3.428 \(\int \frac{1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{16 i \sqrt{a+i a \tan (c+d x)}}{45 a^2 d (e \sec (c+d x))^{3/2}}+\frac{32 i \sqrt{e \sec (c+d x)}}{45 a d e^2 \sqrt{a+i a \tan (c+d x)}}+\frac{4 i}{15 a d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}+\frac{2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}} \]

[Out]

((2*I)/9)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + ((4*I)/15)/(a*d*(e*Sec[c + d*x])^(3/2)*Sqr
t[a + I*a*Tan[c + d*x]]) + (((32*I)/45)*Sqrt[e*Sec[c + d*x]])/(a*d*e^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((16*I)/
45)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d*(e*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.305762, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3502, 3497, 3488} \[ -\frac{16 i \sqrt{a+i a \tan (c+d x)}}{45 a^2 d (e \sec (c+d x))^{3/2}}+\frac{32 i \sqrt{e \sec (c+d x)}}{45 a d e^2 \sqrt{a+i a \tan (c+d x)}}+\frac{4 i}{15 a d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}+\frac{2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((2*I)/9)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + ((4*I)/15)/(a*d*(e*Sec[c + d*x])^(3/2)*Sqr
t[a + I*a*Tan[c + d*x]]) + (((32*I)/45)*Sqrt[e*Sec[c + d*x]])/(a*d*e^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((16*I)/
45)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d*(e*Sec[c + d*x])^(3/2))

Rule 3502

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

Rule 3497

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)/(a*f*m), x] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rule 3488

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)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rubi steps

\begin{align*} \int \frac{1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac{2 \int \frac{1}{(e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}} \, dx}{3 a}\\ &=\frac{2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac{4 i}{15 a d (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{15 a^2}\\ &=\frac{2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac{4 i}{15 a d (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{45 a^2 d (e \sec (c+d x))^{3/2}}+\frac{16 \int \frac{\sqrt{e \sec (c+d x)}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{45 a e^2}\\ &=\frac{2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac{4 i}{15 a d (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}+\frac{32 i \sqrt{e \sec (c+d x)}}{45 a d e^2 \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{45 a^2 d (e \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.438154, size = 100, normalized size = 0.61 \[ -\frac{\sec ^3(c+d x) (-54 i \sin (c+d x)+10 i \sin (3 (c+d x))-81 \cos (c+d x)+5 \cos (3 (c+d x)))}{90 a d (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

-(Sec[c + d*x]^3*(-81*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - (54*I)*Sin[c + d*x] + (10*I)*Sin[3*(c + d*x)]))/(90*
a*d*(e*Sec[c + d*x])^(3/2)*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.304, size = 132, normalized size = 0.8 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( 10\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+10\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+6\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +8\,i\cos \left ( dx+c \right ) +16\,\sin \left ( dx+c \right ) \right ) }{45\,{a}^{2}d{e}^{3}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}\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(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

2/45/d/a^2*cos(d*x+c)^2*(e/cos(d*x+c))^(3/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(10*I*cos(d*x+c)^5
+10*sin(d*x+c)*cos(d*x+c)^4+I*cos(d*x+c)^3+6*cos(d*x+c)^2*sin(d*x+c)+8*I*cos(d*x+c)+16*sin(d*x+c))/e^3

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Maxima [A]  time = 1.97288, size = 240, normalized size = 1.45 \begin{align*} \frac{5 i \, \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 27 i \, \cos \left (\frac{5}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) - 15 i \, \cos \left (\frac{1}{3} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 135 i \, \cos \left (\frac{1}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 27 \, \sin \left (\frac{5}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 15 \, \sin \left (\frac{1}{3} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 135 \, \sin \left (\frac{1}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right )}{180 \, a^{\frac{3}{2}} d e^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/180*(5*I*cos(9/2*d*x + 9/2*c) + 27*I*cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 15*I*cos
(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 135*I*cos(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/
2*d*x + 9/2*c))) + 5*sin(9/2*d*x + 9/2*c) + 27*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) +
15*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 135*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), co
s(9/2*d*x + 9/2*c))))/(a^(3/2)*d*e^(3/2))

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Fricas [A]  time = 1.98072, size = 309, normalized size = 1.87 \begin{align*} \frac{\sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-15 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 120 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 162 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 32 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac{9}{2} i \, d x - \frac{9}{2} i \, c\right )}}{180 \, a^{2} d e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/180*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-15*I*e^(8*I*d*x + 8*I*c) + 120*I*e
^(6*I*d*x + 6*I*c) + 162*I*e^(4*I*d*x + 4*I*c) + 32*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-9/2*I*d*x - 9/2*I*c)/(a^2
*d*e^2)

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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(1/(e*sec(d*x+c))**(3/2)/(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 \frac{1}{\left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^(3/2)), x)