3.333 \(\int \frac{\sec ^8(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=117 \[ \frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{11/2}}{11 a^6 d}+\frac{8 i (a+i a \tan (c+d x))^{9/2}}{3 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{7/2}}{7 a^4 d} \]

[Out]

(((-16*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^4*d) + (((8*I)/3)*(a + I*a*Tan[c + d*x])^(9/2))/(a^5*d) - (((12*
I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^6*d) + (((2*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^7*d)

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Rubi [A]  time = 0.0797237, 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} \[ \frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{11/2}}{11 a^6 d}+\frac{8 i (a+i a \tan (c+d x))^{9/2}}{3 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{7/2}}{7 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

Mathematica [A]  time = 0.466889, size = 95, normalized size = 0.81 \[ \frac{2 \sec ^7(c+d x) (-7 i (26 \sin (c+d x)+59 \sin (3 (c+d x)))+390 \cos (c+d x)+445 \cos (3 (c+d x))) (\sin (4 (c+d x))-i \cos (4 (c+d x)))}{3003 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(2*Sec[c + d*x]^7*(390*Cos[c + d*x] + 445*Cos[3*(c + d*x)] - (7*I)*(26*Sin[c + d*x] + 59*Sin[3*(c + d*x)]))*((
-I)*Cos[4*(c + d*x)] + Sin[4*(c + d*x)]))/(3003*d*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.583, size = 127, normalized size = 1.1 \begin{align*} -{\frac{1024\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-1024\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +128\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-640\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +56\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-504\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +462\,i}{3003\,ad \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)^8/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

-2/3003/d/a*(512*I*cos(d*x+c)^6-512*cos(d*x+c)^5*sin(d*x+c)+64*I*cos(d*x+c)^4-320*cos(d*x+c)^3*sin(d*x+c)+28*I
*cos(d*x+c)^2-252*cos(d*x+c)*sin(d*x+c)+231*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 [B]  time = 1.12363, size = 401, normalized size = 3.43 \begin{align*} -\frac{2 i \,{\left (15015 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} - \frac{3003 \,{\left (3 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} - 10 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} a^{2}\right )}}{a^{2}} + \frac{143 \,{\left (35 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{9}{2}} - 180 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a + 378 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{2} - 420 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{3} + 315 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} a^{4}\right )}}{a^{4}} - \frac{5 \,{\left (231 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{13}{2}} - 1638 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a + 5005 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a^{2} - 8580 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{3} + 9009 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{4} - 6006 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{5} + 3003 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} a^{6}\right )}}{a^{6}}\right )}}{15015 \, a 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))^(1/2),x, algorithm="maxima")

[Out]

-2/15015*I*(15015*sqrt(I*a*tan(d*x + c) + a) - 3003*(3*(I*a*tan(d*x + c) + a)^(5/2) - 10*(I*a*tan(d*x + c) + a
)^(3/2)*a + 15*sqrt(I*a*tan(d*x + c) + a)*a^2)/a^2 + 143*(35*(I*a*tan(d*x + c) + a)^(9/2) - 180*(I*a*tan(d*x +
c) + a)^(7/2)*a + 378*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 420*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 315*sqrt(I*a*
tan(d*x + c) + a)*a^4)/a^4 - 5*(231*(I*a*tan(d*x + c) + a)^(13/2) - 1638*(I*a*tan(d*x + c) + a)^(11/2)*a + 500
5*(I*a*tan(d*x + c) + a)^(9/2)*a^2 - 8580*(I*a*tan(d*x + c) + a)^(7/2)*a^3 + 9009*(I*a*tan(d*x + c) + a)^(5/2)
*a^4 - 6006*(I*a*tan(d*x + c) + a)^(3/2)*a^5 + 3003*sqrt(I*a*tan(d*x + c) + a)*a^6)/a^6)/(a*d)

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Fricas [A]  time = 2.09395, size = 498, normalized size = 4.26 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-2048 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 13312 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 36608 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 54912 i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} e^{\left (i \, d x + i \, c\right )}}{3003 \,{\left (a d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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))^(1/2),x, algorithm="fricas")

[Out]

1/3003*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-2048*I*e^(12*I*d*x + 12*I*c) - 13312*I*e^(10*I*d*x + 10*I*c
) - 36608*I*e^(8*I*d*x + 8*I*c) - 54912*I*e^(6*I*d*x + 6*I*c))*e^(I*d*x + I*c)/(a*d*e^(12*I*d*x + 12*I*c) + 6*
a*d*e^(10*I*d*x + 10*I*c) + 15*a*d*e^(8*I*d*x + 8*I*c) + 20*a*d*e^(6*I*d*x + 6*I*c) + 15*a*d*e^(4*I*d*x + 4*I*
c) + 6*a*d*e^(2*I*d*x + 2*I*c) + a*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))**(1/2),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}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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