\(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [218]
Optimal result
Integrand size = 34, antiderivative size = 167 \[
\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(A-i B) \operatorname {AppellF1}(1+m,1-n,1,2+m,-i \tan (c+d x),i \tan (c+d x)) (1+i \tan (c+d x))^{-n} \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^n}{d (1+m)}+\frac {i B \operatorname {Hypergeometric2F1}(1+m,1-n,2+m,-i \tan (c+d x)) (1+i \tan (c+d x))^{-n} \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^n}{d (1+m)}
\]
[Out]
(A-I*B)*AppellF1(1+m,1-n,1,2+m,-I*tan(d*x+c),I*tan(d*x+c))*tan(d*x+c)^(1+m)*(a+I*a*tan(d*x+c))^n/d/(1+m)/((1+I
*tan(d*x+c))^n)+I*B*hypergeom([1+m, 1-n],[2+m],-I*tan(d*x+c))*tan(d*x+c)^(1+m)*(a+I*a*tan(d*x+c))^n/d/(1+m)/((
1+I*tan(d*x+c))^n)
Rubi [A] (verified)
Time = 0.35 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3682, 3645, 140, 138, 3680, 68,
66} \[
\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(A-i B) \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {AppellF1}(m+1,1-n,1,m+2,-i \tan (c+d x),i \tan (c+d x))}{d (m+1)}+\frac {i B \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}(m+1,1-n,m+2,-i \tan (c+d x))}{d (m+1)}
\]
[In]
Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]
[Out]
((A - I*B)*AppellF1[1 + m, 1 - n, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*(a + I*a*T
an[c + d*x])^n)/(d*(1 + m)*(1 + I*Tan[c + d*x])^n) + (I*B*Hypergeometric2F1[1 + m, 1 - n, 2 + m, (-I)*Tan[c +
d*x]]*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^n)/(d*(1 + m)*(1 + I*Tan[c + d*x])^n)
Rule 66
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Rule 68
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !Int
egerQ[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0] && ((RationalQ[m] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0])) |
| !RationalQ[n])
Rule 138
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Rule 140
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 3645
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dis
t[a*(b/f), Subst[Int[(a + x)^(m - 1)*((c + (d/b)*x)^n/(b^2 + a*x)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b,
c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Rule 3680
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, 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] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]
Rule 3682
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*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Rubi steps \begin{align*}
\text {integral}& = -\left ((-A+i B) \int \tan ^m(c+d x) (a+i a \tan (c+d x))^n \, dx\right )+\frac {(i B) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n \, dx}{a} \\ & = \frac {(i a B) \text {Subst}\left (\int x^m (a+i a x)^{-1+n} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (a^2 (i A+B)\right ) \text {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^m (a+x)^{-1+n}}{-a^2+a x} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = \frac {\left (i B (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \text {Subst}\left (\int (1+i x)^{-1+n} x^m \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (a (i A+B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \text {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^m \left (1+\frac {x}{a}\right )^{-1+n}}{-a^2+a x} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = \frac {(A-i B) \operatorname {AppellF1}(1+m,1-n,1,2+m,-i \tan (c+d x),i \tan (c+d x)) (1+i \tan (c+d x))^{-n} \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^n}{d (1+m)}+\frac {i B \operatorname {Hypergeometric2F1}(1+m,1-n,2+m,-i \tan (c+d x)) (1+i \tan (c+d x))^{-n} \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^n}{d (1+m)} \\
\end{align*}
Mathematica [F]
\[
\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx
\]
[In]
Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]
[Out]
Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]), x]
Maple [F]
\[\int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )d x\]
[In]
int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)
[Out]
int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)
Fricas [F]
\[
\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{m} \,d x }
\]
[In]
integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
integral(((A - I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*((-I*
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m/(e^(2*I*d*x + 2*I*c) + 1), x)
Sympy [F]
\[
\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}\, dx
\]
[In]
integrate(tan(d*x+c)**m*(a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)
[Out]
Integral((I*a*(tan(c + d*x) - I))**n*(A + B*tan(c + d*x))*tan(c + d*x)**m, x)
Maxima [F]
\[
\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{m} \,d x }
\]
[In]
integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*tan(d*x + c)^m, x)
Giac [F]
\[
\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{m} \,d x }
\]
[In]
integrate(tan(d*x+c)^m*(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)^m, x)
Mupad [F(-1)]
Timed out. \[
\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x
\]
[In]
int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n,x)
[Out]
int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n, x)