3.1031 \(\int \frac{1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac{8 \tan (e+f x)}{15 a^2 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \]

[Out]

(I/5)/(f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (4*Tan[e + f*x])/(15*a*f*(a + I*a*Tan[e
+ f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (8*Tan[e + f*x])/(15*a^2*c*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c -
I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.14786, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {3523, 45, 40, 39} \[ \frac{8 \tan (e+f x)}{15 a^2 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

(I/5)/(f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (4*Tan[e + f*x])/(15*a*f*(a + I*a*Tan[e
+ f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (8*Tan[e + f*x])/(15*a^2*c*f*Sqrt[a + I*a*Tan[e + f*x]]*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 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(x*(a + b*x)^(m + 1)*(c + d*x)^(m +
1))/(2*a*c*(m + 1)), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /; F
reeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{7/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac{i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{8 \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a f}\\ &=\frac{i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{8 \tan (e+f x)}{15 a^2 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 4.51353, size = 93, normalized size = 0.63 \[ -\frac{i \sqrt{c-i c \tan (e+f x)} (40 i \sin (2 (e+f x))+4 i \sin (4 (e+f x))+20 \cos (2 (e+f x))+\cos (4 (e+f x))-45)}{120 a^2 c^2 f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

((-I/120)*(-45 + 20*Cos[2*(e + f*x)] + Cos[4*(e + f*x)] + (40*I)*Sin[2*(e + f*x)] + (4*I)*Sin[4*(e + f*x)])*Sq
rt[c - I*c*Tan[e + f*x]])/(a^2*c^2*f*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.041, size = 130, normalized size = 0.9 \begin{align*} -{\frac{8\,i \left ( \tan \left ( fx+e \right ) \right ) ^{5}-8\, \left ( \tan \left ( fx+e \right ) \right ) ^{6}+20\,i \left ( \tan \left ( fx+e \right ) \right ) ^{3}-20\, \left ( \tan \left ( fx+e \right ) \right ) ^{4}+12\,i\tan \left ( fx+e \right ) -15\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}-3}{15\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{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(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/a^3/c^2*(8*I*tan(f*x+e)^5-8*tan(f*x+e)^6+20*I*
tan(f*x+e)^3-20*tan(f*x+e)^4+12*I*tan(f*x+e)-15*tan(f*x+e)^2-3)/(tan(f*x+e)+I)^3/(-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(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.48307, size = 413, normalized size = 2.81 \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 (-5 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 65 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 48 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 30 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 48 i \, e^{\left (5 i \, f x + 5 i \, e\right )} + 110 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 23 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{240 \, a^{3} 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))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/240*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-5*I*e^(10*I*f*x + 10*I*e) - 65*I*e
^(8*I*f*x + 8*I*e) - 48*I*e^(7*I*f*x + 7*I*e) + 30*I*e^(6*I*f*x + 6*I*e) - 48*I*e^(5*I*f*x + 5*I*e) + 110*I*e^
(4*I*f*x + 4*I*e) + 23*I*e^(2*I*f*x + 2*I*e) + 3*I)*e^(-5*I*f*x - 5*I*e)/(a^3*c^2*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))**(5/2)/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}}{\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))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^(5/2)*(-I*c*tan(f*x + e) + c)^(3/2)), x)