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

Optimal. Leaf size=69 \[ \frac{8 i a^2 \sec (c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 i a \sec (c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0643351, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3494, 3493} \[ \frac{8 i a^2 \sec (c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 i a \sec (c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((8*I)/3)*a^2*Sec[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((2*I)/3)*a*Sec[c + d*x]*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 (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=\frac{2 i a \sec (c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{1}{3} (4 a) \int \sec (c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{8 i a^2 \sec (c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 i a \sec (c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.24852, size = 57, normalized size = 0.83 \[ -\frac{2 a (\cos (c)-i \sin (c)) (\tan (c+d x)-5 i) (\cos (d x)-i \sin (d x)) \sqrt{a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(Cos[c] - I*Sin[c])*(Cos[d*x] - I*Sin[d*x])*(-5*I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(3*d)

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Maple [A]  time = 0.237, size = 71, normalized size = 1. \begin{align*}{\frac{2\,a \left ( 4\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +i \right ) }{3\,d\cos \left ( dx+c \right ) }\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)*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

2/3/d*a*(4*I*cos(d*x+c)^2+4*cos(d*x+c)*sin(d*x+c)+I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)

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Maxima [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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.30756, size = 194, normalized size = 2.81 \begin{align*} \frac{\sqrt{2}{\left (12 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 8 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 )}}{3 \,{\left (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)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*(12*I*a*e^(2*I*d*x + 2*I*c) + 8*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(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)*(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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)