3.387 \(\int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=35 \[ \frac{2 i a \sec ^9(c+d x)}{9 d (a+i a \tan (c+d x))^{9/2}} \]

[Out]

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

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Rubi [A]  time = 0.0621879, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {3493} \[ \frac{2 i a \sec ^9(c+d x)}{9 d (a+i a \tan (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((2*I)/9)*a*Sec[c + d*x]^9)/(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]

Rubi steps

\begin{align*} \int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=\frac{2 i a \sec ^9(c+d x)}{9 d (a+i a \tan (c+d x))^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.291017, size = 59, normalized size = 1.69 \[ \frac{2 i (\tan (c+d x)+i) \sec ^7(c+d x)}{9 a^3 d (\tan (c+d x)-i)^3 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((2*I)/9)*Sec[c + d*x]^7*(I + Tan[c + d*x]))/(a^3*d*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.303, size = 115, normalized size = 3.3 \begin{align*}{\frac{32\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+32\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}-40\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-24\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +10\,i\cos \left ( dx+c \right ) +2\,\sin \left ( dx+c \right ) }{9\,{a}^{4}d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}\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))^(7/2),x)

[Out]

2/9/d/a^4*(16*I*cos(d*x+c)^5+16*sin(d*x+c)*cos(d*x+c)^4-20*I*cos(d*x+c)^3-12*cos(d*x+c)^2*sin(d*x+c)+5*I*cos(d
*x+c)+sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^4

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Maxima [B]  time = 1.92944, size = 845, normalized size = 24.14 \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))^(7/2),x, algorithm="maxima")

[Out]

-2/9*(-I*sqrt(a) - 2*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 6*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
- 14*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 42*sqrt
(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 70*sqrt(a)*sin(d*
x + c)^7/(cos(d*x + c) + 1)^7 - 70*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 14*I*sqrt(a)*sin(d*x + c)^10/
(cos(d*x + c) + 1)^10 - 42*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 14*I*sqrt(a)*sin(d*x + c)^12/(cos(d
*x + c) + 1)^12 - 14*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 6*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c)
+ 1)^14 - 2*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 + I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16)*
(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 - 8*a^4*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + 28*a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 56*a^4*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6 + 70*a^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 56*a^4*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a^4*s
in(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a^4*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + a^4*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)^(7/2))

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Fricas [B]  time = 2.09952, size = 281, normalized size = 8.03 \begin{align*} \frac{32 i \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{9 \,{\left (a^{4} d e^{\left (9 i \, d x + 9 i \, c\right )} + 4 \, a^{4} d e^{\left (7 i \, d x + 7 i \, c\right )} + 6 \, a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} 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))^(7/2),x, algorithm="fricas")

[Out]

32/9*I*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(a^4*d*e^(9*I*d*x + 9*I*c) + 4*a^4*d*e^(7*I*d
*x + 7*I*c) + 6*a^4*d*e^(5*I*d*x + 5*I*c) + 4*a^4*d*e^(3*I*d*x + 3*I*c) + a^4*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))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^9/(I*a*tan(d*x + c) + a)^(7/2), x)