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

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, 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 = {2504, 2442, 45} \begin {gather*} -\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} \end {gather*}

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 45

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])

Rule 2442

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 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^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac {1}{2} \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.01, size = 59, normalized size = 1.00 \begin {gather*} \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 {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.33, size = 1190, normalized size = 20.17

method	result	size
risch	\(\text {Expression too large to display}\)	\(1190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 55, normalized size = 0.93 \begin {gather*} \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 {gather*}

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 = 0.37, size = 57, normalized size = 0.97 \begin {gather*} -\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 {gather*}

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 = 0.89, size = 65, normalized size = 1.10 \begin {gather*} \begin {cases} - \frac {a^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4 b^{2}} + \frac {a p x^{2}}{4 b} - \frac {p x^{4}}{8} + \frac {x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4} & \text {for}\: b \neq 0 \\\frac {x^{4} \log {\left (a^{p} c \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

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

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Giac [A]
time = 4.79, size = 97, normalized size = 1.64 \begin {gather*} \frac {2 \, {\left (b x^{2} + a\right )}^{2} p \log \left (b x^{2} + a\right ) - {\left (b x^{2} + a\right )}^{2} p + 2 \, {\left (b x^{2} + a\right )}^{2} \log \left (c\right )}{8 \, b^{2}} + \frac {{\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} a p - {\left (b x^{2} + a\right )} a \log \left (c\right )}{2 \, b^{2}} \end {gather*}

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

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Mupad [B]
time = 0.22, size = 51, normalized size = 0.86 \begin {gather*} \frac {x^4\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{4}-\frac {p\,x^4}{8}-\frac {a^2\,p\,\ln \left (b\,x^2+a\right )}{4\,b^2}+\frac {a\,p\,x^2}{4\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*log(c*(a + b*x^2)^p),x)

[Out]

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

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