3.1028 \(\int \frac{\sqrt{a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=90 \[ -\frac{i \sqrt{a+i a \tan (e+f x)}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{i \sqrt{a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.117379, antiderivative size = 90, 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, 37} \[ -\frac{i \sqrt{a+i a \tan (e+f x)}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{i \sqrt{a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/3)*Sqrt[a + I*a*Tan[e + f*x]])/(f*(c - I*c*Tan[e + f*x])^(3/2)) - ((I/3)*Sqrt[a + I*a*Tan[e + f*x]])/(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{\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}{\sqrt{a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i \sqrt{a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{a \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 \sqrt{a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac{i \sqrt{a+i a \tan (e+f x)}}{3 c f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.04616, size = 89, normalized size = 0.99 \[ \frac{\sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (-i \sin (2 (e+f x))+2 \cos (2 (e+f x))+2) (\sin (2 (e+f x))-i \cos (2 (e+f x)))}{6 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.076, size = 70, normalized size = 0.8 \begin{align*}{\frac{3\,i\tan \left ( fx+e \right ) + \left ( \tan \left ( fx+e \right ) \right ) ^{2}-2}{3\,f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}\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((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)/c^2*(3*I*tan(f*x+e)+tan(f*x+e)^2-2)/(tan(f*x+e)+
I)^3

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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((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 = 1.30412, size = 208, normalized size = 2.31 \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 (4 i \, f x + 4 i \, e\right )} - 4 i \, e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i\right )} e^{\left (i \, f x + i \, e\right )}}{6 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-I*e^(4*I*f*x + 4*I*e) - 4*I*e^(2*I*f
*x + 2*I*e) - 3*I)*e^(I*f*x + I*e)/(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((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{\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((a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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