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

Optimal. Leaf size=167 \[ \frac{i B \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \text{Hypergeometric2F1}(m+1,1-n,m+2,-i \tan (c+d x))}{d (m+1)}+\frac{(A-i B) \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1(m+1;1-n,1;m+2;-i \tan (c+d x),i \tan (c+d x))}{d (m+1)} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.30397, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {3601, 3564, 135, 133, 3599, 66, 64} \[ \frac{(A-i B) \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1(m+1;1-n,1;m+2;-i \tan (c+d x),i \tan (c+d x))}{d (m+1)}+\frac{i B \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1(m+1,1-n;m+2;-i \tan (c+d x))}{d (m+1)} \]

Antiderivative was successfully verified.

[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 3601

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]

Rule 3564

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*x)/b)^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 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])

Rule 3599

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 66

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 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), 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])))

Rubi steps

\begin{align*} \int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=-\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) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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) F_1(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 \, _2F_1(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]  time = 18.1913, size = 0, normalized size = 0. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[c + d*x]^m*(a + 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]  time = 184.135, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( a+ia\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+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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \left (\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(-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+I*a*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 (i \, a \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+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)