3.1053 \(\int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=64 \[ \frac{i c^2 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)}-\frac{2 i c^2 (a+i a \tan (e+f x))^m}{f m} \]

[Out]

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

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Rubi [A]  time = 0.13129, antiderivative size = 64, 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} \[ \frac{i c^2 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)}-\frac{2 i c^2 (a+i a \tan (e+f x))^m}{f m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 41.0072, size = 131, normalized size = 2.05 \[ -\frac{i c^2 2^{m+1} e^{-i (e+f x)} \left (e^{i f x}\right )^m \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{m+1} \left (e^{2 i (e+f x)}+m+1\right ) \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m}{f m (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.261, size = 100, normalized size = 1.6 \begin{align*} -{\frac{{c}^{2}\tan \left ( fx+e \right ){{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+m \right ) }}-{\frac{i{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}{c}^{2}}{f \left ( 1+m \right ) }}-{\frac{2\,i{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}{c}^{2}}{fm \left ( 1+m \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2/f/(1+m)*tan(f*x+e)*exp(m*ln(a+I*a*tan(f*x+e)))-I/f/(1+m)*exp(m*ln(a+I*a*tan(f*x+e)))*c^2-2*I/f/m/(1+m)*ex
p(m*ln(a+I*a*tan(f*x+e)))*c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67163, size = 213, normalized size = 3.33 \begin{align*} \frac{{\left (-2 i \, c^{2} m - 2 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, c^{2}\right )} \left (\frac{2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m}}{f m^{2} + f m +{\left (f m^{2} + f m\right )} e^{\left (2 i \, f x + 2 i \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))**m*(c-I*c*tan(f*x+e))**2,x)

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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