\(\int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx\) [3097]

Optimal result

Integrand size = 26, antiderivative size = 406 \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {2 b (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d)^2 (2+m) (3+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {3 b f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^2 (1+m) (2+m)}+\frac {2 b^2 (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^3 (1+m) (2+m) (3+m)}-\frac {f^3 (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m} \]

[Out]

(-c*f+d*e)^3*(b*x+a)^(1+m)*(d*x+c)^(-3-m)/d^3/(-a*d+b*c)/(3+m)+3*f*(-c*f+d*e)^2*(b*x+a)^(1+m)*(d*x+c)^(-2-m)/d
^3/(-a*d+b*c)/(2+m)+2*b*(-c*f+d*e)^3*(b*x+a)^(1+m)*(d*x+c)^(-2-m)/d^3/(-a*d+b*c)^2/(2+m)/(3+m)+3*f^2*(-c*f+d*e
)*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/d^3/(-a*d+b*c)/(1+m)+3*b*f*(-c*f+d*e)^2*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/d^3/(-a*d+
b*c)^2/(1+m)/(2+m)+2*b^2*(-c*f+d*e)^3*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/d^3/(-a*d+b*c)^3/(1+m)/(2+m)/(3+m)-f^3*(b*x
+a)^m*hypergeom([-m, -m],[1-m],b*(d*x+c)/(-a*d+b*c))/d^4/m/((-d*(b*x+a)/(-a*d+b*c))^m)/((d*x+c)^m)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {128, 47, 37, 72, 71} \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\frac {2 b^2 (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-1}}{d^3 (m+1) (m+2) (m+3) (b c-a d)^3}-\frac {f^3 (a+b x)^m (c+d x)^{-m} \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m}+\frac {3 f^2 (a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1}}{d^3 (m+1) (b c-a d)}+\frac {(a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-3}}{d^3 (m+3) (b c-a d)}+\frac {3 f (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-2}}{d^3 (m+2) (b c-a d)}+\frac {2 b (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-2}}{d^3 (m+2) (m+3) (b c-a d)^2}+\frac {3 b f (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-1}}{d^3 (m+1) (m+2) (b c-a d)^2} \]

[In]

Int[(a + b*x)^m*(c + d*x)^(-4 - m)*(e + f*x)^3,x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 128

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (IGtQ[m, 0] || (
ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(d e-c f)^3 (a+b x)^m (c+d x)^{-4-m}}{d^3}+\frac {3 f (d e-c f)^2 (a+b x)^m (c+d x)^{-3-m}}{d^3}+\frac {3 f^2 (d e-c f) (a+b x)^m (c+d x)^{-2-m}}{d^3}+\frac {f^3 (a+b x)^m (c+d x)^{-1-m}}{d^3}\right ) \, dx \\ & = \frac {f^3 \int (a+b x)^m (c+d x)^{-1-m} \, dx}{d^3}+\frac {\left (3 f^2 (d e-c f)\right ) \int (a+b x)^m (c+d x)^{-2-m} \, dx}{d^3}+\frac {\left (3 f (d e-c f)^2\right ) \int (a+b x)^m (c+d x)^{-3-m} \, dx}{d^3}+\frac {(d e-c f)^3 \int (a+b x)^m (c+d x)^{-4-m} \, dx}{d^3} \\ & = \frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {\left (3 b f (d e-c f)^2\right ) \int (a+b x)^m (c+d x)^{-2-m} \, dx}{d^3 (b c-a d) (2+m)}+\frac {\left (2 b (d e-c f)^3\right ) \int (a+b x)^m (c+d x)^{-3-m} \, dx}{d^3 (b c-a d) (3+m)}+\frac {\left (f^3 (a+b x)^m \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m}\right ) \int (c+d x)^{-1-m} \left (-\frac {a d}{b c-a d}-\frac {b d x}{b c-a d}\right )^m \, dx}{d^3} \\ & = \frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {2 b (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d)^2 (2+m) (3+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {3 b f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^2 (1+m) (2+m)}-\frac {f^3 (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \, _2F_1\left (-m,-m;1-m;\frac {b (c+d x)}{b c-a d}\right )}{d^4 m}+\frac {\left (2 b^2 (d e-c f)^3\right ) \int (a+b x)^m (c+d x)^{-2-m} \, dx}{d^3 (b c-a d)^2 (2+m) (3+m)} \\ & = \frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {2 b (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d)^2 (2+m) (3+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {3 b f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^2 (1+m) (2+m)}+\frac {2 b^2 (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^3 (1+m) (2+m) (3+m)}-\frac {f^3 (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \, _2F_1\left (-m,-m;1-m;\frac {b (c+d x)}{b c-a d}\right )}{d^4 m} \\ \end{align*}

Mathematica [A] (verified)

Time = 12.60 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.69 \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\frac {(a+b x)^m (c+d x)^{-m} \left (\frac {6 f^2 (a+b x)^3 (e+f x)}{(b c-a d)^3 (c+d x)}-\frac {3 f (a+b x)^2 (-a d (1+m)+b c (3+m)+2 b d x) (e+f x)^2}{(b c-a d)^3 (c+d x)^2}+\frac {(a+b x) (e+f x)^3 \left (a^2 d^2 \left (2+3 m+m^2\right )-2 a b d (1+m) (c (3+m)+d x)+b^2 \left (c^2 \left (6+5 m+m^2\right )+2 c d (3+m) x+2 d^2 x^2\right )\right )}{(b c-a d)^3 (c+d x)^3}-\frac {6 f^3 \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-3-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m}\right )}{(1+m) (2+m) (3+m)} \]

[In]

Integrate[(a + b*x)^m*(c + d*x)^(-4 - m)*(e + f*x)^3,x]

[Out]

((a + b*x)^m*((6*f^2*(a + b*x)^3*(e + f*x))/((b*c - a*d)^3*(c + d*x)) - (3*f*(a + b*x)^2*(-(a*d*(1 + m)) + b*c
*(3 + m) + 2*b*d*x)*(e + f*x)^2)/((b*c - a*d)^3*(c + d*x)^2) + ((a + b*x)*(e + f*x)^3*(a^2*d^2*(2 + 3*m + m^2)
- 2*a*b*d*(1 + m)*(c*(3 + m) + d*x) + b^2*(c^2*(6 + 5*m + m^2) + 2*c*d*(3 + m)*x + 2*d^2*x^2)))/((b*c - a*d)^
3*(c + d*x)^3) - (6*f^3*Hypergeometric2F1[-3 - m, -m, 1 - m, (b*(c + d*x))/(b*c - a*d)])/(d^4*m*((d*(a + b*x))
/(-(b*c) + a*d))^m)))/((1 + m)*(2 + m)*(3 + m)*(c + d*x)^m)

Maple [F]

\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-4-m} \left (f x +e \right )^{3}d x\]

[In]

int((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^3,x)

[Out]

int((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^3,x)

Fricas [F]

\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^3,x, algorithm="fricas")

[Out]

integral((f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*x + e^3)*(b*x + a)^m*(d*x + c)^(-m - 4), x)

Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((b*x+a)**m*(d*x+c)**(-4-m)*(f*x+e)**3,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^3,x, algorithm="maxima")

[Out]

integrate((f*x + e)^3*(b*x + a)^m*(d*x + c)^(-m - 4), x)

Giac [F]

\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^3,x, algorithm="giac")

[Out]

integrate((f*x + e)^3*(b*x + a)^m*(d*x + c)^(-m - 4), x)

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int \frac {{\left (e+f\,x\right )}^3\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+4}} \,d x \]

[In]

int(((e + f*x)^3*(a + b*x)^m)/(c + d*x)^(m + 4),x)

[Out]

int(((e + f*x)^3*(a + b*x)^m)/(c + d*x)^(m + 4), x)