3.354 \(\int \frac{\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac{8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac{64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac{256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac{2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(((256*I)/12155)*a^4*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((64*I)/1105)*a^3*Sec[c + d*x]^11)/
(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((8*I)/85)*a^2*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((2*I)
/17)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(5/2))

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Rubi [A]  time = 0.262362, 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} \[ \frac{8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac{64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac{256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac{2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((256*I)/12155)*a^4*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((64*I)/1105)*a^3*Sec[c + d*x]^11)/
(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((8*I)/85)*a^2*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((2*I)
/17)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(5/2))

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 ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{17} (12 a) \int \frac{\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac{8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac{2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{85} \left (32 a^2\right ) \int \frac{\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\\ &=\frac{64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac{8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac{2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac{\left (128 a^3\right ) \int \frac{\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx}{1105}\\ &=\frac{256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac{64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac{8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac{2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.770109, size = 108, normalized size = 0.73 \[ \frac{2 \sec ^9(c+d x) (\sin (4 (c+d x))+i \cos (4 (c+d x))) (-2242 i \cos (2 (c+d x))+374 \tan (c+d x)+1089 \sin (3 (c+d x)) \sec (c+d x)+475 i)}{12155 a d (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*Sec[c + d*x]^9*(I*Cos[4*(c + d*x)] + Sin[4*(c + d*x)])*(475*I - (2242*I)*Cos[2*(c + d*x)] + 1089*Sec[c + d*
x]*Sin[3*(c + d*x)] + 374*Tan[c + d*x]))/(12155*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 3.044, size = 171, normalized size = 1.2 \begin{align*}{\frac{8192\,i \left ( \cos \left ( dx+c \right ) \right ) ^{9}+8192\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{8}-1024\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}+3072\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) -320\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+2240\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}-168\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1848\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -3146\,i\cos \left ( dx+c \right ) -1430\,\sin \left ( dx+c \right ) }{12155\,{a}^{2}d \left ( \cos \left ( dx+c \right ) \right ) ^{8}}\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)^11/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

2/12155/d/a^2*(4096*I*cos(d*x+c)^9+4096*sin(d*x+c)*cos(d*x+c)^8-512*I*cos(d*x+c)^7+1536*cos(d*x+c)^6*sin(d*x+c
)-160*I*cos(d*x+c)^5+1120*sin(d*x+c)*cos(d*x+c)^4-84*I*cos(d*x+c)^3+924*cos(d*x+c)^2*sin(d*x+c)-1573*I*cos(d*x
+c)-715*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

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Maxima [B]  time = 2.40486, size = 1031, normalized size = 7.01 \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)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/12155*(-1767*I*sqrt(a) - 6854*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) + 2088*I*sqrt(a)*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 - 16438*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 5661*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 - 56984*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 13328*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6 - 129336*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7514*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8 - 156468*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 156468*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11
- 7514*I*sqrt(a)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 129336*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13
+ 13328*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 56984*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15
+ 5661*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 16438*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17
- 2088*I*sqrt(a)*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 - 6854*sqrt(a)*sin(d*x + c)^19/(cos(d*x + c) + 1)^19 +
1767*I*sqrt(a)*sin(d*x + c)^20/(cos(d*x + c) + 1)^20)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(sin(d*x + c
)/(cos(d*x + c) + 1) - 1)^(3/2)/((a^2 - 10*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 45*a^2*sin(d*x + c)^4/(co
s(d*x + c) + 1)^4 - 120*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 210*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8
- 252*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 210*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 120*a^2*sin(
d*x + c)^14/(cos(d*x + c) + 1)^14 + 45*a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 10*a^2*sin(d*x + c)^18/(cos
(d*x + c) + 1)^18 + a^2*sin(d*x + c)^20/(cos(d*x + c) + 1)^20)*d*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d
*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(3/2))

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Fricas [A]  time = 2.09942, size = 598, normalized size = 4.07 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (565760 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 261120 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 69632 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8192 i\right )} e^{\left (i \, d x + i \, c\right )}}{12155 \,{\left (a^{2} d e^{\left (17 i \, d x + 17 i \, c\right )} + 8 \, a^{2} d e^{\left (15 i \, d x + 15 i \, c\right )} + 28 \, a^{2} d e^{\left (13 i \, d x + 13 i \, c\right )} + 56 \, a^{2} d e^{\left (11 i \, d x + 11 i \, c\right )} + 70 \, a^{2} d e^{\left (9 i \, d x + 9 i \, c\right )} + 56 \, a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 28 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 8 \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} 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)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/12155*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(565760*I*e^(6*I*d*x + 6*I*c) + 261120*I*e^(4*I*d*x + 4*I*c)
+ 69632*I*e^(2*I*d*x + 2*I*c) + 8192*I)*e^(I*d*x + I*c)/(a^2*d*e^(17*I*d*x + 17*I*c) + 8*a^2*d*e^(15*I*d*x +
15*I*c) + 28*a^2*d*e^(13*I*d*x + 13*I*c) + 56*a^2*d*e^(11*I*d*x + 11*I*c) + 70*a^2*d*e^(9*I*d*x + 9*I*c) + 56*
a^2*d*e^(7*I*d*x + 7*I*c) + 28*a^2*d*e^(5*I*d*x + 5*I*c) + 8*a^2*d*e^(3*I*d*x + 3*I*c) + a^2*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)**11/(a+I*a*tan(d*x+c))**(3/2),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 )^{11}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^11/(I*a*tan(d*x + c) + a)^(3/2), x)