\(\int \tan ^m(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [482]

Optimal result

Integrand size = 29, antiderivative size = 127 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {b B \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {(a A-b B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {(A b+a B) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (2+m)} \]

[Out]

b*B*tan(d*x+c)^(1+m)/d/(1+m)+(A*a-B*b)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(1+m)/d/
(1+m)+(A*b+B*a)*hypergeom([1, 1+1/2*m],[2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(2+m)/d/(2+m)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3673, 3619, 3557, 371} \[ \int \tan ^m(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {(a A-b B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}+\frac {(a B+A b) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}+\frac {b B \tan ^{m+1}(c+d x)}{d (m+1)} \]

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3619

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {b B \tan ^{1+m}(c+d x)}{d (1+m)}+\int \tan ^m(c+d x) (a A-b B+(A b+a B) \tan (c+d x)) \, dx \\ & = \frac {b B \tan ^{1+m}(c+d x)}{d (1+m)}+(A b+a B) \int \tan ^{1+m}(c+d x) \, dx+(a A-b B) \int \tan ^m(c+d x) \, dx \\ & = \frac {b B \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {(A b+a B) \text {Subst}\left (\int \frac {x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac {(a A-b B) \text {Subst}\left (\int \frac {x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b B \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {(a A-b B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {(A b+a B) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (2+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.85 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {\tan ^{1+m}(c+d x) \left (\frac {b B}{1+m}+\frac {(a A-b B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right )}{1+m}+\frac {(A b+a B) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)}{2+m}\right )}{d} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

int(tan(d*x+c)^m*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x)

Fricas [F]

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

[In]

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*b*tan(d*x + c)^2 + A*a + (B*a + A*b)*tan(d*x + c))*tan(d*x + c)^m, x)

Sympy [F]

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

[In]

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

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))*tan(c + d*x)**m, x)

Maxima [F]

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

[In]

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + b*tan(c + d*x)),x)

[Out]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + b*tan(c + d*x)), x)