3.371 \(\int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac{8 i a^2 \sec ^9(c+d x)}{99 d (a+i a \tan (c+d x))^{9/2}}+\frac{2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}} \]

[Out]

(((8*I)/99)*a^2*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((2*I)/11)*a*Sec[c + d*x]^9)/(d*(a + I*a*T
an[c + d*x])^(7/2))

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Rubi [A]  time = 0.125539, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac{8 i a^2 \sec ^9(c+d x)}{99 d (a+i a \tan (c+d x))^{9/2}}+\frac{2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((8*I)/99)*a^2*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((2*I)/11)*a*Sec[c + d*x]^9)/(d*(a + I*a*T
an[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 ^9(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}}+\frac{1}{11} (4 a) \int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\\ &=\frac{8 i a^2 \sec ^9(c+d x)}{99 d (a+i a \tan (c+d x))^{9/2}}+\frac{2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.434218, size = 80, normalized size = 1.1 \[ \frac{2 (9 \tan (c+d x)-13 i) \sec ^7(c+d x) (\cos (2 (c+d x))-i \sin (2 (c+d x)))}{99 a^2 d (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*Sec[c + d*x]^7*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(-13*I + 9*Tan[c + d*x]))/(99*a^2*d*(-I + Tan[c + d*
x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.416, size = 127, normalized size = 1.7 \begin{align*}{\frac{128\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+128\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -16\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+48\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -104\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-64\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +18\,i}{99\,d{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}\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))^(5/2),x)

[Out]

2/99/d/a^3*(64*I*cos(d*x+c)^6+64*cos(d*x+c)^5*sin(d*x+c)-8*I*cos(d*x+c)^4+24*cos(d*x+c)^3*sin(d*x+c)-52*I*cos(
d*x+c)^2-32*cos(d*x+c)*sin(d*x+c)+9*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^5

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Maxima [B]  time = 2.17587, size = 845, normalized size = 11.58 \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))^(5/2),x, algorithm="maxima")

[Out]

-2/99*(-13*I*sqrt(a) - 34*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 46*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c)
+ 1)^2 - 174*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 54*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 -
394*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 22*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 550*sqrt(
a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 550*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 22*I*sqrt(a)*sin(d*
x + c)^10/(cos(d*x + c) + 1)^10 - 394*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 54*I*sqrt(a)*sin(d*x + c
)^12/(cos(d*x + c) + 1)^12 - 174*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 46*I*sqrt(a)*sin(d*x + c)^14/
(cos(d*x + c) + 1)^14 - 34*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 + 13*I*sqrt(a)*sin(d*x + c)^16/(cos(d
*x + c) + 1)^16)*(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
- 8*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 56*a^3*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 + 70*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 56*a^3*sin(d*x + c)^10/(cos(d*x + c) +
1)^10 + 28*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a^3*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + a^3*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)^(5/2))

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Fricas [B]  time = 2.06984, size = 373, normalized size = 5.11 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (704 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 128 i\right )} e^{\left (i \, d x + i \, c\right )}}{99 \,{\left (a^{3} d e^{\left (11 i \, d x + 11 i \, c\right )} + 5 \, a^{3} d e^{\left (9 i \, d x + 9 i \, c\right )} + 10 \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + 10 \, a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )} + 5 \, 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)^9/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/99*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(704*I*e^(2*I*d*x + 2*I*c) + 128*I)*e^(I*d*x + I*c)/(a^3*d*e^(1
1*I*d*x + 11*I*c) + 5*a^3*d*e^(9*I*d*x + 9*I*c) + 10*a^3*d*e^(7*I*d*x + 7*I*c) + 10*a^3*d*e^(5*I*d*x + 5*I*c)
+ 5*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)**9/(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 )^{9}}{{\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)^9/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^9/(I*a*tan(d*x + c) + a)^(5/2), x)