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]

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

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Rubi [A]  time = 0.109789, antiderivative size = 43, normalized size of antiderivative = 1., 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 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 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]

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*}

Mathematica [B]  time = 3.90369, 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]

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

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Maple [B]  time = 0.035, size = 75, normalized size = 1.7 \begin{align*}{\frac{{c}^{2} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( \tan \left ( fx+e \right ) +i \right ) }{5\,f{a}^{3} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{4}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

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 = 1.93079, size = 53, normalized size = 1.23 \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 1.57531, size = 263, normalized size = 6.12 \begin{align*} \frac{{\left (-i \, c^{2} e^{\left (7 i \, f x + 7 i \, e\right )} - i \, c^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + 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} \end{align*}

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^(7*I*f*x + 7*I*e) - I*c^2*e^(5*I*f*x + 5*I*e) + 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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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((c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)