\(\int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [496]

Optimal result

Integrand size = 31, antiderivative size = 219 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {(a B-A b (2+n)) (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {(i A+B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}+\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a-b) d (1+n)}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3688, 3711, 3620, 3618, 70} \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {(a B-A b (n+2)) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+1) (n+2)}+\frac {(B+i A) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}+\frac {(A+i B) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (-b+i a)}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+2)} \]

[In]

Int[Tan[c + d*x]^2*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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 3688

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 3711

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]

Rubi steps \begin{align*} \text {integral}& = \frac {B \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}+\frac {\int (a+b \tan (c+d x))^n \left (-a B-b B (2+n) \tan (c+d x)-(a B-A b (2+n)) \tan ^2(c+d x)\right ) \, dx}{b (2+n)} \\ & = -\frac {(a B-A b (2+n)) (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}+\frac {\int (a+b \tan (c+d x))^n (-A b (2+n)-b B (2+n) \tan (c+d x)) \, dx}{b (2+n)} \\ & = -\frac {(a B-A b (2+n)) (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}+\frac {1}{2} (-A-i B) \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx+\frac {1}{2} (-A+i B) \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx \\ & = -\frac {(a B-A b (2+n)) (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}+\frac {(i A-B) \text {Subst}\left (\int \frac {(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}-\frac {(i A+B) \text {Subst}\left (\int \frac {(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d} \\ & = -\frac {(a B-A b (2+n)) (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {(i A+B) \operatorname {Hypergeometric2F1}\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) \operatorname {Hypergeometric2F1}\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 (i a-b) d (1+n)}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.77 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (\frac {4 A b-2 a B+2 A b n}{b+b n}+\frac {b (i A+B) (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )}{(a-i b) (1+n)}+\frac {b (-i A+B) (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )}{(a+i b) (1+n)}+2 B \tan (c+d x)\right )}{2 b d (2+n)} \]

[In]

Integrate[Tan[c + d*x]^2*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

((a + b*Tan[c + d*x])^(1 + n)*((4*A*b - 2*a*B + 2*A*b*n)/(b + b*n) + (b*(I*A + B)*(2 + n)*Hypergeometric2F1[1,
1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - I*b)])/((a - I*b)*(1 + n)) + (b*((-I)*A + B)*(2 + n)*Hypergeometric2F
1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)])/((a + I*b)*(1 + n)) + 2*B*Tan[c + d*x]))/(2*b*d*(2 + n))

Maple [F]

\[\int \tan \left (d x +c \right )^{2} \left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )d x\]

[In]

int(tan(d*x+c)^2*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^2*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

Fricas [F]

\[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(tan(d*x+c)^2*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*tan(d*x + c)^3 + A*tan(d*x + c)^2)*(b*tan(d*x + c) + a)^n, x)

Sympy [F]

\[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{n} \tan ^{2}{\left (c + d x \right )}\, dx \]

[In]

integrate(tan(d*x+c)**2*(a+b*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**n*tan(c + d*x)**2, x)

Maxima [F]

\[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(tan(d*x+c)^2*(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)^2, x)

Giac [F]

\[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(tan(d*x+c)^2*(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)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^2\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]

[In]

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^n, x)