3.325 \(\int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx\)

Optimal. Leaf size=205 \[ -\frac{i (a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d m}-\frac{i m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d \left (m^2+5 m+6\right )}+\frac{2 i (a+i a \tan (c+d x))^m}{d \left (m^2+5 m+6\right )}+\frac{i \left (m^2+3 m+6\right ) (a+i a \tan (c+d x))^{m+1}}{a d (m+1) (m+2) (m+3)}+\frac{\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)} \]

[Out]

((2*I)*(a + I*a*Tan[c + d*x])^m)/(d*(6 + 5*m + m^2)) - ((I/2)*Hypergeometric2F1[1, m, 1 + m, (1 + I*Tan[c + d*
x])/2]*(a + I*a*Tan[c + d*x])^m)/(d*m) - (I*m*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^m)/(d*(6 + 5*m + m^2)) + (
Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^m)/(d*(3 + m)) + (I*(6 + 3*m + m^2)*(a + I*a*Tan[c + d*x])^(1 + m))/(a*d
*(1 + m)*(2 + m)*(3 + m))

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Rubi [A]  time = 0.358661, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3560, 3597, 3592, 3527, 3481, 68} \[ -\frac{i (a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d m}-\frac{i m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d \left (m^2+5 m+6\right )}+\frac{2 i (a+i a \tan (c+d x))^m}{d \left (m^2+5 m+6\right )}+\frac{i \left (m^2+3 m+6\right ) (a+i a \tan (c+d x))^{m+1}}{a d (m+1) (m+2) (m+3)}+\frac{\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*(a + I*a*Tan[c + d*x])^m)/(d*(6 + 5*m + m^2)) - ((I/2)*Hypergeometric2F1[1, m, 1 + m, (1 + I*Tan[c + d*
x])/2]*(a + I*a*Tan[c + d*x])^m)/(d*m) - (I*m*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^m)/(d*(6 + 5*m + m^2)) + (
Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^m)/(d*(3 + m)) + (I*(6 + 3*m + m^2)*(a + I*a*Tan[c + d*x])^(1 + m))/(a*d
*(1 + m)*(2 + m)*(3 + 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 3597

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

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 ^4(c+d x) (a+i a \tan (c+d x))^m \, dx &=\frac{\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (3+m)}-\frac{\int \tan ^2(c+d x) (a+i a \tan (c+d x))^m (3 a+i a m \tan (c+d x)) \, dx}{a (3+m)}\\ &=-\frac{i m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}+\frac{\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (3+m)}-\frac{\int \tan (c+d x) (a+i a \tan (c+d x))^m \left (-2 i a^2 m+a^2 \left (6+3 m+m^2\right ) \tan (c+d x)\right ) \, dx}{a^2 (2+m) (3+m)}\\ &=-\frac{i m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}+\frac{\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (3+m)}+\frac{i \left (6+3 m+m^2\right ) (a+i a \tan (c+d x))^{1+m}}{a d (1+m) (2+m) (3+m)}-\frac{\int (a+i a \tan (c+d x))^m \left (-a^2 \left (6+3 m+m^2\right )-2 i a^2 m \tan (c+d x)\right ) \, dx}{a^2 (2+m) (3+m)}\\ &=\frac{2 i (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}-\frac{i m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}+\frac{\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (3+m)}+\frac{i \left (6+3 m+m^2\right ) (a+i a \tan (c+d x))^{1+m}}{a d (1+m) (2+m) (3+m)}+\int (a+i a \tan (c+d x))^m \, dx\\ &=\frac{2 i (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}-\frac{i m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}+\frac{\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (3+m)}+\frac{i \left (6+3 m+m^2\right ) (a+i a \tan (c+d x))^{1+m}}{a d (1+m) (2+m) (3+m)}-\frac{(i a) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac{2 i (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}-\frac{i \, _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{i m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}+\frac{\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (3+m)}+\frac{i \left (6+3 m+m^2\right ) (a+i a \tan (c+d x))^{1+m}}{a d (1+m) (2+m) (3+m)}\\ \end{align*}

Mathematica [F]  time = 9.18033, size = 0, normalized size = 0. \[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.316, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{4} \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)^4*(a+I*a*tan(d*x+c))^m,x)

[Out]

int(tan(d*x+c)^4*(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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(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)^4, 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 (e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}{e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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)^4*(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*(e^(8*I*d*x + 8*I*c) - 4*e^(6*I*d*x + 6*I*c) +
6*e^(4*I*d*x + 4*I*c) - 4*e^(2*I*d*x + 2*I*c) + 1)/(e^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x
+ 4*I*c) + 4*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)**4*(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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(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)^4, x)