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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.118939, 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} \[ \frac{i \sqrt{c-i c \tan (e+f x)}}{3 a f \sqrt{a+i a \tan (e+f x)}}+\frac{i \sqrt{c-i c \tan (e+f x)}}{3 f (a+i a \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.46754, size = 68, normalized size = 0.76 \[ \frac{(2+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}{3 a f (\tan (e+f x)-i) \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.08, size = 74, 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{a}^{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((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*(3*I*tan(f*x+e)-tan(f*x+e)^2+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((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 = 1.3365, size = 285, normalized size = 3.17 \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 (-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 )} + 4 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{6 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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