\(\int \tan (c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [497]

Optimal result

Integrand size = 29, antiderivative size = 168 \[ \int \tan (c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {B (a+b \tan (c+d x))^{1+n}}{b d (1+n)}-\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}-\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)} \]

[Out]

B*(a+b*tan(d*x+c))^(1+n)/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)/(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)/(a+I*b)/d/(1+n)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3673, 3620, 3618, 70} \[ \int \tan (c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {(A-i B) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}-\frac {(A+i B) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac {B (a+b \tan (c+d x))^{n+1}}{b d (n+1)} \]

[In]

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

[Out]

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

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 3673

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]

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.74 \[ \int \tan (c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {\left (\frac {2 B}{b}-\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )}{a-i b}-\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 d (1+n)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(tan(d*x+c)*(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), x)

Maxima [F]

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

[In]

integrate(tan(d*x+c)*(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), x)

Giac [F]

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

[In]

integrate(tan(d*x+c)*(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), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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