3.306 \(\int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=117 \[ \frac{2 i (a+i a \tan (c+d x))^{19/2}}{19 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{17/2}}{17 a^6 d}+\frac{8 i (a+i a \tan (c+d x))^{15/2}}{5 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d} \]

[Out]

(((-16*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^4*d) + (((8*I)/5)*(a + I*a*Tan[c + d*x])^(15/2))/(a^5*d) - (((
12*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(a^6*d) + (((2*I)/19)*(a + I*a*Tan[c + d*x])^(19/2))/(a^7*d)

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Rubi [A]  time = 0.0860853, 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} \[ \frac{2 i (a+i a \tan (c+d x))^{19/2}}{19 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{17/2}}{17 a^6 d}+\frac{8 i (a+i a \tan (c+d x))^{15/2}}{5 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((-16*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^4*d) + (((8*I)/5)*(a + I*a*Tan[c + d*x])^(15/2))/(a^5*d) - (((
12*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(a^6*d) + (((2*I)/19)*(a + I*a*Tan[c + d*x])^(19/2))/(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))^{5/2} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^3 (a+x)^{11/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (8 a^3 (a+x)^{11/2}-12 a^2 (a+x)^{13/2}+6 a (a+x)^{15/2}-(a+x)^{17/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{16 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}+\frac{8 i (a+i a \tan (c+d x))^{15/2}}{5 a^5 d}-\frac{12 i (a+i a \tan (c+d x))^{17/2}}{17 a^6 d}+\frac{2 i (a+i a \tan (c+d x))^{19/2}}{19 a^7 d}\\ \end{align*}

Mathematica [A]  time = 1.41642, size = 113, normalized size = 0.97 \[ -\frac{2 a^2 \sec ^8(c+d x) \sqrt{a+i a \tan (c+d x)} (\cos (6 c+8 d x)+i \sin (6 c+8 d x)) (3262 i \cos (2 (c+d x))+494 \tan (c+d x)+1599 \sin (3 (c+d x)) \sec (c+d x)-833 i)}{20995 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(-2*a^2*Sec[c + d*x]^8*(Cos[6*c + 8*d*x] + I*Sin[6*c + 8*d*x])*(-833*I + (3262*I)*Cos[2*(c + d*x)] + 1599*Sec[
c + d*x]*Sin[3*(c + d*x)] + 494*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(20995*d*(Cos[d*x] + I*Sin[d*x])^2)

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Maple [A]  time = 15.969, size = 171, normalized size = 1.5 \begin{align*} -{\frac{2\,{a}^{2} \left ( 4096\,i \left ( \cos \left ( dx+c \right ) \right ) ^{9}-4096\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{8}+512\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}-2560\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) +224\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}-2016\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+132\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-1716\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -2535\,i\cos \left ( dx+c \right ) +1105\,\sin \left ( dx+c \right ) \right ) }{20995\,d \left ( \cos \left ( dx+c \right ) \right ) ^{9}}\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))^(5/2),x)

[Out]

-2/20995/d*a^2*(4096*I*cos(d*x+c)^9-4096*sin(d*x+c)*cos(d*x+c)^8+512*I*cos(d*x+c)^7-2560*cos(d*x+c)^6*sin(d*x+
c)+224*I*cos(d*x+c)^5-2016*sin(d*x+c)*cos(d*x+c)^4+132*I*cos(d*x+c)^3-1716*cos(d*x+c)^2*sin(d*x+c)-2535*I*cos(
d*x+c)+1105*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^9

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Maxima [A]  time = 1.11067, size = 103, normalized size = 0.88 \begin{align*} \frac{2 i \,{\left (1105 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{19}{2}} - 7410 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{17}{2}} a + 16796 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{15}{2}} a^{2} - 12920 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{13}{2}} a^{3}\right )}}{20995 \, 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))^(5/2),x, algorithm="maxima")

[Out]

2/20995*I*(1105*(I*a*tan(d*x + c) + a)^(19/2) - 7410*(I*a*tan(d*x + c) + a)^(17/2)*a + 16796*(I*a*tan(d*x + c)
+ a)^(15/2)*a^2 - 12920*(I*a*tan(d*x + c) + a)^(13/2)*a^3)/(a^7*d)

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Fricas [B]  time = 2.52358, size = 635, normalized size = 5.43 \begin{align*} \frac{\sqrt{2}{\left (-16384 i \, a^{2} e^{\left (18 i \, d x + 18 i \, c\right )} - 155648 i \, a^{2} e^{\left (16 i \, d x + 16 i \, c\right )} - 661504 i \, a^{2} e^{\left (14 i \, d x + 14 i \, c\right )} - 1653760 i \, a^{2} e^{\left (12 i \, d x + 12 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 )}}{20995 \,{\left (d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, 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))^(5/2),x, algorithm="fricas")

[Out]

1/20995*sqrt(2)*(-16384*I*a^2*e^(18*I*d*x + 18*I*c) - 155648*I*a^2*e^(16*I*d*x + 16*I*c) - 661504*I*a^2*e^(14*
I*d*x + 14*I*c) - 1653760*I*a^2*e^(12*I*d*x + 12*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(d*e^
(18*I*d*x + 18*I*c) + 9*d*e^(16*I*d*x + 16*I*c) + 36*d*e^(14*I*d*x + 14*I*c) + 84*d*e^(12*I*d*x + 12*I*c) + 12
6*d*e^(10*I*d*x + 10*I*c) + 126*d*e^(8*I*d*x + 8*I*c) + 84*d*e^(6*I*d*x + 6*I*c) + 36*d*e^(4*I*d*x + 4*I*c) +
9*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))**(5/2),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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*sec(d*x + c)^8, x)