3.369 \(\int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac {256 i a^4 \sec ^{13}(c+d x)}{20995 d (a+i a \tan (c+d x))^{13/2}}+\frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}} \]

[Out]

256/20995*I*a^4*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(13/2)+64/1615*I*a^3*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(11
/2)+24/323*I*a^2*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(9/2)+2/19*I*a*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(7/2)

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Rubi [A]  time = 0.27, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {256 i a^4 \sec ^{13}(c+d x)}{20995 d (a+i a \tan (c+d x))^{13/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((256*I)/20995)*a^4*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(13/2)) + (((64*I)/1615)*a^3*Sec[c + d*x]^13)/
(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((24*I)/323)*a^2*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((2
*I)/19)*a*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(7/2))

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))^{5/2}} \, dx &=\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}+\frac {1}{19} (12 a) \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\\ &=\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}+\frac {1}{323} \left (96 a^2\right ) \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx\\ &=\frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}+\frac {\left (128 a^3\right ) \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{11/2}} \, dx}{1615}\\ &=\frac {256 i a^4 \sec ^{13}(c+d x)}{20995 d (a+i a \tan (c+d x))^{13/2}}+\frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.11, size = 112, normalized size = 0.76 \[ \frac {\sec ^{12}(c+d x) (13 i (38 \sin (c+d x)+123 \sin (3 (c+d x)))+798 \cos (c+d x)+1631 \cos (3 (c+d x))) (-2 \sin (4 (c+d x))-2 i \cos (4 (c+d x)))}{20995 a^2 d (\tan (c+d x)-i)^2 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(Sec[c + d*x]^12*(798*Cos[c + d*x] + 1631*Cos[3*(c + d*x)] + (13*I)*(38*Sin[c + d*x] + 123*Sin[3*(c + d*x)]))*
((-2*I)*Cos[4*(c + d*x)] - 2*Sin[4*(c + d*x)]))/(20995*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 1.68, size = 199, normalized size = 1.35 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (1653760 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 661504 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 155648 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 16384 i\right )}}{20995 \, {\left (a^{3} d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, a^{3} d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/20995*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(1653760*I*e^(6*I*d*x + 6*I*c) + 661504*I*e^(4*I*d*x + 4*I*c
) + 155648*I*e^(2*I*d*x + 2*I*c) + 16384*I)/(a^3*d*e^(18*I*d*x + 18*I*c) + 9*a^3*d*e^(16*I*d*x + 16*I*c) + 36*
a^3*d*e^(14*I*d*x + 14*I*c) + 84*a^3*d*e^(12*I*d*x + 12*I*c) + 126*a^3*d*e^(10*I*d*x + 10*I*c) + 126*a^3*d*e^(
8*I*d*x + 8*I*c) + 84*a^3*d*e^(6*I*d*x + 6*I*c) + 36*a^3*d*e^(4*I*d*x + 4*I*c) + 9*a^3*d*e^(2*I*d*x + 2*I*c) +
a^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{13}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^13/(I*a*tan(d*x + c) + a)^(5/2), x)

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maple [A]  time = 19.23, size = 181, normalized size = 1.23 \[ \frac {2 \left (8192 i \left (\cos ^{10}\left (d x +c \right )\right )+8192 \sin \left (d x +c \right ) \left (\cos ^{9}\left (d x +c \right )\right )-1024 i \left (\cos ^{8}\left (d x +c \right )\right )+3072 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )-320 i \left (\cos ^{6}\left (d x +c \right )\right )+2240 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-168 i \left (\cos ^{4}\left (d x +c \right )\right )+1848 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5356 i \left (\cos ^{2}\left (d x +c \right )\right )-3640 \cos \left (d x +c \right ) \sin \left (d x +c \right )+1105 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{20995 d \cos \left (d x +c \right )^{9} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

2/20995/d*(8192*I*cos(d*x+c)^10+8192*sin(d*x+c)*cos(d*x+c)^9-1024*I*cos(d*x+c)^8+3072*sin(d*x+c)*cos(d*x+c)^7-
320*I*cos(d*x+c)^6+2240*cos(d*x+c)^5*sin(d*x+c)-168*I*cos(d*x+c)^4+1848*cos(d*x+c)^3*sin(d*x+c)-5356*I*cos(d*x
+c)^2-3640*cos(d*x+c)*sin(d*x+c)+1105*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^9/a^3

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maxima [B]  time = 1.14, size = 902, normalized size = 6.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/20995*(-2429*I*sqrt(a) - 8850*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 5122*I*sqrt(a)*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 - 45190*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 12924*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x
+ c) + 1)^4 - 152478*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 40470*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c
) + 1)^6 - 397594*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 50065*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) +
1)^8 - 722228*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 19380*I*sqrt(a)*sin(d*x + c)^10/(cos(d*x + c) + 1
)^10 - 936700*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 936700*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1
)^13 + 19380*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 722228*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) +
1)^15 + 50065*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 397594*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) +
1)^17 + 40470*I*sqrt(a)*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 - 152478*sqrt(a)*sin(d*x + c)^19/(cos(d*x + c)
+ 1)^19 + 12924*I*sqrt(a)*sin(d*x + c)^20/(cos(d*x + c) + 1)^20 - 45190*sqrt(a)*sin(d*x + c)^21/(cos(d*x + c)
+ 1)^21 + 5122*I*sqrt(a)*sin(d*x + c)^22/(cos(d*x + c) + 1)^22 - 8850*sqrt(a)*sin(d*x + c)^23/(cos(d*x + c) +
1)^23 + 2429*I*sqrt(a)*sin(d*x + c)^24/(cos(d*x + c) + 1)^24)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(sin
(d*x + c)/(cos(d*x + c) + 1) - 1)^(5/2)/((a^3 - 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 66*a^3*sin(d*x +
c)^4/(cos(d*x + c) + 1)^4 - 220*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 495*a^3*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 - 792*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 924*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 792*
a^3*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + 495*a^3*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 220*a^3*sin(d*x +
c)^18/(cos(d*x + c) + 1)^18 + 66*a^3*sin(d*x + c)^20/(cos(d*x + c) + 1)^20 - 12*a^3*sin(d*x + c)^22/(cos(d*x +
c) + 1)^22 + a^3*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)^(5/2))

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mupad [B]  time = 11.59, size = 301, normalized size = 2.05 \[ \frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{13\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{5\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,3072{}\mathrm {i}}{17\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{19\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^9} \]

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)^(5/2)),x)

[Out]

(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)*1024i)/(13*a^3*
d*(exp(c*2i + d*x*2i) + 1)^6) - (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)*1024i)/(5*a^3*d*(exp(c*2i + d*x*2i) + 1)^7) + (exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2
i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*3072i)/(17*a^3*d*(exp(c*2i + d*x*2i) + 1)^8) - (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)*1024i)/(19*a^3*d*(exp(c*2i + d
*x*2i) + 1)^9)

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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))**(5/2),x)

[Out]

Timed out

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