3.206 \(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=132 \[ \frac{2 a^2 (A-i B) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}(1,m+1,m+2,i \tan (c+d x))}{d (m+1)}+\frac{i a^2 (B+(m+2) (B+i A)) \tan ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac{i B \left (a^2+i a^2 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.359607, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3594, 3592, 3537, 12, 64} \[ \frac{2 a^2 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}+\frac{i a^2 (B+(m+2) (B+i A)) \tan ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac{i B \left (a^2+i a^2 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

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

Rule 3594

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

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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))^2 (A+B \tan (c+d x)) \, dx &=\frac{i B \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d (2+m)}+\frac{\int \tan ^m(c+d x) (a+i a \tan (c+d x)) (-a (i B (1+m)-A (2+m))+a (B+(i A+B) (2+m)) \tan (c+d x)) \, dx}{2+m}\\ &=\frac{i a^2 (B+(i A+B) (2+m)) \tan ^{1+m}(c+d x)}{d (1+m) (2+m)}+\frac{i B \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d (2+m)}+\frac{\int \tan ^m(c+d x) \left (2 a^2 (A-i B) (2+m)+2 a^2 (i A+B) (2+m) \tan (c+d x)\right ) \, dx}{2+m}\\ &=\frac{i a^2 (B+(i A+B) (2+m)) \tan ^{1+m}(c+d x)}{d (1+m) (2+m)}+\frac{i B \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d (2+m)}+\frac{\left (4 i a^4 (A-i B)^2 (2+m)\right ) \operatorname{Subst}\left (\int \frac{2^{-m} \left (\frac{x}{a^2 (i A+B) (2+m)}\right )^m}{4 a^4 (i A+B)^2 (2+m)^2+2 a^2 (A-i B) (2+m) x} \, dx,x,2 a^2 (i A+B) (2+m) \tan (c+d x)\right )}{d}\\ &=\frac{i a^2 (B+(i A+B) (2+m)) \tan ^{1+m}(c+d x)}{d (1+m) (2+m)}+\frac{i B \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d (2+m)}+\frac{\left (i 2^{2-m} a^4 (A-i B)^2 (2+m)\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{a^2 (i A+B) (2+m)}\right )^m}{4 a^4 (i A+B)^2 (2+m)^2+2 a^2 (A-i B) (2+m) x} \, dx,x,2 a^2 (i A+B) (2+m) \tan (c+d x)\right )}{d}\\ &=\frac{i a^2 (B+(i A+B) (2+m)) \tan ^{1+m}(c+d x)}{d (1+m) (2+m)}+\frac{2 a^2 (A-i B) \, _2F_1(1,1+m;2+m;i \tan (c+d x)) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac{i B \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d (2+m)}\\ \end{align*}

Mathematica [B]  time = 5.58052, size = 323, normalized size = 2.45 \[ -\frac{2 i e^{2 i c} \left (-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \left (\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-m} \cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (\frac{e^{-4 i c} 2^{-m-3} (A-i B) \left (-1+e^{2 i (c+d x)}\right )^{m+3} \text{Hypergeometric2F1}\left (m+3,m+3,m+4,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )}{m+3}+\frac{e^{-4 i c} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{-m-2} \left (A (3 m+4) e^{2 i (c+d x)}-A m-i B (m+1) \left (-1+e^{2 i (c+d x)}\right )\right )}{2 (m+1) (m+2)}\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

((-2*I)*E^((2*I)*c)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c + d*x]^3*(((-1 + E^(
(2*I)*(c + d*x)))^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(-2 - m)*(-(A*m) - I*B*(-1 + E^((2*I)*(c + d*x)))*(1 + m)
+ A*E^((2*I)*(c + d*x))*(4 + 3*m)))/(2*E^((4*I)*c)*(1 + m)*(2 + m)) + (2^(-3 - m)*(A - I*B)*(-1 + E^((2*I)*(c
+ d*x)))^(3 + m)*Hypergeometric2F1[3 + m, 3 + m, 4 + m, (1 - E^((2*I)*(c + d*x)))/2])/(E^((4*I)*c)*(3 + m)))*(
a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(d*((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^m*(Cos
[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [F]  time = 0.346, 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 ) ^{2} \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))^2*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^2*(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 )}^{2} \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))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^2*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{4 \,{\left ({\left (A - i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (A + i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 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 (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(4*((A - I*B)*a^2*e^(6*I*d*x + 6*I*c) + (A + I*B)*a^2*e^(4*I*d*x + 4*I*c))*((-I*e^(2*I*d*x + 2*I*c) +
I)/(e^(2*I*d*x + 2*I*c) + 1))^m/(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3*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^{2} \left (\int A \tan ^{m}{\left (c + d x \right )}\, dx + \int - A \tan ^{2}{\left (c + d x \right )} \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 - B \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int 2 i A \tan{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int 2 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))**2*(A+B*tan(d*x+c)),x)

[Out]

a**2*(Integral(A*tan(c + d*x)**m, x) + Integral(-A*tan(c + d*x)**2*tan(c + d*x)**m, x) + Integral(B*tan(c + d*
x)*tan(c + d*x)**m, x) + Integral(-B*tan(c + d*x)**3*tan(c + d*x)**m, x) + Integral(2*I*A*tan(c + d*x)*tan(c +
d*x)**m, x) + Integral(2*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 )}^{2} \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))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^2*tan(d*x + c)^m, x)