3.4.11 \(\int x^3 (f+g x^2) \log (c (d+e x^2)^p) \, dx\) [311]

Optimal. Leaf size=119 \[ \frac {d (3 e f-2 d g) p x^2}{12 e^2}-\frac {(3 e f-2 d g) p x^4}{24 e}-\frac {1}{18} g p x^6-\frac {d^2 (3 e f-2 d g) p \log \left (d+e x^2\right )}{12 e^3}+\frac {1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right ) \]

[Out]

1/12*d*(-2*d*g+3*e*f)*p*x^2/e^2-1/24*(-2*d*g+3*e*f)*p*x^4/e-1/18*g*p*x^6-1/12*d^2*(-2*d*g+3*e*f)*p*ln(e*x^2+d)
/e^3+1/4*f*x^4*ln(c*(e*x^2+d)^p)+1/6*g*x^6*ln(c*(e*x^2+d)^p)

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Rubi [A]
time = 0.12, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2525, 45, 2461, 12, 78} \begin {gather*} \frac {1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right )-\frac {d^2 p (3 e f-2 d g) \log \left (d+e x^2\right )}{12 e^3}+\frac {d p x^2 (3 e f-2 d g)}{12 e^2}-\frac {p x^4 (3 e f-2 d g)}{24 e}-\frac {1}{18} g p x^6 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(f + g*x^2)*Log[c*(d + e*x^2)^p],x]

[Out]

(d*(3*e*f - 2*d*g)*p*x^2)/(12*e^2) - ((3*e*f - 2*d*g)*p*x^4)/(24*e) - (g*p*x^6)/18 - (d^2*(3*e*f - 2*d*g)*p*Lo
g[d + e*x^2])/(12*e^3) + (f*x^4*Log[c*(d + e*x^2)^p])/4 + (g*x^6*Log[c*(d + e*x^2)^p])/6

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2461

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*(x_)^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = IntHide[x^m*(f + g*x^r)^q, x]}, Dist[a + b*Log[c*(d + e*x)^n], u, x] - Dist[b*e*n, Int[SimplifyI
ntegrand[u/(d + e*x), x], x], x] /; InverseFunctionFreeQ[u, x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, q, r}, x]
&& IntegerQ[m] && IntegerQ[q] && IntegerQ[r]

Rule 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rubi steps

\begin {align*} \int x^3 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx &=\frac {1}{2} \text {Subst}\left (\int x (f+g x) \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right )\\ &=\frac {1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right )-\frac {1}{2} (e p) \text {Subst}\left (\int \frac {x^2 (3 f+2 g x)}{6 (d+e x)} \, dx,x,x^2\right )\\ &=\frac {1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right )-\frac {1}{12} (e p) \text {Subst}\left (\int \frac {x^2 (3 f+2 g x)}{d+e x} \, dx,x,x^2\right )\\ &=\frac {1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right )-\frac {1}{12} (e p) \text {Subst}\left (\int \left (\frac {d (-3 e f+2 d g)}{e^3}+\frac {(3 e f-2 d g) x}{e^2}+\frac {2 g x^2}{e}-\frac {d^2 (-3 e f+2 d g)}{e^3 (d+e x)}\right ) \, dx,x,x^2\right )\\ &=\frac {d (3 e f-2 d g) p x^2}{12 e^2}-\frac {(3 e f-2 d g) p x^4}{24 e}-\frac {1}{18} g p x^6-\frac {d^2 (3 e f-2 d g) p \log \left (d+e x^2\right )}{12 e^3}+\frac {1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 140, normalized size = 1.18 \begin {gather*} \frac {d f p x^2}{4 e}-\frac {d^2 g p x^2}{6 e^2}-\frac {1}{8} f p x^4+\frac {d g p x^4}{12 e}-\frac {1}{18} g p x^6-\frac {d^2 f p \log \left (d+e x^2\right )}{4 e^2}+\frac {d^3 g p \log \left (d+e x^2\right )}{6 e^3}+\frac {1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(f + g*x^2)*Log[c*(d + e*x^2)^p],x]

[Out]

(d*f*p*x^2)/(4*e) - (d^2*g*p*x^2)/(6*e^2) - (f*p*x^4)/8 + (d*g*p*x^4)/(12*e) - (g*p*x^6)/18 - (d^2*f*p*Log[d +
e*x^2])/(4*e^2) + (d^3*g*p*Log[d + e*x^2])/(6*e^3) + (f*x^4*Log[c*(d + e*x^2)^p])/4 + (g*x^6*Log[c*(d + e*x^2
)^p])/6

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.25, size = 387, normalized size = 3.25

method	result	size
risch	\(\left (\frac {1}{6} x^{6} g +\frac {1}{4} x^{4} f \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )+\frac {i \pi g \,x^{6} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{12}-\frac {i \pi g \,x^{6} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{12}-\frac {i \pi f \,x^{4} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{8}-\frac {i \pi f \,x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{8}+\frac {i \pi f \,x^{4} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{8}+\frac {i \pi f \,x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{8}+\frac {i \pi g \,x^{6} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{12}-\frac {i \pi g \,x^{6} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{12}+\frac {x^{6} \ln \left (c \right ) g}{6}-\frac {g p \,x^{6}}{18}+\frac {x^{4} \ln \left (c \right ) f}{4}+\frac {x^{4} d g p}{12 e}-\frac {x^{4} f p}{8}-\frac {x^{2} d^{2} g p}{6 e^{2}}+\frac {x^{2} d f p}{4 e}+\frac {\ln \left (e \,x^{2}+d \right ) d^{3} g p}{6 e^{3}}-\frac {\ln \left (e \,x^{2}+d \right ) d^{2} f p}{4 e^{2}}\)	\(387\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(g*x^2+f)*ln(c*(e*x^2+d)^p),x,method=_RETURNVERBOSE)

[Out]

(1/6*x^6*g+1/4*x^4*f)*ln((e*x^2+d)^p)+1/12*I*Pi*g*x^6*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-1/12*I*Pi*g*
x^6*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-1/8*I*Pi*f*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^
p)*csgn(I*c)-1/8*I*Pi*f*x^4*csgn(I*c*(e*x^2+d)^p)^3+1/8*I*Pi*f*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2
+1/8*I*Pi*f*x^4*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+1/12*I*Pi*g*x^6*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-1/12*I*Pi*
g*x^6*csgn(I*c*(e*x^2+d)^p)^3+1/6*x^6*ln(c)*g-1/18*g*p*x^6+1/4*x^4*ln(c)*f+1/12/e*x^4*d*g*p-1/8*x^4*f*p-1/6/e^
2*x^2*d^2*g*p+1/4/e*x^2*d*f*p+1/6/e^3*ln(e*x^2+d)*d^3*g*p-1/4/e^2*ln(e*x^2+d)*d^2*f*p

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Maxima [A]
time = 0.30, size = 111, normalized size = 0.93 \begin {gather*} \frac {1}{72} \, {\left (6 \, {\left (2 \, d^{3} g - 3 \, d^{2} f e\right )} e^{\left (-4\right )} \log \left (x^{2} e + d\right ) - {\left (4 \, g x^{6} e^{2} - 3 \, {\left (2 \, d g e - 3 \, f e^{2}\right )} x^{4} + 6 \, {\left (2 \, d^{2} g - 3 \, d f e\right )} x^{2}\right )} e^{\left (-3\right )}\right )} p e + \frac {1}{12} \, {\left (2 \, g x^{6} + 3 \, f x^{4}\right )} \log \left ({\left (x^{2} e + d\right )}^{p} c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(g*x^2+f)*log(c*(e*x^2+d)^p),x, algorithm="maxima")

[Out]

1/72*(6*(2*d^3*g - 3*d^2*f*e)*e^(-4)*log(x^2*e + d) - (4*g*x^6*e^2 - 3*(2*d*g*e - 3*f*e^2)*x^4 + 6*(2*d^2*g -
3*d*f*e)*x^2)*e^(-3))*p*e + 1/12*(2*g*x^6 + 3*f*x^4)*log((x^2*e + d)^p*c)

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Fricas [A]
time = 0.34, size = 120, normalized size = 1.01 \begin {gather*} -\frac {1}{72} \, {\left (12 \, d^{2} g p x^{2} e - 6 \, {\left (2 \, g x^{6} + 3 \, f x^{4}\right )} e^{3} \log \left (c\right ) + {\left (4 \, g p x^{6} + 9 \, f p x^{4}\right )} e^{3} - 6 \, {\left (d g p x^{4} + 3 \, d f p x^{2}\right )} e^{2} - 6 \, {\left (2 \, d^{3} g p - 3 \, d^{2} f p e + {\left (2 \, g p x^{6} + 3 \, f p x^{4}\right )} e^{3}\right )} \log \left (x^{2} e + d\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(g*x^2+f)*log(c*(e*x^2+d)^p),x, algorithm="fricas")

[Out]

-1/72*(12*d^2*g*p*x^2*e - 6*(2*g*x^6 + 3*f*x^4)*e^3*log(c) + (4*g*p*x^6 + 9*f*p*x^4)*e^3 - 6*(d*g*p*x^4 + 3*d*
f*p*x^2)*e^2 - 6*(2*d^3*g*p - 3*d^2*f*p*e + (2*g*p*x^6 + 3*f*p*x^4)*e^3)*log(x^2*e + d))*e^(-3)

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Sympy [A]
time = 66.92, size = 156, normalized size = 1.31 \begin {gather*} \begin {cases} \frac {d^{3} g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{6 e^{3}} - \frac {d^{2} f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{4 e^{2}} - \frac {d^{2} g p x^{2}}{6 e^{2}} + \frac {d f p x^{2}}{4 e} + \frac {d g p x^{4}}{12 e} - \frac {f p x^{4}}{8} + \frac {f x^{4} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{4} - \frac {g p x^{6}}{18} + \frac {g x^{6} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{6} & \text {for}\: e \neq 0 \\\left (\frac {f x^{4}}{4} + \frac {g x^{6}}{6}\right ) \log {\left (c d^{p} \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(g*x**2+f)*ln(c*(e*x**2+d)**p),x)

[Out]

Piecewise((d**3*g*log(c*(d + e*x**2)**p)/(6*e**3) - d**2*f*log(c*(d + e*x**2)**p)/(4*e**2) - d**2*g*p*x**2/(6*
e**2) + d*f*p*x**2/(4*e) + d*g*p*x**4/(12*e) - f*p*x**4/8 + f*x**4*log(c*(d + e*x**2)**p)/4 - g*p*x**6/18 + g*
x**6*log(c*(d + e*x**2)**p)/6, Ne(e, 0)), ((f*x**4/4 + g*x**6/6)*log(c*d**p), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (110) = 220\).
time = 5.46, size = 281, normalized size = 2.36 \begin {gather*} \frac {1}{6} \, {\left (x^{2} e + d\right )}^{3} g p e^{\left (-3\right )} \log \left (x^{2} e + d\right ) - \frac {1}{2} \, {\left (x^{2} e + d\right )}^{2} d g p e^{\left (-3\right )} \log \left (x^{2} e + d\right ) - \frac {1}{18} \, {\left (x^{2} e + d\right )}^{3} g p e^{\left (-3\right )} + \frac {1}{4} \, {\left (x^{2} e + d\right )}^{2} d g p e^{\left (-3\right )} + \frac {1}{4} \, {\left (x^{2} e + d\right )}^{2} f p e^{\left (-2\right )} \log \left (x^{2} e + d\right ) + \frac {1}{6} \, {\left (x^{2} e + d\right )}^{3} g e^{\left (-3\right )} \log \left (c\right ) - \frac {1}{2} \, {\left (x^{2} e + d\right )}^{2} d g e^{\left (-3\right )} \log \left (c\right ) - \frac {1}{8} \, {\left (x^{2} e + d\right )}^{2} f p e^{\left (-2\right )} + \frac {1}{4} \, {\left (x^{2} e + d\right )}^{2} f e^{\left (-2\right )} \log \left (c\right ) - \frac {1}{2} \, {\left ({\left (x^{2} e - {\left (x^{2} e + d\right )} \log \left (x^{2} e + d\right ) + d\right )} d^{2} g p - {\left (x^{2} e - {\left (x^{2} e + d\right )} \log \left (x^{2} e + d\right ) + d\right )} d f p e - {\left (x^{2} e + d\right )} d^{2} g \log \left (c\right ) + {\left (x^{2} e + d\right )} d f e \log \left (c\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(g*x^2+f)*log(c*(e*x^2+d)^p),x, algorithm="giac")

[Out]

1/6*(x^2*e + d)^3*g*p*e^(-3)*log(x^2*e + d) - 1/2*(x^2*e + d)^2*d*g*p*e^(-3)*log(x^2*e + d) - 1/18*(x^2*e + d)
^3*g*p*e^(-3) + 1/4*(x^2*e + d)^2*d*g*p*e^(-3) + 1/4*(x^2*e + d)^2*f*p*e^(-2)*log(x^2*e + d) + 1/6*(x^2*e + d)
^3*g*e^(-3)*log(c) - 1/2*(x^2*e + d)^2*d*g*e^(-3)*log(c) - 1/8*(x^2*e + d)^2*f*p*e^(-2) + 1/4*(x^2*e + d)^2*f*
e^(-2)*log(c) - 1/2*((x^2*e - (x^2*e + d)*log(x^2*e + d) + d)*d^2*g*p - (x^2*e - (x^2*e + d)*log(x^2*e + d) +
d)*d*f*p*e - (x^2*e + d)*d^2*g*log(c) + (x^2*e + d)*d*f*e*log(c))*e^(-3)

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Mupad [B]
time = 0.31, size = 103, normalized size = 0.87 \begin {gather*} \ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^6}{6}+\frac {f\,x^4}{4}\right )-x^4\,\left (\frac {f\,p}{8}-\frac {d\,g\,p}{12\,e}\right )-\frac {g\,p\,x^6}{18}+\frac {\ln \left (e\,x^2+d\right )\,\left (2\,d^3\,g\,p-3\,d^2\,e\,f\,p\right )}{12\,e^3}+\frac {d\,x^2\,\left (\frac {f\,p}{2}-\frac {d\,g\,p}{3\,e}\right )}{2\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*log(c*(d + e*x^2)^p)*(f + g*x^2),x)

[Out]

log(c*(d + e*x^2)^p)*((f*x^4)/4 + (g*x^6)/6) - x^4*((f*p)/8 - (d*g*p)/(12*e)) - (g*p*x^6)/18 + (log(d + e*x^2)
*(2*d^3*g*p - 3*d^2*e*f*p))/(12*e^3) + (d*x^2*((f*p)/2 - (d*g*p)/(3*e)))/(2*e)

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