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

Optimal. Leaf size=110 \[ \frac{16 i a^2 \sec ^3(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}+\frac{64 i a^3 \sec ^3(c+d x)}{105 d (a+i a \tan (c+d x))^{3/2}}+\frac{2 i a \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d} \]

[Out]

(((64*I)/105)*a^3*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((16*I)/35)*a^2*Sec[c + d*x]^3)/(d*Sqrt[
a + I*a*Tan[c + d*x]]) + (((2*I)/7)*a*Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d

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Rubi [A]  time = 0.176691, antiderivative size = 110, 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 = {3494, 3493} \[ \frac{16 i a^2 \sec ^3(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}+\frac{64 i a^3 \sec ^3(c+d x)}{105 d (a+i a \tan (c+d x))^{3/2}}+\frac{2 i a \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((64*I)/105)*a^3*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((16*I)/35)*a^2*Sec[c + d*x]^3)/(d*Sqrt[
a + I*a*Tan[c + d*x]]) + (((2*I)/7)*a*Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d

Rule 3494

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rule 3493

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*
(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=\frac{2 i a \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{1}{7} (8 a) \int \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{16 i a^2 \sec ^3(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 i a \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{1}{35} \left (32 a^2\right ) \int \frac{\sec ^3(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{64 i a^3 \sec ^3(c+d x)}{105 d (a+i a \tan (c+d x))^{3/2}}+\frac{16 i a^2 \sec ^3(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 i a \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.414873, size = 91, normalized size = 0.83 \[ \frac{2 a \sec ^3(c+d x) (\cos (d x)-i \sin (d x)) \sqrt{a+i a \tan (c+d x)} (27 i \sin (2 (c+d x))+43 \cos (2 (c+d x))+28) (\sin (2 c+d x)+i \cos (2 c+d x))}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sec[c + d*x]^3*(Cos[d*x] - I*Sin[d*x])*(28 + 43*Cos[2*(c + d*x)] + (27*I)*Sin[2*(c + d*x)])*(I*Cos[2*c +
d*x] + Sin[2*c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(105*d)

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Maple [A]  time = 0.286, size = 98, normalized size = 0.9 \begin{align*}{\frac{2\,a \left ( 64\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+64\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+24\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +15\,i \right ) }{105\,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)^3*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

2/105/d*a*(64*I*cos(d*x+c)^4+64*cos(d*x+c)^3*sin(d*x+c)-8*I*cos(d*x+c)^2+24*cos(d*x+c)*sin(d*x+c)+15*I)*(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 [B]  time = 3.53031, size = 788, normalized size = 7.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-560*I*sqrt(2)*a*cos(4*d*x + 4*c) - 448*I*sqrt(2)*a*cos(2*d*x + 2*c) + 560*sqrt(2)*a*sin(4*d*x + 4*c) + 448*
sqrt(2)*a*sin(2*d*x + 2*c) - 128*I*sqrt(2)*a)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/4)*sqrt(a)/(((210*cos(2*d*x + 2*c)^3 + 105*(2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 210*I*sin(2*d*x
+ 2*c)^3 + 105*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 525*cos(
2*d*x + 2*c)^2 - (-105*I*cos(2*d*x + 2*c)^2 - 105*I*sin(2*d*x + 2*c)^2 - 210*I*cos(2*d*x + 2*c) - 105*I)*sin(4
*d*x + 4*c) - (-210*I*cos(2*d*x + 2*c)^2 - 420*I*cos(2*d*x + 2*c) - 210*I)*sin(2*d*x + 2*c) + 420*cos(2*d*x +
2*c) + 105)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (-210*I*cos(2*d*x + 2*c)^3 + (-210*I*co
s(2*d*x + 2*c) - 105*I)*sin(2*d*x + 2*c)^2 + 210*sin(2*d*x + 2*c)^3 + (-105*I*cos(2*d*x + 2*c)^2 - 105*I*sin(2
*d*x + 2*c)^2 - 210*I*cos(2*d*x + 2*c) - 105*I)*cos(4*d*x + 4*c) - 525*I*cos(2*d*x + 2*c)^2 + 105*(cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) + 210*(cos(2*d*x + 2*c)^2 + 2*cos(2*d*
x + 2*c) + 1)*sin(2*d*x + 2*c) - 420*I*cos(2*d*x + 2*c) - 105*I)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1)))*d)

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Fricas [A]  time = 2.21079, size = 312, normalized size = 2.84 \begin{align*} \frac{\sqrt{2}{\left (560 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 448 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 128 i \, a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{105 \,{\left (d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*sqrt(2)*(560*I*a*e^(4*I*d*x + 4*I*c) + 448*I*a*e^(2*I*d*x + 2*I*c) + 128*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c
) + 1))*e^(I*d*x + I*c)/(d*e^(7*I*d*x + 7*I*c) + 3*d*e^(5*I*d*x + 5*I*c) + 3*d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*
x + I*c))

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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)**3*(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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(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)^3, x)