3.379 \(\int \frac{\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}+\frac{8 i (a+i a \tan (c+d x))^{3/2}}{a^5 d}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{a^4 d} \]

[Out]

((-16*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^4*d) + ((8*I)*(a + I*a*Tan[c + d*x])^(3/2))/(a^5*d) - (((12*I)/5)*(a +
I*a*Tan[c + d*x])^(5/2))/(a^6*d) + (((2*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^7*d)

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Rubi [A]  time = 0.0863583, antiderivative size = 113, 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} \[ \frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}+\frac{8 i (a+i a \tan (c+d x))^{3/2}}{a^5 d}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

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

Mathematica [A]  time = 0.440104, size = 110, normalized size = 0.97 \[ \frac{2 \sec ^7(c+d x) (-i (14 \sin (c+d x)+19 \sin (3 (c+d x)))+126 \cos (c+d x)+51 \cos (3 (c+d x))) (\cos (4 (c+d x))+i \sin (4 (c+d x)))}{35 a^3 d (\tan (c+d x)-i)^3 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(2*Sec[c + d*x]^7*(126*Cos[c + d*x] + 51*Cos[3*(c + d*x)] - I*(14*Sin[c + d*x] + 19*Sin[3*(c + d*x)]))*(Cos[4*
(c + d*x)] + I*Sin[4*(c + d*x)]))/(35*a^3*d*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.282, size = 90, normalized size = 0.8 \begin{align*} -{\frac{408\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+152\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -54\,i\cos \left ( dx+c \right ) -10\,\sin \left ( dx+c \right ) }{35\,{a}^{4}d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}\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))^(7/2),x)

[Out]

-2/35/d/a^4*(204*I*cos(d*x+c)^3+76*cos(d*x+c)^2*sin(d*x+c)-27*I*cos(d*x+c)-5*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(
d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3

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Maxima [A]  time = 0.980628, size = 103, normalized size = 0.91 \begin{align*} \frac{2 i \,{\left (5 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} - 42 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a + 140 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{2} - 280 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} a^{3}\right )}}{35 \, 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))^(7/2),x, algorithm="maxima")

[Out]

2/35*I*(5*(I*a*tan(d*x + c) + a)^(7/2) - 42*(I*a*tan(d*x + c) + a)^(5/2)*a + 140*(I*a*tan(d*x + c) + a)^(3/2)*
a^2 - 280*sqrt(I*a*tan(d*x + c) + a)*a^3)/(a^7*d)

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Fricas [A]  time = 2.15842, size = 343, normalized size = 3.04 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-256 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 896 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 1120 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 560 i\right )} e^{\left (i \, d x + i \, c\right )}}{35 \,{\left (a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^8/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/35*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-256*I*e^(6*I*d*x + 6*I*c) - 896*I*e^(4*I*d*x + 4*I*c) - 1120*
I*e^(2*I*d*x + 2*I*c) - 560*I)*e^(I*d*x + I*c)/(a^4*d*e^(6*I*d*x + 6*I*c) + 3*a^4*d*e^(4*I*d*x + 4*I*c) + 3*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)**8/(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 )^{8}}{{\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)^8/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^8/(I*a*tan(d*x + c) + a)^(7/2), x)