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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.178547, antiderivative size = 119, normalized size of antiderivative = 1., 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 = {2475, 43, 2414, 12, 77} \[ \frac{1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac{1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right )-\frac{d^2 p (3 e f-2 d g) \log \left (d+e x^2\right )}{12 e^3}+\frac{d p x^2 (3 e f-2 d g)}{12 e^2}-\frac{p x^4 (3 e f-2 d g)}{24 e}-\frac{1}{18} g p x^6 \]

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 2475

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])

Rule 43

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 2414

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 12

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

Rule 77

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]))))

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} \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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*}

Mathematica [A]  time = 0.0255616, size = 140, normalized size = 1.18 \[ \frac{1}{4} f x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac{1}{6} g x^6 \log \left (c \left (d+e x^2\right )^p\right )-\frac{d^2 f p \log \left (d+e x^2\right )}{4 e^2}-\frac{d^2 g p x^2}{6 e^2}+\frac{d^3 g p \log \left (d+e x^2\right )}{6 e^3}+\frac{d f p x^2}{4 e}+\frac{d g p x^4}{12 e}-\frac{1}{8} f p x^4-\frac{1}{18} g p x^6 \]

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

________________________________________________________________________________________

Maple [C]  time = 0.573, size = 387, normalized size = 3.3 \begin{align*} \left ({\frac{g{x}^{6}}{6}}+{\frac{f{x}^{4}}{4}} \right ) \ln \left ( \left ( e{x}^{2}+d \right ) ^{p} \right ) +{\frac{i}{12}}\pi \,g{x}^{6} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -{\frac{i}{12}}\pi \,g{x}^{6} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}-{\frac{i}{8}}\pi \,f{x}^{4} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}-{\frac{i}{12}}\pi \,g{x}^{6}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{8}}\pi \,f{x}^{4} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -{\frac{i}{8}}\pi \,f{x}^{4}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{12}}\pi \,g{x}^{6}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}+{\frac{i}{8}}\pi \,f{x}^{4}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}+{\frac{\ln \left ( c \right ) g{x}^{6}}{6}}-{\frac{gp{x}^{6}}{18}}+{\frac{\ln \left ( c \right ) f{x}^{4}}{4}}+{\frac{dgp{x}^{4}}{12\,e}}-{\frac{fp{x}^{4}}{8}}-{\frac{{d}^{2}gp{x}^{2}}{6\,{e}^{2}}}+{\frac{dfp{x}^{2}}{4\,e}}+{\frac{\ln \left ( e{x}^{2}+d \right ){d}^{3}gp}{6\,{e}^{3}}}-{\frac{\ln \left ( e{x}^{2}+d \right ){d}^{2}fp}{4\,{e}^{2}}} \end{align*}

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)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.09469, size = 146, normalized size = 1.23 \begin{align*} -\frac{1}{72} \, e p{\left (\frac{4 \, e^{2} g x^{6} + 3 \,{\left (3 \, e^{2} f - 2 \, d e g\right )} x^{4} - 6 \,{\left (3 \, d e f - 2 \, d^{2} g\right )} x^{2}}{e^{3}} + \frac{6 \,{\left (3 \, d^{2} e f - 2 \, d^{3} g\right )} \log \left (e x^{2} + d\right )}{e^{4}}\right )} + \frac{1}{12} \,{\left (2 \, g x^{6} + 3 \, f x^{4}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \end{align*}

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

________________________________________________________________________________________

Fricas [A]  time = 2.03243, size = 282, normalized size = 2.37 \begin{align*} -\frac{4 \, e^{3} g p x^{6} + 3 \,{\left (3 \, e^{3} f - 2 \, d e^{2} g\right )} p x^{4} - 6 \,{\left (3 \, d e^{2} f - 2 \, d^{2} e g\right )} p x^{2} - 6 \,{\left (2 \, e^{3} g p x^{6} + 3 \, e^{3} f p x^{4} -{\left (3 \, d^{2} e f - 2 \, d^{3} g\right )} p\right )} \log \left (e x^{2} + d\right ) - 6 \,{\left (2 \, e^{3} g x^{6} + 3 \, e^{3} f x^{4}\right )} \log \left (c\right )}{72 \, e^{3}} \end{align*}

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.14927, size = 317, normalized size = 2.66 \begin{align*} \frac{1}{72} \,{\left (12 \, g x^{6} e \log \left (c\right ) + 9 \,{\left (2 \,{\left (x^{2} e + d\right )}^{2} \log \left (x^{2} e + d\right ) - 4 \,{\left (x^{2} e + d\right )} d \log \left (x^{2} e + d\right ) -{\left (x^{2} e + d\right )}^{2} + 4 \,{\left (x^{2} e + d\right )} d\right )} f p e^{\left (-1\right )} + 18 \,{\left ({\left (x^{2} e + d\right )}^{2} - 2 \,{\left (x^{2} e + d\right )} d\right )} f e^{\left (-1\right )} \log \left (c\right ) + 2 \,{\left (6 \,{\left (x^{2} e + d\right )}^{3} e^{\left (-2\right )} \log \left (x^{2} e + d\right ) - 18 \,{\left (x^{2} e + d\right )}^{2} d e^{\left (-2\right )} \log \left (x^{2} e + d\right ) + 18 \,{\left (x^{2} e + d\right )} d^{2} e^{\left (-2\right )} \log \left (x^{2} e + d\right ) - 2 \,{\left (x^{2} e + d\right )}^{3} e^{\left (-2\right )} + 9 \,{\left (x^{2} e + d\right )}^{2} d e^{\left (-2\right )} - 18 \,{\left (x^{2} e + d\right )} d^{2} e^{\left (-2\right )}\right )} g p\right )} e^{\left (-1\right )} \end{align*}

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]

1/72*(12*g*x^6*e*log(c) + 9*(2*(x^2*e + d)^2*log(x^2*e + d) - 4*(x^2*e + d)*d*log(x^2*e + d) - (x^2*e + d)^2 +
4*(x^2*e + d)*d)*f*p*e^(-1) + 18*((x^2*e + d)^2 - 2*(x^2*e + d)*d)*f*e^(-1)*log(c) + 2*(6*(x^2*e + d)^3*e^(-2
)*log(x^2*e + d) - 18*(x^2*e + d)^2*d*e^(-2)*log(x^2*e + d) + 18*(x^2*e + d)*d^2*e^(-2)*log(x^2*e + d) - 2*(x^
2*e + d)^3*e^(-2) + 9*(x^2*e + d)^2*d*e^(-2) - 18*(x^2*e + d)*d^2*e^(-2))*g*p)*e^(-1)