3.1038 \(\int \frac{\sqrt{a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=136 \[ -\frac{2 i \sqrt{a+i a \tan (e+f x)}}{15 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{i \sqrt{a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

((-I/5)*Sqrt[a + I*a*Tan[e + f*x]])/(f*(c - I*c*Tan[e + f*x])^(5/2)) - (((2*I)/15)*Sqrt[a + I*a*Tan[e + f*x]])
/(c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (((2*I)/15)*Sqrt[a + I*a*Tan[e + f*x]])/(c^2*f*Sqrt[c - I*c*Tan[e + f*x]
])

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Rubi [A]  time = 0.133025, antiderivative size = 136, 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} \[ -\frac{2 i \sqrt{a+i a \tan (e+f x)}}{15 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{i \sqrt{a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[e + f*x]]/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((-I/5)*Sqrt[a + I*a*Tan[e + f*x]])/(f*(c - I*c*Tan[e + f*x])^(5/2)) - (((2*I)/15)*Sqrt[a + I*a*Tan[e + f*x]])
/(c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (((2*I)/15)*Sqrt[a + I*a*Tan[e + f*x]])/(c^2*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))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i \sqrt{a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac{i \sqrt{a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}+\frac{(2 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 )}{15 c f}\\ &=-\frac{i \sqrt{a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{15 c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.18302, size = 102, normalized size = 0.75 \[ \frac{\sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} \left (19 \cos (e+f x)+9 \cos (3 (e+f x))-24 i \sin (e+f x) \cos ^2(e+f x)\right ) (\sin (3 (e+f x))-i \cos (3 (e+f x)))}{60 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Tan[e + f*x]]/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((19*Cos[e + f*x] + 9*Cos[3*(e + f*x)] - (24*I)*Cos[e + f*x]^2*Sin[e + f*x])*((-I)*Cos[3*(e + f*x)] + Sin[3*(e
+ f*x)])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(60*c^3*f)

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Maple [A]  time = 0.08, size = 83, normalized size = 0.6 \begin{align*}{\frac{8\,i \left ( \tan \left ( fx+e \right ) \right ) ^{2}+2\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}-7\,i-13\,\tan \left ( fx+e \right ) }{15\,f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}\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))^(5/2),x)

[Out]

1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/c^3*(8*I*tan(f*x+e)^2+2*tan(f*x+e)^3-7*I-13*tan
(f*x+e))/(tan(f*x+e)+I)^4

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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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.38924, size = 251, normalized size = 1.85 \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 )} - 13 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 25 i \, e^{\left (2 i \, f x + 2 i \, e\right )} - 15 i\right )} e^{\left (i \, f x + i \, e\right )}}{60 \, c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/60*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) - 13*I*e^(4
*I*f*x + 4*I*e) - 25*I*e^(2*I*f*x + 2*I*e) - 15*I)*e^(I*f*x + I*e)/(c^3*f)

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Sympy [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))**(1/2)/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(f*x + e) + a)/(-I*c*tan(f*x + e) + c)^(5/2), x)