3.1.37 \(\int x^3 \log (c (a+\frac {b}{x^2})^p) \, dx\) [37]

Optimal. Leaf size=51 \[ \frac {b p x^2}{4 a}+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )-\frac {b^2 p \log \left (b+a x^2\right )}{4 a^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2505, 269, 272, 45} \begin {gather*} -\frac {b^2 p \log \left (a x^2+b\right )}{4 a^2}+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {b p x^2}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Log[c*(a + b/x^2)^p],x]

[Out]

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

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 269

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

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]]

Rule 2505

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m +
1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Dist[b*e*n*(p/(f*(m + 1))), Int[x^(n - 1)*((f*x)^(m + 1)/
(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 56, normalized size = 1.10 \begin {gather*} \frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {1}{4} b p \left (\frac {x^2}{a}-\frac {b \log \left (a+\frac {b}{x^2}\right )}{a^2}-\frac {2 b \log (x)}{a^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Log[c*(a + b/x^2)^p],x]

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int x^{3} \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 44, normalized size = 0.86 \begin {gather*} \frac {1}{4} \, x^{4} \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right ) + \frac {1}{4} \, b p {\left (\frac {x^{2}}{a} - \frac {b \log \left (a x^{2} + b\right )}{a^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*log((a + b/x^2)^p*c) + 1/4*b*p*(x^2/a - b*log(a*x^2 + b)/a^2)

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Fricas [A]
time = 0.38, size = 56, normalized size = 1.10 \begin {gather*} \frac {a^{2} p x^{4} \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + a^{2} x^{4} \log \left (c\right ) + a b p x^{2} - b^{2} p \log \left (a x^{2} + b\right )}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^2*p*x^4*log((a*x^2 + b)/x^2) + a^2*x^4*log(c) + a*b*p*x^2 - b^2*p*log(a*x^2 + b))/a^2

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Sympy [A]
time = 1.65, size = 66, normalized size = 1.29 \begin {gather*} \begin {cases} \frac {x^{4} \log {\left (c \left (a + \frac {b}{x^{2}}\right )^{p} \right )}}{4} + \frac {b p x^{2}}{4 a} - \frac {b^{2} p \log {\left (a x^{2} + b \right )}}{4 a^{2}} & \text {for}\: a \neq 0 \\\frac {p x^{4}}{8} + \frac {x^{4} \log {\left (c \left (\frac {b}{x^{2}}\right )^{p} \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*log(c*(a + b/x**2)**p)/4 + b*p*x**2/(4*a) - b**2*p*log(a*x**2 + b)/(4*a**2), Ne(a, 0)), (p*x**
4/8 + x**4*log(c*(b/x**2)**p)/4, True))

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Giac [A]
time = 6.33, size = 59, normalized size = 1.16 \begin {gather*} \frac {1}{4} \, p x^{4} \log \left (a x^{2} + b\right ) - \frac {1}{4} \, p x^{4} \log \left (x^{2}\right ) + \frac {1}{4} \, x^{4} \log \left (c\right ) + \frac {b p x^{2}}{4 \, a} - \frac {b^{2} p \log \left (a x^{2} + b\right )}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*p*x^4*log(a*x^2 + b) - 1/4*p*x^4*log(x^2) + 1/4*x^4*log(c) + 1/4*b*p*x^2/a - 1/4*b^2*p*log(a*x^2 + b)/a^2

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Mupad [B]
time = 0.24, size = 45, normalized size = 0.88 \begin {gather*} \frac {x^4\,\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{4}-\frac {b^2\,p\,\ln \left (a\,x^2+b\right )}{4\,a^2}+\frac {b\,p\,x^2}{4\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*log(c*(a + b/x^2)^p),x)

[Out]

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

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