\(\int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx\) [260]

Optimal result

Integrand size = 18, antiderivative size = 136 \[ \int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx=-\frac {2 a (b c-2 a d) (b c-a d) x}{b^5}+\frac {(b c-3 a d) (b c-a d) x^2}{2 b^4}+\frac {2 d (b c-a d) x^3}{3 b^3}+\frac {d^2 x^4}{4 b^2}+\frac {a^3 (b c-a d)^2}{b^6 (a+b x)}+\frac {a^2 (3 b c-5 a d) (b c-a d) \log (a+b x)}{b^6} \]

[Out]

-2*a*(-2*a*d+b*c)*(-a*d+b*c)*x/b^5+1/2*(-3*a*d+b*c)*(-a*d+b*c)*x^2/b^4+2/3*d*(-a*d+b*c)*x^3/b^3+1/4*d^2*x^4/b^
2+a^3*(-a*d+b*c)^2/b^6/(b*x+a)+a^2*(-5*a*d+3*b*c)*(-a*d+b*c)*ln(b*x+a)/b^6

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {90} \[ \int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx=\frac {a^3 (b c-a d)^2}{b^6 (a+b x)}+\frac {a^2 (3 b c-5 a d) (b c-a d) \log (a+b x)}{b^6}-\frac {2 a x (b c-2 a d) (b c-a d)}{b^5}+\frac {x^2 (b c-3 a d) (b c-a d)}{2 b^4}+\frac {2 d x^3 (b c-a d)}{3 b^3}+\frac {d^2 x^4}{4 b^2} \]

[In]

Int[(x^3*(c + d*x)^2)/(a + b*x)^2,x]

[Out]

(-2*a*(b*c - 2*a*d)*(b*c - a*d)*x)/b^5 + ((b*c - 3*a*d)*(b*c - a*d)*x^2)/(2*b^4) + (2*d*(b*c - a*d)*x^3)/(3*b^
3) + (d^2*x^4)/(4*b^2) + (a^3*(b*c - a*d)^2)/(b^6*(a + b*x)) + (a^2*(3*b*c - 5*a*d)*(b*c - a*d)*Log[a + b*x])/
b^6

Rule 90

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, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx=\frac {-24 a b \left (b^2 c^2-3 a b c d+2 a^2 d^2\right ) x+6 b^2 \left (b^2 c^2-4 a b c d+3 a^2 d^2\right ) x^2+8 b^3 d (b c-a d) x^3+3 b^4 d^2 x^4+\frac {12 a^3 (b c-a d)^2}{a+b x}+12 a^2 \left (3 b^2 c^2-8 a b c d+5 a^2 d^2\right ) \log (a+b x)}{12 b^6} \]

[In]

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

[Out]

(-24*a*b*(b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x + 6*b^2*(b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x^2 + 8*b^3*d*(b*c - a*
d)*x^3 + 3*b^4*d^2*x^4 + (12*a^3*(b*c - a*d)^2)/(a + b*x) + 12*a^2*(3*b^2*c^2 - 8*a*b*c*d + 5*a^2*d^2)*Log[a +
b*x])/(12*b^6)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {\frac {a \left (5 a^{4} d^{2}-8 a^{3} b c d +3 a^{2} b^{2} c^{2}\right )}{b^{6}}+\frac {d^{2} x^{5}}{4 b}+\frac {\left (5 a^{2} d^{2}-8 a b c d +3 b^{2} c^{2}\right ) x^{3}}{6 b^{3}}-\frac {a \left (5 a^{2} d^{2}-8 a b c d +3 b^{2} c^{2}\right ) x^{2}}{2 b^{4}}-\frac {d \left (5 a d -8 b c \right ) x^{4}}{12 b^{2}}}{b x +a}+\frac {a^{2} \left (5 a^{2} d^{2}-8 a b c d +3 b^{2} c^{2}\right ) \ln \left (b x +a \right )}{b^{6}}\)	\(172\)
default	\(-\frac {-\frac {1}{4} d^{2} x^{4} b^{3}+\frac {2}{3} x^{3} a \,b^{2} d^{2}-\frac {2}{3} x^{3} b^{3} c d -\frac {3}{2} x^{2} a^{2} b \,d^{2}+2 x^{2} a \,b^{2} c d -\frac {1}{2} x^{2} b^{3} c^{2}+4 a^{3} d^{2} x -6 a^{2} c d b x +2 a \,c^{2} b^{2} x}{b^{5}}+\frac {a^{2} \left (5 a^{2} d^{2}-8 a b c d +3 b^{2} c^{2}\right ) \ln \left (b x +a \right )}{b^{6}}+\frac {a^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{b^{6} \left (b x +a \right )}\)	\(174\)
risch	\(\frac {d^{2} x^{4}}{4 b^{2}}-\frac {2 x^{3} a \,d^{2}}{3 b^{3}}+\frac {2 x^{3} c d}{3 b^{2}}+\frac {3 x^{2} a^{2} d^{2}}{2 b^{4}}-\frac {2 x^{2} a c d}{b^{3}}+\frac {x^{2} c^{2}}{2 b^{2}}-\frac {4 a^{3} d^{2} x}{b^{5}}+\frac {6 a^{2} c d x}{b^{4}}-\frac {2 a \,c^{2} x}{b^{3}}+\frac {a^{5} d^{2}}{b^{6} \left (b x +a \right )}-\frac {2 a^{4} c d}{b^{5} \left (b x +a \right )}+\frac {a^{3} c^{2}}{b^{4} \left (b x +a \right )}+\frac {5 a^{4} \ln \left (b x +a \right ) d^{2}}{b^{6}}-\frac {8 a^{3} \ln \left (b x +a \right ) c d}{b^{5}}+\frac {3 a^{2} \ln \left (b x +a \right ) c^{2}}{b^{4}}\)	\(205\)
parallelrisch	\(\frac {3 d^{2} x^{5} b^{5}-5 x^{4} a \,b^{4} d^{2}+8 x^{4} b^{5} c d +10 x^{3} a^{2} b^{3} d^{2}-16 x^{3} a \,b^{4} c d +6 x^{3} b^{5} c^{2}+60 \ln \left (b x +a \right ) x \,a^{4} b \,d^{2}-96 \ln \left (b x +a \right ) x \,a^{3} b^{2} c d +36 \ln \left (b x +a \right ) x \,a^{2} b^{3} c^{2}-30 x^{2} a^{3} b^{2} d^{2}+48 x^{2} a^{2} b^{3} c d -18 x^{2} a \,b^{4} c^{2}+60 \ln \left (b x +a \right ) a^{5} d^{2}-96 \ln \left (b x +a \right ) a^{4} b c d +36 \ln \left (b x +a \right ) a^{3} b^{2} c^{2}+60 a^{5} d^{2}-96 a^{4} b c d +36 a^{3} b^{2} c^{2}}{12 b^{6} \left (b x +a \right )}\)	\(245\)

[In]

int(x^3*(d*x+c)^2/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

(a*(5*a^4*d^2-8*a^3*b*c*d+3*a^2*b^2*c^2)/b^6+1/4*d^2*x^5/b+1/6*(5*a^2*d^2-8*a*b*c*d+3*b^2*c^2)/b^3*x^3-1/2*a*(
5*a^2*d^2-8*a*b*c*d+3*b^2*c^2)/b^4*x^2-1/12*d*(5*a*d-8*b*c)/b^2*x^4)/(b*x+a)+a^2/b^6*(5*a^2*d^2-8*a*b*c*d+3*b^
2*c^2)*ln(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.81 \[ \int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx=\frac {3 \, b^{5} d^{2} x^{5} + 12 \, a^{3} b^{2} c^{2} - 24 \, a^{4} b c d + 12 \, a^{5} d^{2} + {\left (8 \, b^{5} c d - 5 \, a b^{4} d^{2}\right )} x^{4} + 2 \, {\left (3 \, b^{5} c^{2} - 8 \, a b^{4} c d + 5 \, a^{2} b^{3} d^{2}\right )} x^{3} - 6 \, {\left (3 \, a b^{4} c^{2} - 8 \, a^{2} b^{3} c d + 5 \, a^{3} b^{2} d^{2}\right )} x^{2} - 24 \, {\left (a^{2} b^{3} c^{2} - 3 \, a^{3} b^{2} c d + 2 \, a^{4} b d^{2}\right )} x + 12 \, {\left (3 \, a^{3} b^{2} c^{2} - 8 \, a^{4} b c d + 5 \, a^{5} d^{2} + {\left (3 \, a^{2} b^{3} c^{2} - 8 \, a^{3} b^{2} c d + 5 \, a^{4} b d^{2}\right )} x\right )} \log \left (b x + a\right )}{12 \, {\left (b^{7} x + a b^{6}\right )}} \]

[In]

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

[Out]

1/12*(3*b^5*d^2*x^5 + 12*a^3*b^2*c^2 - 24*a^4*b*c*d + 12*a^5*d^2 + (8*b^5*c*d - 5*a*b^4*d^2)*x^4 + 2*(3*b^5*c^
2 - 8*a*b^4*c*d + 5*a^2*b^3*d^2)*x^3 - 6*(3*a*b^4*c^2 - 8*a^2*b^3*c*d + 5*a^3*b^2*d^2)*x^2 - 24*(a^2*b^3*c^2 -
3*a^3*b^2*c*d + 2*a^4*b*d^2)*x + 12*(3*a^3*b^2*c^2 - 8*a^4*b*c*d + 5*a^5*d^2 + (3*a^2*b^3*c^2 - 8*a^3*b^2*c*d
+ 5*a^4*b*d^2)*x)*log(b*x + a))/(b^7*x + a*b^6)

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.29 \[ \int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx=\frac {a^{2} \left (a d - b c\right ) \left (5 a d - 3 b c\right ) \log {\left (a + b x \right )}}{b^{6}} + x^{3} \left (- \frac {2 a d^{2}}{3 b^{3}} + \frac {2 c d}{3 b^{2}}\right ) + x^{2} \cdot \left (\frac {3 a^{2} d^{2}}{2 b^{4}} - \frac {2 a c d}{b^{3}} + \frac {c^{2}}{2 b^{2}}\right ) + x \left (- \frac {4 a^{3} d^{2}}{b^{5}} + \frac {6 a^{2} c d}{b^{4}} - \frac {2 a c^{2}}{b^{3}}\right ) + \frac {a^{5} d^{2} - 2 a^{4} b c d + a^{3} b^{2} c^{2}}{a b^{6} + b^{7} x} + \frac {d^{2} x^{4}}{4 b^{2}} \]

[In]

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

[Out]

a**2*(a*d - b*c)*(5*a*d - 3*b*c)*log(a + b*x)/b**6 + x**3*(-2*a*d**2/(3*b**3) + 2*c*d/(3*b**2)) + x**2*(3*a**2
*d**2/(2*b**4) - 2*a*c*d/b**3 + c**2/(2*b**2)) + x*(-4*a**3*d**2/b**5 + 6*a**2*c*d/b**4 - 2*a*c**2/b**3) + (a*
*5*d**2 - 2*a**4*b*c*d + a**3*b**2*c**2)/(a*b**6 + b**7*x) + d**2*x**4/(4*b**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.29 \[ \int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx=\frac {a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}}{b^{7} x + a b^{6}} + \frac {3 \, b^{3} d^{2} x^{4} + 8 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{3} + 6 \, {\left (b^{3} c^{2} - 4 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x^{2} - 24 \, {\left (a b^{2} c^{2} - 3 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} x}{12 \, b^{5}} + \frac {{\left (3 \, a^{2} b^{2} c^{2} - 8 \, a^{3} b c d + 5 \, a^{4} d^{2}\right )} \log \left (b x + a\right )}{b^{6}} \]

[In]

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

[Out]

(a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2)/(b^7*x + a*b^6) + 1/12*(3*b^3*d^2*x^4 + 8*(b^3*c*d - a*b^2*d^2)*x^3 + 6*
(b^3*c^2 - 4*a*b^2*c*d + 3*a^2*b*d^2)*x^2 - 24*(a*b^2*c^2 - 3*a^2*b*c*d + 2*a^3*d^2)*x)/b^5 + (3*a^2*b^2*c^2 -
8*a^3*b*c*d + 5*a^4*d^2)*log(b*x + a)/b^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.74 \[ \int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx=\frac {{\left (3 \, d^{2} + \frac {4 \, {\left (2 \, b^{2} c d - 5 \, a b d^{2}\right )}}{{\left (b x + a\right )} b} + \frac {6 \, {\left (b^{4} c^{2} - 8 \, a b^{3} c d + 10 \, a^{2} b^{2} d^{2}\right )}}{{\left (b x + a\right )}^{2} b^{2}} - \frac {12 \, {\left (3 \, a b^{5} c^{2} - 12 \, a^{2} b^{4} c d + 10 \, a^{3} b^{3} d^{2}\right )}}{{\left (b x + a\right )}^{3} b^{3}}\right )} {\left (b x + a\right )}^{4}}{12 \, b^{6}} - \frac {{\left (3 \, a^{2} b^{2} c^{2} - 8 \, a^{3} b c d + 5 \, a^{4} d^{2}\right )} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{6}} + \frac {\frac {a^{3} b^{6} c^{2}}{b x + a} - \frac {2 \, a^{4} b^{5} c d}{b x + a} + \frac {a^{5} b^{4} d^{2}}{b x + a}}{b^{10}} \]

[In]

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

[Out]

1/12*(3*d^2 + 4*(2*b^2*c*d - 5*a*b*d^2)/((b*x + a)*b) + 6*(b^4*c^2 - 8*a*b^3*c*d + 10*a^2*b^2*d^2)/((b*x + a)^
2*b^2) - 12*(3*a*b^5*c^2 - 12*a^2*b^4*c*d + 10*a^3*b^3*d^2)/((b*x + a)^3*b^3))*(b*x + a)^4/b^6 - (3*a^2*b^2*c^
2 - 8*a^3*b*c*d + 5*a^4*d^2)*log(abs(b*x + a)/((b*x + a)^2*abs(b)))/b^6 + (a^3*b^6*c^2/(b*x + a) - 2*a^4*b^5*c
*d/(b*x + a) + a^5*b^4*d^2/(b*x + a))/b^10

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.74 \[ \int \frac {x^3 (c+d x)^2}{(a+b x)^2} \, dx=x\,\left (\frac {a^2\,\left (\frac {2\,a\,d^2}{b^3}-\frac {2\,c\,d}{b^2}\right )}{b^2}-\frac {2\,a\,\left (\frac {c^2}{b^2}+\frac {2\,a\,\left (\frac {2\,a\,d^2}{b^3}-\frac {2\,c\,d}{b^2}\right )}{b}-\frac {a^2\,d^2}{b^4}\right )}{b}\right )+x^2\,\left (\frac {c^2}{2\,b^2}+\frac {a\,\left (\frac {2\,a\,d^2}{b^3}-\frac {2\,c\,d}{b^2}\right )}{b}-\frac {a^2\,d^2}{2\,b^4}\right )-x^3\,\left (\frac {2\,a\,d^2}{3\,b^3}-\frac {2\,c\,d}{3\,b^2}\right )+\frac {a^5\,d^2-2\,a^4\,b\,c\,d+a^3\,b^2\,c^2}{b\,\left (x\,b^6+a\,b^5\right )}+\frac {\ln \left (a+b\,x\right )\,\left (5\,a^4\,d^2-8\,a^3\,b\,c\,d+3\,a^2\,b^2\,c^2\right )}{b^6}+\frac {d^2\,x^4}{4\,b^2} \]

[In]

int((x^3*(c + d*x)^2)/(a + b*x)^2,x)

[Out]

x*((a^2*((2*a*d^2)/b^3 - (2*c*d)/b^2))/b^2 - (2*a*(c^2/b^2 + (2*a*((2*a*d^2)/b^3 - (2*c*d)/b^2))/b - (a^2*d^2)
/b^4))/b) + x^2*(c^2/(2*b^2) + (a*((2*a*d^2)/b^3 - (2*c*d)/b^2))/b - (a^2*d^2)/(2*b^4)) - x^3*((2*a*d^2)/(3*b^
3) - (2*c*d)/(3*b^2)) + (a^5*d^2 + a^3*b^2*c^2 - 2*a^4*b*c*d)/(b*(a*b^5 + b^6*x)) + (log(a + b*x)*(5*a^4*d^2 +
3*a^2*b^2*c^2 - 8*a^3*b*c*d))/b^6 + (d^2*x^4)/(4*b^2)