\(\int x (f+g x^2)^2 \log (c (d+e x^2)^p) \, dx\) [325]

Optimal result

Integrand size = 23, antiderivative size = 124 \[ \int x \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {(e f-d g)^2 p x^2}{6 e^2}-\frac {(e f-d g) p \left (f+g x^2\right )^2}{12 e g}-\frac {p \left (f+g x^2\right )^3}{18 g}-\frac {(e f-d g)^3 p \log \left (d+e x^2\right )}{6 e^3 g}+\frac {\left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right )}{6 g} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2525, 2442, 45} \[ \int x \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {\left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right )}{6 g}-\frac {p (e f-d g)^3 \log \left (d+e x^2\right )}{6 e^3 g}-\frac {p x^2 (e f-d g)^2}{6 e^2}-\frac {p \left (f+g x^2\right )^2 (e f-d g)}{12 e g}-\frac {p \left (f+g x^2\right )^3}{18 g} \]

[In]

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

[Out]

-1/6*((e*f - d*g)^2*p*x^2)/e^2 - ((e*f - d*g)*p*(f + g*x^2)^2)/(12*e*g) - (p*(f + g*x^2)^3)/(18*g) - ((e*f - d
*g)^3*p*Log[d + e*x^2])/(6*e^3*g) + ((f + g*x^2)^3*Log[c*(d + e*x^2)^p])/(6*g)

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09 \[ \int x \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {6 d^2 g (-3 e f+d g) p \log \left (d+e x^2\right )+e \left (-p x^2 \left (6 d^2 g^2-3 d e g \left (6 f+g x^2\right )+e^2 \left (18 f^2+9 f g x^2+2 g^2 x^4\right )\right )+6 e \left (3 d f^2+e x^2 \left (3 f^2+3 f g x^2+g^2 x^4\right )\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )}{36 e^3} \]

[In]

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

[Out]

(6*d^2*g*(-3*e*f + d*g)*p*Log[d + e*x^2] + e*(-(p*x^2*(6*d^2*g^2 - 3*d*e*g*(6*f + g*x^2) + e^2*(18*f^2 + 9*f*g
*x^2 + 2*g^2*x^4))) + 6*e*(3*d*f^2 + e*x^2*(3*f^2 + 3*f*g*x^2 + g^2*x^4))*Log[c*(d + e*x^2)^p]))/(36*e^3)

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.66

method	result	size
parts	\(\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) g^{2} x^{6}}{6}+\frac {f g \,x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{2}+\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) f^{2} x^{2}}{2}+\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) f^{3}}{6 g}-\frac {p e \left (\frac {g \left (\frac {1}{3} e^{2} g^{2} x^{6}-\frac {1}{2} d e \,g^{2} x^{4}+\frac {3}{2} e^{2} f g \,x^{4}+d^{2} g^{2} x^{2}-3 d e f g \,x^{2}+3 e^{2} f^{2} x^{2}\right )}{2 e^{3}}+\frac {\left (-d^{3} g^{3}+3 d^{2} e f \,g^{2}-3 d \,e^{2} f^{2} g +e^{3} f^{3}\right ) \ln \left (e \,x^{2}+d \right )}{2 e^{4}}\right )}{3 g}\)	\(206\)
parallelrisch	\(\frac {6 x^{6} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{3} g^{2}-2 x^{6} e^{3} g^{2} p +18 x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{3} f g +3 x^{4} d \,e^{2} g^{2} p -9 x^{4} e^{3} f g p +18 x^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{3} f^{2}-6 x^{2} d^{2} e \,g^{2} p +18 x^{2} d \,e^{2} f g p -18 x^{2} e^{3} f^{2} p +6 \ln \left (e \,x^{2}+d \right ) d^{3} g^{2} p -18 \ln \left (e \,x^{2}+d \right ) d^{2} e f g p +36 \ln \left (e \,x^{2}+d \right ) d \,e^{2} f^{2} p -18 \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d \,e^{2} f^{2}+6 d^{3} g^{2} p -18 d^{2} e f g p +18 d \,e^{2} f^{2} p}{36 e^{3}}\)	\(249\)
risch	\(\frac {\ln \left (c \right ) f^{2} x^{2}}{2}+\frac {g^{2} \ln \left (c \right ) x^{6}}{6}+\frac {g^{2} d p \,x^{4}}{12 e}+\frac {i g^{2} \pi \,x^{6} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{12}+\frac {i g^{2} \pi \,x^{6} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{12}-\frac {i g \pi f \,x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{4}+\frac {i \pi \,f^{2} x^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{4}+\frac {i \pi \,f^{2} x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{4}-\frac {i \pi \,f^{2} x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{4}-\frac {i g^{2} \pi \,x^{6} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{12}+\frac {\left (g \,x^{2}+f \right )^{3} \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{6 g}-\frac {g^{2} d^{2} p \,x^{2}}{6 e^{2}}-\frac {f^{2} p \,x^{2}}{2}+\frac {\ln \left (e \,x^{2}+d \right ) d \,f^{2} p}{2 e}+\frac {g^{2} \ln \left (e \,x^{2}+d \right ) d^{3} p}{6 e^{3}}-\frac {g^{2} p \,x^{6}}{18}+\frac {g \ln \left (c \right ) f \,x^{4}}{2}-\frac {\ln \left (e \,x^{2}+d \right ) f^{3} p}{6 g}-\frac {i g \pi f \,x^{4} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{4}-\frac {i g^{2} \pi \,x^{6} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{12}+\frac {i g \pi f \,x^{4} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{4}+\frac {i g \pi f \,x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{4}-\frac {i \pi \,f^{2} x^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{4}+\frac {d f g p \,x^{2}}{2 e}-\frac {d^{2} f g p \ln \left (e \,x^{2}+d \right )}{2 e^{2}}-\frac {f g p \,x^{4}}{4}\)	\(605\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.45 \[ \int x \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {2 \, e^{3} g^{2} p x^{6} + 3 \, {\left (3 \, e^{3} f g - d e^{2} g^{2}\right )} p x^{4} + 6 \, {\left (3 \, e^{3} f^{2} - 3 \, d e^{2} f g + d^{2} e g^{2}\right )} p x^{2} - 6 \, {\left (e^{3} g^{2} p x^{6} + 3 \, e^{3} f g p x^{4} + 3 \, e^{3} f^{2} p x^{2} + {\left (3 \, d e^{2} f^{2} - 3 \, d^{2} e f g + d^{3} g^{2}\right )} p\right )} \log \left (e x^{2} + d\right ) - 6 \, {\left (e^{3} g^{2} x^{6} + 3 \, e^{3} f g x^{4} + 3 \, e^{3} f^{2} x^{2}\right )} \log \left (c\right )}{36 \, e^{3}} \]

[In]

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

[Out]

-1/36*(2*e^3*g^2*p*x^6 + 3*(3*e^3*f*g - d*e^2*g^2)*p*x^4 + 6*(3*e^3*f^2 - 3*d*e^2*f*g + d^2*e*g^2)*p*x^2 - 6*(
e^3*g^2*p*x^6 + 3*e^3*f*g*p*x^4 + 3*e^3*f^2*p*x^2 + (3*d*e^2*f^2 - 3*d^2*e*f*g + d^3*g^2)*p)*log(e*x^2 + d) -
6*(e^3*g^2*x^6 + 3*e^3*f*g*x^4 + 3*e^3*f^2*x^2)*log(c))/e^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (104) = 208\).

Time = 60.08 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.90 \[ \int x \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} \frac {d^{3} g^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{6 e^{3}} - \frac {d^{2} f g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{2 e^{2}} - \frac {d^{2} g^{2} p x^{2}}{6 e^{2}} + \frac {d f^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{2 e} + \frac {d f g p x^{2}}{2 e} + \frac {d g^{2} p x^{4}}{12 e} - \frac {f^{2} p x^{2}}{2} + \frac {f^{2} x^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{2} - \frac {f g p x^{4}}{4} + \frac {f g x^{4} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{2} - \frac {g^{2} p x^{6}}{18} + \frac {g^{2} x^{6} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{6} & \text {for}\: e \neq 0 \\\left (\frac {f^{2} x^{2}}{2} + \frac {f g x^{4}}{2} + \frac {g^{2} x^{6}}{6}\right ) \log {\left (c d^{p} \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.23 \[ \int x \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {{\left (g x^{2} + f\right )}^{3} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{6 \, g} - \frac {e p {\left (\frac {2 \, e^{2} g^{3} x^{6} + 3 \, {\left (3 \, e^{2} f g^{2} - d e g^{3}\right )} x^{4} + 6 \, {\left (3 \, e^{2} f^{2} g - 3 \, d e f g^{2} + d^{2} g^{3}\right )} x^{2}}{e^{3}} + \frac {6 \, {\left (e^{3} f^{3} - 3 \, d e^{2} f^{2} g + 3 \, d^{2} e f g^{2} - d^{3} g^{3}\right )} \log \left (e x^{2} + d\right )}{e^{4}}\right )}}{36 \, g} \]

[In]

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

[Out]

1/6*(g*x^2 + f)^3*log((e*x^2 + d)^p*c)/g - 1/36*e*p*((2*e^2*g^3*x^6 + 3*(3*e^2*f*g^2 - d*e*g^3)*x^4 + 6*(3*e^2
*f^2*g - 3*d*e*f*g^2 + d^2*g^3)*x^2)/e^3 + 6*(e^3*f^3 - 3*d*e^2*f^2*g + 3*d^2*e*f*g^2 - d^3*g^3)*log(e*x^2 + d
)/e^4)/g

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (114) = 228\).

Time = 0.30 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.74 \[ \int x \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {{\left (e x^{2} + d\right )}^{2} f g p \log \left (e x^{2} + d\right )}{2 \, e^{2}} + \frac {{\left (e x^{2} + d\right )}^{3} g^{2} p \log \left (e x^{2} + d\right )}{6 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{2} d g^{2} p \log \left (e x^{2} + d\right )}{2 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{2} f g p}{4 \, e^{2}} - \frac {{\left (e x^{2} + d\right )}^{3} g^{2} p}{18 \, e^{3}} + \frac {{\left (e x^{2} + d\right )}^{2} d g^{2} p}{4 \, e^{3}} + \frac {{\left (e x^{2} + d\right )}^{2} f g \log \left (c\right )}{2 \, e^{2}} + \frac {{\left (e x^{2} + d\right )}^{3} g^{2} \log \left (c\right )}{6 \, e^{3}} - \frac {{\left (e x^{2} + d\right )}^{2} d g^{2} \log \left (c\right )}{2 \, e^{3}} - \frac {{\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} e^{2} f^{2} p - 2 \, {\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} d e f g p + {\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} d^{2} g^{2} p - {\left (e x^{2} + d\right )} e^{2} f^{2} \log \left (c\right ) + 2 \, {\left (e x^{2} + d\right )} d e f g \log \left (c\right ) - {\left (e x^{2} + d\right )} d^{2} g^{2} \log \left (c\right )}{2 \, e^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.64 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15 \[ \int x \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2\,x^2}{2}+\frac {f\,g\,x^4}{2}+\frac {g^2\,x^6}{6}\right )-x^2\,\left (\frac {f^2\,p}{2}-\frac {d\,\left (f\,g\,p-\frac {d\,g^2\,p}{3\,e}\right )}{2\,e}\right )-x^4\,\left (\frac {f\,g\,p}{4}-\frac {d\,g^2\,p}{12\,e}\right )+\frac {\ln \left (e\,x^2+d\right )\,\left (p\,d^3\,g^2-3\,p\,d^2\,e\,f\,g+3\,p\,d\,e^2\,f^2\right )}{6\,e^3}-\frac {g^2\,p\,x^6}{18} \]

[In]

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

[Out]

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