3.942 \(\int \frac{(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=58 \[ \frac{i a^2}{2 c f (c-i c \tan (e+f x))^2}-\frac{2 i a^2}{3 f (c-i c \tan (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.113296, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ \frac{i a^2}{2 c f (c-i c \tan (e+f x))^2}-\frac{2 i a^2}{3 f (c-i c \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.3485, size = 53, normalized size = 0.91 \[ \frac{a^2 (5 \cos (e+f x)-i \sin (e+f x)) (\sin (5 (e+f x))-i \cos (5 (e+f x)))}{24 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(5*Cos[e + f*x] - I*Sin[e + f*x])*((-I)*Cos[5*(e + f*x)] + Sin[5*(e + f*x)]))/(24*c^3*f)

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Maple [A]  time = 0.027, size = 39, normalized size = 0.7 \begin{align*}{\frac{{a}^{2}}{f{c}^{3}} \left ({\frac{-{\frac{i}{2}}}{ \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{2}{3\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}} \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))^3,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.26518, size = 101, normalized size = 1.74 \begin{align*} \frac{-2 i \, a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )}}{24 \, 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))^3,x, algorithm="fricas")

[Out]

1/24*(-2*I*a^2*e^(6*I*f*x + 6*I*e) - 3*I*a^2*e^(4*I*f*x + 4*I*e))/(c^3*f)

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Sympy [A]  time = 0.447423, size = 97, normalized size = 1.67 \begin{align*} \begin{cases} \frac{- 8 i a^{2} c^{3} f e^{6 i e} e^{6 i f x} - 12 i a^{2} c^{3} f e^{4 i e} e^{4 i f x}}{96 c^{6} f^{2}} & \text{for}\: 96 c^{6} f^{2} \neq 0 \\\frac{x \left (a^{2} e^{6 i e} + a^{2} e^{4 i e}\right )}{2 c^{3}} & \text{otherwise} \end{cases} \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))**3,x)

[Out]

Piecewise(((-8*I*a**2*c**3*f*exp(6*I*e)*exp(6*I*f*x) - 12*I*a**2*c**3*f*exp(4*I*e)*exp(4*I*f*x))/(96*c**6*f**2
), Ne(96*c**6*f**2, 0)), (x*(a**2*exp(6*I*e) + a**2*exp(4*I*e))/(2*c**3), True))

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Giac [B]  time = 1.48326, size = 143, normalized size = 2.47 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 3 i \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 8 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 i \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{3 \, c^{3} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{6}} \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))^3,x, algorithm="giac")

[Out]

-2/3*(3*a^2*tan(1/2*f*x + 1/2*e)^5 + 3*I*a^2*tan(1/2*f*x + 1/2*e)^4 - 8*a^2*tan(1/2*f*x + 1/2*e)^3 - 3*I*a^2*t
an(1/2*f*x + 1/2*e)^2 + 3*a^2*tan(1/2*f*x + 1/2*e))/(c^3*f*(tan(1/2*f*x + 1/2*e) + I)^6)