\(\int \frac {x^3 (a c+b c x^2)}{(a+b x^2)^2} \, dx\) [127]

Optimal result

Integrand size = 23, antiderivative size = 29 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {c x^2}{2 b}-\frac {a c \log \left (a+b x^2\right )}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {21, 272, 45} \[ \int \frac {x^3 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {c x^2}{2 b}-\frac {a c \log \left (a+b x^2\right )}{2 b^2} \]

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^2} \, dx=c \left (\frac {x^2}{2 b}-\frac {a \log \left (a+b x^2\right )}{2 b^2}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\(-\frac {-c b \,x^{2}+a c \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(25\)
default	\(c \left (\frac {x^{2}}{2 b}-\frac {a \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\right )\)	\(26\)
risch	\(\frac {c \,x^{2}}{2 b}-\frac {a c \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(26\)
norman	\(\frac {\frac {x^{4} c}{2}-\frac {a^{2} c}{2 b^{2}}}{b \,x^{2}+a}-\frac {a c \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {b c x^{2} - a c \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

1/2*(b*c*x^2 - a*c*log(b*x^2 + a))/b^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^2} \, dx=c \left (- \frac {a \log {\left (a + b x^{2} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b}\right ) \]

[In]

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

[Out]

c*(-a*log(a + b*x**2)/(2*b**2) + x**2/(2*b))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {c x^{2}}{2 \, b} - \frac {a c \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

1/2*c*x^2/b - 1/2*a*c*log(b*x^2 + a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {\frac {a c \log \left (\frac {{\left | b x^{2} + a \right |}}{{\left (b x^{2} + a\right )}^{2} {\left | b \right |}}\right )}{b} + \frac {{\left (b x^{2} + a\right )} c}{b}}{2 \, b} \]

[In]

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

[Out]

1/2*(a*c*log(abs(b*x^2 + a)/((b*x^2 + a)^2*abs(b)))/b + (b*x^2 + a)*c/b)/b

Mupad [B] (verification not implemented)

Time = 5.43 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {c\,\left (a\,\ln \left (b\,x^2+a\right )-b\,x^2\right )}{2\,b^2} \]

[In]

int((x^3*(a*c + b*c*x^2))/(a + b*x^2)^2,x)

[Out]

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