3.385 \(\int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=110 \[ \frac {64 i a^3 \sec ^{13}(c+d x)}{3315 d (a+i a \tan (c+d x))^{13/2}}+\frac {16 i a^2 \sec ^{13}(c+d x)}{255 d (a+i a \tan (c+d x))^{11/2}}+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}} \]

[Out]

64/3315*I*a^3*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(13/2)+16/255*I*a^2*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(11/2)
+2/17*I*a*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(9/2)

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Rubi [A]  time = 0.19, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac {16 i a^2 \sec ^{13}(c+d x)}{255 d (a+i a \tan (c+d x))^{11/2}}+\frac {64 i a^3 \sec ^{13}(c+d x)}{3315 d (a+i a \tan (c+d x))^{13/2}}+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((64*I)/3315)*a^3*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(13/2)) + (((16*I)/255)*a^2*Sec[c + d*x]^13)/(d*
(a + I*a*Tan[c + d*x])^(11/2)) + (((2*I)/17)*a*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(9/2))

Rule 3493

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*
(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rule 3494

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}}+\frac {1}{17} (8 a) \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx\\ &=\frac {16 i a^2 \sec ^{13}(c+d x)}{255 d (a+i a \tan (c+d x))^{11/2}}+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}}+\frac {1}{255} \left (32 a^2\right ) \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{11/2}} \, dx\\ &=\frac {64 i a^3 \sec ^{13}(c+d x)}{3315 d (a+i a \tan (c+d x))^{13/2}}+\frac {16 i a^2 \sec ^{13}(c+d x)}{255 d (a+i a \tan (c+d x))^{11/2}}+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 1.10, size = 92, normalized size = 0.84 \[ -\frac {2 \sec ^{12}(c+d x) (247 i \sin (2 (c+d x))+263 \cos (2 (c+d x))+68) (\cos (3 (c+d x))-i \sin (3 (c+d x)))}{3315 a^3 d (\tan (c+d x)-i)^3 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(-2*Sec[c + d*x]^12*(68 + 263*Cos[2*(c + d*x)] + (247*I)*Sin[2*(c + d*x)])*(Cos[3*(c + d*x)] - I*Sin[3*(c + d*
x)]))/(3315*a^3*d*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.87, size = 173, normalized size = 1.57 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (130560 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 34816 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 4096 i\right )}}{3315 \, {\left (a^{4} d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, a^{4} d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, a^{4} d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/3315*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(130560*I*e^(4*I*d*x + 4*I*c) + 34816*I*e^(2*I*d*x + 2*I*c) +
4096*I)/(a^4*d*e^(16*I*d*x + 16*I*c) + 8*a^4*d*e^(14*I*d*x + 14*I*c) + 28*a^4*d*e^(12*I*d*x + 12*I*c) + 56*a^
4*d*e^(10*I*d*x + 10*I*c) + 70*a^4*d*e^(8*I*d*x + 8*I*c) + 56*a^4*d*e^(6*I*d*x + 6*I*c) + 28*a^4*d*e^(4*I*d*x
+ 4*I*c) + 8*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{13}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^13/(I*a*tan(d*x + c) + a)^(7/2), x)

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maple [A]  time = 6.86, size = 171, normalized size = 1.55 \[ \frac {2 \left (2048 i \left (\cos ^{9}\left (d x +c \right )\right )+2048 \sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )-256 i \left (\cos ^{7}\left (d x +c \right )\right )+768 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )-80 i \left (\cos ^{5}\left (d x +c \right )\right )+560 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-2252 i \left (\cos ^{3}\left (d x +c \right )\right )-1748 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+871 i \cos \left (d x +c \right )+195 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{3315 d \cos \left (d x +c \right )^{8} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

2/3315/d*(2048*I*cos(d*x+c)^9+2048*sin(d*x+c)*cos(d*x+c)^8-256*I*cos(d*x+c)^7+768*sin(d*x+c)*cos(d*x+c)^6-80*I
*cos(d*x+c)^5+560*sin(d*x+c)*cos(d*x+c)^4-2252*I*cos(d*x+c)^3-1748*cos(d*x+c)^2*sin(d*x+c)+871*I*cos(d*x+c)+19
5*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^8/a^4

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maxima [B]  time = 1.08, size = 902, normalized size = 8.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-2/3315*(-331*I*sqrt(a) - 998*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 1838*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 - 7522*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 4836*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c)
+ 1)^4 - 27882*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 8954*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^
6 - 68926*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 12631*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 -
125052*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 10540*I*sqrt(a)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 1
68980*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 168980*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 1
0540*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 125052*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 +
12631*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 68926*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17 +
8954*I*sqrt(a)*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 - 27882*sqrt(a)*sin(d*x + c)^19/(cos(d*x + c) + 1)^19 + 4
836*I*sqrt(a)*sin(d*x + c)^20/(cos(d*x + c) + 1)^20 - 7522*sqrt(a)*sin(d*x + c)^21/(cos(d*x + c) + 1)^21 + 183
8*I*sqrt(a)*sin(d*x + c)^22/(cos(d*x + c) + 1)^22 - 998*sqrt(a)*sin(d*x + c)^23/(cos(d*x + c) + 1)^23 + 331*I*
sqrt(a)*sin(d*x + c)^24/(cos(d*x + c) + 1)^24)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(sin(d*x + c)/(cos(
d*x + c) + 1) - 1)^(7/2)/((a^4 - 12*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 66*a^4*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 - 220*a^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 495*a^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 792*a
^4*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 924*a^4*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 792*a^4*sin(d*x + c
)^14/(cos(d*x + c) + 1)^14 + 495*a^4*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 220*a^4*sin(d*x + c)^18/(cos(d*x
+ c) + 1)^18 + 66*a^4*sin(d*x + c)^20/(cos(d*x + c) + 1)^20 - 12*a^4*sin(d*x + c)^22/(cos(d*x + c) + 1)^22 + a
^4*sin(d*x + c)^24/(cos(d*x + c) + 1)^24)*d*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 - 1)^(7/2))

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mupad [B]  time = 9.28, size = 105, normalized size = 0.95 \[ \frac {512\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,68{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,255{}\mathrm {i}+8{}\mathrm {i}\right )}{3315\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^13*(a + a*tan(c + d*x)*1i)^(7/2)),x)

[Out]

(512*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*(exp(c*2i +
d*x*2i)*68i + exp(c*4i + d*x*4i)*255i + 8i))/(3315*a^4*d*(exp(c*2i + d*x*2i) + 1)^8)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**13/(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

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