\(\int \tan ^m(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [207]

Optimal result

Integrand size = 32, antiderivative size = 70 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {i a B \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {a (A-i B) \operatorname {Hypergeometric2F1}(1,1+m,2+m,i \tan (c+d x)) \tan ^{1+m}(c+d x)}{d (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3673, 3618, 66} \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {a (A-i B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}(1,m+1,m+2,i \tan (c+d x))}{d (m+1)}+\frac {i a B \tan ^{m+1}(c+d x)}{d (m+1)} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {a (i B+(A-i B) \operatorname {Hypergeometric2F1}(1,1+m,2+m,i \tan (c+d x))) \tan ^{1+m}(c+d x)}{d (1+m)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

Fricas [F]

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

[In]

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

[Out]

integral(2*((A - I*B)*a*e^(4*I*d*x + 4*I*c) + (A + I*B)*a*e^(2*I*d*x + 2*I*c))*((-I*e^(2*I*d*x + 2*I*c) + I)/(
e^(2*I*d*x + 2*I*c) + 1))^m/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1), x)

Sympy [F]

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

[In]

integrate(tan(d*x+c)**m*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

I*a*(Integral(-I*A*tan(c + d*x)**m, x) + Integral(A*tan(c + d*x)*tan(c + d*x)**m, x) + Integral(B*tan(c + d*x)
**2*tan(c + d*x)**m, x) + Integral(-I*B*tan(c + d*x)*tan(c + d*x)**m, x))

Maxima [F]

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

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)*tan(d*x + c)^m, x)

Giac [F]

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

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)*tan(d*x + c)^m, x)

Mupad [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right ) \,d x \]

[In]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i),x)

[Out]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i), x)