3.386 \(\int \frac{\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac{8 i a^2 \sec ^{11}(c+d x)}{143 d (a+i a \tan (c+d x))^{11/2}}+\frac{2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}} \]

[Out]

(((8*I)/143)*a^2*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((2*I)/13)*a*Sec[c + d*x]^11)/(d*(a + I
*a*Tan[c + d*x])^(9/2))

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Rubi [A]  time = 0.127292, 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} \[ \frac{8 i a^2 \sec ^{11}(c+d x)}{143 d (a+i a \tan (c+d x))^{11/2}}+\frac{2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((8*I)/143)*a^2*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((2*I)/13)*a*Sec[c + d*x]^11)/(d*(a + I
*a*Tan[c + d*x])^(9/2))

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))^{7/2}} \, dx &=\frac{2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}+\frac{1}{13} (4 a) \int \frac{\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx\\ &=\frac{8 i a^2 \sec ^{11}(c+d x)}{143 d (a+i a \tan (c+d x))^{11/2}}+\frac{2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.418194, size = 82, normalized size = 1.12 \[ -\frac{2 i (11 \tan (c+d x)-15 i) \sec ^9(c+d x) (\cos (2 (c+d x))-i \sin (2 (c+d x)))}{143 a^3 d (\tan (c+d x)-i)^3 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((-2*I)/143)*Sec[c + d*x]^9*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(-15*I + 11*Tan[c + d*x]))/(a^3*d*(-I + T
an[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.37, size = 144, normalized size = 2. \begin{align*}{\frac{256\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}+256\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) -32\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+96\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}-296\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-216\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +102\,i\cos \left ( dx+c \right ) +22\,\sin \left ( dx+c \right ) }{143\,{a}^{4}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)^11/(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

2/143/d/a^4*(128*I*cos(d*x+c)^7+128*cos(d*x+c)^6*sin(d*x+c)-16*I*cos(d*x+c)^5+48*sin(d*x+c)*cos(d*x+c)^4-148*I
*cos(d*x+c)^3-108*cos(d*x+c)^2*sin(d*x+c)+51*I*cos(d*x+c)+11*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 = 2.1739, size = 1031, normalized size = 14.12 \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))^(7/2),x, algorithm="maxima")

[Out]

-2/143*(-15*I*sqrt(a) - 38*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 88*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c)
+ 1)^2 - 278*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 213*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- 920*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 272*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 1848*s
qrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 182*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2548*sqrt(a)*
sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 2548*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 182*I*sqrt(a)*sin(d
*x + c)^12/(cos(d*x + c) + 1)^12 - 1848*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 272*I*sqrt(a)*sin(d*x
+ c)^14/(cos(d*x + c) + 1)^14 - 920*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 + 213*I*sqrt(a)*sin(d*x + c)
^16/(cos(d*x + c) + 1)^16 - 278*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17 + 88*I*sqrt(a)*sin(d*x + c)^18/(
cos(d*x + c) + 1)^18 - 38*sqrt(a)*sin(d*x + c)^19/(cos(d*x + c) + 1)^19 + 15*I*sqrt(a)*sin(d*x + c)^20/(cos(d*
x + c) + 1)^20)*(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
- 10*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 45*a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 120*a^4*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 + 210*a^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 252*a^4*sin(d*x + c)^10/(cos(d*x + c
) + 1)^10 + 210*a^4*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 120*a^4*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + 45
*a^4*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 10*a^4*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 + a^4*sin(d*x + c)^2
0/(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)^(7
/2))

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Fricas [B]  time = 2.08415, size = 420, normalized size = 5.75 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (1664 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 256 i\right )} e^{\left (i \, d x + i \, c\right )}}{143 \,{\left (a^{4} d e^{\left (13 i \, d x + 13 i \, c\right )} + 6 \, a^{4} d e^{\left (11 i \, d x + 11 i \, c\right )} + 15 \, a^{4} d e^{\left (9 i \, d x + 9 i \, c\right )} + 20 \, a^{4} d e^{\left (7 i \, d x + 7 i \, c\right )} + 15 \, a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 6 \, 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)^11/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/143*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(1664*I*e^(2*I*d*x + 2*I*c) + 256*I)*e^(I*d*x + I*c)/(a^4*d*e^
(13*I*d*x + 13*I*c) + 6*a^4*d*e^(11*I*d*x + 11*I*c) + 15*a^4*d*e^(9*I*d*x + 9*I*c) + 20*a^4*d*e^(7*I*d*x + 7*I
*c) + 15*a^4*d*e^(5*I*d*x + 5*I*c) + 6*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)**11/(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 )^{11}}{{\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)^11/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^11/(I*a*tan(d*x + c) + a)^(7/2), x)