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

Optimal. Leaf size=62 \[ \frac{2 i a^2}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{4 i a^2}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

(((-4*I)/5)*a^2)/(f*(c - I*c*Tan[e + f*x])^(5/2)) + (((2*I)/3)*a^2)/(c*f*(c - I*c*Tan[e + f*x])^(3/2))

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Rubi [A]  time = 0.153213, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3522, 3487, 43} \[ \frac{2 i a^2}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{4 i a^2}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)/5)*a^2)/(f*(c - I*c*Tan[e + f*x])^(5/2)) + (((2*I)/3)*a^2)/(c*f*(c - I*c*Tan[e + f*x])^(3/2))

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\left (a^2 c^2\right ) \int \frac{\sec ^4(e+f x)}{(c-i c \tan (e+f x))^{9/2}} \, dx\\ &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{c-x}{(c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \left (\frac{2 c}{(c+x)^{7/2}}-\frac{1}{(c+x)^{5/2}}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=-\frac{4 i a^2}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 i a^2}{3 c f (c-i c \tan (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 5.76541, size = 95, normalized size = 1.53 \[ \frac{2 a^2 \cos ^2(e+f x) \sqrt{c-i c \tan (e+f x)} (5 \sin (e+f x)-i \cos (e+f x)) (\cos (3 e+5 f x)+i \sin (3 e+5 f x))}{15 c^3 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Cos[e + f*x]^2*((-I)*Cos[e + f*x] + 5*Sin[e + f*x])*(Cos[3*e + 5*f*x] + I*Sin[3*e + 5*f*x])*Sqrt[c - I*
c*Tan[e + f*x]])/(15*c^3*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.022, size = 47, normalized size = 0.8 \begin{align*}{\frac{-2\,i{a}^{2}}{cf} \left ( -{\frac{1}{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,c}{5} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06826, size = 59, normalized size = 0.95 \begin{align*} \frac{2 i \,{\left (5 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} - 6 \, a^{2} c\right )}}{15 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*I*(5*(-I*c*tan(f*x + e) + c)*a^2 - 6*a^2*c)/((-I*c*tan(f*x + e) + c)^(5/2)*c*f)

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Fricas [A]  time = 1.44041, size = 209, normalized size = 3.37 \begin{align*} \frac{\sqrt{2}{\left (-3 i \, a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 4 i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(2)*(-3*I*a^2*e^(6*I*f*x + 6*I*e) - 4*I*a^2*e^(4*I*f*x + 4*I*e) + I*a^2*e^(2*I*f*x + 2*I*e) + 2*I*a^2
)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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