\(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [205]

Optimal result

Integrand size = 34, antiderivative size = 205 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) (2+m) (3+m)}+\frac {4 a^3 (A-i B) \operatorname {Hypergeometric2F1}(1,1+m,2+m,i \tan (c+d x)) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac {(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3675, 3673, 3618, 12, 66} \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {4 a^3 (A-i B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}(1,m+1,m+2,i \tan (c+d x))}{d (m+1)}-\frac {a^3 \left (A \left (2 m^2+11 m+15\right )-i B \left (2 m^2+11 m+17\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3)}-\frac {(A (m+3)-i B (m+5)) \left (a^3+i a^3 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac {i a B (a+i a \tan (c+d x))^2 \tan ^{m+1}(c+d x)}{d (m+3)} \]

[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 12

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

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]

Rule 3675

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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) \operatorname {Hypergeometric2F1}(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] (verified)

Time = 2.55 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.58 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^3 \tan ^{1+m}(c+d x) \left (\frac {(A-i B) (-3 (2+m)+4 (2+m) \operatorname {Hypergeometric2F1}(1,1+m,2+m,i \tan (c+d x))-i (1+m) \tan (c+d x))}{(1+m) (2+m)}+i B \left (\frac {1}{1+m}+\frac {2 i \tan (c+d x)}{2+m}-\frac {\tan ^2(c+d x)}{3+m}\right )\right )}{d} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{3} \left (A +B \tan \left (d x +c \right )\right )d x\]

[In]

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

Fricas [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{m} \,d x } \]

[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)

Sympy [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=- i a^{3} \left (\int i A \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 3 A \tan {\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 3 B \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{4}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 3 i A \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int i B \tan {\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 3 i B \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

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

[Out]

-I*a**3*(Integral(I*A*tan(c + d*x)**m, x) + Integral(-3*A*tan(c + d*x)*tan(c + d*x)**m, x) + Integral(A*tan(c
+ d*x)**3*tan(c + d*x)**m, x) + Integral(-3*B*tan(c + d*x)**2*tan(c + d*x)**m, x) + Integral(B*tan(c + d*x)**4
*tan(c + d*x)**m, x) + Integral(-3*I*A*tan(c + d*x)**2*tan(c + d*x)**m, x) + Integral(I*B*tan(c + d*x)*tan(c +
d*x)**m, x) + Integral(-3*I*B*tan(c + d*x)**3*tan(c + d*x)**m, x))

Maxima [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{m} \,d x } \]

[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)

Giac [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{m} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

[In]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3, x)