\(\int \frac {x^3}{(a+b x)^4} \, dx\) [198]

Optimal result

Integrand size = 11, antiderivative size = 58 \[ \int \frac {x^3}{(a+b x)^4} \, dx=\frac {a^3}{3 b^4 (a+b x)^3}-\frac {3 a^2}{2 b^4 (a+b x)^2}+\frac {3 a}{b^4 (a+b x)}+\frac {\log (a+b x)}{b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {x^3}{(a+b x)^4} \, dx=\frac {a^3}{3 b^4 (a+b x)^3}-\frac {3 a^2}{2 b^4 (a+b x)^2}+\frac {3 a}{b^4 (a+b x)}+\frac {\log (a+b x)}{b^4} \]

[In]

Int[x^3/(a + b*x)^4,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {a^3}{b^3 (a+b x)^4}+\frac {3 a^2}{b^3 (a+b x)^3}-\frac {3 a}{b^3 (a+b x)^2}+\frac {1}{b^3 (a+b x)}\right ) \, dx \\ & = \frac {a^3}{3 b^4 (a+b x)^3}-\frac {3 a^2}{2 b^4 (a+b x)^2}+\frac {3 a}{b^4 (a+b x)}+\frac {\log (a+b x)}{b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.76 \[ \int \frac {x^3}{(a+b x)^4} \, dx=\frac {\frac {a \left (11 a^2+27 a b x+18 b^2 x^2\right )}{(a+b x)^3}+6 \log (a+b x)}{6 b^4} \]

[In]

Integrate[x^3/(a + b*x)^4,x]

[Out]

((a*(11*a^2 + 27*a*b*x + 18*b^2*x^2))/(a + b*x)^3 + 6*Log[a + b*x])/(6*b^4)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81

method	result	size
norman	\(\frac {\frac {11 a^{3}}{6 b^{4}}+\frac {3 a \,x^{2}}{b^{2}}+\frac {9 a^{2} x}{2 b^{3}}}{\left (b x +a \right )^{3}}+\frac {\ln \left (b x +a \right )}{b^{4}}\)	\(47\)
risch	\(\frac {\frac {11 a^{3}}{6 b^{4}}+\frac {3 a \,x^{2}}{b^{2}}+\frac {9 a^{2} x}{2 b^{3}}}{\left (b x +a \right )^{3}}+\frac {\ln \left (b x +a \right )}{b^{4}}\)	\(47\)
default	\(\frac {a^{3}}{3 b^{4} \left (b x +a \right )^{3}}-\frac {3 a^{2}}{2 b^{4} \left (b x +a \right )^{2}}+\frac {3 a}{b^{4} \left (b x +a \right )}+\frac {\ln \left (b x +a \right )}{b^{4}}\)	\(55\)
parallelrisch	\(\frac {6 b^{3} \ln \left (b x +a \right ) x^{3}+18 \ln \left (b x +a \right ) x^{2} a \,b^{2}+18 \ln \left (b x +a \right ) x \,a^{2} b +18 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )+27 a^{2} b x +11 a^{3}}{6 b^{4} \left (b x +a \right )^{3}}\)	\(88\)

[In]

int(x^3/(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

(11/6*a^3/b^4+3*a*x^2/b^2+9/2*a^2*x/b^3)/(b*x+a)^3+ln(b*x+a)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.62 \[ \int \frac {x^3}{(a+b x)^4} \, dx=\frac {18 \, a b^{2} x^{2} + 27 \, a^{2} b x + 11 \, a^{3} + 6 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{6 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} \]

[In]

integrate(x^3/(b*x+a)^4,x, algorithm="fricas")

[Out]

1/6*(18*a*b^2*x^2 + 27*a^2*b*x + 11*a^3 + 6*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*log(b*x + a))/(b^7*x^3 +
3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.21 \[ \int \frac {x^3}{(a+b x)^4} \, dx=\frac {11 a^{3} + 27 a^{2} b x + 18 a b^{2} x^{2}}{6 a^{3} b^{4} + 18 a^{2} b^{5} x + 18 a b^{6} x^{2} + 6 b^{7} x^{3}} + \frac {\log {\left (a + b x \right )}}{b^{4}} \]

[In]

integrate(x**3/(b*x+a)**4,x)

[Out]

(11*a**3 + 27*a**2*b*x + 18*a*b**2*x**2)/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + log(a
+ b*x)/b**4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.21 \[ \int \frac {x^3}{(a+b x)^4} \, dx=\frac {18 \, a b^{2} x^{2} + 27 \, a^{2} b x + 11 \, a^{3}}{6 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} + \frac {\log \left (b x + a\right )}{b^{4}} \]

[In]

integrate(x^3/(b*x+a)^4,x, algorithm="maxima")

[Out]

1/6*(18*a*b^2*x^2 + 27*a^2*b*x + 11*a^3)/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4) + log(b*x + a)/b^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79 \[ \int \frac {x^3}{(a+b x)^4} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac {18 \, a b x^{2} + 27 \, a^{2} x + \frac {11 \, a^{3}}{b}}{6 \, {\left (b x + a\right )}^{3} b^{3}} \]

[In]

integrate(x^3/(b*x+a)^4,x, algorithm="giac")

[Out]

log(abs(b*x + a))/b^4 + 1/6*(18*a*b*x^2 + 27*a^2*x + 11*a^3/b)/((b*x + a)^3*b^3)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{(a+b x)^4} \, dx=\frac {\ln \left (a+b\,x\right )+\frac {3\,a}{a+b\,x}-\frac {3\,a^2}{2\,{\left (a+b\,x\right )}^2}+\frac {a^3}{3\,{\left (a+b\,x\right )}^3}}{b^4} \]

[In]

int(x^3/(a + b*x)^4,x)

[Out]

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