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

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

[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))

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Rubi [A]  time = 0.265408, antiderivative size = 147, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.952615, 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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Maple [A]  time = 15.287, size = 181, normalized size = 1.2 \begin{align*}{\frac{16384\,i \left ( \cos \left ( dx+c \right ) \right ) ^{10}+16384\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{9}-2048\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+6144\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}-640\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+4480\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -336\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+3696\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -10712\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-7280\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +2210\,i}{20995\,d{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{9}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

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/a^3*(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

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Maxima [B]  time = 2.87335, size = 1218, normalized size = 8.29 \begin{align*} \text{result too large to display} \end{align*}

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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Fricas [A]  time = 2.24708, size = 649, normalized size = 4.41 \begin{align*} \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 )} e^{\left (i \, d x + i \, c\right )}}{20995 \,{\left (a^{3} d e^{\left (19 i \, d x + 19 i \, c\right )} + 9 \, a^{3} d e^{\left (17 i \, d x + 17 i \, c\right )} + 36 \, a^{3} d e^{\left (15 i \, d x + 15 i \, c\right )} + 84 \, a^{3} d e^{\left (13 i \, d x + 13 i \, c\right )} + 126 \, a^{3} d e^{\left (11 i \, d x + 11 i \, c\right )} + 126 \, a^{3} d e^{\left (9 i \, d x + 9 i \, c\right )} + 84 \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + 36 \, a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )} + 9 \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

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)*e^(I*d*x + I*c)/(a^3*d*e^(19*I*d*x + 19*I*c) + 9*a^3*d*e^(17*I*d*x
+ 17*I*c) + 36*a^3*d*e^(15*I*d*x + 15*I*c) + 84*a^3*d*e^(13*I*d*x + 13*I*c) + 126*a^3*d*e^(11*I*d*x + 11*I*c)
+ 126*a^3*d*e^(9*I*d*x + 9*I*c) + 84*a^3*d*e^(7*I*d*x + 7*I*c) + 36*a^3*d*e^(5*I*d*x + 5*I*c) + 9*a^3*d*e^(3*
I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)