3.1029 \(\int \frac{1}{\sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=137 \[ -\frac{2 i \sqrt{a+i a \tan (e+f x)}}{3 a c f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{3 a f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.137222, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {3523, 45, 37} \[ -\frac{2 i \sqrt{a+i a \tan (e+f x)}}{3 a c f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{3 a f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

I/(f*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2)) - (((2*I)/3)*Sqrt[a + I*a*Tan[e + f*x]])/(a*f*(c
- I*c*Tan[e + f*x])^(3/2)) - (((2*I)/3)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*f*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 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 2.40179, size = 89, normalized size = 0.65 \[ \frac{i \sqrt{c-i c \tan (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) (-2 i \sin (2 (e+f x))+\cos (2 (e+f x))-3)}{6 c^2 f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.082, size = 109, normalized size = 0.8 \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\,fa{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{2}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.31013, size = 313, normalized size = 2.28 \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 (-i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 7 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, e^{\left (i \, f x + i \, e\right )} + 3 i\right )} e^{\left (-i \, f x - i \, e\right )}}{12 \, a c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*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))*(-I*e^(6*I*f*x + 6*I*e) - 7*I*e^(4*I*
f*x + 4*I*e) + 4*I*e^(3*I*f*x + 3*I*e) - 3*I*e^(2*I*f*x + 2*I*e) + 4*I*e^(I*f*x + I*e) + 3*I)*e^(-I*f*x - I*e)
/(a*c^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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