3.1012 \(\int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.11, 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 = {3523, 37} \[ \frac {i (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [B]  time = 4.47, size = 90, normalized size = 2.09 \[ -\frac {i c^2 \sec ^2(e+f x) \sqrt {c-i c \tan (e+f x)} (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{5 a^2 f (\tan (e+f x)-i)^2 \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 71, normalized size = 1.65 \[ \frac {{\left (i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{5 \, a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.24, size = 75, normalized size = 1.74 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (\tan \left (f x +e \right )+i\right )}{5 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a^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 = 0.66, size = 39, normalized size = 0.91 \[ \frac {{\left (i \, c^{2} \cos \left (5 \, f x + 5 \, e\right ) + c^{2} \sin \left (5 \, f x + 5 \, e\right )\right )} \sqrt {c}}{5 \, a^{\frac {5}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.76, size = 65, normalized size = 1.51 \[ \frac {c^2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^5\,\sqrt {a\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}\,\sqrt {-c\,\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}}{5\,a^3\,f\,{\left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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