\(\int \frac {x^3 (c+d x)}{a+b x} \, dx\) [205]

Optimal result

Integrand size = 16, antiderivative size = 87 \[ \int \frac {x^3 (c+d x)}{a+b x} \, dx=\frac {a^2 (b c-a d) x}{b^4}-\frac {a (b c-a d) x^2}{2 b^3}+\frac {(b c-a d) x^3}{3 b^2}+\frac {d x^4}{4 b}-\frac {a^3 (b c-a d) \log (a+b x)}{b^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 87, 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} \[ \int \frac {x^3 (c+d x)}{a+b x} \, dx=-\frac {a^3 (b c-a d) \log (a+b x)}{b^5}+\frac {a^2 x (b c-a d)}{b^4}-\frac {a x^2 (b c-a d)}{2 b^3}+\frac {x^3 (b c-a d)}{3 b^2}+\frac {d x^4}{4 b} \]

[In]

Int[(x^3*(c + d*x))/(a + b*x),x]

[Out]

(a^2*(b*c - a*d)*x)/b^4 - (a*(b*c - a*d)*x^2)/(2*b^3) + ((b*c - a*d)*x^3)/(3*b^2) + (d*x^4)/(4*b) - (a^3*(b*c
- a*d)*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 {a^2 (-b c+a d)}{b^4}+\frac {a (-b c+a d) x}{b^3}+\frac {(b c-a d) x^2}{b^2}+\frac {d x^3}{b}+\frac {a^3 (-b c+a d)}{b^4 (a+b x)}\right ) \, dx \\ & = \frac {a^2 (b c-a d) x}{b^4}-\frac {a (b c-a d) x^2}{2 b^3}+\frac {(b c-a d) x^3}{3 b^2}+\frac {d x^4}{4 b}-\frac {a^3 (b c-a d) \log (a+b x)}{b^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 (c+d x)}{a+b x} \, dx=\frac {b x \left (-12 a^3 d+6 a^2 b (2 c+d x)-2 a b^2 x (3 c+2 d x)+b^3 x^2 (4 c+3 d x)\right )+12 a^3 (-b c+a d) \log (a+b x)}{12 b^5} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94

method	result	size
norman	\(\frac {d \,x^{4}}{4 b}-\frac {\left (a d -b c \right ) x^{3}}{3 b^{2}}+\frac {a \left (a d -b c \right ) x^{2}}{2 b^{3}}-\frac {a^{2} \left (a d -b c \right ) x}{b^{4}}+\frac {a^{3} \left (a d -b c \right ) \ln \left (b x +a \right )}{b^{5}}\)	\(82\)
default	\(-\frac {-\frac {1}{4} d \,x^{4} b^{3}+\frac {1}{3} a \,b^{2} d \,x^{3}-\frac {1}{3} b^{3} c \,x^{3}-\frac {1}{2} a^{2} b d \,x^{2}+\frac {1}{2} a \,b^{2} c \,x^{2}+a^{3} d x -a^{2} b c x}{b^{4}}+\frac {a^{3} \left (a d -b c \right ) \ln \left (b x +a \right )}{b^{5}}\)	\(91\)
risch	\(\frac {d \,x^{4}}{4 b}-\frac {a d \,x^{3}}{3 b^{2}}+\frac {c \,x^{3}}{3 b}+\frac {a^{2} d \,x^{2}}{2 b^{3}}-\frac {a c \,x^{2}}{2 b^{2}}-\frac {a^{3} d x}{b^{4}}+\frac {a^{2} c x}{b^{3}}+\frac {a^{4} \ln \left (b x +a \right ) d}{b^{5}}-\frac {a^{3} \ln \left (b x +a \right ) c}{b^{4}}\)	\(100\)
parallelrisch	\(\frac {3 d \,x^{4} b^{4}-4 x^{3} a \,b^{3} d +4 x^{3} b^{4} c +6 x^{2} a^{2} b^{2} d -6 x^{2} a \,b^{3} c +12 \ln \left (b x +a \right ) a^{4} d -12 \ln \left (b x +a \right ) a^{3} b c -12 x \,a^{3} b d +12 x \,a^{2} b^{2} c}{12 b^{5}}\)	\(100\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.08 \[ \int \frac {x^3 (c+d x)}{a+b x} \, dx=\frac {3 \, b^{4} d x^{4} + 4 \, {\left (b^{4} c - a b^{3} d\right )} x^{3} - 6 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} x^{2} + 12 \, {\left (a^{2} b^{2} c - a^{3} b d\right )} x - 12 \, {\left (a^{3} b c - a^{4} d\right )} \log \left (b x + a\right )}{12 \, b^{5}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \frac {x^3 (c+d x)}{a+b x} \, dx=\frac {a^{3} \left (a d - b c\right ) \log {\left (a + b x \right )}}{b^{5}} + x^{3} \left (- \frac {a d}{3 b^{2}} + \frac {c}{3 b}\right ) + x^{2} \left (\frac {a^{2} d}{2 b^{3}} - \frac {a c}{2 b^{2}}\right ) + x \left (- \frac {a^{3} d}{b^{4}} + \frac {a^{2} c}{b^{3}}\right ) + \frac {d x^{4}}{4 b} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07 \[ \int \frac {x^3 (c+d x)}{a+b x} \, dx=\frac {3 \, b^{3} d x^{4} + 4 \, {\left (b^{3} c - a b^{2} d\right )} x^{3} - 6 \, {\left (a b^{2} c - a^{2} b d\right )} x^{2} + 12 \, {\left (a^{2} b c - a^{3} d\right )} x}{12 \, b^{4}} - \frac {{\left (a^{3} b c - a^{4} d\right )} \log \left (b x + a\right )}{b^{5}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.09 \[ \int \frac {x^3 (c+d x)}{a+b x} \, dx=\frac {3 \, b^{3} d x^{4} + 4 \, b^{3} c x^{3} - 4 \, a b^{2} d x^{3} - 6 \, a b^{2} c x^{2} + 6 \, a^{2} b d x^{2} + 12 \, a^{2} b c x - 12 \, a^{3} d x}{12 \, b^{4}} - \frac {{\left (a^{3} b c - a^{4} d\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.08 \[ \int \frac {x^3 (c+d x)}{a+b x} \, dx=x^3\,\left (\frac {c}{3\,b}-\frac {a\,d}{3\,b^2}\right )+\frac {\ln \left (a+b\,x\right )\,\left (a^4\,d-a^3\,b\,c\right )}{b^5}+\frac {d\,x^4}{4\,b}-\frac {a\,x^2\,\left (\frac {c}{b}-\frac {a\,d}{b^2}\right )}{2\,b}+\frac {a^2\,x\,\left (\frac {c}{b}-\frac {a\,d}{b^2}\right )}{b^2} \]

[In]

int((x^3*(c + d*x))/(a + b*x),x)

[Out]

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