3.294 \(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=88 \[ -\frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}+\frac{8 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d}-\frac{8 i (a+i a \tan (c+d x))^{9/2}}{9 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.0768244, antiderivative size = 88, 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} \[ -\frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}+\frac{8 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d}-\frac{8 i (a+i a \tan (c+d x))^{9/2}}{9 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.491337, size = 93, normalized size = 1.06 \[ \frac{2 a \sec ^6(c+d x) (\cos (d x)-i \sin (d x)) \sqrt{a+i a \tan (c+d x)} (-135 i \sin (2 (c+d x))+151 \cos (2 (c+d x))+52) (\sin (4 c+5 d x)-i \cos (4 c+5 d x))}{1287 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*a*Sec[c + d*x]^6*(Cos[d*x] - I*Sin[d*x])*(52 + 151*Cos[2*(c + d*x)] - (135*I)*Sin[2*(c + d*x)])*((-I)*Cos[4
*c + 5*d*x] + Sin[4*c + 5*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(1287*d)

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Maple [A]  time = 0.335, size = 125, normalized size = 1.4 \begin{align*} -{\frac{2\,a \left ( 256\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-256\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +32\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-160\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +14\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-126\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -99\,i \right ) }{1287\,d \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)^6*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-2/1287/d*a*(256*I*cos(d*x+c)^6-256*cos(d*x+c)^5*sin(d*x+c)+32*I*cos(d*x+c)^4-160*cos(d*x+c)^3*sin(d*x+c)+14*I
*cos(d*x+c)^2-126*cos(d*x+c)*sin(d*x+c)-99*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 = 1.00145, size = 78, normalized size = 0.89 \begin{align*} -\frac{2 i \,{\left (99 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{13}{2}} - 468 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a + 572 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a^{2}\right )}}{1287 \, a^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/1287*I*(99*(I*a*tan(d*x + c) + a)^(13/2) - 468*(I*a*tan(d*x + c) + a)^(11/2)*a + 572*(I*a*tan(d*x + c) + a)
^(9/2)*a^2)/(a^5*d)

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Fricas [B]  time = 2.3263, size = 446, normalized size = 5.07 \begin{align*} \frac{\sqrt{2}{\left (-1024 i \, a e^{\left (12 i \, d x + 12 i \, c\right )} - 6656 i \, a e^{\left (10 i \, d x + 10 i \, c\right )} - 18304 i \, a e^{\left (8 i \, d x + 8 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 )}}{1287 \,{\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, 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)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/1287*sqrt(2)*(-1024*I*a*e^(12*I*d*x + 12*I*c) - 6656*I*a*e^(10*I*d*x + 10*I*c) - 18304*I*a*e^(8*I*d*x + 8*I*
c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 1
5*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*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)**6*(a+I*a*tan(d*x+c))**(3/2),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{3}{2}} \sec \left (d x + c\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*sec(d*x + c)^6, x)