3.1021 \(\int \frac{1}{(a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=94 \[ \frac{2 \tan (e+f x)}{3 a f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12086, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {3523, 45, 39} \[ \frac{2 \tan (e+f x)}{3 a f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

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

Rule 3523

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{5/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{(2 c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{2 \tan (e+f x)}{3 a f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.74399, size = 73, normalized size = 0.78 \[ \frac{\sqrt{c-i c \tan (e+f x)} (2 \sin (2 (e+f x))-i \cos (2 (e+f x))+3 i)}{6 a c f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

((3*I - I*Cos[2*(e + f*x)] + 2*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(6*a*c*f*Sqrt[a + I*a*Tan[e + f*x
]])

________________________________________________________________________________________

Maple [A]  time = 0.084, size = 109, normalized size = 1.2 \begin{align*}{\frac{2\,i \left ( \tan \left ( fx+e \right ) \right ) ^{3}-2\, \left ( \tan \left ( fx+e \right ) \right ) ^{4}+2\,i\tan \left ( fx+e \right ) -3\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}-1}{3\,f{a}^{2}c \left ( \tan \left ( fx+e \right ) +i \right ) ^{2} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{3}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x)

[Out]

1/3/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2/c*(2*I*tan(f*x+e)^3-2*tan(f*x+e)^4+2*I*tan(f
*x+e)-3*tan(f*x+e)^2-1)/(tan(f*x+e)+I)^2/(-tan(f*x+e)+I)^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 1.36922, size = 324, normalized size = 3.45 \begin{align*} \frac{\sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-3 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 4 i \, e^{\left (5 i \, f x + 5 i \, e\right )} + 3 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, e^{\left (3 i \, f x + 3 i \, e\right )} + 7 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, a^{2} c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-3*I*e^(6*I*f*x + 6*I*e) - 4*I*e^(5*
I*f*x + 5*I*e) + 3*I*e^(4*I*f*x + 4*I*e) - 4*I*e^(3*I*f*x + 3*I*e) + 7*I*e^(2*I*f*x + 2*I*e) + I)*e^(-3*I*f*x
- 3*I*e)/(a^2*c*f)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^(3/2)*sqrt(-I*c*tan(f*x + e) + c)), x)