3.1039 \(\int \frac{1}{\sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=186 \[ -\frac{2 i \sqrt{a+i a \tan (e+f x)}}{5 a c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{5 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{3 i \sqrt{a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.161175, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {3523, 45, 37} \[ -\frac{2 i \sqrt{a+i a \tan (e+f x)}}{5 a c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{5 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{3 i \sqrt{a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

I/(f*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2)) - (((3*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a*f*(c
- I*c*Tan[e + f*x])^(5/2)) - (((2*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((
(2*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a*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{1}{\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}{(a+i a x)^{3/2} (c-i c x)^{7/2}} \, 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))^{5/2}}+\frac{(3 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}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac{3 i \sqrt{a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac{6 \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}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac{3 i \sqrt{a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{5 a c 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 )}{5 c f}\\ &=\frac{i}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac{3 i \sqrt{a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{5 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i \sqrt{a+i a \tan (e+f x)}}{5 a c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.7411, size = 106, normalized size = 0.57 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) (-5 \sin (e+f x)+3 \sin (3 (e+f x))-10 i \cos (e+f x)+2 i \cos (3 (e+f x)))}{20 c^3 f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

((Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*((-10*I)*Cos[e + f*x] + (2*I)*Cos[3*(e + f*x)] - 5*Sin[e + f*x] + 3*S
in[3*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(20*c^3*f*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.085, size = 118, normalized size = 0.6 \begin{align*}{\frac{4\,i \left ( \tan \left ( fx+e \right ) \right ) ^{4}+2\, \left ( \tan \left ( fx+e \right ) \right ) ^{5}+6\,i \left ( \tan \left ( fx+e \right ) \right ) ^{2}+ \left ( \tan \left ( fx+e \right ) \right ) ^{3}+2\,i-\tan \left ( fx+e \right ) }{5\,fa{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4} \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))^(5/2),x)

[Out]

1/5/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/a/c^3*(4*I*tan(f*x+e)^4+2*tan(f*x+e)^5+6*I*tan(f
*x+e)^2+tan(f*x+e)^3+2*I-tan(f*x+e))/(tan(f*x+e)+I)^4/(-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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.38921, size = 354, normalized size = 1.9 \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 (8 i \, f x + 8 i \, e\right )} - 6 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 20 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 16 i \, e^{\left (3 i \, f x + 3 i \, e\right )} - 10 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 16 i \, e^{\left (i \, f x + i \, e\right )} + 5 i\right )} e^{\left (-i \, f x - i \, e\right )}}{40 \, a c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/40*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-I*e^(8*I*f*x + 8*I*e) - 6*I*e^(6*I*
f*x + 6*I*e) - 20*I*e^(4*I*f*x + 4*I*e) + 16*I*e^(3*I*f*x + 3*I*e) - 10*I*e^(2*I*f*x + 2*I*e) + 16*I*e^(I*f*x
+ I*e) + 5*I)*e^(-I*f*x - I*e)/(a*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(1/(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{1}{\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(1/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(I*a*tan(f*x + e) + a)*(-I*c*tan(f*x + e) + c)^(5/2)), x)