\(\int x \log ^3(c (a+b x^2)^p) \, dx\) [93]

Optimal result

Integrand size = 16, antiderivative size = 93 \[ \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=-3 p^3 x^2+\frac {3 p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}-\frac {3 p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b}+\frac {\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b} \]

[Out]

-3*p^3*x^2+3*p^2*(b*x^2+a)*ln(c*(b*x^2+a)^p)/b-3/2*p*(b*x^2+a)*ln(c*(b*x^2+a)^p)^2/b+1/2*(b*x^2+a)*ln(c*(b*x^2
+a)^p)^3/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2504, 2436, 2333, 2332} \[ \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {3 p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}+\frac {\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b}-\frac {3 p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b}-3 p^3 x^2 \]

[In]

Int[x*Log[c*(a + b*x^2)^p]^3,x]

[Out]

-3*p^3*x^2 + (3*p^2*(a + b*x^2)*Log[c*(a + b*x^2)^p])/b - (3*p*(a + b*x^2)*Log[c*(a + b*x^2)^p]^2)/(2*b) + ((a
+ b*x^2)*Log[c*(a + b*x^2)^p]^3)/(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 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94 \[ \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {-6 b p^3 x^2+6 p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )-3 p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )+\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b} \]

[In]

Integrate[x*Log[c*(a + b*x^2)^p]^3,x]

[Out]

(-6*b*p^3*x^2 + 6*p^2*(a + b*x^2)*Log[c*(a + b*x^2)^p] - 3*p*(a + b*x^2)*Log[c*(a + b*x^2)^p]^2 + (a + b*x^2)*
Log[c*(a + b*x^2)^p]^3)/(2*b)

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.62

method	result	size
parallelrisch	\(\frac {x^{2} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{3} a b p -3 x^{2} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a b \,p^{2}+6 x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a b \,p^{3}-6 x^{2} a b \,p^{4}+{\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{3} a^{2} p -3 {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a^{2} p^{2}+6 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{2} p^{3}}{2 a b p}\)	\(151\)
risch	\(\text {Expression too large to display}\)	\(3925\)

[In]

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

[Out]

1/2*(x^2*ln(c*(b*x^2+a)^p)^3*a*b*p-3*x^2*ln(c*(b*x^2+a)^p)^2*a*b*p^2+6*x^2*ln(c*(b*x^2+a)^p)*a*b*p^3-6*x^2*a*b
*p^4+ln(c*(b*x^2+a)^p)^3*a^2*p-3*ln(c*(b*x^2+a)^p)^2*a^2*p^2+6*ln(c*(b*x^2+a)^p)*a^2*p^3)/a/b/p

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.89 \[ \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {6 \, b p^{3} x^{2} - 6 \, b p^{2} x^{2} \log \left (c\right ) + 3 \, b p x^{2} \log \left (c\right )^{2} - b x^{2} \log \left (c\right )^{3} - {\left (b p^{3} x^{2} + a p^{3}\right )} \log \left (b x^{2} + a\right )^{3} + 3 \, {\left (b p^{3} x^{2} + a p^{3} - {\left (b p^{2} x^{2} + a p^{2}\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right )^{2} - 3 \, {\left (2 \, b p^{3} x^{2} + 2 \, a p^{3} + {\left (b p x^{2} + a p\right )} \log \left (c\right )^{2} - 2 \, {\left (b p^{2} x^{2} + a p^{2}\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*(6*b*p^3*x^2 - 6*b*p^2*x^2*log(c) + 3*b*p*x^2*log(c)^2 - b*x^2*log(c)^3 - (b*p^3*x^2 + a*p^3)*log(b*x^2 +
a)^3 + 3*(b*p^3*x^2 + a*p^3 - (b*p^2*x^2 + a*p^2)*log(c))*log(b*x^2 + a)^2 - 3*(2*b*p^3*x^2 + 2*a*p^3 + (b*p*
x^2 + a*p)*log(c)^2 - 2*(b*p^2*x^2 + a*p^2)*log(c))*log(b*x^2 + a))/b

Sympy [A] (verification not implemented)

Time = 0.79 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.54 \[ \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \frac {3 a p^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b} - \frac {3 a p \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{2 b} + \frac {a \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{2 b} - 3 p^{3} x^{2} + 3 p^{2} x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {3 p x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{2} + \frac {x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{2} & \text {for}\: b \neq 0 \\\frac {x^{2} \log {\left (a^{p} c \right )}^{3}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((3*a*p**2*log(c*(a + b*x**2)**p)/b - 3*a*p*log(c*(a + b*x**2)**p)**2/(2*b) + a*log(c*(a + b*x**2)**p
)**3/(2*b) - 3*p**3*x**2 + 3*p**2*x**2*log(c*(a + b*x**2)**p) - 3*p*x**2*log(c*(a + b*x**2)**p)**2/2 + x**2*lo
g(c*(a + b*x**2)**p)**3/2, Ne(b, 0)), (x**2*log(a**p*c)**3/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.76 \[ \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {3}{2} \, b p {\left (\frac {x^{2}}{b} - \frac {a \log \left (b x^{2} + a\right )}{b^{2}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} + \frac {1}{2} \, x^{2} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3} + \frac {1}{2} \, b p {\left (\frac {{\left (a \log \left (b x^{2} + a\right )^{3} - 6 \, b x^{2} + 3 \, a \log \left (b x^{2} + a\right )^{2} + 6 \, a \log \left (b x^{2} + a\right )\right )} p^{2}}{b^{2}} + \frac {3 \, {\left (2 \, b x^{2} - a \log \left (b x^{2} + a\right )^{2} - 2 \, a \log \left (b x^{2} + a\right )\right )} p \log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{b^{2}}\right )} \]

[In]

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

[Out]

-3/2*b*p*(x^2/b - a*log(b*x^2 + a)/b^2)*log((b*x^2 + a)^p*c)^2 + 1/2*x^2*log((b*x^2 + a)^p*c)^3 + 1/2*b*p*((a*
log(b*x^2 + a)^3 - 6*b*x^2 + 3*a*log(b*x^2 + a)^2 + 6*a*log(b*x^2 + a))*p^2/b^2 + 3*(2*b*x^2 - a*log(b*x^2 + a
)^2 - 2*a*log(b*x^2 + a))*p*log((b*x^2 + a)^p*c)/b^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.82 \[ \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {{\left ({\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{3} - 6 \, b x^{2} - 3 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} + 6 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) - 6 \, a\right )} p^{3} + 3 \, {\left (2 \, b x^{2} + {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + 2 \, a\right )} p^{2} \log \left (c\right ) - 3 \, {\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} p \log \left (c\right )^{2} + {\left (b x^{2} + a\right )} \log \left (c\right )^{3}}{2 \, b} \]

[In]

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

[Out]

1/2*(((b*x^2 + a)*log(b*x^2 + a)^3 - 6*b*x^2 - 3*(b*x^2 + a)*log(b*x^2 + a)^2 + 6*(b*x^2 + a)*log(b*x^2 + a) -
6*a)*p^3 + 3*(2*b*x^2 + (b*x^2 + a)*log(b*x^2 + a)^2 - 2*(b*x^2 + a)*log(b*x^2 + a) + 2*a)*p^2*log(c) - 3*(b*
x^2 - (b*x^2 + a)*log(b*x^2 + a) + a)*p*log(c)^2 + (b*x^2 + a)*log(c)^3)/b

Mupad [B] (verification not implemented)

Time = 1.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11 \[ \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx={\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^3\,\left (\frac {a}{2\,b}+\frac {x^2}{2}\right )-{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {3\,p\,x^2}{2}+\frac {3\,a\,p}{2\,b}\right )-3\,p^3\,x^2+3\,p^2\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )+\frac {3\,a\,p^3\,\ln \left (b\,x^2+a\right )}{b} \]

[In]

int(x*log(c*(a + b*x^2)^p)^3,x)

[Out]

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