\(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [206]

Optimal result

Integrand size = 34, antiderivative size = 132 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {i a^2 (B+(i A+B) (2+m)) \tan ^{1+m}(c+d x)}{d (1+m) (2+m)}+\frac {2 a^2 (A-i B) \operatorname {Hypergeometric2F1}(1,1+m,2+m,i \tan (c+d x)) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {i B \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d (2+m)} \]

[Out]

I*a^2*(B+(I*A+B)*(2+m))*tan(d*x+c)^(1+m)/d/(1+m)/(2+m)+2*a^2*(A-I*B)*hypergeom([1, 1+m],[2+m],I*tan(d*x+c))*ta
n(d*x+c)^(1+m)/d/(1+m)+I*B*tan(d*x+c)^(1+m)*(a^2+I*a^2*tan(d*x+c))/d/(2+m)

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3675, 3673, 3618, 12, 66} \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {2 a^2 (A-i B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}(1,m+1,m+2,i \tan (c+d x))}{d (m+1)}+\frac {i a^2 (B+(m+2) (B+i A)) \tan ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac {i B \left (a^2+i a^2 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2)} \]

[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 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 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 ) \text {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 ) \text {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) \operatorname {Hypergeometric2F1}(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 [A] (verified)

Time = 1.57 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {a^2 \tan ^{1+m}(c+d x) ((A-2 i B) (2+m)-2 (A-i B) (2+m) \operatorname {Hypergeometric2F1}(1,1+m,2+m,i \tan (c+d x))+B (1+m) \tan (c+d x))}{d (1+m) (2+m)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{2} \left (A +B \tan \left (d x +c \right )\right )d x\]

[In]

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

Fricas [F]

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

[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)

Sympy [F]

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

[In]

integrate(tan(d*x+c)**m*(a+I*a*tan(d*x+c))**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))

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]

[In]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2, x)