3.129 \(\int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=179 \[ \frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {3 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^4 d}-\frac {a^2+b^2}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}-\frac {2 a \sec (c+d x)}{b^3 d}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3106, 3094, 3770, 3074, 206, 3768, 3104} \[ \frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac {a^2+b^2}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {3 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^4 d}-\frac {2 a \sec (c+d x)}{b^3 d}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

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

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Mathematica [C]  time = 6.12, size = 709, normalized size = 3.96 \[ -\frac {3 \left (2 a^2+b^2\right ) \sec ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{2 b^4 d (a+b \tan (c+d x))^2}+\frac {3 \left (2 a^2+b^2\right ) \sec ^2(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{2 b^4 d (a+b \tan (c+d x))^2}-\frac {6 a \sqrt {a^2+b^2} \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \tanh ^{-1}\left (\frac {\sqrt {a^2+b^2} \left (a \sin \left (\frac {1}{2} (c+d x)\right )-b \cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \cos \left (\frac {1}{2} (c+d x)\right )+b^2 \cos \left (\frac {1}{2} (c+d x)\right )}\right )}{b^4 d (a+b \tan (c+d x))^2}-\frac {2 a \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}-\frac {2 a \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d (a+b \tan (c+d x))^2}+\frac {2 a \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}-\frac {(a-i b) (a+i b) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))}{b^3 d (a+b \tan (c+d x))^2}+\frac {\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{4 b^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a+b \tan (c+d x))^2}-\frac {\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{4 b^2 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

-(((a - I*b)*(a + I*b)*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(b^3*d*(a + b*Tan[c + d*x])^2)) - (2*
a*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(a + b*Tan[c + d*x])^2) - (6*a*Sqrt[a^2 + b^2]*Ar
cTanh[(Sqrt[a^2 + b^2]*(-(b*Cos[(c + d*x)/2]) + a*Sin[(c + d*x)/2]))/(a^2*Cos[(c + d*x)/2] + b^2*Cos[(c + d*x)
/2])]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^4*d*(a + b*Tan[c + d*x])^2) - (3*(2*a^2 + b^2)*Lo
g[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(2*b^4*d*(a + b*Tan
[c + d*x])^2) + (3*(2*a^2 + b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*S
in[c + d*x])^2)/(2*b^4*d*(a + b*Tan[c + d*x])^2) + (Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(4*b^2
*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a + b*Tan[c + d*x])^2) - (2*a*Sec[c + d*x]^2*Sin[(c + d*x)/2]*(a*C
os[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^2) - (Sec[c
+ d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(4*b^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a + b*Tan[c +
d*x])^2) + (2*a*Sec[c + d*x]^2*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^2)

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fricas [B]  time = 0.61, size = 355, normalized size = 1.98 \[ -\frac {6 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, b^{3} + 6 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 3 \, {\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \, {\left (a b^{4} d \cos \left (d x + c\right )^{3} + b^{5} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

-1/4*(6*a*b^2*cos(d*x + c)*sin(d*x + c) - 2*b^3 + 6*(2*a^2*b + b^3)*cos(d*x + c)^2 - 6*(a^2*cos(d*x + c)^3 + a
*b*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*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)) - 3*((2*a^3 + a*b^2)*cos(d*x + c)^3 + (2*a^2*b + b^3)*cos(d*x + c)^2*sin(d*x
+ c))*log(sin(d*x + c) + 1) + 3*((2*a^3 + a*b^2)*cos(d*x + c)^3 + (2*a^2*b + b^3)*cos(d*x + c)^2*sin(d*x + c))
*log(-sin(d*x + c) + 1))/(a*b^4*d*cos(d*x + c)^3 + b^5*d*cos(d*x + c)^2*sin(d*x + c))

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giac [A]  time = 0.34, size = 280, normalized size = 1.56 \[ \frac {\frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} + \frac {6 \, {\left (a^{3} + a b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{3}} + \frac {4 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} + a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )} a b^{3}}}{2 \, d} \]

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))^2,x, algorithm="giac")

[Out]

1/2*(3*(2*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 3*(2*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1
))/b^4 + 6*(a^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^4) + 2*(b*tan(1/2*d*x + 1/2*c)^3 + 4*a*tan(1/2*d*x + 1/2*c)
^2 + b*tan(1/2*d*x + 1/2*c) - 4*a)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^3) + 4*(a^2*b*tan(1/2*d*x + 1/2*c) + b^3*
tan(1/2*d*x + 1/2*c) + a^3 + a*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)*a*b^3))/d

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maple [B]  time = 0.32, size = 440, normalized size = 2.46 \[ \frac {1}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 a}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}{d \,b^{4}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,b^{2}}-\frac {1}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}}{d \,b^{4}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,b^{2}}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right ) a}+\frac {2 a^{2}}{d \,b^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}+\frac {2}{d b \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}-\frac {6 a \sqrt {a^{2}+b^{2}}\, \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{4}} \]

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))^2,x)

[Out]

1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)^2+2/d/b^3/(tan(1/2*d*x+1/2*c)-1)*a+1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)-3/d/b^4*l
n(tan(1/2*d*x+1/2*c)-1)*a^2-3/2/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)-1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2-2/d/b^3/(tan
(1/2*d*x+1/2*c)+1)*a+1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)+3/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)*a^2+3/2/d/b^2*ln(tan(1/
2*d*x+1/2*c)+1)+2/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)*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)/a*tan(1/2*d*x+1/2*c)+2/d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/
2*c)-a)*a^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)-6/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))

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maxima [B]  time = 0.44, size = 471, normalized size = 2.63 \[ -\frac {\frac {2 \, {\left (6 \, a^{3} + 2 \, a b^{2} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (9 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {6 \, {\left (2 \, a^{3} + a b^{2}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} b^{3} + \frac {2 \, a b^{4} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, a^{2} b^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, a b^{4} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{2} b^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {2 \, a b^{4} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{2} b^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {6 \, \sqrt {a^{2} + b^{2}} a \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{b^{4}} - \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{4}} + \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{4}}}{2 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

-1/2*(2*(6*a^3 + 2*a*b^2 + 6*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (9*a^2*b + 2*b^3)*sin(d*x + c)/(cos(d*x
+ c) + 1) - 6*(2*a^3 + a*b^2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*(3*a^2*b + b^3)*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + (3*a^2*b + 2*b^3)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^2*b^3 + 2*a*b^4*sin(d*x + c)/(cos(d*x
+ c) + 1) - 3*a^2*b^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*a*b^4*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a
^2*b^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 2*a*b^4*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - a^2*b^3*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6) - 6*sqrt(a^2 + 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)))/b^4 - 3*(2*a^2 + b^2)*log(sin(d*x + c)/(cos(d*x + c
) + 1) + 1)/b^4 + 3*(2*a^2 + b^2)*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b^4)/d

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mupad [B]  time = 1.90, size = 585, normalized size = 3.27 \[ \frac {\mathrm {atanh}\left (\frac {648\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{216\,a\,b^2+648\,a^3+\frac {432\,a^5}{b^2}}+\frac {432\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{432\,a^5+648\,a^3\,b^2+216\,a\,b^4}+\frac {216\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{216\,a+\frac {648\,a^3}{b^2}+\frac {432\,a^5}{b^4}}\right )\,\left (6\,a^2+3\,b^2\right )}{b^4\,d}-\frac {\frac {2\,\left (3\,a^2+b^2\right )}{b^3}+\frac {6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{b^3}-\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+b^2\right )}{b^3}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^2+2\,b^2\right )}{a\,b^2}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2+b^2\right )}{a\,b^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,a^2+2\,b^2\right )}{a\,b^2}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {6\,a\,\mathrm {atanh}\left (\frac {432\,a^3\,\sqrt {a^2+b^2}}{432\,a^3\,b+\frac {432\,a^5}{b}+864\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+864\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {864\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}}{432\,a^3+\frac {432\,a^5}{b^2}+864\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {864\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b}}+\frac {432\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2+b^2}}{432\,a^5+864\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b+432\,a^3\,b^2+864\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3}\right )\,\sqrt {a^2+b^2}}{b^4\,d} \]

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))^2),x)

[Out]

(atanh((648*a^3*tan(c/2 + (d*x)/2))/(216*a*b^2 + 648*a^3 + (432*a^5)/b^2) + (432*a^5*tan(c/2 + (d*x)/2))/(216*
a*b^4 + 432*a^5 + 648*a^3*b^2) + (216*a*tan(c/2 + (d*x)/2))/(216*a + (648*a^3)/b^2 + (432*a^5)/b^4))*(6*a^2 +
3*b^2))/(b^4*d) - ((2*(3*a^2 + b^2))/b^3 + (6*a^2*tan(c/2 + (d*x)/2)^4)/b^3 - (6*tan(c/2 + (d*x)/2)^2*(2*a^2 +
b^2))/b^3 + (tan(c/2 + (d*x)/2)*(9*a^2 + 2*b^2))/(a*b^2) - (4*tan(c/2 + (d*x)/2)^3*(3*a^2 + b^2))/(a*b^2) + (
tan(c/2 + (d*x)/2)^5*(3*a^2 + 2*b^2))/(a*b^2))/(d*(a + 2*b*tan(c/2 + (d*x)/2) - 3*a*tan(c/2 + (d*x)/2)^2 + 3*a
*tan(c/2 + (d*x)/2)^4 - a*tan(c/2 + (d*x)/2)^6 - 4*b*tan(c/2 + (d*x)/2)^3 + 2*b*tan(c/2 + (d*x)/2)^5)) - (6*a*
atanh((432*a^3*(a^2 + b^2)^(1/2))/(432*a^3*b + (432*a^5)/b + 864*a^4*tan(c/2 + (d*x)/2) + 864*a^2*b^2*tan(c/2
+ (d*x)/2)) + (864*a^2*tan(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2))/(432*a^3 + (432*a^5)/b^2 + 864*a^2*b*tan(c/2 + (d
*x)/2) + (864*a^4*tan(c/2 + (d*x)/2))/b) + (432*a^4*tan(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2))/(432*a^5 + 432*a^3*b
^2 + 864*a^4*b*tan(c/2 + (d*x)/2) + 864*a^2*b^3*tan(c/2 + (d*x)/2)))*(a^2 + b^2)^(1/2))/(b^4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

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))**2,x)

[Out]

Integral(sec(c + d*x)**3/(a*cos(c + d*x) + b*sin(c + d*x))**2, x)

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