3.2 \(\int x^3 \log (c (a+b x^2)^p) \, dx\)

Optimal. Leaf size=59 \[ -\frac{a^2 p \log \left (a+b x^2\right )}{4 b^2}+\frac{1}{4} x^4 \log \left (c \left (a+b x^2\right )^p\right )+\frac{a p x^2}{4 b}-\frac{p x^4}{8} \]

[Out]

(a*p*x^2)/(4*b) - (p*x^4)/8 - (a^2*p*Log[a + b*x^2])/(4*b^2) + (x^4*Log[c*(a + b*x^2)^p])/4

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Rubi [A]  time = 0.0492964, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2454, 2395, 43} \[ -\frac{a^2 p \log \left (a+b x^2\right )}{4 b^2}+\frac{1}{4} x^4 \log \left (c \left (a+b x^2\right )^p\right )+\frac{a p x^2}{4 b}-\frac{p x^4}{8} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Log[c*(a + b*x^2)^p],x]

[Out]

(a*p*x^2)/(4*b) - (p*x^4)/8 - (a^2*p*Log[a + b*x^2])/(4*b^2) + (x^4*Log[c*(a + b*x^2)^p])/4

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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+b x^2\right )^p\right ) \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x \log \left (c (a+b x)^p\right ) \, dx,x,x^2\right )\\ &=\frac{1}{4} x^4 \log \left (c \left (a+b x^2\right )^p\right )-\frac{1}{4} (b p) \operatorname{Subst}\left (\int \frac{x^2}{a+b x} \, dx,x,x^2\right )\\ &=\frac{1}{4} x^4 \log \left (c \left (a+b x^2\right )^p\right )-\frac{1}{4} (b p) \operatorname{Subst}\left (\int \left (-\frac{a}{b^2}+\frac{x}{b}+\frac{a^2}{b^2 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac{a p x^2}{4 b}-\frac{p x^4}{8}-\frac{a^2 p \log \left (a+b x^2\right )}{4 b^2}+\frac{1}{4} x^4 \log \left (c \left (a+b x^2\right )^p\right )\\ \end{align*}

Mathematica [A]  time = 0.0145955, size = 59, normalized size = 1. \[ -\frac{a^2 p \log \left (a+b x^2\right )}{4 b^2}+\frac{1}{4} x^4 \log \left (c \left (a+b x^2\right )^p\right )+\frac{a p x^2}{4 b}-\frac{p x^4}{8} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Log[c*(a + b*x^2)^p],x]

[Out]

(a*p*x^2)/(4*b) - (p*x^4)/8 - (a^2*p*Log[a + b*x^2])/(4*b^2) + (x^4*Log[c*(a + b*x^2)^p])/4

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Maple [C]  time = 0.414, size = 183, normalized size = 3.1 \begin{align*}{\frac{{x}^{4}\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{4}}-{\frac{i}{8}}\pi \,{x}^{4}{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{8}}\pi \,{x}^{4} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{i}{8}}\pi \,{x}^{4}{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}-{\frac{i}{8}}\pi \,{x}^{4} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+{\frac{\ln \left ( c \right ){x}^{4}}{4}}-{\frac{p{x}^{4}}{8}}+{\frac{ap{x}^{2}}{4\,b}}-{\frac{{a}^{2}p\ln \left ( b{x}^{2}+a \right ) }{4\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*ln((b*x^2+a)^p)-1/8*I*Pi*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+1/8*I*Pi*x^4*csgn(I*c
*(b*x^2+a)^p)^2*csgn(I*c)+1/8*I*Pi*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/8*I*Pi*x^4*csgn(I*c*(b*x^
2+a)^p)^3+1/4*ln(c)*x^4-1/8*p*x^4+1/4*a*p*x^2/b-1/4*a^2*p*ln(b*x^2+a)/b^2

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Maxima [A]  time = 1.10431, size = 74, normalized size = 1.25 \begin{align*} \frac{1}{4} \, x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) - \frac{1}{8} \, b p{\left (\frac{2 \, a^{2} \log \left (b x^{2} + a\right )}{b^{3}} + \frac{b x^{4} - 2 \, a x^{2}}{b^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*log((b*x^2 + a)^p*c) - 1/8*b*p*(2*a^2*log(b*x^2 + a)/b^3 + (b*x^4 - 2*a*x^2)/b^2)

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Fricas [A]  time = 2.0214, size = 127, normalized size = 2.15 \begin{align*} -\frac{b^{2} p x^{4} - 2 \, b^{2} x^{4} \log \left (c\right ) - 2 \, a b p x^{2} - 2 \,{\left (b^{2} p x^{4} - a^{2} p\right )} \log \left (b x^{2} + a\right )}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b^2*p*x^4 - 2*b^2*x^4*log(c) - 2*a*b*p*x^2 - 2*(b^2*p*x^4 - a^2*p)*log(b*x^2 + a))/b^2

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Sympy [A]  time = 8.44226, size = 70, normalized size = 1.19 \begin{align*} \begin{cases} - \frac{a^{2} p \log{\left (a + b x^{2} \right )}}{4 b^{2}} + \frac{a p x^{2}}{4 b} + \frac{p x^{4} \log{\left (a + b x^{2} \right )}}{4} - \frac{p x^{4}}{8} + \frac{x^{4} \log{\left (c \right )}}{4} & \text{for}\: b \neq 0 \\\frac{x^{4} \log{\left (a^{p} 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*(b*x**2+a)**p),x)

[Out]

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

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Giac [A]  time = 1.22004, size = 131, normalized size = 2.22 \begin{align*} \frac{\frac{{\left (2 \,{\left (b x^{2} + a\right )}^{2} \log \left (b x^{2} + a\right ) - 4 \,{\left (b x^{2} + a\right )} a \log \left (b x^{2} + a\right ) -{\left (b x^{2} + a\right )}^{2} + 4 \,{\left (b x^{2} + a\right )} a\right )} p}{b} + \frac{2 \,{\left ({\left (b x^{2} + a\right )}^{2} - 2 \,{\left (b x^{2} + a\right )} a\right )} \log \left (c\right )}{b}}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((2*(b*x^2 + a)^2*log(b*x^2 + a) - 4*(b*x^2 + a)*a*log(b*x^2 + a) - (b*x^2 + a)^2 + 4*(b*x^2 + a)*a)*p/b +
2*((b*x^2 + a)^2 - 2*(b*x^2 + a)*a)*log(c)/b)/b