3.4.26 \(\int \frac {(f+g x^2)^2 \log (c (d+e x^2)^p)}{x} \, dx\) [326]

Optimal. Leaf size=153 \[ -f g p x^2+\frac {d g^2 p x^2}{4 e}-\frac {1}{8} g^2 p x^4-\frac {d^2 g^2 p \log \left (d+e x^2\right )}{4 e^2}+\frac {1}{4} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {f g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac {1}{2} f^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f^2 p \text {Li}_2\left (1+\frac {e x^2}{d}\right ) \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2525, 45, 2463, 2436, 2332, 2441, 2352, 2442} \begin {gather*} \frac {1}{2} f^2 p \text {PolyLog}\left (2,\frac {e x^2}{d}+1\right )+\frac {1}{2} f^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {f g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac {1}{4} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )-\frac {d^2 g^2 p \log \left (d+e x^2\right )}{4 e^2}+\frac {d g^2 p x^2}{4 e}-f g p x^2-\frac {1}{8} g^2 p x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(f*g*p*x^2) + (d*g^2*p*x^2)/(4*e) - (g^2*p*x^4)/8 - (d^2*g^2*p*Log[d + e*x^2])/(4*e^2) + (g^2*x^4*Log[c*(d +
e*x^2)^p])/4 + (f*g*(d + e*x^2)*Log[c*(d + e*x^2)^p])/e + (f^2*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p])/2 + (f^
2*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 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 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 )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(f+g x)^2 \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (2 f g \log \left (c (d+e x)^p\right )+\frac {f^2 \log \left (c (d+e x)^p\right )}{x}+g^2 x \log \left (c (d+e x)^p\right )\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2} f^2 \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^2\right )+(f g) \text {Subst}\left (\int \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right )+\frac {1}{2} g^2 \text {Subst}\left (\int x \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right )\\ &=\frac {1}{4} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {(f g) \text {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^2\right )}{e}-\frac {1}{2} \left (e f^2 p\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^2\right )-\frac {1}{4} \left (e g^2 p\right ) \text {Subst}\left (\int \frac {x^2}{d+e x} \, dx,x,x^2\right )\\ &=-f g p x^2+\frac {1}{4} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {f g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac {1}{2} f^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f^2 p \text {Li}_2\left (1+\frac {e x^2}{d}\right )-\frac {1}{4} \left (e g^2 p\right ) \text {Subst}\left (\int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx,x,x^2\right )\\ &=-f g p x^2+\frac {d g^2 p x^2}{4 e}-\frac {1}{8} g^2 p x^4-\frac {d^2 g^2 p \log \left (d+e x^2\right )}{4 e^2}+\frac {1}{4} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {f g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac {1}{2} f^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f^2 p \text {Li}_2\left (1+\frac {e x^2}{d}\right )\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 121, normalized size = 0.79 \begin {gather*} \frac {-e g p x^2 \left (8 e f-2 d g+e g x^2\right )-2 d^2 g^2 p \log \left (d+e x^2\right )+2 e \left (g \left (4 d f+4 e f x^2+e g x^4\right )+2 e f^2 \log \left (-\frac {e x^2}{d}\right )\right ) \log \left (c \left (d+e x^2\right )^p\right )+4 e^2 f^2 p \text {Li}_2\left (1+\frac {e x^2}{d}\right )}{8 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(e*g*p*x^2*(8*e*f - 2*d*g + e*g*x^2)) - 2*d^2*g^2*p*Log[d + e*x^2] + 2*e*(g*(4*d*f + 4*e*f*x^2 + e*g*x^4) +
2*e*f^2*Log[-((e*x^2)/d)])*Log[c*(d + e*x^2)^p] + 4*e^2*f^2*p*PolyLog[2, 1 + (e*x^2)/d])/(8*e^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.53, size = 652, normalized size = 4.26

method	result	size
risch	\(-\frac {d^{2} g^{2} p \ln \left (e \,x^{2}+d \right )}{4 e^{2}}-\frac {g^{2} p \,x^{4}}{8}-f g p \,x^{2}+\frac {d \,g^{2} p \,x^{2}}{4 e}-p \,f^{2} \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )-p \,f^{2} \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )+\frac {\ln \left (\left (e \,x^{2}+d \right )^{p}\right ) g^{2} x^{4}}{4}+\ln \left (\left (e \,x^{2}+d \right )^{p}\right ) f^{2} \ln \left (x \right )+\ln \left (c \right ) f^{2} \ln \left (x \right )+\ln \left (\left (e \,x^{2}+d \right )^{p}\right ) f g \,x^{2}-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )+\ln \left (c \right ) f g \,x^{2}-\frac {i \pi \,g^{2} x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{8}-\frac {i \pi \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3} f^{2} \ln \left (x \right )}{2}-\frac {i \pi f g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{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^{2} \ln \left (x \right )}{2}+\frac {i \pi \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) f^{2} \ln \left (x \right )}{2}+\frac {i \pi \,g^{2} 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 \,g^{2} x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{8}+\frac {p g d \ln \left (e \,x^{2}+d \right ) f}{e}-\frac {i \pi f 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 )}{2}+\frac {\ln \left (c \right ) g^{2} x^{4}}{4}+\frac {i \pi f 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}}{2}+\frac {i \pi f g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \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 ) \mathrm {csgn}\left (i c \right ) f^{2} \ln \left (x \right )}{2}-\frac {i \pi \,g^{2} 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}\)	\(652\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 157, normalized size = 1.03 \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^{2} p + f^{2} \log \left (c\right ) \log \left (x\right ) - \frac {1}{8} \, {\left ({\left (g^{2} p - 2 \, g^{2} \log \left (c\right )\right )} x^{4} e^{2} - 2 \, {\left (d g^{2} p e - 4 \, {\left (f g p - f g \log \left (c\right )\right )} e^{2}\right )} x^{2} - 2 \, {\left (g^{2} p x^{4} e^{2} + 4 \, f g p x^{2} e^{2} - d^{2} g^{2} p + 4 \, d f g p e\right )} \log \left (x^{2} e + d\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

integral((g^2*x^4 + 2*f*g*x^2 + f^2)*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 )^{2} \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)**2*ln(c*(e*x**2+d)**p)/x,x)

[Out]

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

[Out]

integrate((g*x^2 + f)^2*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 )}^2}{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)^2)/x,x)

[Out]

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

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