\(\int x \log (c (a+b x^2)^p) \, dx\) [4]

Optimal result

Integrand size = 14, antiderivative size = 35 \[ \int x \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {p x^2}{2}+\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2504, 2436, 2332} \[ \int x \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 b}-\frac {p x^2}{2} \]

[In]

Int[x*Log[c*(a + b*x^2)^p],x]

[Out]

-1/2*(p*x^2) + ((a + b*x^2)*Log[c*(a + b*x^2)^p])/(2*b)

Rule 2332

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

Rule 2436

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 2504

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])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \log \left (c (a+b x)^p\right ) \, dx,x,x^2\right ) \\ & = \frac {\text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int x \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{2} \left (-p x^2+\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}\right ) \]

[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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) \left (b \,x^{2}+a \right )-\left (b \,x^{2}+a \right ) p}{2 b}\)	\(37\)
default	\(\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) \left (b \,x^{2}+a \right )-\left (b \,x^{2}+a \right ) p}{2 b}\)	\(37\)
norman	\(-\frac {p \,x^{2}}{2}+\frac {x^{2} \ln \left (c \,{\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}\right )}{2}+\frac {p a \ln \left (b \,x^{2}+a \right )}{2 b}\)	\(42\)
parts	\(\frac {x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{2}-p b \left (\frac {x^{2}}{2 b}-\frac {a \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\right )\)	\(46\)
parallelrisch	\(\frac {x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b p -x^{2} b \,p^{2}+\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a p +a \,p^{2}}{2 p b}\)	\(57\)
risch	\(\frac {x^{2} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{4}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{4}-\frac {i \pi \,x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{4}+\frac {i \pi \,x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{4}+\frac {\ln \left (c \right ) x^{2}}{2}-\frac {p \,x^{2}}{2}+\frac {p a \ln \left (b \,x^{2}+a \right )}{2 b}\)	\(171\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14 \[ \int x \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {b p x^{2} - b x^{2} \log \left (c\right ) - {\left (b p x^{2} + a p\right )} \log \left (b x^{2} + a\right )}{2 \, b} \]

[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

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.46 \[ \int x \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \frac {a \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 b} - \frac {p x^{2}}{2} + \frac {x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2} & \text {for}\: b \neq 0 \\\frac {x^{2} \log {\left (a^{p} c \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int x \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {1}{2} \, b p {\left (\frac {x^{2}}{b} - \frac {a \log \left (b x^{2} + a\right )}{b^{2}}\right )} + \frac {1}{2} \, x^{2} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int x \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {{\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} p - {\left (b x^{2} + a\right )} \log \left (c\right )}{2 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 1.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int x \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2}-\frac {p\,x^2}{2}+\frac {a\,p\,\ln \left (b\,x^2+a\right )}{2\,b} \]

[In]

int(x*log(c*(a + b*x^2)^p),x)

[Out]

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