3.1180 \(\int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=192 \[ -\frac{2 d \left (c^2 (-(m+3))+i c d m+d^2\right ) (a+i a \tan (e+f x))^m}{f m (m+2)}-\frac{d^2 (d m+i c (m+4)) (a+i a \tan (e+f x))^{m+1}}{a f (m+1) (m+2)}+\frac{(d+i c)^3 (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m}+\frac{d (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^2}{f (m+2)} \]

[Out]

(-2*d*(d^2 + I*c*d*m - c^2*(3 + m))*(a + I*a*Tan[e + f*x])^m)/(f*m*(2 + m)) + ((I*c + d)^3*Hypergeometric2F1[1
, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(2*f*m) - (d^2*(d*m + I*c*(4 + m))*(a + I*a*Tan[
e + f*x])^(1 + m))/(a*f*(1 + m)*(2 + m)) + (d*(a + I*a*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^2)/(f*(2 + m))

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Rubi [A]  time = 0.462147, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3560, 3592, 3527, 3481, 68} \[ -\frac{2 d \left (c^2 (-(m+3))+i c d m+d^2\right ) (a+i a \tan (e+f x))^m}{f m (m+2)}-\frac{d^2 (d m+i c (m+4)) (a+i a \tan (e+f x))^{m+1}}{a f (m+1) (m+2)}+\frac{(d+i c)^3 (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m}+\frac{d (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^2}{f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(d^2 + I*c*d*m - c^2*(3 + m))*(a + I*a*Tan[e + f*x])^m)/(f*m*(2 + m)) + ((I*c + d)^3*Hypergeometric2F1[1
, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(2*f*m) - (d^2*(d*m + I*c*(4 + m))*(a + I*a*Tan[
e + f*x])^(1 + m))/(a*f*(1 + m)*(2 + m)) + (d*(a + I*a*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^2)/(f*(2 + m))

Rule 3560

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[
a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

Rule 3592

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rule 3481

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^3 \, dx &=\frac{d (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^2}{f (2+m)}-\frac{\int (a+i a \tan (e+f x))^m (c+d \tan (e+f x)) \left (-a \left (c^2 (2+m)-d (2 d+i c m)\right )+a d (i d m-c (4+m)) \tan (e+f x)\right ) \, dx}{a (2+m)}\\ &=-\frac{d^2 (d m+i c (4+m)) (a+i a \tan (e+f x))^{1+m}}{a f (1+m) (2+m)}+\frac{d (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^2}{f (2+m)}-\frac{\int (a+i a \tan (e+f x))^m \left (a \left (i c^2 d m-i d^3 m-c^3 (2+m)+c d^2 (6+m)\right )+2 a d \left (d^2+i c d m-c^2 (3+m)\right ) \tan (e+f x)\right ) \, dx}{a (2+m)}\\ &=-\frac{2 d \left (d^2+i c d m-c^2 (3+m)\right ) (a+i a \tan (e+f x))^m}{f m (2+m)}-\frac{d^2 (d m+i c (4+m)) (a+i a \tan (e+f x))^{1+m}}{a f (1+m) (2+m)}+\frac{d (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^2}{f (2+m)}+(c-i d)^3 \int (a+i a \tan (e+f x))^m \, dx\\ &=-\frac{2 d \left (d^2+i c d m-c^2 (3+m)\right ) (a+i a \tan (e+f x))^m}{f m (2+m)}-\frac{d^2 (d m+i c (4+m)) (a+i a \tan (e+f x))^{1+m}}{a f (1+m) (2+m)}+\frac{d (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^2}{f (2+m)}+\frac{\left (a (i c+d)^3\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=-\frac{2 d \left (d^2+i c d m-c^2 (3+m)\right ) (a+i a \tan (e+f x))^m}{f m (2+m)}+\frac{(i c+d)^3 \, _2F_1\left (1,m;1+m;\frac{1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 f m}-\frac{d^2 (d m+i c (4+m)) (a+i a \tan (e+f x))^{1+m}}{a f (1+m) (2+m)}+\frac{d (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^2}{f (2+m)}\\ \end{align*}

Mathematica [F]  time = 45.1208, size = 0, normalized size = 0. \[ \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^3 \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.765, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^m*(c+d*tan(f*x+e))^3,x)

[Out]

int((a+I*a*tan(f*x+e))^m*(c+d*tan(f*x+e))^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \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+d*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{3} + 3 i \, c^{2} d - 3 \, c d^{2} - i \, d^{3} +{\left (c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (3 \, c^{3} - 3 i \, c^{2} d + 3 \, c d^{2} - 3 i \, d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (3 \, c^{3} + 3 i \, c^{2} d + 3 \, c d^{2} + 3 i \, d^{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}}{e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^m*(c+d*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

integral((c^3 + 3*I*c^2*d - 3*c*d^2 - I*d^3 + (c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)*e^(6*I*f*x + 6*I*e) + (3*c^3
- 3*I*c^2*d + 3*c*d^2 - 3*I*d^3)*e^(4*I*f*x + 4*I*e) + (3*c^3 + 3*I*c^2*d + 3*c*d^2 + 3*I*d^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/(e^(6*I*f*x + 6*I*e) + 3*e^(4*I*f*x + 4*I*e) + 3*
e^(2*I*f*x + 2*I*e) + 1), x)

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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+d*tan(f*x+e))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \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+d*tan(f*x+e))^3,x, algorithm="giac")

[Out]

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