3.1052 \(\int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=99 \[ -\frac{i c^3 (a+i a \tan (e+f x))^{m+2}}{a^2 f (m+2)}-\frac{4 i c^3 (a+i a \tan (e+f x))^m}{f m}+\frac{4 i c^3 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)} \]

[Out]

((-4*I)*c^3*(a + I*a*Tan[e + f*x])^m)/(f*m) + ((4*I)*c^3*(a + I*a*Tan[e + f*x])^(1 + m))/(a*f*(1 + m)) - (I*c^
3*(a + I*a*Tan[e + f*x])^(2 + m))/(a^2*f*(2 + m))

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Rubi [A]  time = 0.145773, antiderivative size = 99, 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} \[ -\frac{i c^3 (a+i a \tan (e+f x))^{m+2}}{a^2 f (m+2)}-\frac{4 i c^3 (a+i a \tan (e+f x))^m}{f m}+\frac{4 i c^3 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*c^3*(a + I*a*Tan[e + f*x])^m)/(f*m) + ((4*I)*c^3*(a + I*a*Tan[e + f*x])^(1 + m))/(a*f*(1 + m)) - (I*c^
3*(a + I*a*Tan[e + f*x])^(2 + m))/(a^2*f*(2 + 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))^3 \, dx &=\left (a^3 c^3\right ) \int \sec ^6(e+f x) (a+i a \tan (e+f x))^{-3+m} \, dx\\ &=-\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int (a-x)^2 (a+x)^{-1+m} \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=-\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \left (4 a^2 (a+x)^{-1+m}-4 a (a+x)^m+(a+x)^{1+m}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=-\frac{4 i c^3 (a+i a \tan (e+f x))^m}{f m}+\frac{4 i c^3 (a+i a \tan (e+f x))^{1+m}}{a f (1+m)}-\frac{i c^3 (a+i a \tan (e+f x))^{2+m}}{a^2 f (2+m)}\\ \end{align*}

Mathematica [F]  time = 180.002, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [B]  time = 0.357, size = 192, normalized size = 1.9 \begin{align*}{\frac{i{c}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{2}{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 2+m \right ) }}-2\,{\frac{{c}^{3} \left ( 3+m \right ) \tan \left ( fx+e \right ){{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+m \right ) \left ( 2+m \right ) }}-{\frac{i{c}^{3}m{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+m \right ) \left ( 2+m \right ) }}-{\frac{5\,i{c}^{3}{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+m \right ) \left ( 2+m \right ) }}-{\frac{8\,i{c}^{3}{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{fm \left ( 1+m \right ) \left ( 2+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))^3,x)

[Out]

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

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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 )}^{3}{\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))^3,x, algorithm="maxima")

[Out]

integrate((-I*c*tan(f*x + e) + c)^3*(I*a*tan(f*x + e) + a)^m, x)

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Fricas [A]  time = 1.63905, size = 396, normalized size = 4. \begin{align*} \frac{{\left (-4 i \, c^{3} m^{2} - 12 i \, c^{3} m - 8 i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 8 i \, c^{3} +{\left (-8 i \, c^{3} m - 16 i \, c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\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^{3} + 3 \, f m^{2} + 2 \, f m +{\left (f m^{3} + 3 \, f m^{2} + 2 \, f m\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \,{\left (f m^{3} + 3 \, f m^{2} + 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))^3,x, algorithm="fricas")

[Out]

(-4*I*c^3*m^2 - 12*I*c^3*m - 8*I*c^3*e^(4*I*f*x + 4*I*e) - 8*I*c^3 + (-8*I*c^3*m - 16*I*c^3)*e^(2*I*f*x + 2*I*
e))*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m/(f*m^3 + 3*f*m^2 + 2*f*m + (f*m^3 + 3*f*m^2 + 2*f*m)
*e^(4*I*f*x + 4*I*e) + 2*(f*m^3 + 3*f*m^2 + 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))**3,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 )}^{3}{\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))^3,x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^3*(I*a*tan(f*x + e) + a)^m, x)