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

Optimal. Leaf size=70 \[ \frac{a (A-i B) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}(1,m+1,m+2,i \tan (c+d x))}{d (m+1)}+\frac{i a B \tan ^{m+1}(c+d x)}{d (m+1)} \]

[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))

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Rubi [A]  time = 0.116656, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3592, 3537, 64} \[ \frac{a (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}+\frac{i a B \tan ^{m+1}(c+d x)}{d (m+1)} \]

Antiderivative was successfully verified.

[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 3592

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]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^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 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)) (A+B \tan (c+d x)) \, dx &=\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 ) \operatorname{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) \, _2F_1(1,1+m;2+m;i \tan (c+d x)) \tan ^{1+m}(c+d x)}{d (1+m)}\\ \end{align*}

Mathematica [B]  time = 2.20329, size = 190, normalized size = 2.71 \[ -\frac{i a e^{-i c} 2^{-m-1} \left (-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^{m+1} \cos ^2(c+d x) (1+i \tan (c+d x)) (A+B \tan (c+d x)) \left (-B 2^{m+1}+(B+i A) \left (1+e^{2 i (c+d x)}\right )^{m+1} \text{Hypergeometric2F1}\left (m+1,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right )}{d (m+1) (\cos (d x)+i \sin (d x)) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.821, 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 ) \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))*(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)

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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 )} \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))*(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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{2 \,{\left ({\left (A - i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (A + i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \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 (4 i \, d x + 4 i \, c\right )} + 2 \, 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))*(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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int A \tan ^{m}{\left (c + d x \right )}\, dx + \int B \tan{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int i A \tan{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int i B \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx\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))*(A+B*tan(d*x+c)),x)

[Out]

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

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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 )} \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))*(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)