\(\int \frac {x^m (c+d x)^3}{a+b x} \, dx\) [529]

Optimal result

Integrand size = 18, antiderivative size = 127 \[ \int \frac {x^m (c+d x)^3}{a+b x} \, dx=\frac {d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{1+m}}{b^3 (1+m)}+\frac {d^2 (3 b c-a d) x^{2+m}}{b^2 (2+m)}+\frac {d^3 x^{3+m}}{b (3+m)}+\frac {(b c-a d)^3 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,1,1-m,\frac {a}{a+b x}\right )}{b^3 m (a+b x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.93 \[ \int \frac {x^m (c+d x)^3}{a+b x} \, dx=\frac {x^{1+m} \left (d \left (\frac {a^2 d^2}{1+m}+a b d \left (-\frac {3 c}{1+m}-\frac {d x}{2+m}\right )+b^2 \left (\frac {3 c^2}{1+m}+\frac {3 c d x}{2+m}+\frac {d^2 x^2}{3+m}\right )\right )+\frac {(b c-a d)^3 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )}{a (1+m)}\right )}{b^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.35, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {99, 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.

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.90 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.30 \[ \int \frac {x^m (c+d x)^3}{a+b x} \, dx=\frac {c^{3} m x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac {c^{3} x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac {3 c^{2} d m x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac {6 c^{2} d x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac {3 c d^{2} m x^{m + 3} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} + \frac {9 c d^{2} x^{m + 3} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} + \frac {d^{3} m x^{m + 4} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 4\right ) \Gamma \left (m + 4\right )}{a \Gamma \left (m + 5\right )} + \frac {4 d^{3} x^{m + 4} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 4\right ) \Gamma \left (m + 4\right )}{a \Gamma \left (m + 5\right )} \] Input:

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

Output:

c**3*m*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/( 
a*gamma(m + 2)) + c**3*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1 
)*gamma(m + 1)/(a*gamma(m + 2)) + 3*c**2*d*m*x**(m + 2)*lerchphi(b*x*exp_p 
olar(I*pi)/a, 1, m + 2)*gamma(m + 2)/(a*gamma(m + 3)) + 6*c**2*d*x**(m + 2 
)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m + 2)/(a*gamma(m + 3)) 
+ 3*c*d**2*m*x**(m + 3)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 3)*gamma(m 
+ 3)/(a*gamma(m + 4)) + 9*c*d**2*x**(m + 3)*lerchphi(b*x*exp_polar(I*pi)/a 
, 1, m + 3)*gamma(m + 3)/(a*gamma(m + 4)) + d**3*m*x**(m + 4)*lerchphi(b*x 
*exp_polar(I*pi)/a, 1, m + 4)*gamma(m + 4)/(a*gamma(m + 5)) + 4*d**3*x**(m 
 + 4)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 4)*gamma(m + 4)/(a*gamma(m + 
5))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - x**m*a**3*d**3*m**3 - 6*x**m*a**3*d**3*m**2 - 11*x**m*a**3*d**3*m - 6* 
x**m*a**3*d**3 + 3*x**m*a**2*b*c*d**2*m**3 + 18*x**m*a**2*b*c*d**2*m**2 + 
33*x**m*a**2*b*c*d**2*m + 18*x**m*a**2*b*c*d**2 + x**m*a**2*b*d**3*m**3*x 
+ 5*x**m*a**2*b*d**3*m**2*x + 6*x**m*a**2*b*d**3*m*x - 3*x**m*a*b**2*c**2* 
d*m**3 - 18*x**m*a*b**2*c**2*d*m**2 - 33*x**m*a*b**2*c**2*d*m - 18*x**m*a* 
b**2*c**2*d - 3*x**m*a*b**2*c*d**2*m**3*x - 15*x**m*a*b**2*c*d**2*m**2*x - 
 18*x**m*a*b**2*c*d**2*m*x - x**m*a*b**2*d**3*m**3*x**2 - 4*x**m*a*b**2*d* 
*3*m**2*x**2 - 3*x**m*a*b**2*d**3*m*x**2 + x**m*b**3*c**3*m**3 + 6*x**m*b* 
*3*c**3*m**2 + 11*x**m*b**3*c**3*m + 6*x**m*b**3*c**3 + 3*x**m*b**3*c**2*d 
*m**3*x + 15*x**m*b**3*c**2*d*m**2*x + 18*x**m*b**3*c**2*d*m*x + 3*x**m*b* 
*3*c*d**2*m**3*x**2 + 12*x**m*b**3*c*d**2*m**2*x**2 + 9*x**m*b**3*c*d**2*m 
*x**2 + x**m*b**3*d**3*m**3*x**3 + 3*x**m*b**3*d**3*m**2*x**3 + 2*x**m*b** 
3*d**3*m*x**3 + int(x**m/(a*x + b*x**2),x)*a**4*d**3*m**4 + 6*int(x**m/(a* 
x + b*x**2),x)*a**4*d**3*m**3 + 11*int(x**m/(a*x + b*x**2),x)*a**4*d**3*m* 
*2 + 6*int(x**m/(a*x + b*x**2),x)*a**4*d**3*m - 3*int(x**m/(a*x + b*x**2), 
x)*a**3*b*c*d**2*m**4 - 18*int(x**m/(a*x + b*x**2),x)*a**3*b*c*d**2*m**3 - 
 33*int(x**m/(a*x + b*x**2),x)*a**3*b*c*d**2*m**2 - 18*int(x**m/(a*x + b*x 
**2),x)*a**3*b*c*d**2*m + 3*int(x**m/(a*x + b*x**2),x)*a**2*b**2*c**2*d*m* 
*4 + 18*int(x**m/(a*x + b*x**2),x)*a**2*b**2*c**2*d*m**3 + 33*int(x**m/(a* 
x + b*x**2),x)*a**2*b**2*c**2*d*m**2 + 18*int(x**m/(a*x + b*x**2),x)*a*...