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

Optimal. Leaf size=383 \[ -\frac {4 a^3 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {4 a^2 \sec (c+d x)}{b^5 d}-\frac {6 a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {8 a^2 \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}-\frac {2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}+\frac {2 \left (a^2+b^2\right ) \sec (c+d x)}{b^5 d}+\frac {4 a \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac {\left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {\sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^4 d}-\frac {3 a \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b^4 d}+\frac {\sec ^3(c+d x)}{3 b^3 d} \]

[Out]

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

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Rubi [A]  time = 0.79, antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 31, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3106, 3076, 3074, 206, 3104, 3770, 3094, 3768} \[ \frac {4 a^2 \sec (c+d x)}{b^5 d}+\frac {2 \left (a^2+b^2\right ) \sec (c+d x)}{b^5 d}-\frac {4 a^3 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {6 a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {4 a \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac {\left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {8 a^2 \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}-\frac {2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}-\frac {\sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^4 d}-\frac {3 a \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {3 a \tan (c+d x) \sec (c+d x)}{2 b^4 d}+\frac {\sec ^3(c+d x)}{3 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^3*ArcTanh[Sin[c + d*x]])/(b^6*d) - (3*a*ArcTanh[Sin[c + d*x]])/(2*b^4*d) - (6*a*(a^2 + b^2)*ArcTanh[Sin[
c + d*x]])/(b^6*d) - (8*a^2*Sqrt[a^2 + b^2]*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^6*d
) - (Sqrt[a^2 + b^2]*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(2*b^4*d) - (2*(a^2 + b^2)^(3
/2)*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^6*d) + (4*a^2*Sec[c + d*x])/(b^5*d) + (2*(a
^2 + b^2)*Sec[c + d*x])/(b^5*d) + Sec[c + d*x]^3/(3*b^3*d) - ((a^2 + b^2)*(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*(a^2 + b^2))/(b^5*d*(a*Cos[c + d*x] + b*Sin[c + d*x])) - (
3*a*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 ^4(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx &=\frac {\int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^2}-\frac {(2 a) \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^2}+\frac {\left (a^2+b^2\right ) \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^2}\\ &=\frac {\sec ^3(c+d x)}{3 b^3 d}-\frac {a \int \sec ^3(c+d x) \, dx}{b^4}-\frac {(2 a) \int \sec ^3(c+d x) \, dx}{b^4}+\frac {\left (4 a^2\right ) \int \frac {\sec ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^4}+2 \frac {\left (a^2+b^2\right ) \int \frac {\sec ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^4}-2 \frac {\left (2 a \left (a^2+b^2\right )\right ) \int \frac {\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^4}+\frac {\left (a^2+b^2\right )^2 \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^4}\\ &=\frac {4 a^2 \sec (c+d x)}{b^5 d}+\frac {\sec ^3(c+d x)}{3 b^3 d}-\frac {\left (a^2+b^2\right ) (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 {3 a \sec (c+d x) \tan (c+d x)}{2 b^4 d}-\frac {\left (4 a^3\right ) \int \sec (c+d x) \, dx}{b^6}-\frac {a \int \sec (c+d x) \, dx}{2 b^4}-\frac {a \int \sec (c+d x) \, dx}{b^4}-2 \left (-\frac {2 a \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {\left (2 a \left (a^2+b^2\right )\right ) \int \sec (c+d x) \, dx}{b^6}-\frac {\left (2 a^2 \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 {\left (4 a^2 \left (a^2+b^2\right )\right ) \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^6}+\frac {\left (a^2+b^2\right ) \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{2 b^4}+2 \left (\frac {\left (a^2+b^2\right ) \sec (c+d x)}{b^5 d}-\frac {\left (a \left (a^2+b^2\right )\right ) \int \sec (c+d x) \, dx}{b^6}+\frac {\left (a^2+b^2\right )^2 \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^6}\right )\\ &=-\frac {4 a^3 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {3 a \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac {4 a^2 \sec (c+d x)}{b^5 d}+\frac {\sec ^3(c+d x)}{3 b^3 d}-\frac {\left (a^2+b^2\right ) (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 {3 a \sec (c+d x) \tan (c+d x)}{2 b^4 d}-2 \left (\frac {2 a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {2 a \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {\left (2 a^2 \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 {\left (4 a^2 \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}-\frac {\left (a^2+b^2\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 )}{2 b^4 d}+2 \left (-\frac {a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {\left (a^2+b^2\right ) \sec (c+d x)}{b^5 d}-\frac {\left (a^2+b^2\right )^2 \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^3 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {3 a \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {4 a^2 \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 {\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 )}{2 b^4 d}+\frac {4 a^2 \sec (c+d x)}{b^5 d}+\frac {\sec ^3(c+d x)}{3 b^3 d}+2 \left (-\frac {a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac {\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^6 d}+\frac {\left (a^2+b^2\right ) \sec (c+d x)}{b^5 d}\right )-\frac {\left (a^2+b^2\right ) (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}-2 \left (\frac {2 a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac {2 a^2 \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 \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}\right )-\frac {3 a \sec (c+d x) \tan (c+d x)}{2 b^4 d}\\ \end {align*}

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Mathematica [C]  time = 2.46, size = 688, normalized size = 1.80 \[ \frac {\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac {6 b^2 \left (a^2+b^2\right )^2 \sin (c+d x)}{a}+2 b \left (36 a^2+13 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2+\frac {2 b \left (36 a^2+13 b^2\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {2 b \left (36 a^2+13 b^2\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+\frac {6 b (a-i b) (a+i b) \left (8 a^2-b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}{a}+30 a \left (4 a^2+3 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2-30 a \left (4 a^2+3 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2+60 \sqrt {a^2+b^2} \left (4 a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )+\frac {2 b^3 \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {2 b^3 \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {b^2 (b-9 a) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b^2 (9 a+b) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{12 b^6 d (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])*((6*b^2*(a^2 + b^2)^2*Sin[c + d*x])/a + (6*(a - I*b)*(a + I*
b)*b*(8*a^2 - b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x]))/a + 2*b*(36*a^2 + 13*b^2)*(a*Cos[c + d*x] + b*Sin[c + d*
x])^2 + 60*Sqrt[a^2 + b^2]*(4*a^2 + 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])^2 + 30*a*(4*a^2 + 3*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c +
d*x])^2 - 30*a*(4*a^2 + 3*b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 +
(b^2*(-9*a + b)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^3*Sin[(c +
d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (2*b*(36*a^2 + 13*b^2)
*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (2*b^3*Sin[(c +
d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (b^2*(9*a + b)*(a*Cos[
c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (2*b*(36*a^2 + 13*b^2)*Sin[(c + d*x)/2
]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(12*b^6*d*(a + b*Tan[c + d*x])^
3)

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fricas [A]  time = 0.89, size = 564, normalized size = 1.47 \[ \frac {4 \, b^{5} + 30 \, {\left (4 \, a^{4} b + a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} + 20 \, {\left (2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left ({\left (4 \, a^{4} - 3 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (4 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{3}\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^{5} - a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left ({\left (4 \, a^{5} - a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (a b^{4} \cos \left (d x + c\right ) - 6 \, {\left (3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (2 \, a b^{7} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + b^{8} d \cos \left (d x + c\right )^{3} + {\left (a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.07, size = 510, normalized size = 1.33 \[ -\frac {\frac {15 \, {\left (4 \, a^{3} + 3 \, a 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 ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{6}} + \frac {15 \, {\left (4 \, a^{4} + 5 \, a^{2} b^{2} + b^{4}\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 {2 \, {\left (9 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 72 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a^{2} + 14 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} b^{5}} + \frac {6 \, {\left (7 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 25 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 23 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{6} - 7 \, a^{4} b^{2} + a^{2} 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 )}^{2} a^{2} b^{5}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(15*(4*a^3 + 3*a*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(tan(1/2*d*x +
1/2*c) - 1))/b^6 + 15*(4*a^4 + 5*a^2*b^2 + b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/a
bs(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) + 2*(9*a*b*tan(1/2*d*x + 1/2*c)^
5 + 36*a^2*tan(1/2*d*x + 1/2*c)^4 + 18*b^2*tan(1/2*d*x + 1/2*c)^4 - 72*a^2*tan(1/2*d*x + 1/2*c)^2 - 24*b^2*tan
(1/2*d*x + 1/2*c)^2 - 9*a*b*tan(1/2*d*x + 1/2*c) + 36*a^2 + 14*b^2)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*b^5) + 6*(
7*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 5*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 8*a^6*tan
(1/2*d*x + 1/2*c)^2 - 9*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^6*tan(1/2*d*x
+ 1/2*c)^2 - 25*a^5*b*tan(1/2*d*x + 1/2*c) - 23*a^3*b^3*tan(1/2*d*x + 1/2*c) + 2*a*b^5*tan(1/2*d*x + 1/2*c) -
8*a^6 - 7*a^4*b^2 + a^2*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2*a^2*b^5))/d

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maple [B]  time = 0.40, size = 1125, normalized size = 2.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/2/d*a/b^4*ln(tan(1/2*d*x+1/2*c)-1)-15/2/d*a/b^4*ln(tan(1/2*d*x+1/2*c)+1)+2/d/(tan(1/2*d*x+1/2*c)^2*a-2*b*ta
n(1/2*d*x+1/2*c)-a)^2/a*tan(1/2*d*x+1/2*c)^3+15/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2*tan(1/
2*d*x+1/2*c)^2-2/d/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2/a*tan(1/2*d*x+1/2*c)+7/d/b^3/(tan(1/2*d
*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2*a^2+5/d/b^2/(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))-5/2/d/b^3/(tan(1/2*d*x+1/2*c)-1)+5/2/d/b^3/(tan(1/2*d*x+1/2*c)+1)-1/d/b/(tan(1/2*d*x+1/2*c)^
2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2+8/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2*a^4-3/2/d/b^4/(tan
(1/2*d*x+1/2*c)-1)^2*a-6/d/b^5/(tan(1/2*d*x+1/2*c)-1)*a^2-3/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)*a+10/d*a^3/b^6*ln(t
an(1/2*d*x+1/2*c)-1)+3/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)^2*a+6/d/b^5/(tan(1/2*d*x+1/2*c)+1)*a^2-3/2/d/b^4/(tan(1/
2*d*x+1/2*c)+1)*a-10/d*a^3/b^6*ln(tan(1/2*d*x+1/2*c)+1)-1/3/d/b^3/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/b^3/(tan(1/2*
d*x+1/2*c)-1)^2+1/3/d/b^3/(tan(1/2*d*x+1/2*c)+1)^3-1/2/d/b^3/(tan(1/2*d*x+1/2*c)+1)^2-5/d/b^2/(tan(1/2*d*x+1/2
*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2*a*tan(1/2*d*x+1/2*c)^3+9/d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2
*c)-a)^2*a^2*tan(1/2*d*x+1/2*c)^2-2/d*b/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2/a^2*tan(1/2*d*x+1/
2*c)^2+23/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2*a*tan(1/2*d*x+1/2*c)+25/d/b^4/(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))*a^2+25/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2
*d*x+1/2*c)-a)^2*a^3*tan(1/2*d*x+1/2*c)+20/d/b^6/(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))*a^4-7/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2*a^3*tan(1/2*d*x+1/2*c)^3-8/d/b^5/
(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2*a^4*tan(1/2*d*x+1/2*c)^2

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maxima [B]  time = 0.52, size = 902, normalized size = 2.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(60*a^6 + 35*a^4*b^2 - 3*a^2*b^4 + (210*a^5*b + 125*a^3*b^3 - 6*a*b^5)*sin(d*x + c)/(cos(d*x + c) + 1)
- 2*(120*a^6 - 10*a^4*b^2 - 55*a^2*b^4 + 3*b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(330*a^5*b + 205*a^3*b
^3 - 12*a*b^5)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*(180*a^6 - 95*a^4*b^2 - 120*a^2*b^4 + 9*b^6)*sin(d*x +
c)^4/(cos(d*x + c) + 1)^4 + 12*(60*a^5*b + 35*a^3*b^3 - 3*a*b^5)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 6*(40*a
^6 - 30*a^4*b^2 - 35*a^2*b^4 + 3*b^6)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6*(50*a^5*b + 25*a^3*b^3 - 4*a*b^5
)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3*(20*a^6 - 15*a^4*b^2 - 15*a^2*b^4 + 2*b^6)*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8 + 3*(10*a^5*b + 5*a^3*b^3 - 2*a*b^5)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^4*b^5 + 4*a^3*b^6*sin(
d*x + c)/(cos(d*x + c) + 1) - 16*a^3*b^6*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 24*a^3*b^6*sin(d*x + c)^5/(cos(
d*x + c) + 1)^5 - 16*a^3*b^6*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 4*a^3*b^6*sin(d*x + c)^9/(cos(d*x + c) + 1)
^9 - a^4*b^5*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - (5*a^4*b^5 - 4*a^2*b^7)*sin(d*x + c)^2/(cos(d*x + c) + 1)
^2 + 2*(5*a^4*b^5 - 6*a^2*b^7)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2*(5*a^4*b^5 - 6*a^2*b^7)*sin(d*x + c)^6/
(cos(d*x + c) + 1)^6 + (5*a^4*b^5 - 4*a^2*b^7)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) - 15*(4*a^3 + 3*a*b^2)*log
(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b^6 + 15*(4*a^3 + 3*a*b^2)*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b^6
- 15*(4*a^4 + 5*a^2*b^2 + b^4)*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))/d

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mupad [B]  time = 3.82, size = 1203, normalized size = 3.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((60*a^4 - 3*b^4 + 35*a^2*b^2)/(3*b^5) + (tan(c/2 + (d*x)/2)*(210*a^4 - 6*b^4 + 125*a^2*b^2))/(3*a*b^4) + (tan
(c/2 + (d*x)/2)^8*(20*a^6 + 2*b^6 - 15*a^2*b^4 - 15*a^4*b^2))/(a^2*b^5) - (2*tan(c/2 + (d*x)/2)^6*(40*a^6 + 3*
b^6 - 35*a^2*b^4 - 30*a^4*b^2))/(a^2*b^5) - (2*tan(c/2 + (d*x)/2)^2*(120*a^6 + 3*b^6 - 55*a^2*b^4 - 10*a^4*b^2
))/(3*a^2*b^5) + (2*tan(c/2 + (d*x)/2)^4*(180*a^6 + 9*b^6 - 120*a^2*b^4 - 95*a^4*b^2))/(3*a^2*b^5) + (tan(c/2
+ (d*x)/2)^9*(10*a^4 - 2*b^4 + 5*a^2*b^2))/(a*b^4) - (2*tan(c/2 + (d*x)/2)^7*(50*a^4 - 4*b^4 + 25*a^2*b^2))/(a
*b^4) + (4*tan(c/2 + (d*x)/2)^5*(60*a^4 - 3*b^4 + 35*a^2*b^2))/(a*b^4) - (2*tan(c/2 + (d*x)/2)^3*(330*a^4 - 12
*b^4 + 205*a^2*b^2))/(3*a*b^4))/(d*(tan(c/2 + (d*x)/2)^8*(5*a^2 - 4*b^2) - tan(c/2 + (d*x)/2)^2*(5*a^2 - 4*b^2
) + tan(c/2 + (d*x)/2)^4*(10*a^2 - 12*b^2) - tan(c/2 + (d*x)/2)^6*(10*a^2 - 12*b^2) - a^2*tan(c/2 + (d*x)/2)^1
0 + a^2 - 16*a*b*tan(c/2 + (d*x)/2)^3 + 24*a*b*tan(c/2 + (d*x)/2)^5 - 16*a*b*tan(c/2 + (d*x)/2)^7 + 4*a*b*tan(
c/2 + (d*x)/2)^9 + 4*a*b*tan(c/2 + (d*x)/2))) - (atanh((3000*a^2*tan(c/2 + (d*x)/2))/(3000*a^2 + (7000*a^4)/b^
2 + (4000*a^6)/b^4) + (7000*a^4*tan(c/2 + (d*x)/2))/(7000*a^4 + 3000*a^2*b^2 + (4000*a^6)/b^2) + (4000*a^6*tan
(c/2 + (d*x)/2))/(4000*a^6 + 3000*a^2*b^4 + 7000*a^4*b^2))*(15*a*b^2 + 20*a^3))/(b^6*d) + (5*atanh((1000*a^2*(
a^2 + b^2)^(1/2))/(1000*a^2*b + (5000*a^4)/b + (4000*a^6)/b^3 + 10000*a^3*tan(c/2 + (d*x)/2) + 2000*a*b^2*tan(
c/2 + (d*x)/2) + (8000*a^5*tan(c/2 + (d*x)/2))/b^2) + (4000*a^4*(a^2 + b^2)^(1/2))/(5000*a^4*b + 1000*a^2*b^3
+ (4000*a^6)/b + 8000*a^5*tan(c/2 + (d*x)/2) + 2000*a*b^4*tan(c/2 + (d*x)/2) + 10000*a^3*b^2*tan(c/2 + (d*x)/2
)) + (9000*a^3*tan(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2))/(5000*a^4 + 1000*a^2*b^2 + (4000*a^6)/b^2 + 2000*a*b^3*ta
n(c/2 + (d*x)/2) + 10000*a^3*b*tan(c/2 + (d*x)/2) + (8000*a^5*tan(c/2 + (d*x)/2))/b) + (4000*a^5*tan(c/2 + (d*
x)/2)*(a^2 + b^2)^(1/2))/(4000*a^6 + 1000*a^2*b^4 + 5000*a^4*b^2 + 2000*a*b^5*tan(c/2 + (d*x)/2) + 8000*a^5*b*
tan(c/2 + (d*x)/2) + 10000*a^3*b^3*tan(c/2 + (d*x)/2)) + (2000*a*tan(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2))/(1000*a
^2 + (5000*a^4)/b^2 + (4000*a^6)/b^4 + (10000*a^3*tan(c/2 + (d*x)/2))/b + (8000*a^5*tan(c/2 + (d*x)/2))/b^3 +
2000*a*b*tan(c/2 + (d*x)/2)))*(4*a^2 + b^2)*(a^2 + b^2)^(1/2))/(b^6*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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