3.2.34 \(\int \frac {x^3 (a c+b c x^2)}{(a+b x^2)^3} \, dx\) [134]

Optimal. Leaf size=35 \[ \frac {a c}{2 b^2 \left (a+b x^2\right )}+\frac {c \log \left (a+b x^2\right )}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {21, 272, 45} \begin {gather*} \frac {a c}{2 b^2 \left (a+b x^2\right )}+\frac {c \log \left (a+b x^2\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 28, normalized size = 0.80 \begin {gather*} \frac {c \left (\frac {a}{a+b x^2}+\log \left (a+b x^2\right )\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 32, normalized size = 0.91

method	result	size
default	\(c \left (\frac {a}{2 b^{2} \left (b \,x^{2}+a \right )}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{2}}\right )\)	\(32\)
risch	\(\frac {a c}{2 b^{2} \left (b \,x^{2}+a \right )}+\frac {c \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(32\)
norman	\(\frac {\frac {a^{2} c}{2 b^{2}}+\frac {c a \,x^{2}}{2 b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {c \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 34, normalized size = 0.97 \begin {gather*} \frac {a c}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} + \frac {c \log \left (b x^{2} + a\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*c/(b^3*x^2 + a*b^2) + 1/2*c*log(b*x^2 + a)/b^2

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Fricas [A]
time = 1.12, size = 40, normalized size = 1.14 \begin {gather*} \frac {a c + {\left (b c x^{2} + a c\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*c + (b*c*x^2 + a*c)*log(b*x^2 + a))/(b^3*x^2 + a*b^2)

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Sympy [A]
time = 0.09, size = 31, normalized size = 0.89 \begin {gather*} c \left (\frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {\log {\left (a + b x^{2} \right )}}{2 b^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(a/(2*a*b**2 + 2*b**3*x**2) + log(a + b*x**2)/(2*b**2))

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Giac [A]
time = 0.79, size = 32, normalized size = 0.91 \begin {gather*} \frac {c \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{2}} + \frac {a c}{2 \, {\left (b x^{2} + a\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 31, normalized size = 0.89 \begin {gather*} \frac {c\,\ln \left (b\,x^2+a\right )}{2\,b^2}+\frac {a\,c}{2\,b^2\,\left (b\,x^2+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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