3.11.27 \(\int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [1027]

Optimal. Leaf size=43 \[ -\frac {i (a+i a \tan (e+f x))^{3/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {3604, 37} \begin {gather*} -\frac {i (a+i a \tan (e+f x))^{3/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(87\) vs. \(2(43)=86\).
time = 1.52, size = 87, normalized size = 2.02 \begin {gather*} \frac {a \cos (e+f x) (\cos (f x)-i \sin (f x)) (-i \cos (3 e+4 f x)+\sin (3 e+4 f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{3 c^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.34, size = 62, normalized size = 1.44

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan ^{2}\left (f x +e \right )\right )}{3 f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(62\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan ^{2}\left (f x +e \right )\right )}{3 f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(62\)
risch	\(-\frac {i a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{3 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]
time = 0.54, size = 37, normalized size = 0.86 \begin {gather*} \frac {{\left (-i \, a \cos \left (3 \, f x + 3 \, e\right ) + a \sin \left (3 \, f x + 3 \, e\right )\right )} \sqrt {a}}{3 \, c^{\frac {3}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (33) = 66\).
time = 0.74, size = 71, normalized size = 1.65 \begin {gather*} \frac {{\left (-i \, a e^{\left (5 i \, f x + 5 i \, e\right )} - i \, a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \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}}}{3 \, c^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\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((a+I*a*tan(f*x+e))**(3/2)/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

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

[Out]

integrate((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 = 0.77, size = 108, normalized size = 2.51 \begin {gather*} -\frac {\sqrt {2}\,a\,\left (\cos \left (2\,f\,x\right )+\sin \left (2\,f\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,e\right )+\sin \left (2\,e\right )\,1{}\mathrm {i}\right )\,\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}}\,1{}\mathrm {i}}{6\,c\,f\,\sqrt {\frac {c}{\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2^(1/2)*a*(cos(2*f*x) + sin(2*f*x)*1i)*(cos(2*e) + sin(2*e)*1i)*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i
+ 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*1i)/(6*c*f*(c/(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))^(1/2))

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