\(\int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [480]

Optimal result

Integrand size = 31, antiderivative size = 267 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac {\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (2+m)}+\frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)} \]

[Out]

b*(3*a*A*b*(3+m)-b^2*B*(3+m)+2*a^2*B*(4+m))*tan(d*x+c)^(1+m)/d/(1+m)/(3+m)+(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*h
ypergeom([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)+b^2*(A*b*(3+m)+a*B*(5+m))*tan(d*x+
c)^(2+m)/d/(2+m)/(3+m)+(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b^2)*hypergeom([1, 1+1/2*m],[2+1/2*m],-tan(d*x+c)^2)*tan(d
*x+c)^(2+m)/d/(2+m)+b*B*tan(d*x+c)^(1+m)*(a+b*tan(d*x+c))^2/d/(3+m)

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3688, 3718, 3711, 3619, 3557, 371} \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {b \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+3)}+\frac {\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}+\frac {\left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}+\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2) (m+3)}+\frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)} \]

[In]

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

[Out]

(b*(3*a*A*b*(3 + m) - b^2*B*(3 + m) + 2*a^2*B*(4 + m))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(3 + m)) + ((a^3*A - 3
*a*A*b^2 - 3*a^2*b*B + b^3*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m)
)/(d*(1 + m)) + (b^2*(A*b*(3 + m) + a*B*(5 + m))*Tan[c + d*x]^(2 + m))/(d*(2 + m)*(3 + m)) + ((3*a^2*A*b - A*b
^3 + a^3*B - 3*a*b^2*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/(d*(
2 + m)) + (b*B*Tan[c + d*x]^(1 + m)*(a + b*Tan[c + d*x])^2)/(d*(3 + 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 3688

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 - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3711

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[C*((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*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3718

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

Rubi steps \begin{align*} \text {integral}& = \frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}+\frac {\int \tan ^m(c+d x) (a+b \tan (c+d x)) \left (-a (b B (1+m)-a A (3+m))+\left (2 a A b+a^2 B-b^2 B\right ) (3+m) \tan (c+d x)+b (A b (3+m)+a B (5+m)) \tan ^2(c+d x)\right ) \, dx}{3+m} \\ & = \frac {b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}-\frac {\int \tan ^m(c+d x) \left (a^2 (2+m) (b B (1+m)-a A (3+m))-\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) (2+m) (3+m) \tan (c+d x)-b (2+m) \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^2(c+d x)\right ) \, dx}{6+5 m+m^2} \\ & = \frac {b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac {b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}-\frac {\int \tan ^m(c+d x) \left (-\left (\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) (2+m) (3+m)\right )-\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) (2+m) (3+m) \tan (c+d x)\right ) \, dx}{6+5 m+m^2} \\ & = \frac {b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac {b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \int \tan ^{1+m}(c+d x) \, dx+\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \int \tan ^m(c+d x) \, dx \\ & = \frac {b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac {b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \text {Subst}\left (\int \frac {x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \text {Subst}\left (\int \frac {x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac {\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \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 {b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \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)}+\frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.87 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {\tan ^{1+m}(c+d x) \left (b (2+m) \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right )+\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) (2+m) (3+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right )+b^2 (1+m) (A b (3+m)+a B (5+m)) \tan (c+d x)+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) (1+m) (3+m) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)+b B (1+m) (2+m) (a+b \tan (c+d x))^2\right )}{d (1+m) (2+m) (3+m)} \]

[In]

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

[Out]

(Tan[c + d*x]^(1 + m)*(b*(2 + m)*(3*a*A*b*(3 + m) - b^2*B*(3 + m) + 2*a^2*B*(4 + m)) + (a^3*A - 3*a*A*b^2 - 3*
a^2*b*B + b^3*B)*(2 + m)*(3 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2] + b^2*(1 + m)*(A*
b*(3 + m) + a*B*(5 + m))*Tan[c + d*x] + (3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*(1 + m)*(3 + m)*Hypergeometric
2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x] + b*B*(1 + m)*(2 + m)*(a + b*Tan[c + d*x])^2))/(d*(
1 + m)*(2 + m)*(3 + m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*b^3*tan(d*x + c)^4 + A*a^3 + (3*B*a*b^2 + A*b^3)*tan(d*x + c)^3 + 3*(B*a^2*b + A*a*b^2)*tan(d*x +
c)^2 + (B*a^3 + 3*A*a^2*b)*tan(d*x + c))*tan(d*x + c)^m, x)

Sympy [F]

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

[In]

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

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3*tan(c + d*x)**m, x)

Maxima [F]

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

[In]

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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