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

Optimal. Leaf size=291 \[ \frac{(B+i A) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}+\frac{(A+i B) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac{\left (2 a^2 B-a A b (n+3)-b^2 B \left (n^2+5 n+6\right )\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+1) (n+2) (n+3)}-\frac{\tan (c+d x) (2 a B-A b (n+3)) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}+\frac{B \tan ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+3)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.583964, antiderivative size = 289, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3607, 3647, 3630, 3539, 3537, 68} \[ \frac{\left (2 a^2 B-a A b (n+3)-b^2 B (n+2) (n+3)\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+1) (n+2) (n+3)}-\frac{\tan (c+d x) (2 a B-A b (n+3)) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}+\frac{(B+i A) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}+\frac{(A+i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac{B \tan ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3607

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*
f*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3647

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

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(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*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3539

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

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^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+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac{B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac{\int \tan (c+d x) (a+b \tan (c+d x))^n \left (-2 a B-b B (3+n) \tan (c+d x)-(2 a B-A b (3+n)) \tan ^2(c+d x)\right ) \, dx}{b (3+n)}\\ &=-\frac{(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac{\int (a+b \tan (c+d x))^n \left (a (2 a B-A b (3+n))-A b^2 (2+n) (3+n) \tan (c+d x)+\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) \tan ^2(c+d x)\right ) \, dx}{b^2 (2+n) (3+n)}\\ &=\frac{\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac{(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac{\int (a+b \tan (c+d x))^n \left (b^2 B (2+n) (3+n)-A b^2 (2+n) (3+n) \tan (c+d x)\right ) \, dx}{b^2 (2+n) (3+n)}\\ &=\frac{\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac{(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac{1}{2} (-i A+B) \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx+\frac{1}{2} (i A+B) \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx\\ &=\frac{\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac{(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}-\frac{(A-i B) \operatorname{Subst}\left (\int \frac{(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac{(A+i B) \operatorname{Subst}\left (\int \frac{(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=\frac{\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}+\frac{(A-i B) \, _2F_1\left (1,1+n;2+n;\frac{a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}+\frac{(A+i B) \, _2F_1\left (1,1+n;2+n;\frac{a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)}-\frac{(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}\\ \end{align*}

Mathematica [A]  time = 2.24764, size = 281, normalized size = 0.97 \[ \frac{(a+b \tan (c+d x))^{n+1} \left (b^3 \left (n^2+5 n+6\right ) (a+i b) (A-i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )+b^3 \left (n^2+5 n+6\right ) (a-i b) (A+i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )+2 (a-i b) (a+i b) \left (2 a^2 B-a A b (n+3)-b^2 B (n+2) (n+3)\right )-2 b (n+1) (a-i b) (a+i b) \tan (c+d x) (2 a B-A b (n+3))+2 b^2 B (n+1) (n+2) (a-i b) (a+i b) \tan ^2(c+d x)\right )}{2 b^3 d (n+1) (n+2) (n+3) (a-i b) (a+i b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.355, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{3} \left ( a+b\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+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \tan \left (d x + c\right )^{4} + A \tan \left (d x + c\right )^{3}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*tan(d*x + c)^4 + A*tan(d*x + c)^3)*(b*tan(d*x + c) + a)^n, x)

________________________________________________________________________________________

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+b*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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