\(\int \frac {x^3 (d+e x)^m}{a+b x+c x^2} \, dx\) [125]

Optimal result

Integrand size = 23, antiderivative size = 290 \[ \int \frac {x^3 (d+e x)^m}{a+b x+c x^2} \, dx=-\frac {(c d+b e) (d+e x)^{1+m}}{c^2 e^2 (1+m)}+\frac {(d+e x)^{2+m}}{c e^2 (2+m)}+\frac {\left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{c \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}+\frac {\left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.10 \[ \int \frac {x^3 (d+e x)^m}{a+b x+c x^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {\left (b^3-3 a b c-b^2 \sqrt {b^2-4 a c}+a c \sqrt {b^2-4 a c}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}+\frac {-\left (\left (b \left (b+\sqrt {b^2-4 a c}\right ) e-2 c \left (\sqrt {b^2-4 a c} d+2 a e\right )\right ) (b e (2+m)+c (d-e (1+m) x))\right )+\left (b^3-3 a b c+b^2 \sqrt {b^2-4 a c}-a c \sqrt {b^2-4 a c}\right ) e^2 (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e^2 \left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (2+m)}\right )}{c^2 \sqrt {b^2-4 a c} (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.25 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1200, 2009}

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.

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {x^3 (d+e x)^m}{a+b x+c x^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 (d+e x)^m}{a+b x+c x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((d + e*x)**m*a*b*e**2*m**2 + 3*(d + e*x)**m*a*b*e**2*m + 2*(d + e*x)**m*a 
*b*e**2 - (d + e*x)**m*a*c*d*e*m**2 - 3*(d + e*x)**m*a*c*d*e*m - 2*(d + e* 
x)**m*a*c*d*e + (d + e*x)**m*b**2*d*e*m + 2*(d + e*x)**m*b**2*d*e - (d + e 
*x)**m*b**2*e**2*m**2*x - 2*(d + e*x)**m*b**2*e**2*m*x - (d + e*x)**m*b*c* 
d**2*m + (d + e*x)**m*b*c*d*e*m**2*x + (d + e*x)**m*b*c*e**2*m**2*x**2 + ( 
d + e*x)**m*b*c*e**2*m*x**2 - int((d + e*x)**m/(a*d + a*e*x + b*d*x + b*e* 
x**2 + c*d*x**2 + c*e*x**3),x)*a**2*b*e**3*m**3 - 3*int((d + e*x)**m/(a*d 
+ a*e*x + b*d*x + b*e*x**2 + c*d*x**2 + c*e*x**3),x)*a**2*b*e**3*m**2 - 2* 
int((d + e*x)**m/(a*d + a*e*x + b*d*x + b*e*x**2 + c*d*x**2 + c*e*x**3),x) 
*a**2*b*e**3*m + int((d + e*x)**m/(a*d + a*e*x + b*d*x + b*e*x**2 + c*d*x* 
*2 + c*e*x**3),x)*a**2*c*d*e**2*m**3 + 3*int((d + e*x)**m/(a*d + a*e*x + b 
*d*x + b*e*x**2 + c*d*x**2 + c*e*x**3),x)*a**2*c*d*e**2*m**2 + 2*int((d + 
e*x)**m/(a*d + a*e*x + b*d*x + b*e*x**2 + c*d*x**2 + c*e*x**3),x)*a**2*c*d 
*e**2*m - 2*int(((d + e*x)**m*x**2)/(a*d + a*e*x + b*d*x + b*e*x**2 + c*d* 
x**2 + c*e*x**3),x)*a*b*c*e**3*m**3 - 6*int(((d + e*x)**m*x**2)/(a*d + a*e 
*x + b*d*x + b*e*x**2 + c*d*x**2 + c*e*x**3),x)*a*b*c*e**3*m**2 - 4*int((( 
d + e*x)**m*x**2)/(a*d + a*e*x + b*d*x + b*e*x**2 + c*d*x**2 + c*e*x**3),x 
)*a*b*c*e**3*m + int(((d + e*x)**m*x**2)/(a*d + a*e*x + b*d*x + b*e*x**2 + 
 c*d*x**2 + c*e*x**3),x)*a*c**2*d*e**2*m**3 + 3*int(((d + e*x)**m*x**2)/(a 
*d + a*e*x + b*d*x + b*e*x**2 + c*d*x**2 + c*e*x**3),x)*a*c**2*d*e**2*m...