3.4.13 \(\int \frac {(f+g x^2) \log (c (d+e x^2)^p)}{x} \, dx\) [313]

Optimal. Leaf size=82 \[ -\frac {1}{2} g p x^2+\frac {g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e}+\frac {1}{2} f \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f p \text {Li}_2\left (1+\frac {e x^2}{d}\right ) \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2525, 45, 2463, 2436, 2332, 2441, 2352} \begin {gather*} \frac {1}{2} f p \text {PolyLog}\left (2,\frac {e x^2}{d}+1\right )+\frac {1}{2} f \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e}-\frac {1}{2} g p x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(g*p*x^2) + (g*(d + e*x^2)*Log[c*(d + e*x^2)^p])/(2*e) + (f*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p])/2 + (
f*p*PolyLog[2, 1 + (e*x^2)/d])/2

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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

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 \frac {\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(f+g x) \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (g \log \left (c (d+e x)^p\right )+\frac {f \log \left (c (d+e x)^p\right )}{x}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2} f \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^2\right )+\frac {1}{2} g \text {Subst}\left (\int \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2} f \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {g \text {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^2\right )}{2 e}-\frac {1}{2} (e f p) \text {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^2\right )\\ &=-\frac {1}{2} g p x^2+\frac {g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e}+\frac {1}{2} f \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f p \text {Li}_2\left (1+\frac {e x^2}{d}\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 80, normalized size = 0.98 \begin {gather*} \frac {1}{2} g \left (-p x^2+\frac {\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}\right )+\frac {1}{2} f \left (\log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+p \text {Li}_2\left (\frac {d+e x^2}{d}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(g*(-(p*x^2) + ((d + e*x^2)*Log[c*(d + e*x^2)^p])/e))/2 + (f*(Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p] + p*PolyL
og[2, (d + e*x^2)/d]))/2

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.16, size = 419, normalized size = 5.11

method	result	size
risch	\(\frac {\ln \left (\left (e \,x^{2}+d \right )^{p}\right ) g \,x^{2}}{2}+\ln \left (\left (e \,x^{2}+d \right )^{p}\right ) f \ln \left (x \right )-\frac {g p \,x^{2}}{2}+\frac {p g d \ln \left (e \,x^{2}+d \right )}{2 e}-p f \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )-p f \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )-p f \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )-p f \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )+\frac {i \pi \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) f \ln \left (x \right )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} f \ln \left (x \right )}{2}-\frac {i \pi \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3} f \ln \left (x \right )}{2}-\frac {i \pi g \,x^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{4}+\frac {i \pi g \,x^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{4}+\frac {i \pi g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{4}-\frac {i \pi \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) f \ln \left (x \right )}{2}-\frac {i \pi g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{4}+\frac {\ln \left (c \right ) g \,x^{2}}{2}+\ln \left (c \right ) f \ln \left (x \right )\)	\(419\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.59, size = 95, normalized size = 1.16 \begin {gather*} \frac {1}{2} \, {\left (\log \left (x^{2} e + d\right ) \log \left (-\frac {x^{2} e + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {x^{2} e + d}{d}\right )\right )} f p + f \log \left (c\right ) \log \left (x\right ) - \frac {1}{2} \, {\left ({\left (g p - g \log \left (c\right )\right )} x^{2} e - {\left (g p x^{2} e + d g p\right )} \log \left (x^{2} e + d\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (f + g x^{2}\right ) \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((f + g*x**2)*log(c*(d + e*x**2)**p)/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (g\,x^2+f\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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