3.313 \(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=147 \[ \frac{256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac{64 i a^3 \sec ^3(c+d x)}{105 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 i a^2 \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

[Out]

(((256*I)/315)*a^4*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((64*I)/105)*a^3*Sec[c + d*x]^3)/(d*Sqr
t[a + I*a*Tan[c + d*x]]) + (((8*I)/21)*a^2*Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + (((2*I)/9)*a*Sec[c +
d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/d

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Rubi [A]  time = 0.240449, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac{256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac{64 i a^3 \sec ^3(c+d x)}{105 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 i a^2 \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((256*I)/315)*a^4*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((64*I)/105)*a^3*Sec[c + d*x]^3)/(d*Sqr
t[a + I*a*Tan[c + d*x]]) + (((8*I)/21)*a^2*Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + (((2*I)/9)*a*Sec[c +
d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/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))^{5/2} \, dx &=\frac{2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac{1}{3} (4 a) \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac{8 i a^2 \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac{1}{21} \left (32 a^2\right ) \int \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 \sqrt{a+i a \tan (c+d x)}}+\frac{8 i a^2 \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac{1}{105} \left (128 a^3\right ) \int \frac{\sec ^3(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac{64 i a^3 \sec ^3(c+d x)}{105 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 i a^2 \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.680422, size = 103, normalized size = 0.7 \[ \frac{2 a^2 (\sin (2 c)+i \cos (2 c)) \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)} (242 \cos (2 (c+d x))+54 i \tan (c+d x)+89 i \sin (3 (c+d x)) \sec (c+d x)+77)}{315 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*a^2*Sec[c + d*x]^3*(I*Cos[2*c] + Sin[2*c])*(77 + 242*Cos[2*(c + d*x)] + (89*I)*Sec[c + d*x]*Sin[3*(c + d*x)
] + (54*I)*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*(Cos[d*x] + I*Sin[d*x])^2)

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Maple [A]  time = 0.293, size = 117, normalized size = 0.8 \begin{align*}{\frac{2\,{a}^{2} \left ( 256\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+256\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}-32\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+96\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +95\,i\cos \left ( dx+c \right ) -35\,\sin \left ( dx+c \right ) \right ) }{315\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}\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))^(5/2),x)

[Out]

2/315/d*a^2*(256*I*cos(d*x+c)^5+256*sin(d*x+c)*cos(d*x+c)^4-32*I*cos(d*x+c)^3+96*cos(d*x+c)^2*sin(d*x+c)+95*I*
cos(d*x+c)-35*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^4

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Maxima [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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.48567, size = 402, normalized size = 2.73 \begin{align*} \frac{\sqrt{2}{\left (3360 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 4032 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2304 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 512 i \, a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{315 \,{\left (d e^{\left (9 i \, d x + 9 i \, c\right )} + 4 \, d e^{\left (7 i \, d x + 7 i \, c\right )} + 6 \, d e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, 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))^(5/2),x, algorithm="fricas")

[Out]

1/315*sqrt(2)*(3360*I*a^2*e^(6*I*d*x + 6*I*c) + 4032*I*a^2*e^(4*I*d*x + 4*I*c) + 2304*I*a^2*e^(2*I*d*x + 2*I*c
) + 512*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(d*e^(9*I*d*x + 9*I*c) + 4*d*e^(7*I*d*x + 7*I
*c) + 6*d*e^(5*I*d*x + 5*I*c) + 4*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))**(5/2),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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*sec(d*x + c)^3, x)