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

Optimal. Leaf size=117 \[ \frac{2 i (a+i a \tan (c+d x))^{17/2}}{17 a^7 d}-\frac{4 i (a+i a \tan (c+d x))^{15/2}}{5 a^6 d}+\frac{24 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d} \]

[Out]

(((-16*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^4*d) + (((24*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^5*d) - (
((4*I)/5)*(a + I*a*Tan[c + d*x])^(15/2))/(a^6*d) + (((2*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(a^7*d)

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Rubi [A]  time = 0.0893523, 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} \[ \frac{2 i (a+i a \tan (c+d x))^{17/2}}{17 a^7 d}-\frac{4 i (a+i a \tan (c+d x))^{15/2}}{5 a^6 d}+\frac{24 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-16*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^4*d) + (((24*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^5*d) - (
((4*I)/5)*(a + I*a*Tan[c + d*x])^(15/2))/(a^6*d) + (((2*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(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 \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)^{9/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (8 a^3 (a+x)^{9/2}-12 a^2 (a+x)^{11/2}+6 a (a+x)^{13/2}-(a+x)^{15/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{16 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d}+\frac{24 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}-\frac{4 i (a+i a \tan (c+d x))^{15/2}}{5 a^6 d}+\frac{2 i (a+i a \tan (c+d x))^{17/2}}{17 a^7 d}\\ \end{align*}

Mathematica [A]  time = 1.12643, size = 111, normalized size = 0.95 \[ \frac{2 a \sec ^8(c+d x) (\cos (d x)-i \sin (d x)) \sqrt{a+i a \tan (c+d x)} (-11 i (34 \sin (c+d x)+99 \sin (3 (c+d x)))+646 \cos (c+d x)+1121 \cos (3 (c+d x))) (\sin (5 c+6 d x)-i \cos (5 c+6 d x))}{12155 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sec[c + d*x]^8*(Cos[d*x] - I*Sin[d*x])*(646*Cos[c + d*x] + 1121*Cos[3*(c + d*x)] - (11*I)*(34*Sin[c + d*x
] + 99*Sin[3*(c + d*x)]))*((-I)*Cos[5*c + 6*d*x] + Sin[5*c + 6*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(12155*d)

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Maple [A]  time = 5.216, size = 152, normalized size = 1.3 \begin{align*} -{\frac{2\,a \left ( 2048\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}-2048\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+256\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-1280\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +112\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-1008\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +66\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-858\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -715\,i \right ) }{12155\,d \left ( \cos \left ( dx+c \right ) \right ) ^{8}}\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]

-2/12155/d*a*(2048*I*cos(d*x+c)^8-2048*sin(d*x+c)*cos(d*x+c)^7+256*I*cos(d*x+c)^6-1280*cos(d*x+c)^5*sin(d*x+c)
+112*I*cos(d*x+c)^4-1008*cos(d*x+c)^3*sin(d*x+c)+66*I*cos(d*x+c)^2-858*cos(d*x+c)*sin(d*x+c)-715*I)*(a*(I*sin(
d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^8

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Maxima [A]  time = 1.04075, size = 103, normalized size = 0.88 \begin{align*} \frac{2 i \,{\left (715 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{17}{2}} - 4862 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{15}{2}} a + 11220 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{13}{2}} a^{2} - 8840 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a^{3}\right )}}{12155 \, 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]

2/12155*I*(715*(I*a*tan(d*x + c) + a)^(17/2) - 4862*(I*a*tan(d*x + c) + a)^(15/2)*a + 11220*(I*a*tan(d*x + c)
+ a)^(13/2)*a^2 - 8840*(I*a*tan(d*x + c) + a)^(11/2)*a^3)/(a^7*d)

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Fricas [B]  time = 2.59432, size = 578, normalized size = 4.94 \begin{align*} \frac{\sqrt{2}{\left (-8192 i \, a e^{\left (16 i \, d x + 16 i \, c\right )} - 69632 i \, a e^{\left (14 i \, d x + 14 i \, c\right )} - 261120 i \, a e^{\left (12 i \, d x + 12 i \, c\right )} - 565760 i \, a e^{\left (10 i \, d x + 10 i \, c\right )}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{12155 \,{\left (d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 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]

1/12155*sqrt(2)*(-8192*I*a*e^(16*I*d*x + 16*I*c) - 69632*I*a*e^(14*I*d*x + 14*I*c) - 261120*I*a*e^(12*I*d*x +
12*I*c) - 565760*I*a*e^(10*I*d*x + 10*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(d*e^(16*I*d*x +
16*I*c) + 8*d*e^(14*I*d*x + 14*I*c) + 28*d*e^(12*I*d*x + 12*I*c) + 56*d*e^(10*I*d*x + 10*I*c) + 70*d*e^(8*I*d
*x + 8*I*c) + 56*d*e^(6*I*d*x + 6*I*c) + 28*d*e^(4*I*d*x + 4*I*c) + 8*d*e^(2*I*d*x + 2*I*c) + 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{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )^{8}\,{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((I*a*tan(d*x + c) + a)^(3/2)*sec(d*x + c)^8, x)