3.347 \(\int \frac{\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^7 d}-\frac{4 i (a+i a \tan (c+d x))^{9/2}}{3 a^6 d}+\frac{24 i (a+i a \tan (c+d x))^{7/2}}{7 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d} \]

[Out]

(((-16*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^4*d) + (((24*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^5*d) - (((4*
I)/3)*(a + I*a*Tan[c + d*x])^(9/2))/(a^6*d) + (((2*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^7*d)

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Rubi [A]  time = 0.0869304, antiderivative size = 117, 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 = {3487, 43} \[ \frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^7 d}-\frac{4 i (a+i a \tan (c+d x))^{9/2}}{3 a^6 d}+\frac{24 i (a+i a \tan (c+d x))^{7/2}}{7 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((-16*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^4*d) + (((24*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^5*d) - (((4*
I)/3)*(a + I*a*Tan[c + d*x])^(9/2))/(a^6*d) + (((2*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^7*d)

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^3 (a+x)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (8 a^3 (a+x)^{3/2}-12 a^2 (a+x)^{5/2}+6 a (a+x)^{7/2}-(a+x)^{9/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{16 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d}+\frac{24 i (a+i a \tan (c+d x))^{7/2}}{7 a^5 d}-\frac{4 i (a+i a \tan (c+d x))^{9/2}}{3 a^6 d}+\frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^7 d}\\ \end{align*}

Mathematica [A]  time = 0.669403, size = 110, normalized size = 0.94 \[ \frac{2 i \sec ^6(c+d x) (\cos (4 (c+d x))+i \sin (4 (c+d x))) (494 i \cos (2 (c+d x))+110 \tan (c+d x)+215 \sin (3 (c+d x)) \sec (c+d x)+39 i)}{1155 a d (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((2*I)/1155)*Sec[c + d*x]^6*(Cos[4*(c + d*x)] + I*Sin[4*(c + d*x)])*(39*I + (494*I)*Cos[2*(c + d*x)] + 215*Se
c[c + d*x]*Sin[3*(c + d*x)] + 110*Tan[c + d*x]))/(a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.318, size = 117, normalized size = 1. \begin{align*} -{\frac{512\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}-512\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+64\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-320\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +490\,i\cos \left ( dx+c \right ) +210\,\sin \left ( dx+c \right ) }{1155\,{a}^{2}d \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)^8/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-2/1155/d/a^2*(256*I*cos(d*x+c)^5-256*sin(d*x+c)*cos(d*x+c)^4+32*I*cos(d*x+c)^3-160*cos(d*x+c)^2*sin(d*x+c)+24
5*I*cos(d*x+c)+105*sin(d*x+c))*(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 [A]  time = 1.10234, size = 103, normalized size = 0.88 \begin{align*} \frac{2 i \,{\left (105 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} - 770 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a + 1980 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{2} - 1848 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{3}\right )}}{1155 \, a^{7} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

2/1155*I*(105*(I*a*tan(d*x + c) + a)^(11/2) - 770*(I*a*tan(d*x + c) + a)^(9/2)*a + 1980*(I*a*tan(d*x + c) + a)
^(7/2)*a^2 - 1848*(I*a*tan(d*x + c) + a)^(5/2)*a^3)/(a^7*d)

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Fricas [A]  time = 2.06482, size = 468, normalized size = 4. \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-1024 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 5632 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 12672 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 14784 i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} e^{\left (i \, d x + i \, c\right )}}{1155 \,{\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/1155*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-1024*I*e^(10*I*d*x + 10*I*c) - 5632*I*e^(8*I*d*x + 8*I*c) -
12672*I*e^(6*I*d*x + 6*I*c) - 14784*I*e^(4*I*d*x + 4*I*c))*e^(I*d*x + I*c)/(a^2*d*e^(10*I*d*x + 10*I*c) + 5*a
^2*d*e^(8*I*d*x + 8*I*c) + 10*a^2*d*e^(6*I*d*x + 6*I*c) + 10*a^2*d*e^(4*I*d*x + 4*I*c) + 5*a^2*d*e^(2*I*d*x +
2*I*c) + a^2*d)

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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)**8/(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 )^{8}}{{\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)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^8/(I*a*tan(d*x + c) + a)^(3/2), x)