\(\int x (f+g x^2) \log (c (d+e x^2)^p) \, dx\) [312]

Optimal result

Integrand size = 21, antiderivative size = 94 \[ \int x \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {(e f-d g) p x^2}{4 e}-\frac {p \left (f+g x^2\right )^2}{8 g}-\frac {(e f-d g)^2 p \log \left (d+e x^2\right )}{4 e^2 g}+\frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 g} \]

[Out]

-1/4*(-d*g+e*f)*p*x^2/e-1/8*p*(g*x^2+f)^2/g-1/4*(-d*g+e*f)^2*p*ln(e*x^2+d)/e^2/g+1/4*(g*x^2+f)^2*ln(c*(e*x^2+d
)^p)/g

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2525, 2442, 45} \[ \int x \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 g}-\frac {p (e f-d g)^2 \log \left (d+e x^2\right )}{4 e^2 g}-\frac {p x^2 (e f-d g)}{4 e}-\frac {p \left (f+g x^2\right )^2}{8 g} \]

[In]

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

[Out]

-1/4*((e*f - d*g)*p*x^2)/e - (p*(f + g*x^2)^2)/(8*g) - ((e*f - d*g)^2*p*Log[d + e*x^2])/(4*e^2*g) + ((f + g*x^
2)^2*Log[c*(d + e*x^2)^p])/(4*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) \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right ) \\ & = \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 g}-\frac {(e p) \text {Subst}\left (\int \frac {(f+g x)^2}{d+e x} \, dx,x,x^2\right )}{4 g} \\ & = \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 g}-\frac {(e p) \text {Subst}\left (\int \left (\frac {g (e f-d g)}{e^2}+\frac {(e f-d g)^2}{e^2 (d+e x)}+\frac {g (f+g x)}{e}\right ) \, dx,x,x^2\right )}{4 g} \\ & = -\frac {(e f-d g) p x^2}{4 e}-\frac {p \left (f+g x^2\right )^2}{8 g}-\frac {(e f-d g)^2 p \log \left (d+e x^2\right )}{4 e^2 g}+\frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04 \[ \int x \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {d g p x^2}{4 e}-\frac {1}{8} g p x^4-\frac {d^2 g p \log \left (d+e x^2\right )}{4 e^2}+\frac {1}{4} g x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f \left (-p x^2+\frac {\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.35

method	result	size
parts	\(\frac {g \,x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{4}+\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) f \,x^{2}}{2}+\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) f^{2}}{4 g}-\frac {p e \left (-\frac {g \left (-\frac {1}{2} e g \,x^{4}+d g \,x^{2}-2 f e \,x^{2}\right )}{2 e^{2}}+\frac {\left (g^{2} d^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}\right )}{2 g}\)	\(127\)
parallelrisch	\(-\frac {-2 x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{2} g +g p \,x^{4} e^{2}-4 x^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{2} f -2 d g p \,x^{2} e +4 x^{2} e^{2} f p +2 \ln \left (e \,x^{2}+d \right ) d^{2} g p -8 \ln \left (e \,x^{2}+d \right ) d e f p +4 \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d e f +2 d^{2} g p -4 d e f p}{8 e^{2}}\)	\(136\)
risch	\(\text {Expression too large to display}\)	\(3275\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int x \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {e^{2} g p x^{4} + 2 \, {\left (2 \, e^{2} f - d e g\right )} p x^{2} - 2 \, {\left (e^{2} g p x^{4} + 2 \, e^{2} f p x^{2} + {\left (2 \, d e f - d^{2} g\right )} p\right )} \log \left (e x^{2} + d\right ) - 2 \, {\left (e^{2} g x^{4} + 2 \, e^{2} f x^{2}\right )} \log \left (c\right )}{8 \, e^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 15.77 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.34 \[ \int x \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} - \frac {d^{2} g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{4 e^{2}} + \frac {d f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{2 e} + \frac {d g p x^{2}}{4 e} - \frac {f p x^{2}}{2} + \frac {f x^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{2} - \frac {g p x^{4}}{8} + \frac {g x^{4} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{4} & \text {for}\: e \neq 0 \\\left (\frac {f x^{2}}{2} + \frac {g x^{4}}{4}\right ) \log {\left (c d^{p} \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int x \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {e p {\left (\frac {e g^{2} x^{4} + 2 \, {\left (2 \, e f g - d g^{2}\right )} x^{2}}{e^{2}} + \frac {2 \, {\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x^{2} + d\right )}{e^{3}}\right )}}{8 \, g} + \frac {{\left (g x^{2} + f\right )}^{2} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{4 \, g} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.52 \[ \int x \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {2 \, {\left (e x^{2} + d\right )}^{2} g p \log \left (e x^{2} + d\right ) - {\left (e x^{2} + d\right )}^{2} g p + 2 \, {\left (e x^{2} + d\right )}^{2} g \log \left (c\right )}{8 \, e^{2}} - \frac {{\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} e f p - {\left (e x^{2} - {\left (e x^{2} + d\right )} \log \left (e x^{2} + d\right ) + d\right )} d g p - {\left (e x^{2} + d\right )} e f \log \left (c\right ) + {\left (e x^{2} + d\right )} d g \log \left (c\right )}{2 \, e^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.49 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int x \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^4}{4}+\frac {f\,x^2}{2}\right )-x^2\,\left (\frac {f\,p}{2}-\frac {d\,g\,p}{4\,e}\right )-\frac {g\,p\,x^4}{8}-\frac {\ln \left (e\,x^2+d\right )\,\left (d^2\,g\,p-2\,d\,e\,f\,p\right )}{4\,e^2} \]

[In]

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

[Out]

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