3.378 \(\int \frac{\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=146 \[ -\frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^9 d}+\frac{16 i (a+i a \tan (c+d x))^{9/2}}{9 a^8 d}-\frac{48 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d}+\frac{64 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}-\frac{32 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d} \]

[Out]

(((-32*I)/3)*(a + I*a*Tan[c + d*x])^(3/2))/(a^5*d) + (((64*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^6*d) - (((48
*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^7*d) + (((16*I)/9)*(a + I*a*Tan[c + d*x])^(9/2))/(a^8*d) - (((2*I)/11)
*(a + I*a*Tan[c + d*x])^(11/2))/(a^9*d)

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Rubi [A]  time = 0.0931584, 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} \[ -\frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^9 d}+\frac{16 i (a+i a \tan (c+d x))^{9/2}}{9 a^8 d}-\frac{48 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d}+\frac{64 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}-\frac{32 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((-32*I)/3)*(a + I*a*Tan[c + d*x])^(3/2))/(a^5*d) + (((64*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^6*d) - (((48
*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^7*d) + (((16*I)/9)*(a + I*a*Tan[c + d*x])^(9/2))/(a^8*d) - (((2*I)/11)
*(a + I*a*Tan[c + d*x])^(11/2))/(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))^{7/2}} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^4 \sqrt{a+x} \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (16 a^4 \sqrt{a+x}-32 a^3 (a+x)^{3/2}+24 a^2 (a+x)^{5/2}-8 a (a+x)^{7/2}+(a+x)^{9/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac{32 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d}+\frac{64 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}-\frac{48 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d}+\frac{16 i (a+i a \tan (c+d x))^{9/2}}{9 a^8 d}-\frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^9 d}\\ \end{align*}

Mathematica [A]  time = 0.852603, size = 114, normalized size = 0.78 \[ \frac{2 \sec ^9(c+d x) (-1144 i \sin (2 (c+d x))-1027 i \sin (4 (c+d x))+2552 \cos (2 (c+d x))+1283 \cos (4 (c+d x))+1584) (\cos (5 (c+d x))+i \sin (5 (c+d x)))}{3465 a^3 d (\tan (c+d x)-i)^3 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(2*Sec[c + d*x]^9*(1584 + 2552*Cos[2*(c + d*x)] + 1283*Cos[4*(c + d*x)] - (1144*I)*Sin[2*(c + d*x)] - (1027*I)
*Sin[4*(c + d*x)])*(Cos[5*(c + d*x)] + I*Sin[5*(c + d*x)]))/(3465*a^3*d*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan
[c + d*x]])

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Maple [A]  time = 0.317, size = 117, normalized size = 0.8 \begin{align*} -{\frac{4096\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}-4096\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+9752\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+6680\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -3010\,i\cos \left ( dx+c \right ) -630\,\sin \left ( dx+c \right ) }{3465\,{a}^{4}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)^10/(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

-2/3465/d/a^4*(2048*I*cos(d*x+c)^5-2048*sin(d*x+c)*cos(d*x+c)^4+4876*I*cos(d*x+c)^3+3340*cos(d*x+c)^2*sin(d*x+
c)-1505*I*cos(d*x+c)-315*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 = 0.993697, size = 127, normalized size = 0.87 \begin{align*} -\frac{2 i \,{\left (315 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} - 3080 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a + 11880 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{2} - 22176 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{3} + 18480 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{4}\right )}}{3465 \, 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))^(7/2),x, algorithm="maxima")

[Out]

-2/3465*I*(315*(I*a*tan(d*x + c) + a)^(11/2) - 3080*(I*a*tan(d*x + c) + a)^(9/2)*a + 11880*(I*a*tan(d*x + c) +
a)^(7/2)*a^2 - 22176*(I*a*tan(d*x + c) + a)^(5/2)*a^3 + 18480*(I*a*tan(d*x + c) + a)^(3/2)*a^4)/(a^9*d)

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Fricas [A]  time = 2.15303, size = 513, normalized size = 3.51 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-8192 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 45056 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 101376 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 118272 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 73920 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (i \, d x + i \, c\right )}}{3465 \,{\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} 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))^(7/2),x, algorithm="fricas")

[Out]

1/3465*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-8192*I*e^(10*I*d*x + 10*I*c) - 45056*I*e^(8*I*d*x + 8*I*c)
- 101376*I*e^(6*I*d*x + 6*I*c) - 118272*I*e^(4*I*d*x + 4*I*c) - 73920*I*e^(2*I*d*x + 2*I*c))*e^(I*d*x + I*c)/(
a^4*d*e^(10*I*d*x + 10*I*c) + 5*a^4*d*e^(8*I*d*x + 8*I*c) + 10*a^4*d*e^(6*I*d*x + 6*I*c) + 10*a^4*d*e^(4*I*d*x
+ 4*I*c) + 5*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*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))**(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 )^{10}}{{\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)^10/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^10/(I*a*tan(d*x + c) + a)^(7/2), x)