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

Optimal. Leaf size=183 \[ \frac{(A+i B) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1)}+\frac{(A-i B) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1)} \]

[Out]

((A + I*B)*AppellF1[1 + m, -n, 1, 2 + m, -((b*Tan[c + d*x])/a), (-I)*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*(a + b
*Tan[c + d*x])^n)/(2*d*(1 + m)*(1 + (b*Tan[c + d*x])/a)^n) + ((A - I*B)*AppellF1[1 + m, -n, 1, 2 + m, -((b*Tan
[c + d*x])/a), I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*(a + b*Tan[c + d*x])^n)/(2*d*(1 + m)*(1 + (b*Tan[c + d*x])
/a)^n)

________________________________________________________________________________________

Rubi [A]  time = 0.313777, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3603, 3602, 135, 133} \[ \frac{(A+i B) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1)}+\frac{(A-i B) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + I*B)*AppellF1[1 + m, -n, 1, 2 + m, -((b*Tan[c + d*x])/a), (-I)*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*(a + b
*Tan[c + d*x])^n)/(2*d*(1 + m)*(1 + (b*Tan[c + d*x])/a)^n) + ((A - I*B)*AppellF1[1 + m, -n, 1, 2 + m, -((b*Tan
[c + d*x])/a), I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*(a + b*Tan[c + d*x])^n)/(2*d*(1 + m)*(1 + (b*Tan[c + d*x])
/a)^n)

Rule 3603

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

Rule 3602

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

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac{1}{2} (A-i B) \int (1+i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^n \, dx+\frac{1}{2} (A+i B) \int (1-i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^n \, dx\\ &=\frac{(A-i B) \operatorname{Subst}\left (\int \frac{x^m (a+b x)^n}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{(A+i B) \operatorname{Subst}\left (\int \frac{x^m (a+b x)^n}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\left ((A-i B) (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1+\frac{b x}{a}\right )^n}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left ((A+i B) (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1+\frac{b x}{a}\right )^n}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{(A+i B) F_1\left (1+m;-n,1;2+m;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}}{2 d (1+m)}+\frac{(A-i B) F_1\left (1+m;-n,1;2+m;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}}{2 d (1+m)}\\ \end{align*}

Mathematica [F]  time = 2.10371, size = 0, normalized size = 0. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.415, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{m} \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)^m*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(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)^m, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\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 )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^n*tan(d*x + c)^m, 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)**m*(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 )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(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)^m, x)