3.205 \(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=205 \[ \frac{4 a^3 (A-i B) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}(1,m+1,m+2,i \tan (c+d x))}{d (m+1)}-\frac{a^3 \left (A \left (2 m^2+11 m+15\right )-i B \left (2 m^2+11 m+17\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3)}-\frac{(A (m+3)-i B (m+5)) \left (a^3+i a^3 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac{i a B (a+i a \tan (c+d x))^2 \tan ^{m+1}(c+d x)}{d (m+3)} \]

[Out]

-((a^3*(A*(15 + 11*m + 2*m^2) - I*B*(17 + 11*m + 2*m^2))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(2 + m)*(3 + m))) +
(4*a^3*(A - I*B)*Hypergeometric2F1[1, 1 + m, 2 + m, I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m))/(d*(1 + m)) + (I*a*B
*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^2)/(d*(3 + m)) - ((A*(3 + m) - I*B*(5 + m))*Tan[c + d*x]^(1 + m)*
(a^3 + I*a^3*Tan[c + d*x]))/(d*(2 + m)*(3 + m))

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Rubi [A]  time = 0.642078, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{4 a^3 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}-\frac{a^3 \left (A \left (2 m^2+11 m+15\right )-i B \left (2 m^2+11 m+17\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3)}-\frac{(A (m+3)-i B (m+5)) \left (a^3+i a^3 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac{i a B (a+i a \tan (c+d x))^2 \tan ^{m+1}(c+d x)}{d (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-((a^3*(A*(15 + 11*m + 2*m^2) - I*B*(17 + 11*m + 2*m^2))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(2 + m)*(3 + m))) +
(4*a^3*(A - I*B)*Hypergeometric2F1[1, 1 + m, 2 + m, I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m))/(d*(1 + m)) + (I*a*B
*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^2)/(d*(3 + m)) - ((A*(3 + m) - I*B*(5 + m))*Tan[c + d*x]^(1 + m)*
(a^3 + I*a^3*Tan[c + d*x]))/(d*(2 + m)*(3 + 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))^3 (A+B \tan (c+d x)) \, dx &=\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}+\frac{\int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (-a (i B (1+m)-A (3+m))+a (i A (3+m)+B (5+m)) \tan (c+d x)) \, dx}{3+m}\\ &=\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac{(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}+\frac{\int \tan ^m(c+d x) (a+i a \tan (c+d x)) \left (-a^2 \left (i B \left (7+9 m+2 m^2\right )-A \left (9+9 m+2 m^2\right )\right )+a^2 \left (i A \left (15+11 m+2 m^2\right )+B \left (17+11 m+2 m^2\right )\right ) \tan (c+d x)\right ) \, dx}{6+5 m+m^2}\\ &=-\frac{a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (6+5 m+m^2\right )}+\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac{(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}+\frac{\int \tan ^m(c+d x) \left (4 a^3 (A-i B) (2+m) (3+m)+4 a^3 (i A+B) (2+m) (3+m) \tan (c+d x)\right ) \, dx}{6+5 m+m^2}\\ &=-\frac{a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (6+5 m+m^2\right )}+\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac{(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}+\frac{\left (16 i a^6 (A-i B)^2 (2+m) (3+m)\right ) \operatorname{Subst}\left (\int \frac{4^{-m} \left (\frac{x}{a^3 (i A+B) (2+m) (3+m)}\right )^m}{16 a^6 (i A+B)^2 (2+m)^2 (3+m)^2+4 a^3 (A-i B) (2+m) (3+m) x} \, dx,x,4 a^3 (i A+B) (2+m) (3+m) \tan (c+d x)\right )}{d}\\ &=-\frac{a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (6+5 m+m^2\right )}+\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac{(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}+\frac{\left (i 4^{2-m} a^6 (A-i B)^2 (2+m) (3+m)\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{a^3 (i A+B) (2+m) (3+m)}\right )^m}{16 a^6 (i A+B)^2 (2+m)^2 (3+m)^2+4 a^3 (A-i B) (2+m) (3+m) x} \, dx,x,4 a^3 (i A+B) (2+m) (3+m) \tan (c+d x)\right )}{d}\\ &=-\frac{a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (6+5 m+m^2\right )}+\frac{4 a^3 (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 a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac{(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}\\ \end{align*}

Mathematica [A]  time = 10.8678, size = 374, normalized size = 1.82 \[ -\frac{i e^{i c} \left (-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \left (\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-m} \cos ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (2 e^{-4 i c} (m+3) (A-i B) \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (2^{1-m} (m+2) \text{Hypergeometric2F1}\left (m+1,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+\left (-(4 m+7) e^{2 i (c+d x)}-2 m-5\right ) \left (1+e^{2 i (c+d x)}\right )^{-m-2}\right )+4 i B e^{-4 i c} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (\left (2 m^2+8 m+7\right ) e^{4 i (c+d x)}+2 (m+2) e^{2 i (c+d x)}+1\right ) \left (1+e^{2 i (c+d x)}\right )^{-m-3}\right )}{2 d (m+1) (m+2) (m+3) (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

((-I/2)*E^(I*c)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c + d*x]^4*(((4*I)*B*(-1 +
E^((2*I)*(c + d*x)))^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(-3 - m)*(1 + 2*E^((2*I)*(c + d*x))*(2 + m) + E^((4*I)
*(c + d*x))*(7 + 8*m + 2*m^2)))/E^((4*I)*c) + (2*(A - I*B)*(-1 + E^((2*I)*(c + d*x)))^(1 + m)*(3 + m)*((1 + E^
((2*I)*(c + d*x)))^(-2 - m)*(-5 - 2*m - E^((2*I)*(c + d*x))*(7 + 4*m)) + 2^(1 - m)*(2 + m)*Hypergeometric2F1[1
+ m, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/2]))/E^((4*I)*c))*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))
/(d*((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^m*(1 + m)*(2 + m)*(3 + m)*(Cos[d*x] + I*Sin[d*x])^3
*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [F]  time = 0.383, 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 ) ^{3} \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))^3*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^3*(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 )}^{3} \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))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*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{8 \,{\left ({\left (A - i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (A + i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 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 (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(8*((A - I*B)*a^3*e^(8*I*d*x + 8*I*c) + (A + I*B)*a^3*e^(6*I*d*x + 6*I*c))*((-I*e^(2*I*d*x + 2*I*c) +
I)/(e^(2*I*d*x + 2*I*c) + 1))^m/(e^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*c) + 4*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^{3} \left (\int A \tan ^{m}{\left (c + d x \right )}\, dx + \int - 3 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 - 3 B \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int 3 i A \tan{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int - i A \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int 3 i B \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int - i B \tan ^{4}{\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))**3*(A+B*tan(d*x+c)),x)

[Out]

a**3*(Integral(A*tan(c + d*x)**m, x) + Integral(-3*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(-3*B*tan(c + d*x)**3*tan(c + d*x)**m, x) + Integral(3*I*A*tan(c + d*x)*tan
(c + d*x)**m, x) + Integral(-I*A*tan(c + d*x)**3*tan(c + d*x)**m, x) + Integral(3*I*B*tan(c + d*x)**2*tan(c +
d*x)**m, x) + Integral(-I*B*tan(c + d*x)**4*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 )}^{3} \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))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*tan(d*x + c)^m, x)