3.362 \(\int \frac{\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=146 \[ -\frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^9 d}+\frac{16 i (a+i a \tan (c+d x))^{11/2}}{11 a^8 d}-\frac{16 i (a+i a \tan (c+d x))^{9/2}}{3 a^7 d}+\frac{64 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}-\frac{32 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d} \]

[Out]

(((-32*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^5*d) + (((64*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^6*d) - (((16
*I)/3)*(a + I*a*Tan[c + d*x])^(9/2))/(a^7*d) + (((16*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^8*d) - (((2*I)/1
3)*(a + I*a*Tan[c + d*x])^(13/2))/(a^9*d)

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Rubi [A]  time = 0.0931937, antiderivative size = 146, 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} \[ -\frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^9 d}+\frac{16 i (a+i a \tan (c+d x))^{11/2}}{11 a^8 d}-\frac{16 i (a+i a \tan (c+d x))^{9/2}}{3 a^7 d}+\frac{64 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}-\frac{32 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((-32*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^5*d) + (((64*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^6*d) - (((16
*I)/3)*(a + I*a*Tan[c + d*x])^(9/2))/(a^7*d) + (((16*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^8*d) - (((2*I)/1
3)*(a + I*a*Tan[c + d*x])^(13/2))/(a^9*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 ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^4 (a+x)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (16 a^4 (a+x)^{3/2}-32 a^3 (a+x)^{5/2}+24 a^2 (a+x)^{7/2}-8 a (a+x)^{9/2}+(a+x)^{11/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac{32 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d}+\frac{64 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}-\frac{16 i (a+i a \tan (c+d x))^{9/2}}{3 a^7 d}+\frac{16 i (a+i a \tan (c+d x))^{11/2}}{11 a^8 d}-\frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^9 d}\\ \end{align*}

Mathematica [A]  time = 0.692974, size = 116, normalized size = 0.79 \[ \frac{2 \sec ^9(c+d x) (2600 \sin (2 (c+d x))+2875 \sin (4 (c+d x))+4264 i \cos (2 (c+d x))+3131 i \cos (4 (c+d x))+2288 i) (\cos (5 (c+d x))+i \sin (5 (c+d x)))}{15015 a^2 d (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*Sec[c + d*x]^9*(2288*I + (4264*I)*Cos[2*(c + d*x)] + (3131*I)*Cos[4*(c + d*x)] + 2600*Sin[2*(c + d*x)] + 28
75*Sin[4*(c + d*x)])*(Cos[5*(c + d*x)] + I*Sin[5*(c + d*x)]))/(15015*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*
Tan[c + d*x]])

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Maple [A]  time = 0.349, size = 127, normalized size = 0.9 \begin{align*}{\frac{-8192\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+8192\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -1024\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+5120\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -12460\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-7980\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +2310\,i}{15015\,d{a}^{3} \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)^10/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

2/15015/d/a^3*(-4096*I*cos(d*x+c)^6+4096*cos(d*x+c)^5*sin(d*x+c)-512*I*cos(d*x+c)^4+2560*cos(d*x+c)^3*sin(d*x+
c)-6230*I*cos(d*x+c)^2-3990*cos(d*x+c)*sin(d*x+c)+1155*I)*(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 [A]  time = 0.97875, size = 127, normalized size = 0.87 \begin{align*} -\frac{2 i \,{\left (1155 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{13}{2}} - 10920 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a + 40040 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a^{2} - 68640 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{3} + 48048 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{4}\right )}}{15015 \, a^{9} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/15015*I*(1155*(I*a*tan(d*x + c) + a)^(13/2) - 10920*(I*a*tan(d*x + c) + a)^(11/2)*a + 40040*(I*a*tan(d*x +
c) + a)^(9/2)*a^2 - 68640*(I*a*tan(d*x + c) + a)^(7/2)*a^3 + 48048*(I*a*tan(d*x + c) + a)^(5/2)*a^4)/(a^9*d)

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Fricas [A]  time = 2.27935, size = 566, normalized size = 3.88 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-16384 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 106496 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 292864 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 439296 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 384384 i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} e^{\left (i \, d x + i \, c\right )}}{15015 \,{\left (a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/15015*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-16384*I*e^(12*I*d*x + 12*I*c) - 106496*I*e^(10*I*d*x + 10*
I*c) - 292864*I*e^(8*I*d*x + 8*I*c) - 439296*I*e^(6*I*d*x + 6*I*c) - 384384*I*e^(4*I*d*x + 4*I*c))*e^(I*d*x +
I*c)/(a^3*d*e^(12*I*d*x + 12*I*c) + 6*a^3*d*e^(10*I*d*x + 10*I*c) + 15*a^3*d*e^(8*I*d*x + 8*I*c) + 20*a^3*d*e^
(6*I*d*x + 6*I*c) + 15*a^3*d*e^(4*I*d*x + 4*I*c) + 6*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*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)**10/(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 )^{10}}{{\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)^10/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^10/(I*a*tan(d*x + c) + a)^(5/2), x)