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

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

[Out]

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

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Rubi [A]  time = 0.0244695, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2454, 2389, 2295} \[ \frac{\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 b}-\frac{p x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*Log[c*(a + b*x^2)^p],x]

[Out]

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

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 2389

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

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

Mathematica [A]  time = 0.0088682, size = 34, normalized size = 0.97 \[ \frac{1}{2} \left (\frac{\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}-p x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[c*(a + b*x^2)^p],x]

[Out]

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

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Maple [A]  time = 0.097, size = 50, normalized size = 1.4 \begin{align*}{\frac{{x}^{2}\ln \left ( c \left ( b{x}^{2}+a \right ) ^{p} \right ) }{2}}-{\frac{p{x}^{2}}{2}}+{\frac{\ln \left ( c \left ( b{x}^{2}+a \right ) ^{p} \right ) a}{2\,b}}-{\frac{ap}{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09696, size = 59, normalized size = 1.69 \begin{align*} -\frac{1}{2} \, b p{\left (\frac{x^{2}}{b} - \frac{a \log \left (b x^{2} + a\right )}{b^{2}}\right )} + \frac{1}{2} \, x^{2} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.94739, size = 89, normalized size = 2.54 \begin{align*} -\frac{b p x^{2} - b x^{2} \log \left (c\right ) -{\left (b p x^{2} + a p\right )} \log \left (b x^{2} + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.69366, size = 56, normalized size = 1.6 \begin{align*} \begin{cases} \frac{a p \log{\left (a + b x^{2} \right )}}{2 b} + \frac{p x^{2} \log{\left (a + b x^{2} \right )}}{2} - \frac{p x^{2}}{2} + \frac{x^{2} \log{\left (c \right )}}{2} & \text{for}\: b \neq 0 \\\frac{x^{2} \log{\left (a^{p} c \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18202, size = 58, normalized size = 1.66 \begin{align*} -\frac{{\left (b x^{2} -{\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} p -{\left (b x^{2} + a\right )} \log \left (c\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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