3.1051 \(\int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^4 \, dx\)

Optimal. Leaf size=134 \[ -\frac{6 i c^4 (a+i a \tan (e+f x))^{m+2}}{a^2 f (m+2)}+\frac{i c^4 (a+i a \tan (e+f x))^{m+3}}{a^3 f (m+3)}-\frac{8 i c^4 (a+i a \tan (e+f x))^m}{f m}+\frac{12 i c^4 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)} \]

[Out]

((-8*I)*c^4*(a + I*a*Tan[e + f*x])^m)/(f*m) + ((12*I)*c^4*(a + I*a*Tan[e + f*x])^(1 + m))/(a*f*(1 + m)) - ((6*
I)*c^4*(a + I*a*Tan[e + f*x])^(2 + m))/(a^2*f*(2 + m)) + (I*c^4*(a + I*a*Tan[e + f*x])^(3 + m))/(a^3*f*(3 + m)
)

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Rubi [A]  time = 0.170084, antiderivative size = 134, 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} \[ -\frac{6 i c^4 (a+i a \tan (e+f x))^{m+2}}{a^2 f (m+2)}+\frac{i c^4 (a+i a \tan (e+f x))^{m+3}}{a^3 f (m+3)}-\frac{8 i c^4 (a+i a \tan (e+f x))^m}{f m}+\frac{12 i c^4 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^m*(c - I*c*Tan[e + f*x])^4,x]

[Out]

((-8*I)*c^4*(a + I*a*Tan[e + f*x])^m)/(f*m) + ((12*I)*c^4*(a + I*a*Tan[e + f*x])^(1 + m))/(a*f*(1 + m)) - ((6*
I)*c^4*(a + I*a*Tan[e + f*x])^(2 + m))/(a^2*f*(2 + m)) + (I*c^4*(a + I*a*Tan[e + f*x])^(3 + m))/(a^3*f*(3 + 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))^4 \, dx &=\left (a^4 c^4\right ) \int \sec ^8(e+f x) (a+i a \tan (e+f x))^{-4+m} \, dx\\ &=-\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int (a-x)^3 (a+x)^{-1+m} \, dx,x,i a \tan (e+f x)\right )}{a^3 f}\\ &=-\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \left (8 a^3 (a+x)^{-1+m}-12 a^2 (a+x)^m+6 a (a+x)^{1+m}-(a+x)^{2+m}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^3 f}\\ &=-\frac{8 i c^4 (a+i a \tan (e+f x))^m}{f m}+\frac{12 i c^4 (a+i a \tan (e+f x))^{1+m}}{a f (1+m)}-\frac{6 i c^4 (a+i a \tan (e+f x))^{2+m}}{a^2 f (2+m)}+\frac{i c^4 (a+i a \tan (e+f x))^{3+m}}{a^3 f (3+m)}\\ \end{align*}

Mathematica [F]  time = 108.492, size = 0, normalized size = 0. \[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^4 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + I*a*Tan[e + f*x])^m*(c - I*c*Tan[e + f*x])^4,x]

[Out]

Integrate[(a + I*a*Tan[e + f*x])^m*(c - I*c*Tan[e + f*x])^4, x]

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Maple [C]  time = 0.605, size = 5385, normalized size = 40.2 \begin{align*} \text{output too large to display} \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))^4,x)

[Out]

result too large to display

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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 )}^{4}{\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))^4,x, algorithm="maxima")

[Out]

integrate((-I*c*tan(f*x + e) + c)^4*(I*a*tan(f*x + e) + a)^m, x)

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Fricas [B]  time = 1.68052, size = 635, normalized size = 4.74 \begin{align*} \frac{{\left (-8 i \, c^{4} m^{3} - 48 i \, c^{4} m^{2} - 88 i \, c^{4} m - 48 i \, c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} - 48 i \, c^{4} +{\left (-48 i \, c^{4} m - 144 i \, c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-24 i \, c^{4} m^{2} - 120 i \, c^{4} m - 144 i \, c^{4}\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^{4} + 6 \, f m^{3} + 11 \, f m^{2} + 6 \, f m +{\left (f m^{4} + 6 \, f m^{3} + 11 \, f m^{2} + 6 \, f m\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \,{\left (f m^{4} + 6 \, f m^{3} + 11 \, f m^{2} + 6 \, f m\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \,{\left (f m^{4} + 6 \, f m^{3} + 11 \, f m^{2} + 6 \, 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))^4,x, algorithm="fricas")

[Out]

(-8*I*c^4*m^3 - 48*I*c^4*m^2 - 88*I*c^4*m - 48*I*c^4*e^(6*I*f*x + 6*I*e) - 48*I*c^4 + (-48*I*c^4*m - 144*I*c^4
)*e^(4*I*f*x + 4*I*e) + (-24*I*c^4*m^2 - 120*I*c^4*m - 144*I*c^4)*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^4 + 6*f*m^3 + 11*f*m^2 + 6*f*m + (f*m^4 + 6*f*m^3 + 11*f*m^2 + 6*f*m)*e^(6
*I*f*x + 6*I*e) + 3*(f*m^4 + 6*f*m^3 + 11*f*m^2 + 6*f*m)*e^(4*I*f*x + 4*I*e) + 3*(f*m^4 + 6*f*m^3 + 11*f*m^2 +
6*f*m)*e^(2*I*f*x + 2*I*e))

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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((a+I*a*tan(f*x+e))**m*(c-I*c*tan(f*x+e))**4,x)

[Out]

Timed out

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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 )}^{4}{\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))^4,x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^4*(I*a*tan(f*x + e) + a)^m, x)