3.1.79 \(\int x \log ^2(c (a+b x^2)^p) \, dx\) [79]

Optimal. Leaf size=61 \[ p^2 x^2-\frac {p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}+\frac {\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2504, 2436, 2333, 2332} \begin {gather*} \frac {\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b}-\frac {p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}+p^2 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Log[c*(a + b*x^2)^p]^2,x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 63, normalized size = 1.03 \begin {gather*} \frac {1}{2} \left (\frac {\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{b}-2 p \left (-p x^2+\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Log[c*(a + b*x^2)^p]^2,x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.69, size = 1034, normalized size = 16.95

method	result	size
risch	\(\text {Expression too large to display}\)	\(1034\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*p^2/b*ln(b*x^2+a)+1/2*x^2*ln((b*x^2+a)^p)^2+1/b*ln(c)*ln(b*x^2+a)*a*p+1/4*Pi^2*x^2*csgn(I*(b*x^2+a)^p)^2*cs
gn(I*c*(b*x^2+a)^p)^3*csgn(I*c)-1/8*Pi^2*x^2*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)^2-1/2*Pi^
2*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c)+1/4*Pi^2*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a
)^p)^3*csgn(I*c)^2-1/2*I/b*Pi*ln(b*x^2+a)*a*p*csgn(I*c*(b*x^2+a)^p)^3-1/2*I*ln(c)*Pi*x^2*csgn(I*(b*x^2+a)^p)*c
sgn(I*c*(b*x^2+a)^p)*csgn(I*c)+1/2*I*Pi*p*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-ln(c)*p*x^2-
1/8*Pi^2*x^2*csgn(I*c*(b*x^2+a)^p)^6+1/2*ln(c)^2*x^2+1/2*I/b*Pi*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(
b*x^2+a)^p)^2+1/2*I/b*Pi*ln(b*x^2+a)*a*p*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-1/2*I*ln(c)*Pi*x^2*csgn(I*c*(b*x^2+
a)^p)^3+1/2*I*Pi*p*x^2*csgn(I*c*(b*x^2+a)^p)^3+p^2*x^2+1/2*I*ln(c)*Pi*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+
a)^p)^2+1/2*I*ln(c)*Pi*x^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-1/2*I*Pi*p*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^
2+a)^p)^2-1/2*I*Pi*p*x^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+1/2*(I*Pi*b*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2
+a)^p)^2-I*Pi*b*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*b*x^2*csgn(I*c*(b*x^2+a)^p)^3+I*P
i*b*x^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*ln(c)*b*x^2-2*x^2*p*b+2*p*a*ln(b*x^2+a))/b*ln((b*x^2+a)^p)-1/2*I/b
*Pi*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+1/4*Pi^2*x^2*csgn(I*c*(b*x^2+a)^p)^5*c
sgn(I*c)-1/8*Pi^2*x^2*csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c)^2-1/8*Pi^2*x^2*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a
)^p)^4+1/4*Pi^2*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^5-1/2/b*a*p^2*ln(b*x^2+a)^2

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Maxima [A]
time = 0.28, size = 97, normalized size = 1.59 \begin {gather*} -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 ) + \frac {1}{2} \, x^{2} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} + \frac {{\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^{2}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 96, normalized size = 1.57 \begin {gather*} \frac {2 \, b p^{2} x^{2} - 2 \, b p x^{2} \log \left (c\right ) + b x^{2} \log \left (c\right )^{2} + {\left (b p^{2} x^{2} + a p^{2}\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b p^{2} x^{2} + a p^{2} - {\left (b p x^{2} + a p\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.59, size = 90, normalized size = 1.48 \begin {gather*} \begin {cases} - \frac {a p \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b} + \frac {a \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{2 b} + p^{2} x^{2} - p x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )} + \frac {x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{2} & \text {for}\: b \neq 0 \\\frac {x^{2} \log {\left (a^{p} c \right )}^{2}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.77, size = 96, normalized size = 1.57 \begin {gather*} \frac {{\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} - 2 \, {\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} p \log \left (c\right ) + {\left (b x^{2} + a\right )} \log \left (c\right )^{2}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.23, size = 70, normalized size = 1.15 \begin {gather*} p^2\,x^2+{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {a}{2\,b}+\frac {x^2}{2}\right )-p\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-\frac {a\,p^2\,\ln \left (b\,x^2+a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(c*(a + b*x^2)^p)^2,x)

[Out]

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

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