3.37 \(\int x^3 \log (c (a+\frac{b}{x^2})^p) \, dx\)

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

[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)

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Rubi [A]  time = 0.0335074, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2455, 263, 266, 43} \[ -\frac{b^2 p \log \left (a x^2+b\right )}{4 a^2}+\frac{1}{4} x^4 \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )+\frac{b p x^2}{4 a} \]

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 2455

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]

Rule 263

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 266

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 43

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*} \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) \operatorname{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) \operatorname{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*}

Mathematica [A]  time = 0.0202259, size = 56, normalized size = 1.1 \[ \frac{1}{4} b p \left (-\frac{b \log \left (a+\frac{b}{x^2}\right )}{a^2}-\frac{2 b \log (x)}{a^2}+\frac{x^2}{a}\right )+\frac{1}{4} x^4 \log \left (c \left (a+\frac{b}{x^2}\right )^p\right ) \]

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 = 0.245, size = 0, normalized size = 0. \begin{align*} \int{x}^{3}\ln \left ( c \left ( a+{\frac{b}{{x}^{2}}} \right ) ^{p} \right ) \, dx \end{align*}

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 = 1.08037, size = 59, normalized size = 1.16 \begin{align*} \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{align*}

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 = 2.32064, size = 127, normalized size = 2.49 \begin{align*} \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{align*}

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 = 29.7322, size = 87, normalized size = 1.71 \begin{align*} \begin{cases} \frac{p x^{4} \log{\left (a + \frac{b}{x^{2}} \right )}}{4} + \frac{x^{4} \log{\left (c \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} \log{\left (b \right )}}{4} - \frac{p x^{4} \log{\left (x \right )}}{2} + \frac{p x^{4}}{8} + \frac{x^{4} \log{\left (c \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21072, size = 80, normalized size = 1.57 \begin{align*} \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{align*}

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