3.219 \(\int \tan ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=245 \[ \frac{(A-i B) (a+i a \tan (c+d x))^n \text{Hypergeometric2F1}\left (1,n,n+1,\frac{1}{2} (1+i \tan (c+d x))\right )}{2 d n}-\frac{\left (A n (n+3)-i B \left (n^2+3 n+6\right )\right ) (a+i a \tan (c+d x))^{n+1}}{a d (n+1) (n+2) (n+3)}-\frac{(-A (n+3)+i B n) \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (n+2) (n+3)}+\frac{2 (-A (n+3)+i B n) (a+i a \tan (c+d x))^n}{d n (n+2) (n+3)}+\frac{B \tan ^3(c+d x) (a+i a \tan (c+d x))^n}{d (n+3)} \]

[Out]

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

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Rubi [A]  time = 0.649361, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3597, 3592, 3527, 3481, 68} \[ \frac{(A-i B) (a+i a \tan (c+d x))^n \, _2F_1\left (1,n;n+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d n}-\frac{\left (A n (n+3)-i B \left (n^2+3 n+6\right )\right ) (a+i a \tan (c+d x))^{n+1}}{a d (n+1) (n+2) (n+3)}-\frac{(-A (n+3)+i B n) \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (n+2) (n+3)}+\frac{2 (-A (n+3)+i B n) (a+i a \tan (c+d x))^n}{d n (n+2) (n+3)}+\frac{B \tan ^3(c+d x) (a+i a \tan (c+d x))^n}{d (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(I*B*n - A*(3 + n))*(a + I*a*Tan[c + d*x])^n)/(d*n*(2 + n)*(3 + n)) + ((A - I*B)*Hypergeometric2F1[1, n, 1
+ n, (1 + I*Tan[c + d*x])/2]*(a + I*a*Tan[c + d*x])^n)/(2*d*n) - ((I*B*n - A*(3 + n))*Tan[c + d*x]^2*(a + I*a*
Tan[c + d*x])^n)/(d*(2 + n)*(3 + n)) + (B*Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^n)/(d*(3 + n)) - ((A*n*(3 + n)
- I*B*(6 + 3*n + n^2))*(a + I*a*Tan[c + d*x])^(1 + n))/(a*d*(1 + n)*(2 + n)*(3 + 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 ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac{B \tan ^3(c+d x) (a+i a \tan (c+d x))^n}{d (3+n)}+\frac{\int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (-3 a B-a (i B n-A (3+n)) \tan (c+d x)) \, dx}{a (3+n)}\\ &=-\frac{(i B n-A (3+n)) \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n) (3+n)}+\frac{B \tan ^3(c+d x) (a+i a \tan (c+d x))^n}{d (3+n)}+\frac{\int \tan (c+d x) (a+i a \tan (c+d x))^n \left (2 a^2 (i B n-A (3+n))-a^2 \left (i A n (3+n)+B \left (6+3 n+n^2\right )\right ) \tan (c+d x)\right ) \, dx}{a^2 (2+n) (3+n)}\\ &=-\frac{(i B n-A (3+n)) \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n) (3+n)}+\frac{B \tan ^3(c+d x) (a+i a \tan (c+d x))^n}{d (3+n)}-\frac{\left (A n (3+n)-i B \left (6+3 n+n^2\right )\right ) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n) (3+n)}+\frac{\int (a+i a \tan (c+d x))^n \left (a^2 \left (i A n (3+n)+B \left (6+3 n+n^2\right )\right )+2 a^2 (i B n-A (3+n)) \tan (c+d x)\right ) \, dx}{a^2 (2+n) (3+n)}\\ &=\frac{2 (i B n-A (3+n)) (a+i a \tan (c+d x))^n}{d n (2+n) (3+n)}-\frac{(i B n-A (3+n)) \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n) (3+n)}+\frac{B \tan ^3(c+d x) (a+i a \tan (c+d x))^n}{d (3+n)}-\frac{\left (A n (3+n)-i B \left (6+3 n+n^2\right )\right ) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n) (3+n)}-(-i A-B) \int (a+i a \tan (c+d x))^n \, dx\\ &=\frac{2 (i B n-A (3+n)) (a+i a \tan (c+d x))^n}{d n (2+n) (3+n)}-\frac{(i B n-A (3+n)) \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n) (3+n)}+\frac{B \tan ^3(c+d x) (a+i a \tan (c+d x))^n}{d (3+n)}-\frac{\left (A n (3+n)-i B \left (6+3 n+n^2\right )\right ) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n) (3+n)}+\frac{(a (A-i B)) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+n}}{a-x} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac{2 (i B n-A (3+n)) (a+i a \tan (c+d x))^n}{d n (2+n) (3+n)}+\frac{(A-i B) \, _2F_1\left (1,n;1+n;\frac{1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^n}{2 d n}-\frac{(i B n-A (3+n)) \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n) (3+n)}+\frac{B \tan ^3(c+d x) (a+i a \tan (c+d x))^n}{d (3+n)}-\frac{\left (A n (3+n)-i B \left (6+3 n+n^2\right )\right ) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n) (3+n)}\\ \end{align*}

Mathematica [F]  time = 21.1936, size = 0, normalized size = 0. \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.647, 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 ) ^{n} \left ( A+B\tan \left ( dx+c \right ) \right ) \, 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))^n*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

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Maxima [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))^n*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (i \, A + B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (-2 i \, A - 4 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, B e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (2 i \, A - 4 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \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 )^{n}}{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)^3*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(((I*A + B)*e^(8*I*d*x + 8*I*c) + (-2*I*A - 4*B)*e^(6*I*d*x + 6*I*c) + 6*B*e^(4*I*d*x + 4*I*c) + (2*I*
A - 4*B)*e^(2*I*d*x + 2*I*c) - I*A + B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n/(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)**3*(a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \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))^n*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*tan(d*x + c)^3, x)