\(\int \frac {x (c+d x)^3}{a+b x} \, dx\) [224]

Optimal result

Integrand size = 16, antiderivative size = 106 \[ \int \frac {x (c+d x)^3}{a+b x} \, dx=\frac {(b c-a d)^3 x}{b^4}+\frac {d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^2}{2 b^3}+\frac {d^2 (3 b c-a d) x^3}{3 b^2}+\frac {d^3 x^4}{4 b}-\frac {a (b c-a d)^3 \log (a+b x)}{b^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {x (c+d x)^3}{a+b x} \, dx=\frac {d x^2 \left (a^2 d^2-3 a b c d+3 b^2 c^2\right )}{2 b^3}-\frac {a (b c-a d)^3 \log (a+b x)}{b^5}+\frac {x (b c-a d)^3}{b^4}+\frac {d^2 x^3 (3 b c-a d)}{3 b^2}+\frac {d^3 x^4}{4 b} \]

[In]

Int[(x*(c + d*x)^3)/(a + b*x),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.08 \[ \int \frac {x (c+d x)^3}{a+b x} \, dx=\frac {b x \left (-12 a^3 d^3+6 a^2 b d^2 (6 c+d x)-2 a b^2 d \left (18 c^2+9 c d x+2 d^2 x^2\right )+3 b^3 \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )\right )+12 a (-b c+a d)^3 \log (a+b x)}{12 b^5} \]

[In]

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

[Out]

(b*x*(-12*a^3*d^3 + 6*a^2*b*d^2*(6*c + d*x) - 2*a*b^2*d*(18*c^2 + 9*c*d*x + 2*d^2*x^2) + 3*b^3*(4*c^3 + 6*c^2*
d*x + 4*c*d^2*x^2 + d^3*x^3)) + 12*a*(-(b*c) + a*d)^3*Log[a + b*x])/(12*b^5)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.43

method	result	size
norman	\(-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{b^{4}}+\frac {d^{3} x^{4}}{4 b}-\frac {d^{2} \left (a d -3 b c \right ) x^{3}}{3 b^{2}}+\frac {d \left (a^{2} d^{2}-3 a b c d +3 b^{2} c^{2}\right ) x^{2}}{2 b^{3}}+\frac {a \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (b x +a \right )}{b^{5}}\)	\(152\)
default	\(-\frac {-\frac {1}{4} d^{3} x^{4} b^{3}+\frac {1}{3} x^{3} a \,b^{2} d^{3}-x^{3} b^{3} c \,d^{2}-\frac {1}{2} x^{2} a^{2} b \,d^{3}+\frac {3}{2} x^{2} a \,b^{2} c \,d^{2}-\frac {3}{2} x^{2} b^{3} c^{2} d +a^{3} d^{3} x -3 a^{2} b c \,d^{2} x +3 a \,b^{2} c^{2} d x -b^{3} c^{3} x}{b^{4}}+\frac {a \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (b x +a \right )}{b^{5}}\)	\(166\)
risch	\(\frac {d^{3} x^{4}}{4 b}-\frac {x^{3} a \,d^{3}}{3 b^{2}}+\frac {x^{3} c \,d^{2}}{b}+\frac {x^{2} a^{2} d^{3}}{2 b^{3}}-\frac {3 x^{2} a c \,d^{2}}{2 b^{2}}+\frac {3 x^{2} c^{2} d}{2 b}-\frac {a^{3} d^{3} x}{b^{4}}+\frac {3 a^{2} c \,d^{2} x}{b^{3}}-\frac {3 a \,c^{2} d x}{b^{2}}+\frac {c^{3} x}{b}+\frac {a^{4} \ln \left (b x +a \right ) d^{3}}{b^{5}}-\frac {3 a^{3} \ln \left (b x +a \right ) c \,d^{2}}{b^{4}}+\frac {3 a^{2} \ln \left (b x +a \right ) c^{2} d}{b^{3}}-\frac {a \ln \left (b x +a \right ) c^{3}}{b^{2}}\)	\(186\)
parallelrisch	\(\frac {3 d^{3} x^{4} b^{4}-4 x^{3} a \,b^{3} d^{3}+12 x^{3} b^{4} c \,d^{2}+6 x^{2} a^{2} b^{2} d^{3}-18 x^{2} a \,b^{3} c \,d^{2}+18 x^{2} b^{4} c^{2} d +12 \ln \left (b x +a \right ) a^{4} d^{3}-36 \ln \left (b x +a \right ) a^{3} b c \,d^{2}+36 \ln \left (b x +a \right ) a^{2} b^{2} c^{2} d -12 \ln \left (b x +a \right ) a \,b^{3} c^{3}-12 x \,a^{3} b \,d^{3}+36 x \,a^{2} b^{2} c \,d^{2}-36 x a \,b^{3} c^{2} d +12 b^{4} c^{3} x}{12 b^{5}}\)	\(187\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.56 \[ \int \frac {x (c+d x)^3}{a+b x} \, dx=\frac {3 \, b^{4} d^{3} x^{4} + 4 \, {\left (3 \, b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{3} + 6 \, {\left (3 \, b^{4} c^{2} d - 3 \, a b^{3} c d^{2} + a^{2} b^{2} d^{3}\right )} x^{2} + 12 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x - 12 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \log \left (b x + a\right )}{12 \, b^{5}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.24 \[ \int \frac {x (c+d x)^3}{a+b x} \, dx=\frac {a \left (a d - b c\right )^{3} \log {\left (a + b x \right )}}{b^{5}} + x^{3} \left (- \frac {a d^{3}}{3 b^{2}} + \frac {c d^{2}}{b}\right ) + x^{2} \left (\frac {a^{2} d^{3}}{2 b^{3}} - \frac {3 a c d^{2}}{2 b^{2}} + \frac {3 c^{2} d}{2 b}\right ) + x \left (- \frac {a^{3} d^{3}}{b^{4}} + \frac {3 a^{2} c d^{2}}{b^{3}} - \frac {3 a c^{2} d}{b^{2}} + \frac {c^{3}}{b}\right ) + \frac {d^{3} x^{4}}{4 b} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.55 \[ \int \frac {x (c+d x)^3}{a+b x} \, dx=\frac {3 \, b^{3} d^{3} x^{4} + 4 \, {\left (3 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{3} + 6 \, {\left (3 \, b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + 12 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x}{12 \, b^{4}} - \frac {{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \log \left (b x + a\right )}{b^{5}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.60 \[ \int \frac {x (c+d x)^3}{a+b x} \, dx=\frac {3 \, b^{3} d^{3} x^{4} + 12 \, b^{3} c d^{2} x^{3} - 4 \, a b^{2} d^{3} x^{3} + 18 \, b^{3} c^{2} d x^{2} - 18 \, a b^{2} c d^{2} x^{2} + 6 \, a^{2} b d^{3} x^{2} + 12 \, b^{3} c^{3} x - 36 \, a b^{2} c^{2} d x + 36 \, a^{2} b c d^{2} x - 12 \, a^{3} d^{3} x}{12 \, b^{4}} - \frac {{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \]

[In]

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

[Out]

1/12*(3*b^3*d^3*x^4 + 12*b^3*c*d^2*x^3 - 4*a*b^2*d^3*x^3 + 18*b^3*c^2*d*x^2 - 18*a*b^2*c*d^2*x^2 + 6*a^2*b*d^3
*x^2 + 12*b^3*c^3*x - 36*a*b^2*c^2*d*x + 36*a^2*b*c*d^2*x - 12*a^3*d^3*x)/b^4 - (a*b^3*c^3 - 3*a^2*b^2*c^2*d +
3*a^3*b*c*d^2 - a^4*d^3)*log(abs(b*x + a))/b^5

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.62 \[ \int \frac {x (c+d x)^3}{a+b x} \, dx=x^2\,\left (\frac {3\,c^2\,d}{2\,b}+\frac {a\,\left (\frac {a\,d^3}{b^2}-\frac {3\,c\,d^2}{b}\right )}{2\,b}\right )-x^3\,\left (\frac {a\,d^3}{3\,b^2}-\frac {c\,d^2}{b}\right )+x\,\left (\frac {c^3}{b}-\frac {a\,\left (\frac {3\,c^2\,d}{b}+\frac {a\,\left (\frac {a\,d^3}{b^2}-\frac {3\,c\,d^2}{b}\right )}{b}\right )}{b}\right )+\frac {d^3\,x^4}{4\,b}+\frac {\ln \left (a+b\,x\right )\,\left (a^4\,d^3-3\,a^3\,b\,c\,d^2+3\,a^2\,b^2\,c^2\,d-a\,b^3\,c^3\right )}{b^5} \]

[In]

int((x*(c + d*x)^3)/(a + b*x),x)

[Out]

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