3.148 \(\int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=400 \[ \frac {8 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {2 \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {6 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}-\frac {2 \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {3 a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^4 d \sqrt {a^2+b^2}}-\frac {a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {4 a^3 \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d \sqrt {a^2+b^2}}-\frac {4 a \sec (c+d x)}{b^5 d}+\frac {3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b^4 d} \]

[Out]

8*a^2*arctanh(sin(d*x+c))/b^6/d+1/2*arctanh(sin(d*x+c))/b^4/d+2*(a^2+b^2)*arctanh(sin(d*x+c))/b^6/d-4*a*sec(d*
x+c)/b^5/d+1/3*(-a^2-b^2)/b^3/d/(a*cos(d*x+c)+b*sin(d*x+c))^3+3/2*a*(b*cos(d*x+c)-a*sin(d*x+c))/b^4/d/(a*cos(d
*x+c)+b*sin(d*x+c))^2-4*a^2/b^5/d/(a*cos(d*x+c)+b*sin(d*x+c))-2*(a^2+b^2)/b^5/d/(a*cos(d*x+c)+b*sin(d*x+c))+4*
a^3*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/b^6/d/(a^2+b^2)^(1/2)+3/2*a*arctanh((b*cos(d*x+c)-a*s
in(d*x+c))/(a^2+b^2)^(1/2))/b^4/d/(a^2+b^2)^(1/2)+6*a*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))*(a^
2+b^2)^(1/2)/b^6/d+1/2*sec(d*x+c)*tan(d*x+c)/b^4/d

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Rubi [A]  time = 0.80, antiderivative size = 400, normalized size of antiderivative = 1.00, number of steps used = 32, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3106, 3094, 3770, 3074, 206, 3076, 3768, 3104} \[ \frac {8 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {2 \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac {2 \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac {a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {4 a^3 \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d \sqrt {a^2+b^2}}+\frac {6 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}+\frac {3 a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^4 d \sqrt {a^2+b^2}}-\frac {4 a \sec (c+d x)}{b^5 d}+\frac {3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(8*a^2*ArcTanh[Sin[c + d*x]])/(b^6*d) + ArcTanh[Sin[c + d*x]]/(2*b^4*d) + (2*(a^2 + b^2)*ArcTanh[Sin[c + d*x]]
)/(b^6*d) + (4*a^3*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^6*Sqrt[a^2 + b^2]*d) + (3*a*
ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(2*b^4*Sqrt[a^2 + b^2]*d) + (6*a*Sqrt[a^2 + b^2]*A
rcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^6*d) - (4*a*Sec[c + d*x])/(b^5*d) - (a^2 + b^2)/
(3*b^3*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*a*(b*Cos[c + d*x] - a*Sin[c + d*x]))/(2*b^4*d*(a*Cos[c + d*
x] + b*Sin[c + d*x])^2) - (4*a^2)/(b^5*d*(a*Cos[c + d*x] + b*Sin[c + d*x])) - (2*(a^2 + b^2))/(b^5*d*(a*Cos[c
+ d*x] + b*Sin[c + d*x])) + (Sec[c + d*x]*Tan[c + d*x])/(2*b^4*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3076

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*Cos[c + d*x] -
a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1))/(d*(n + 1)*(a^2 + b^2)), x] + Dist[(n + 2)/((n + 1
)*(a^2 + b^2)), Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && LtQ[n, -1] && NeQ[n, -2]

Rule 3094

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_)/cos[(c_.) + (d_.)*(x_)], x_Symbol] :>
Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)), x] + (Dist[1/b^2, Int[(a*Cos[c + d*x] + b*Sin[c
+ d*x])^(n + 2)/Cos[c + d*x], x], x] - Dist[a/b^2, Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1), x], x]) /;
FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3104

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[Cos[c + d*x]^(m + 1)/(b*d*(m + 1)), x] + (-Dist[a/b^2, Int[Cos[c + d*x]^(m + 1), x], x] + Dist[(a^2 + b
^2)/b^2, Int[Cos[c + d*x]^(m + 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[
a^2 + b^2, 0] && LtQ[m, -1]

Rule 3106

Int[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbo
l] :> Dist[(a^2 + b^2)/b^2, Int[Cos[c + d*x]^(m + 2)*(a*Cos[c + d*x] + b*Sin[c + d*x])^n, x], x] + (Dist[1/b^2
, Int[Cos[c + d*x]^m*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] - Dist[(2*a)/b^2, Int[Cos[c + d*x]^(m +
1)*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n
, -1] && LtQ[m, -1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx &=\frac {\int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^2}-\frac {(2 a) \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^2}+\frac {\left (a^2+b^2\right ) \int \frac {\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx}{b^2}\\ &=-\frac {a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\int \sec ^3(c+d x) \, dx}{b^4}-2 \frac {(2 a) \int \frac {\sec ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^4}+\frac {\left (4 a^2\right ) \int \frac {\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^4}+2 \frac {\left (a^2+b^2\right ) \int \frac {\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^4}-\frac {\left (a \left (a^2+b^2\right )\right ) \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^4}-\frac {\left (2 a \left (a^2+b^2\right )\right ) \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^4}\\ &=-\frac {a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {\sec (c+d x) \tan (c+d x)}{2 b^4 d}+\frac {\left (4 a^2\right ) \int \sec (c+d x) \, dx}{b^6}-\frac {\left (4 a^3\right ) \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^6}+\frac {\int \sec (c+d x) \, dx}{2 b^4}-\frac {a \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{2 b^4}-\frac {a \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^4}+2 \left (-\frac {a^2+b^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {\left (a^2+b^2\right ) \int \sec (c+d x) \, dx}{b^6}-\frac {\left (a \left (a^2+b^2\right )\right ) \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^6}\right )-2 \left (\frac {2 a \sec (c+d x)}{b^5 d}-\frac {\left (2 a^2\right ) \int \sec (c+d x) \, dx}{b^6}+\frac {\left (2 a \left (a^2+b^2\right )\right ) \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^6}\right )\\ &=\frac {4 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {\sec (c+d x) \tan (c+d x)}{2 b^4 d}+\frac {\left (4 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{b^6 d}+\frac {a \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{2 b^4 d}+\frac {a \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{b^4 d}+2 \left (\frac {\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {a^2+b^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {\left (a \left (a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{b^6 d}\right )-2 \left (-\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {2 a \sec (c+d x)}{b^5 d}-\frac {\left (2 a \left (a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{b^6 d}\right )\\ &=\frac {4 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac {4 a^3 \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 \sqrt {a^2+b^2} d}+\frac {3 a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^4 \sqrt {a^2+b^2} d}-2 \left (-\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {2 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}+\frac {2 a \sec (c+d x)}{b^5 d}\right )-\frac {a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+2 \left (\frac {\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}-\frac {a^2+b^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}\right )+\frac {\sec (c+d x) \tan (c+d x)}{2 b^4 d}\\ \end {align*}

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Mathematica [A]  time = 3.47, size = 538, normalized size = 1.34 \[ -\frac {\sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (18 b^2 \left (a^2+b^2\right ) \sin (c+d x) (a \cos (c+d x)+b \sin (c+d x))+6 b \left (12 a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2+30 \left (4 a^2+b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3-30 \left (4 a^2+b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3+\frac {60 a \left (4 a^2+3 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3 \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+4 b^3 \left (a^2+b^2\right )-\frac {3 b^2 (a \cos (c+d x)+b \sin (c+d x))^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {3 b^2 (a \cos (c+d x)+b \sin (c+d x))^3}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+48 a b (a \cos (c+d x)+b \sin (c+d x))^3+\frac {48 a b \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^3}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {48 a b \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^3}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}\right )}{12 b^6 d (a+b \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

-1/12*(Sec[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])*(4*b^3*(a^2 + b^2) + 18*b^2*(a^2 + b^2)*Sin[c + d*x]*(
a*Cos[c + d*x] + b*Sin[c + d*x]) + 6*b*(12*a^2 + b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 + 48*a*b*(a*Cos[c +
d*x] + b*Sin[c + d*x])^3 + (60*a*(4*a^2 + 3*b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]]*(a*Cos[c +
d*x] + b*Sin[c + d*x])^3)/Sqrt[a^2 + b^2] + 30*(4*a^2 + b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a*Cos[
c + d*x] + b*Sin[c + d*x])^3 - 30*(4*a^2 + b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*S
in[c + d*x])^3 - (3*b^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (48*a*b
*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (3*b^2*(a*Cos[c
+ d*x] + b*Sin[c + d*x])^3)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (48*a*b*Sin[(c + d*x)/2]*(a*Cos[c + d*x
] + b*Sin[c + d*x])^3)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(b^6*d*(a + b*Tan[c + d*x])^4)

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fricas [B]  time = 0.81, size = 820, normalized size = 2.05 \[ \frac {6 \, a^{2} b^{5} + 6 \, b^{7} - 30 \, {\left (4 \, a^{6} b - 3 \, a^{4} b^{3} - 8 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} - 20 \, {\left (11 \, a^{4} b^{3} + 13 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left ({\left (4 \, a^{6} - 9 \, a^{4} b^{2} - 9 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (4 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left ({\left (12 \, a^{5} b + 5 \, a^{3} b^{3} - 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 15 \, {\left ({\left (4 \, a^{7} - 7 \, a^{5} b^{2} - 14 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (4 \, a^{5} b^{2} + 5 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} + {\left ({\left (12 \, a^{6} b + 11 \, a^{4} b^{3} - 2 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{4} b^{3} + 5 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (4 \, a^{7} - 7 \, a^{5} b^{2} - 14 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (4 \, a^{5} b^{2} + 5 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} + {\left ({\left (12 \, a^{6} b + 11 \, a^{4} b^{3} - 2 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{4} b^{3} + 5 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 30 \, {\left (10 \, {\left (a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{5} b^{6} - 2 \, a^{3} b^{8} - 3 \, a b^{10}\right )} d \cos \left (d x + c\right )^{5} + 3 \, {\left (a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{3} + {\left ({\left (3 \, a^{4} b^{7} + 2 \, a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{4} + {\left (a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/12*(6*a^2*b^5 + 6*b^7 - 30*(4*a^6*b - 3*a^4*b^3 - 8*a^2*b^5 - b^7)*cos(d*x + c)^4 - 20*(11*a^4*b^3 + 13*a^2*
b^5 + 2*b^7)*cos(d*x + c)^2 + 15*((4*a^6 - 9*a^4*b^2 - 9*a^2*b^4)*cos(d*x + c)^5 + 3*(4*a^4*b^2 + 3*a^2*b^4)*c
os(d*x + c)^3 + ((12*a^5*b + 5*a^3*b^3 - 3*a*b^5)*cos(d*x + c)^4 + (4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^2)*sin(d
*x + c))*sqrt(a^2 + b^2)*log((2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*s
qrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^
2 + b^2)) + 15*((4*a^7 - 7*a^5*b^2 - 14*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^5 + 3*(4*a^5*b^2 + 5*a^3*b^4 + a*b^6)*
cos(d*x + c)^3 + ((12*a^6*b + 11*a^4*b^3 - 2*a^2*b^5 - b^7)*cos(d*x + c)^4 + (4*a^4*b^3 + 5*a^2*b^5 + b^7)*cos
(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 15*((4*a^7 - 7*a^5*b^2 - 14*a^3*b^4 - 3*a*b^6)*cos(d*x + c)
^5 + 3*(4*a^5*b^2 + 5*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + ((12*a^6*b + 11*a^4*b^3 - 2*a^2*b^5 - b^7)*cos(d*x + c
)^4 + (4*a^4*b^3 + 5*a^2*b^5 + b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 30*(10*(a^5*b^2 + a
^3*b^4)*cos(d*x + c)^3 + (a^3*b^4 + a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^5*b^6 - 2*a^3*b^8 - 3*a*b^10)*d*cos
(d*x + c)^5 + 3*(a^3*b^8 + a*b^10)*d*cos(d*x + c)^3 + ((3*a^4*b^7 + 2*a^2*b^9 - b^11)*d*cos(d*x + c)^4 + (a^2*
b^9 + b^11)*d*cos(d*x + c)^2)*sin(d*x + c))

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giac [A]  time = 3.10, size = 548, normalized size = 1.37 \[ \frac {\frac {15 \, {\left (4 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{6}} - \frac {15 \, {\left (4 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{6}} + \frac {15 \, {\left (4 \, a^{3} + 3 \, a b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{6}} + \frac {6 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{5}} + \frac {2 \, {\left (27 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, a^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 117 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 216 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 114 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, a^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 300 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 54 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 189 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a^{7} + 5 \, a^{5} b^{2} + 2 \, a^{3} b^{4}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{3} a^{3} b^{5}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/6*(15*(4*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^6 - 15*(4*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/b^6 + 15*(4*a^3 + 3*a*b^2)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*
x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^6) + 6*(b*tan(1/2*d*x + 1/2*c)^3 + 8*a*tan(1/2*d*x +
1/2*c)^2 + b*tan(1/2*d*x + 1/2*c) - 8*a)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^5) + 2*(27*a^6*b*tan(1/2*d*x + 1/2
*c)^5 + 6*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 + 36*a^7*tan(1/2*d*x + 1/2*c)^4 - 117*a^5*b^2*tan(1/2*d*x + 1/2*c)^4
- 12*a*b^6*tan(1/2*d*x + 1/2*c)^4 - 216*a^6*b*tan(1/2*d*x + 1/2*c)^3 + 114*a^4*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*
a^2*b^5*tan(1/2*d*x + 1/2*c)^3 + 8*b^7*tan(1/2*d*x + 1/2*c)^3 - 72*a^7*tan(1/2*d*x + 1/2*c)^2 + 300*a^5*b^2*ta
n(1/2*d*x + 1/2*c)^2 + 54*a^3*b^4*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^6*tan(1/2*d*x + 1/2*c)^2 + 189*a^6*b*tan(1/2
*d*x + 1/2*c) + 30*a^4*b^3*tan(1/2*d*x + 1/2*c) + 6*a^2*b^5*tan(1/2*d*x + 1/2*c) + 36*a^7 + 5*a^5*b^2 + 2*a^3*
b^4)/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^3*a^3*b^5))/d

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maple [B]  time = 0.47, size = 1255, normalized size = 3.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

2/3/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3+1/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)^2+1/2/d/b^4/(tan(
1/2*d*x+1/2*c)-1)-1/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)^2+1/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)+18/d/b/(tan(1/2*d*x+1/2*
c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*tan(1/2*d*x+1/2*c)^2+12/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c
)-a)^3*a^4+5/3/d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^2+2/d/(tan(1/2*d*x+1/2*c)^2*a-2*b*t
an(1/2*d*x+1/2*c)-a)^3/a*tan(1/2*d*x+1/2*c)+2/d/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/a*tan(1/2*
d*x+1/2*c)^5+8/3/d/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/a*tan(1/2*d*x+1/2*c)^3+4/d/b^5/(tan(1/2
*d*x+1/2*c)-1)*a-10/d/b^6*ln(tan(1/2*d*x+1/2*c)-1)*a^2-4/d/b^5/(tan(1/2*d*x+1/2*c)+1)*a+10/d/b^6*ln(tan(1/2*d*
x+1/2*c)+1)*a^2+8/3/d*b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/a^3*tan(1/2*d*x+1/2*c)^3-24/d/b^
5/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^4*tan(1/2*d*x+1/2*c)^2+100/d/b^3/(tan(1/2*d*x+1/2*c)^2
*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^2*tan(1/2*d*x+1/2*c)^2+4/d*b/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a
)^3/a^2*tan(1/2*d*x+1/2*c)^2+63/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^3*tan(1/2*d*x+1/2*
c)+10/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a*tan(1/2*d*x+1/2*c)-20/d/b^6*a^3/(a^2+b^2)^(1
/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-15/d/b^4*a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(
1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))+9/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^3*tan(1/2*d
*x+1/2*c)^5+12/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^4*tan(1/2*d*x+1/2*c)^4-39/d/b^3/(ta
n(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^2*tan(1/2*d*x+1/2*c)^4-4/d*b/(tan(1/2*d*x+1/2*c)^2*a-2*b*ta
n(1/2*d*x+1/2*c)-a)^3/a^2*tan(1/2*d*x+1/2*c)^4-72/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^
3*tan(1/2*d*x+1/2*c)^3+38/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a*tan(1/2*d*x+1/2*c)^3-5/2
/d/b^4*ln(tan(1/2*d*x+1/2*c)-1)+5/2/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.48, size = 936, normalized size = 2.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/6*(2*(60*a^7 + 5*a^5*b^2 + 2*a^3*b^4 + 6*(55*a^6*b + 5*a^4*b^3 + a^2*b^5)*sin(d*x + c)/(cos(d*x + c) + 1) -
2*(120*a^7 - 280*a^5*b^2 - 25*a^3*b^4 - 6*a*b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(510*a^6*b - 105*a^4
*b^3 + 2*a^2*b^5 - 4*b^7)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*(180*a^7 - 635*a^5*b^2 - 65*a^3*b^4 - 18*a*b
^6)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 2*(540*a^6*b - 195*a^4*b^3 - 2*a^2*b^5 - 8*b^7)*sin(d*x + c)^5/(cos(
d*x + c) + 1)^5 - 6*(40*a^7 - 140*a^5*b^2 - 5*a^3*b^4 - 6*a*b^6)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*(210*
a^6*b - 75*a^4*b^3 + 2*a^2*b^5 - 4*b^7)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3*(20*a^7 - 45*a^5*b^2 - 4*a*b^6
)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 6*(5*a^6*b + a^2*b^5)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^6*b^5 +
6*a^5*b^6*sin(d*x + c)/(cos(d*x + c) + 1) + 6*a^5*b^6*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - a^6*b^5*sin(d*x +
c)^10/(cos(d*x + c) + 1)^10 - (5*a^6*b^5 - 12*a^4*b^7)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*(3*a^5*b^6 - a^
3*b^8)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*(5*a^6*b^5 - 18*a^4*b^7)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 +
4*(9*a^5*b^6 - 4*a^3*b^8)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2*(5*a^6*b^5 - 18*a^4*b^7)*sin(d*x + c)^6/(cos
(d*x + c) + 1)^6 - 8*(3*a^5*b^6 - a^3*b^8)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + (5*a^6*b^5 - 12*a^4*b^7)*sin(
d*x + c)^8/(cos(d*x + c) + 1)^8) - 15*(4*a^2 + 3*b^2)*a*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2
+ b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^6) - 15*(4*a^2 + b^2)*lo
g(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b^6 + 15*(4*a^2 + b^2)*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b^6)/d

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mupad [B]  time = 4.59, size = 1961, normalized size = 4.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a*cos(c + d*x) + b*sin(c + d*x))^4),x)

[Out]

(atanh((4000*a^3*tan(c/2 + (d*x)/2))/(1000*a*b^2 + 4000*a^3) + (1000*a*tan(c/2 + (d*x)/2))/(1000*a + (4000*a^3
)/b^2))*(20*a^2 + 5*b^2))/(b^6*d) - ((60*a^4 + 2*b^4 + 5*a^2*b^2)/(3*b^5) + (2*tan(c/2 + (d*x)/2)^9*(5*a^4 + b
^4))/(a*b^4) + (2*tan(c/2 + (d*x)/2)^6*(6*b^6 - 40*a^6 + 5*a^2*b^4 + 140*a^4*b^2))/(a^2*b^5) - (2*tan(c/2 + (d
*x)/2)^7*(210*a^6 - 4*b^6 + 2*a^2*b^4 - 75*a^4*b^2))/(3*a^3*b^4) + (2*tan(c/2 + (d*x)/2)^2*(6*b^6 - 120*a^6 +
25*a^2*b^4 + 280*a^4*b^2))/(3*a^2*b^5) - (2*tan(c/2 + (d*x)/2)^3*(510*a^6 - 4*b^6 + 2*a^2*b^4 - 105*a^4*b^2))/
(3*a^3*b^4) - (2*tan(c/2 + (d*x)/2)^4*(18*b^6 - 180*a^6 + 65*a^2*b^4 + 635*a^4*b^2))/(3*a^2*b^5) - (tan(c/2 +
(d*x)/2)^8*(4*b^6 - 20*a^6 + 45*a^4*b^2))/(a^2*b^5) + (2*tan(c/2 + (d*x)/2)*(55*a^4 + b^4 + 5*a^2*b^2))/(a*b^4
) + (2*tan(c/2 + (d*x)/2)^5*(9*a^2 - 4*b^2)*(60*a^4 + 2*b^4 + 5*a^2*b^2))/(3*a^3*b^4))/(d*(tan(c/2 + (d*x)/2)^
2*(12*a*b^2 - 5*a^3) - a^3*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x)/2)^8*(12*a*b^2 - 5*a^3) - tan(c/2 + (d*x)/2
)^4*(36*a*b^2 - 10*a^3) + tan(c/2 + (d*x)/2)^6*(36*a*b^2 - 10*a^3) - tan(c/2 + (d*x)/2)^3*(24*a^2*b - 8*b^3) -
tan(c/2 + (d*x)/2)^7*(24*a^2*b - 8*b^3) + tan(c/2 + (d*x)/2)^5*(36*a^2*b - 16*b^3) + a^3 + 6*a^2*b*tan(c/2 +
(d*x)/2) + 6*a^2*b*tan(c/2 + (d*x)/2)^9)) - (a*atan(((a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*((8*(25*a^2*b^9 + 20
0*a^4*b^7 + 400*a^6*b^5))/b^14 + (8*tan(c/2 + (d*x)/2)*(50*a*b^11 + 650*a^3*b^9 + 1600*a^5*b^7 + 800*a^7*b^5))
/b^15 - (5*a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*((8*tan(c/2 + (d*x)/2)*(60*a^2*b^14 + 80*a^4*b^12))/b^15 - (8*(
10*a*b^14 + 20*a^3*b^12))/b^14 + (5*a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(1
2*a*b^19 + 8*a^3*b^17))/b^15))/(2*(b^8 + a^2*b^6))))/(2*(b^8 + a^2*b^6)))*5i)/(2*(b^8 + a^2*b^6)) + (a*(4*a^2
+ 3*b^2)*(a^2 + b^2)^(1/2)*((8*(25*a^2*b^9 + 200*a^4*b^7 + 400*a^6*b^5))/b^14 + (8*tan(c/2 + (d*x)/2)*(50*a*b^
11 + 650*a^3*b^9 + 1600*a^5*b^7 + 800*a^7*b^5))/b^15 - (5*a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*((8*(10*a*b^14 +
20*a^3*b^12))/b^14 - (8*tan(c/2 + (d*x)/2)*(60*a^2*b^14 + 80*a^4*b^12))/b^15 + (5*a*(4*a^2 + 3*b^2)*(a^2 + b^
2)^(1/2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^19 + 8*a^3*b^17))/b^15))/(2*(b^8 + a^2*b^6))))/(2*(b^8 +
a^2*b^6)))*5i)/(2*(b^8 + a^2*b^6)))/((16*(2000*a^7 + 375*a^3*b^4 + 2000*a^5*b^2))/b^14 - (16*tan(c/2 + (d*x)/2
)*(8000*a^8 + 375*a^2*b^6 + 3500*a^4*b^4 + 10000*a^6*b^2))/b^15 - (5*a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*((8*(
25*a^2*b^9 + 200*a^4*b^7 + 400*a^6*b^5))/b^14 + (8*tan(c/2 + (d*x)/2)*(50*a*b^11 + 650*a^3*b^9 + 1600*a^5*b^7
+ 800*a^7*b^5))/b^15 - (5*a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*((8*tan(c/2 + (d*x)/2)*(60*a^2*b^14 + 80*a^4*b^1
2))/b^15 - (8*(10*a*b^14 + 20*a^3*b^12))/b^14 + (5*a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*(32*a^2*b^3 + (8*tan(c/
2 + (d*x)/2)*(12*a*b^19 + 8*a^3*b^17))/b^15))/(2*(b^8 + a^2*b^6))))/(2*(b^8 + a^2*b^6))))/(2*(b^8 + a^2*b^6))
+ (5*a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*((8*(25*a^2*b^9 + 200*a^4*b^7 + 400*a^6*b^5))/b^14 + (8*tan(c/2 + (d*
x)/2)*(50*a*b^11 + 650*a^3*b^9 + 1600*a^5*b^7 + 800*a^7*b^5))/b^15 - (5*a*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*((
8*(10*a*b^14 + 20*a^3*b^12))/b^14 - (8*tan(c/2 + (d*x)/2)*(60*a^2*b^14 + 80*a^4*b^12))/b^15 + (5*a*(4*a^2 + 3*
b^2)*(a^2 + b^2)^(1/2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^19 + 8*a^3*b^17))/b^15))/(2*(b^8 + a^2*b^6)
)))/(2*(b^8 + a^2*b^6))))/(2*(b^8 + a^2*b^6))))*(4*a^2 + 3*b^2)*(a^2 + b^2)^(1/2)*5i)/(d*(b^8 + a^2*b^6))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**3/(a*cos(c + d*x) + b*sin(c + d*x))**4, x)

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