3.10.24 \(\int \frac {(d+e x)^m (a+b x+c x^2)}{(f+g x)^3} \, dx\) [924]

3.10.24.1 Optimal result

Integrand size = 25, antiderivative size = 245 \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(f+g x)^3} \, dx=\frac {\left (a+\frac {f (c f-b g)}{g^2}\right ) (d+e x)^{1+m}}{2 (e f-d g) (f+g x)^2}+\frac {(c f (4 d g-e f (3+m))+g (a e g (1-m)-b (2 d g-e f (1+m)))) (d+e x)^{1+m}}{2 g^2 (e f-d g)^2 (f+g x)}+\frac {\left (c \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-e g m (a e g (1-m)-b (2 d g-e f (1+m)))\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {g (d+e x)}{e f-d g}\right )}{2 g^2 (e f-d g)^3 (1+m)} \]

output

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

3.10.24.2 Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.64 \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(f+g x)^3} \, dx=-\frac {(d+e x)^{1+m} \left (c (e f-d g)^2 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )+e \left (-\left ((2 c f-b g) (e f-d g) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )\right )+e \left (c f^2+g (-b f+a g)\right ) \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )\right )\right )}{g^2 (-e f+d g)^3 (1+m)} \]

input

Integrate[((d + e*x)^m*(a + b*x + c*x^2))/(f + g*x)^3,x]
 

output

-(((d + e*x)^(1 + m)*(c*(e*f - d*g)^2*Hypergeometric2F1[1, 1 + m, 2 + m, ( 
g*(d + e*x))/(-(e*f) + d*g)] + e*(-((2*c*f - b*g)*(e*f - d*g)*Hypergeometr 
ic2F1[2, 1 + m, 2 + m, (g*(d + e*x))/(-(e*f) + d*g)]) + e*(c*f^2 + g*(-(b* 
f) + a*g))*Hypergeometric2F1[3, 1 + m, 2 + m, (g*(d + e*x))/(-(e*f) + d*g) 
])))/(g^2*(-(e*f) + d*g)^3*(1 + m)))
 

3.10.24.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1193, 27, 87, 78}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right ) (d+e x)^m}{(f+g x)^3} \, dx\)

\(\Big \downarrow \) 1193

\(\displaystyle \frac {\int \frac {(d+e x)^m \left (-2 c \left (d-\frac {e f}{g}\right ) x g^2-(2 b d g-a e (1-m) g-b e f (m+1)) g+c f (2 d g-e f (m+1))\right )}{g^2 (f+g x)^2}dx}{2 (e f-d g)}+\frac {(d+e x)^{m+1} \left (a+\frac {f (c f-b g)}{g^2}\right )}{2 (f+g x)^2 (e f-d g)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d+e x)^m (c f (2 d g-e f (m+1))-g (2 b d g-a e (1-m) g-b e f (m+1))+2 c g (e f-d g) x)}{(f+g x)^2}dx}{2 g^2 (e f-d g)}+\frac {(d+e x)^{m+1} \left (a+\frac {f (c f-b g)}{g^2}\right )}{2 (f+g x)^2 (e f-d g)}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\frac {\left (e g m (-a e g (1-m)+2 b d g-b e f (m+1))+c \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )\right ) \int \frac {(d+e x)^m}{f+g x}dx}{e f-d g}-\frac {(d+e x)^{m+1} (g (-a e g (1-m)+2 b d g-b e f (m+1))-c f (4 d g-e f (m+3)))}{(f+g x) (e f-d g)}}{2 g^2 (e f-d g)}+\frac {(d+e x)^{m+1} \left (a+\frac {f (c f-b g)}{g^2}\right )}{2 (f+g x)^2 (e f-d g)}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {(d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {g (d+e x)}{e f-d g}\right ) \left (e g m (-a e g (1-m)+2 b d g-b e f (m+1))+c \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )\right )}{(m+1) (e f-d g)^2}-\frac {(d+e x)^{m+1} (g (-a e g (1-m)+2 b d g-b e f (m+1))-c f (4 d g-e f (m+3)))}{(f+g x) (e f-d g)}}{2 g^2 (e f-d g)}+\frac {(d+e x)^{m+1} \left (a+\frac {f (c f-b g)}{g^2}\right )}{2 (f+g x)^2 (e f-d g)}\)

input

Int[((d + e*x)^m*(a + b*x + c*x^2))/(f + g*x)^3,x]
 

output

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

3.10.24.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 78

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

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 1193

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x 
 + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + 
e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) 
), x] + Simp[1/((m + 1)*(e*f - d*g))   Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex 
pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a 
, b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] &&  !IntegerQ[n] 
&&  !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
 

3.10.24.4 Maple [F]

\[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )}{\left (g x +f \right )^{3}}d x\]

input

int((e*x+d)^m*(c*x^2+b*x+a)/(g*x+f)^3,x)
 

output

int((e*x+d)^m*(c*x^2+b*x+a)/(g*x+f)^3,x)
 

3.10.24.5 Fricas [F]

\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]

input

integrate((e*x+d)^m*(c*x^2+b*x+a)/(g*x+f)^3,x, algorithm="fricas")
 

output

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

3.10.24.6 Sympy [F]

\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(f+g x)^3} \, dx=\int \frac {\left (d + e x\right )^{m} \left (a + b x + c x^{2}\right )}{\left (f + g x\right )^{3}}\, dx \]

input

integrate((e*x+d)**m*(c*x**2+b*x+a)/(g*x+f)**3,x)
 

output

Integral((d + e*x)**m*(a + b*x + c*x**2)/(f + g*x)**3, x)
 

3.10.24.7 Maxima [F]

\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]

input

integrate((e*x+d)^m*(c*x^2+b*x+a)/(g*x+f)^3,x, algorithm="maxima")
 

output

integrate((c*x^2 + b*x + a)*(e*x + d)^m/(g*x + f)^3, x)
 

3.10.24.8 Giac [F]

\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]

input

integrate((e*x+d)^m*(c*x^2+b*x+a)/(g*x+f)^3,x, algorithm="giac")
 

output

integrate((c*x^2 + b*x + a)*(e*x + d)^m/(g*x + f)^3, x)
 

3.10.24.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(f+g x)^3} \, dx=\int \frac {{\left (d+e\,x\right )}^m\,\left (c\,x^2+b\,x+a\right )}{{\left (f+g\,x\right )}^3} \,d x \]

input

int(((d + e*x)^m*(a + b*x + c*x^2))/(f + g*x)^3,x)
 

output

int(((d + e*x)^m*(a + b*x + c*x^2))/(f + g*x)^3, x)