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

Optimal. Leaf size=144 \[ \frac{(a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d m}-\frac{m (a+i a \tan (c+d x))^{m+1}}{a d \left (m^2+3 m+2\right )}+\frac{\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac{2 (a+i a \tan (c+d x))^m}{d m (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.191903, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3560, 3592, 3527, 3481, 68} \[ \frac{(a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d m}-\frac{m (a+i a \tan (c+d x))^{m+1}}{a d \left (m^2+3 m+2\right )}+\frac{\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac{2 (a+i a \tan (c+d x))^m}{d m (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [F]  time = 6.54585, size = 0, normalized size = 0. \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.622, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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