3.11.36 \(\int \frac {(a+i a \tan (e+f x))^{5/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [1036]

Optimal. Leaf size=43 \[ -\frac {i (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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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))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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))^{5/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}\\ \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. \(91\) vs. \(2(43)=86\).
time = 2.61, size = 91, normalized size = 2.12 \begin {gather*} \frac {a^2 \cos (e+f x) (-i \cos (5 e+7 f x)+\sin (5 e+7 f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{5 c^3 f (\cos (f x)+i \sin (f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[e + f*x]*((-I)*Cos[5*e + 7*f*x] + Sin[5*e + 7*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f
*x]])/(5*c^3*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (34 ) = 68\).
time = 0.38, size = 75, normalized size = 1.74

method	result	size
risch	\(-\frac {i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{4 i \left (f x +e \right )}}{5 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(65\)
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (-\tan \left (f x +e \right )+i\right )}{5 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}}\)	\(75\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (-\tan \left (f x +e \right )+i\right )}{5 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/5/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2/c^3*(1+tan(f*x+e)^2)*(-tan(f*x+e)+I)/(tan(f*
x+e)+I)^4

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Maxima [A]
time = 0.56, size = 41, normalized size = 0.95 \begin {gather*} \frac {{\left (-i \, a^{2} \cos \left (5 \, f x + 5 \, e\right ) + a^{2} \sin \left (5 \, f x + 5 \, e\right )\right )} \sqrt {a}}{5 \, c^{\frac {5}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(-I*a^2*cos(5*f*x + 5*e) + a^2*sin(5*f*x + 5*e))*sqrt(a)/(c^(5/2)*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. 75 vs. \(2 (33) = 66\).
time = 0.97, size = 75, normalized size = 1.74 \begin {gather*} \frac {{\left (-i \, a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} - i \, a^{2} e^{\left (5 i \, f x + 5 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}}}{5 \, c^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/5*(-I*a^2*e^(7*I*f*x + 7*I*e) - I*a^2*e^(5*I*f*x + 5*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^3*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 {5}{2}}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*(tan(e + f*x) - I))**(5/2)/(-I*c*(tan(e + f*x) + I))**(5/2), 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))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.29, size = 114, normalized size = 2.65 \begin {gather*} -\frac {a^2\,\left (\cos \left (4\,e+4\,f\,x\right )+\sin \left (4\,e+4\,f\,x\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}}{5\,c^2\,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((a + a*tan(e + f*x)*1i)^(5/2)/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

-(a^2*(cos(4*e + 4*f*x) + sin(4*e + 4*f*x)*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)/(5*c^2*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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