3.1162 \(\int \frac{(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=129 \[ -\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}-\frac{2 a \sqrt{a+i a \tan (e+f x)}}{f (d+i c) \sqrt{c+d \tan (e+f x)}} \]

[Out]

((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e +
f*x]])])/((c - I*d)^(3/2)*f) - (2*a*Sqrt[a + I*a*Tan[e + f*x]])/((I*c + d)*f*Sqrt[c + d*Tan[e + f*x]])

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Rubi [A]  time = 0.220858, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3545, 3544, 208} \[ -\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}-\frac{2 a \sqrt{a+i a \tan (e+f x)}}{f (d+i c) \sqrt{c+d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e +
f*x]])])/((c - I*d)^(3/2)*f) - (2*a*Sqrt[a + I*a*Tan[e + f*x]])/((I*c + d)*f*Sqrt[c + d*Tan[e + f*x]])

Rule 3545

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{2 a \sqrt{a+i a \tan (e+f x)}}{(i c+d) f \sqrt{c+d \tan (e+f x)}}+\frac{(2 a) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{c-i d}\\ &=-\frac{2 a \sqrt{a+i a \tan (e+f x)}}{(i c+d) f \sqrt{c+d \tan (e+f x)}}+\frac{\left (4 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{a+i a \tan (e+f x)}}\right )}{(i c+d) f}\\ &=-\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}-\frac{2 a \sqrt{a+i a \tan (e+f x)}}{(i c+d) f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 4.39455, size = 209, normalized size = 1.62 \[ \frac{2 i a e^{-i (e+f x)} \sqrt{a+i a \tan (e+f x)} \left (\sqrt{c-i d} e^{i (e+f x)}-\sqrt{1+e^{2 i (e+f x)}} \log \left (2 e^{-i e} \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right ) \sqrt{c+d \tan (e+f x)}\right )}{f (c-i d)^{3/2} \sqrt{c+d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*a*Sqrt[a + I*a*Tan[e + f*x]]*(Sqrt[c - I*d]*E^(I*(e + f*x)) - Sqrt[1 + E^((2*I)*(e + f*x))]*Log[(2*(Sqr
t[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((
2*I)*(e + f*x)))]))/E^(I*e)]*Sqrt[c + d*Tan[e + f*x]]))/((c - I*d)^(3/2)*E^(I*(e + f*x))*f*Sqrt[c + d*Tan[e +
f*x]])

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Maple [B]  time = 0.059, size = 1840, normalized size = 14.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2),x)

[Out]

1/2/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(-I*2^(1/2)*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1
+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d*(-a*(I*d-c))^(1/2)+I*ln((3*a*c+I*a*tan(f*x+e)*c-
I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I
))*a*c*d*(I*a*d)^(1/2)-I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+
d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*c^2*(I*a*d)^(1/2)+I*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+
2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c
^2+I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*t
an(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)*a*d^2*(I*a*d)^(1/2)+2*I*2^(1/2)*c^3*(a*(c+d*tan(f*x+e))*(1+I*tan
(f*x+e)))^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)-I*2^(1/2)*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+
e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*tan(f*x+e)*a*d^2*(-a*(I*d-c))^(1/2)+2*I*2^(1/2)*
c*d^2*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+2^(1/2)*ln(1/2*(2*I*a*tan(f
*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*tan(f*x+e)*a*c*d
*(-a*(I*d-c))^(1/2)+2^(1/2)*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*
a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*tan(f*x+e)*a*d^2*(-a*(I*d-c))^(1/2)-2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/
2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d-2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*
2^(1/2)*(-a*(I*d-c))^(1/2)*d^3-I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2
)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)*a*c*d*(I*a*d)^(1/2)+I*2^(1/2)*ln(1/2
*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*tan
(f*x+e)*a*c*d*(-a*(I*d-c))^(1/2)+ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2
)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2+ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+
d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d-ln((3
*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))
)^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)*a*c*d*(I*a*d)^(1/2)-ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^
(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)*a*d^2*(I*a*d)
^(1/2)-ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I
*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*c^2*(I*a*d)^(1/2)-ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2
^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*c*d)*a/
(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(c^2+d^2)^2/(I*a*d)^(1/2)/(-a*(I*d-c))^(1/2)/(c+d*tan(f*x+e))^(1/2
)

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

[Out]

Timed out

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Fricas [B]  time = 1.4628, size = 1463, normalized size = 11.34 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(2)*(-4*I*a*e^(2*I*f*x + 2*I*e) - 4*I*a)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x
+ 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - ((c^2 - 2*I*c*d - d^2)*f*e^(2*I*f*x + 2*I*e
) + (c^2 + d^2)*f)*sqrt(8*I*a^3/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/2*((I*c^2 + 2*c*d - I*d^2)*f
*sqrt(8*I*a^3/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*e^(2*I*f*x + 2*I*e) + 2*sqrt(2)*(a*e^(2*I*f*x + 2*I*
e) + a)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e)
+ 1))*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/a) + ((c^2 - 2*I*c*d - d^2)*f*e^(2*I*f*x + 2*I*e) + (c^2 + d^2)*f)
*sqrt(8*I*a^3/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/2*((-I*c^2 - 2*c*d + I*d^2)*f*sqrt(8*I*a^3/((-
I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*e^(2*I*f*x + 2*I*e) + 2*sqrt(2)*(a*e^(2*I*f*x + 2*I*e) + a)*sqrt(((c
- I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x +
I*e))*e^(-2*I*f*x - 2*I*e)/a))/((c^2 - 2*I*c*d - d^2)*f*e^(2*I*f*x + 2*I*e) + (c^2 + d^2)*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \left (i \tan{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(I*tan(e + f*x) + 1))**(3/2)/(c + d*tan(e + f*x))**(3/2), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError