3.355 \(\int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=110 \[ \frac{16 i a^2 \sec ^9(c+d x)}{143 d (a+i a \tan (c+d x))^{7/2}}+\frac{64 i a^3 \sec ^9(c+d x)}{1287 d (a+i a \tan (c+d x))^{9/2}}+\frac{2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(((64*I)/1287)*a^3*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((16*I)/143)*a^2*Sec[c + d*x]^9)/(d*(a
+ I*a*Tan[c + d*x])^(7/2)) + (((2*I)/13)*a*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(5/2))

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Rubi [A]  time = 0.191073, antiderivative size = 110, normalized size of antiderivative = 1., 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} \[ \frac{16 i a^2 \sec ^9(c+d x)}{143 d (a+i a \tan (c+d x))^{7/2}}+\frac{64 i a^3 \sec ^9(c+d x)}{1287 d (a+i a \tan (c+d x))^{9/2}}+\frac{2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((64*I)/1287)*a^3*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((16*I)/143)*a^2*Sec[c + d*x]^9)/(d*(a
+ I*a*Tan[c + d*x])^(7/2)) + (((2*I)/13)*a*Sec[c + d*x]^9)/(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 ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{13} (8 a) \int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac{16 i a^2 \sec ^9(c+d x)}{143 d (a+i a \tan (c+d x))^{7/2}}+\frac{2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{143} \left (32 a^2\right ) \int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\\ &=\frac{64 i a^3 \sec ^9(c+d x)}{1287 d (a+i a \tan (c+d x))^{9/2}}+\frac{16 i a^2 \sec ^9(c+d x)}{143 d (a+i a \tan (c+d x))^{7/2}}+\frac{2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.404781, size = 92, normalized size = 0.84 \[ \frac{2 \sec ^8(c+d x) (135 i \sin (2 (c+d x))+151 \cos (2 (c+d x))+52) (\cos (3 (c+d x))-i \sin (3 (c+d x)))}{1287 a d (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*Sec[c + d*x]^8*(52 + 151*Cos[2*(c + d*x)] + (135*I)*Sin[2*(c + d*x)])*(Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)
]))/(1287*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.342, size = 144, normalized size = 1.3 \begin{align*}{\frac{1024\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}+1024\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) -128\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+384\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}-40\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+280\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -450\,i\cos \left ( dx+c \right ) -198\,\sin \left ( dx+c \right ) }{1287\,{a}^{2}d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}\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)^9/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

2/1287/d/a^2*(512*I*cos(d*x+c)^7+512*cos(d*x+c)^6*sin(d*x+c)-64*I*cos(d*x+c)^5+192*sin(d*x+c)*cos(d*x+c)^4-20*
I*cos(d*x+c)^3+140*cos(d*x+c)^2*sin(d*x+c)-225*I*cos(d*x+c)-99*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*
x+c))^(1/2)/cos(d*x+c)^6

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Maxima [B]  time = 2.36937, size = 845, normalized size = 7.68 \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)^9/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/1287*(-203*I*sqrt(a) - 678*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 - 1802*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 26*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^
4 - 3614*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 858*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 657
8*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 6578*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 858*I*sqrt(
a)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 3614*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 26*I*sqrt(a)*s
in(d*x + c)^12/(cos(d*x + c) + 1)^12 - 1802*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 2*I*sqrt(a)*sin(d*
x + c)^14/(cos(d*x + c) + 1)^14 - 678*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 + 203*I*sqrt(a)*sin(d*x +
c)^16/(cos(d*x + c) + 1)^16)*(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 - 8*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 56*a^2
*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 56*a^2*sin(d*x + c)^10/(co
s(d*x + c) + 1)^10 + 28*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a^2*sin(d*x + c)^14/(cos(d*x + c) + 1)^1
4 + a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16)*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 = 2.31033, size = 463, normalized size = 4.21 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (18304 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 6656 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1024 i\right )} e^{\left (i \, d x + i \, c\right )}}{1287 \,{\left (a^{2} d e^{\left (13 i \, d x + 13 i \, c\right )} + 6 \, a^{2} d e^{\left (11 i \, d x + 11 i \, c\right )} + 15 \, a^{2} d e^{\left (9 i \, d x + 9 i \, c\right )} + 20 \, a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 15 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 6 \, 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)^9/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/1287*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(18304*I*e^(4*I*d*x + 4*I*c) + 6656*I*e^(2*I*d*x + 2*I*c) + 1
024*I)*e^(I*d*x + I*c)/(a^2*d*e^(13*I*d*x + 13*I*c) + 6*a^2*d*e^(11*I*d*x + 11*I*c) + 15*a^2*d*e^(9*I*d*x + 9*
I*c) + 20*a^2*d*e^(7*I*d*x + 7*I*c) + 15*a^2*d*e^(5*I*d*x + 5*I*c) + 6*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)**9/(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 )^{9}}{{\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)^9/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^9/(I*a*tan(d*x + c) + a)^(3/2), x)