3.11.30 \(\int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx\) [1030]

Optimal. Leaf size=101 \[ \frac {\tan (e+f x)}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {2 \tan (e+f x)}{3 a c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 101, normalized size of antiderivative = 1.00, 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 = {3604, 40, 39} \begin {gather*} \frac {2 \tan (e+f x)}{3 a c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {\tan (e+f x)}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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] /;
FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rule 3604

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]

Rubi steps

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

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Mathematica [A]
time = 1.71, size = 103, normalized size = 1.02 \begin {gather*} \frac {\sec (e+f x) (-i \cos (2 (e+f x))+\sin (2 (e+f x))) (9 \sin (e+f x)+\sin (3 (e+f x))) \sqrt {c-i c \tan (e+f x)}}{12 a c^2 f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)])*(9*Sin[e + f*x] + Sin[3*(e + f*x)])*Sqrt[c - I*c*Tan[
e + f*x]])/(12*a*c^2*f*(-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]
time = 0.37, size = 95, normalized size = 0.94

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right ) \left (2 \left (\tan ^{2}\left (f x +e \right )\right )+3\right )}{3 f \,a^{2} c^{2} \left (-\tan \left (f x +e \right )+i\right )^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(95\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right ) \left (2 \left (\tan ^{2}\left (f x +e \right )\right )+3\right )}{3 f \,a^{2} c^{2} \left (-\tan \left (f x +e \right )+i\right )^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.62, size = 48, normalized size = 0.48 \begin {gather*} \frac {\sin \left (3 \, f x + 3 \, e\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, f x + 3 \, e\right ), \cos \left (3 \, f x + 3 \, e\right )\right )\right )}{12 \, a^{\frac {3}{2}} c^{\frac {3}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(sin(3*f*x + 3*e) + 9*sin(1/3*arctan2(sin(3*f*x + 3*e), cos(3*f*x + 3*e))))/(a^(3/2)*c^(3/2)*f)

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Fricas [A]
time = 0.82, size = 95, normalized size = 0.94 \begin {gather*} \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 )} - 10 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 10 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{24 \, a^{2} c^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*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) - 10*I*e^(6*I
*f*x + 6*I*e) + 10*I*e^(2*I*f*x + 2*I*e) + I)*e^(-3*I*f*x - 3*I*e)/(a^2*c^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.54, size = 138, normalized size = 1.37 \begin {gather*} \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,8{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+10\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )-9{}\mathrm {i}\right )}{24\,a^2\,c\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(3/2)),x)

[Out]

(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(cos(2*e + 2*f*x)*8i + cos(4*
e + 4*f*x)*1i + 10*sin(2*e + 2*f*x) + sin(4*e + 4*f*x) - 9i))/(24*a^2*c*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*
f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))

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