3.3.32 \(\int \frac {\log (c (a+b x^2)^p)}{x^3 (d+e x)} \, dx\) [232]

Optimal. Leaf size=371 \[ -\frac {2 \sqrt {b} e p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d^2}+\frac {b p \log (x)}{a d}+\frac {e^2 p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{d^3}-\frac {b p \log \left (a+b x^2\right )}{2 a d}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 d x^2}+\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}+\frac {e^2 \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^3} \]

[Out]

b*p*ln(x)/a/d-1/2*b*p*ln(b*x^2+a)/a/d-1/2*ln(c*(b*x^2+a)^p)/d/x^2+e*ln(c*(b*x^2+a)^p)/d^2/x+1/2*e^2*ln(-b*x^2/
a)*ln(c*(b*x^2+a)^p)/d^3-e^2*ln(e*x+d)*ln(c*(b*x^2+a)^p)/d^3+e^2*p*ln(e*x+d)*ln(e*((-a)^(1/2)-x*b^(1/2))/(e*(-
a)^(1/2)+d*b^(1/2)))/d^3+e^2*p*ln(e*x+d)*ln(-e*((-a)^(1/2)+x*b^(1/2))/(-e*(-a)^(1/2)+d*b^(1/2)))/d^3+1/2*e^2*p
*polylog(2,1+b*x^2/a)/d^3+e^2*p*polylog(2,(e*x+d)*b^(1/2)/(-e*(-a)^(1/2)+d*b^(1/2)))/d^3+e^2*p*polylog(2,(e*x+
d)*b^(1/2)/(e*(-a)^(1/2)+d*b^(1/2)))/d^3-2*e*p*arctan(x*b^(1/2)/a^(1/2))*b^(1/2)/d^2/a^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 15, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {2516, 2504, 2442, 36, 29, 31, 2505, 211, 2441, 2352, 2512, 266, 2463, 2440, 2438} \begin {gather*} \frac {e^2 p \text {PolyLog}\left (2,\frac {b x^2}{a}+1\right )}{2 d^3}+\frac {e^2 p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^3}+\frac {e^2 p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {-a} e+\sqrt {b} d}\right )}{d^3}-\frac {2 \sqrt {b} e p \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d^2}+\frac {e^2 \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^3}+\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 d x^2}+\frac {e^2 p \log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )}{d^3}+\frac {e^2 p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^3}-\frac {b p \log \left (a+b x^2\right )}{2 a d}+\frac {b p \log (x)}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/(x^3*(d + e*x)),x]

[Out]

(-2*Sqrt[b]*e*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*d^2) + (b*p*Log[x])/(a*d) + (e^2*p*Log[(e*(Sqrt[-a] - Sq
rt[b]*x))/(Sqrt[b]*d + Sqrt[-a]*e)]*Log[d + e*x])/d^3 + (e^2*p*Log[-((e*(Sqrt[-a] + Sqrt[b]*x))/(Sqrt[b]*d - S
qrt[-a]*e))]*Log[d + e*x])/d^3 - (b*p*Log[a + b*x^2])/(2*a*d) - Log[c*(a + b*x^2)^p]/(2*d*x^2) + (e*Log[c*(a +
b*x^2)^p])/(d^2*x) + (e^2*Log[-((b*x^2)/a)]*Log[c*(a + b*x^2)^p])/(2*d^3) - (e^2*Log[d + e*x]*Log[c*(a + b*x^
2)^p])/d^3 + (e^2*p*PolyLog[2, (Sqrt[b]*(d + e*x))/(Sqrt[b]*d - Sqrt[-a]*e)])/d^3 + (e^2*p*PolyLog[2, (Sqrt[b]
*(d + e*x))/(Sqrt[b]*d + Sqrt[-a]*e)])/d^3 + (e^2*p*PolyLog[2, 1 + (b*x^2)/a])/(2*d^3)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 2438

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

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

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 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2505

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

Rule 2512

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

Rule 2516

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

Rubi steps

\begin {align*} \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3 (d+e x)} \, dx &=\int \left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x^3}-\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x^2}+\frac {e^2 \log \left (c \left (a+b x^2\right )^p\right )}{d^3 x}-\frac {e^3 \log \left (c \left (a+b x^2\right )^p\right )}{d^3 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx}{d}-\frac {e \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx}{d^2}+\frac {e^2 \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx}{d^3}-\frac {e^3 \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{d+e x} \, dx}{d^3}\\ &=\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}-\frac {e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^3}+\frac {\text {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x^2} \, dx,x,x^2\right )}{2 d}+\frac {e^2 \text {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x} \, dx,x,x^2\right )}{2 d^3}-\frac {(2 b e p) \int \frac {1}{a+b x^2} \, dx}{d^2}+\frac {\left (2 b e^2 p\right ) \int \frac {x \log (d+e x)}{a+b x^2} \, dx}{d^3}\\ &=-\frac {2 \sqrt {b} e p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 d x^2}+\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}+\frac {e^2 \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^3}+\frac {(b p) \text {Subst}\left (\int \frac {1}{x (a+b x)} \, dx,x,x^2\right )}{2 d}-\frac {\left (b e^2 p\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {b x}{a}\right )}{a+b x} \, dx,x,x^2\right )}{2 d^3}+\frac {\left (2 b e^2 p\right ) \int \left (-\frac {\log (d+e x)}{2 \sqrt {b} \left (\sqrt {-a}-\sqrt {b} x\right )}+\frac {\log (d+e x)}{2 \sqrt {b} \left (\sqrt {-a}+\sqrt {b} x\right )}\right ) \, dx}{d^3}\\ &=-\frac {2 \sqrt {b} e p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 d x^2}+\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}+\frac {e^2 \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^3}+\frac {(b p) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 a d}-\frac {\left (b^2 p\right ) \text {Subst}\left (\int \frac {1}{a+b x} \, dx,x,x^2\right )}{2 a d}-\frac {\left (\sqrt {b} e^2 p\right ) \int \frac {\log (d+e x)}{\sqrt {-a}-\sqrt {b} x} \, dx}{d^3}+\frac {\left (\sqrt {b} e^2 p\right ) \int \frac {\log (d+e x)}{\sqrt {-a}+\sqrt {b} x} \, dx}{d^3}\\ &=-\frac {2 \sqrt {b} e p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d^2}+\frac {b p \log (x)}{a d}+\frac {e^2 p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{d^3}-\frac {b p \log \left (a+b x^2\right )}{2 a d}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 d x^2}+\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}+\frac {e^2 \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^3}-\frac {\left (e^3 p\right ) \int \frac {\log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right )}{d+e x} \, dx}{d^3}-\frac {\left (e^3 p\right ) \int \frac {\log \left (\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{-\sqrt {b} d+\sqrt {-a} e}\right )}{d+e x} \, dx}{d^3}\\ &=-\frac {2 \sqrt {b} e p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d^2}+\frac {b p \log (x)}{a d}+\frac {e^2 p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{d^3}-\frac {b p \log \left (a+b x^2\right )}{2 a d}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 d x^2}+\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}+\frac {e^2 \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^3}-\frac {\left (e^2 p\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt {b} x}{-\sqrt {b} d+\sqrt {-a} e}\right )}{x} \, dx,x,d+e x\right )}{d^3}-\frac {\left (e^2 p\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {\sqrt {b} x}{\sqrt {b} d+\sqrt {-a} e}\right )}{x} \, dx,x,d+e x\right )}{d^3}\\ &=-\frac {2 \sqrt {b} e p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d^2}+\frac {b p \log (x)}{a d}+\frac {e^2 p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{d^3}-\frac {b p \log \left (a+b x^2\right )}{2 a d}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 d x^2}+\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}+\frac {e^2 \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 320, normalized size = 0.86 \begin {gather*} \frac {-\frac {4 \sqrt {b} d e p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {b d^2 p \left (2 \log (x)-\log \left (a+b x^2\right )\right )}{a}-\frac {d^2 \log \left (c \left (a+b x^2\right )^p\right )}{x^2}+\frac {2 d e \log \left (c \left (a+b x^2\right )^p\right )}{x}-2 e^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )+2 e^2 p \left (\left (\log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right )+\log \left (\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{-\sqrt {b} d+\sqrt {-a} e}\right )\right ) \log (d+e x)+\text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )+\text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )\right )+e^2 \left (\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+p \text {Li}_2\left (1+\frac {b x^2}{a}\right )\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]/(x^3*(d + e*x)),x]

[Out]

((-4*Sqrt[b]*d*e*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[a] + (b*d^2*p*(2*Log[x] - Log[a + b*x^2]))/a - (d^2*Log[c
*(a + b*x^2)^p])/x^2 + (2*d*e*Log[c*(a + b*x^2)^p])/x - 2*e^2*Log[d + e*x]*Log[c*(a + b*x^2)^p] + 2*e^2*p*((Lo
g[(e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*d + Sqrt[-a]*e)] + Log[(e*(Sqrt[-a] + Sqrt[b]*x))/(-(Sqrt[b]*d) + Sqrt[-
a]*e)])*Log[d + e*x] + PolyLog[2, (Sqrt[b]*(d + e*x))/(Sqrt[b]*d - Sqrt[-a]*e)] + PolyLog[2, (Sqrt[b]*(d + e*x
))/(Sqrt[b]*d + Sqrt[-a]*e)]) + e^2*(Log[-((b*x^2)/a)]*Log[c*(a + b*x^2)^p] + p*PolyLog[2, 1 + (b*x^2)/a]))/(2
*d^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.40, size = 1071, normalized size = 2.89

method	result	size
risch	\(\text {Expression too large to display}\)	\(1071\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^2+a)^p)/x^3/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

1/4*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)/d/x^2-1/2*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^
2+a)^p)*csgn(I*c)*e/d^2/x-p*e^2/d^3*dilog((-b*x+(-b*a)^(1/2))/(-b*a)^(1/2))-p*e^2/d^3*dilog((b*x+(-b*a)^(1/2))
/(-b*a)^(1/2))+p*e^2/d^3*dilog((e*(-b*a)^(1/2)-(e*x+d)*b+b*d)/(e*(-b*a)^(1/2)+b*d))+p*e^2/d^3*dilog((e*(-b*a)^
(1/2)+(e*x+d)*b-b*d)/(e*(-b*a)^(1/2)-b*d))+1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)*e^2/d^3*ln(x)+1/2*I*Pi*c
sgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)*e^2/d^3*ln(e*x+d)-1/2*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*
x^2+a)^p)*csgn(I*c)*e^2/d^3*ln(x)-ln(c)*e^2/d^3*ln(e*x+d)+ln(c)*e^2/d^3*ln(x)+ln(c)*e/d^2/x-1/2*ln(c)/d/x^2+1/
2*I*Pi*csgn(I*c*(b*x^2+a)^p)^3*e^2/d^3*ln(e*x+d)-ln((b*x^2+a)^p)*e^2/d^3*ln(e*x+d)+ln((b*x^2+a)^p)*e^2/d^3*ln(
x)+ln((b*x^2+a)^p)*e/d^2/x+p*e^2/d^3*ln(e*x+d)*ln((e*(-b*a)^(1/2)+(e*x+d)*b-b*d)/(e*(-b*a)^(1/2)-b*d))-1/2*b*p
*ln(b*x^2+a)/a/d-1/2*ln((b*x^2+a)^p)/d/x^2+1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)*e/d^2/x+1/2*I*Pi*csgn(I*
(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2*e^2/d^3*ln(x)-1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)^3*e^2/d^3*ln(x)-1/2*I*Pi*csg
n(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2*e^2/d^3*ln(e*x+d)+b*p*ln(x)/a/d-1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)^3*e/d^
2/x+1/4*I*Pi*csgn(I*c*(b*x^2+a)^p)^3/d/x^2-p*e^2/d^3*ln(x)*ln((-b*x+(-b*a)^(1/2))/(-b*a)^(1/2))-p*e^2/d^3*ln(x
)*ln((b*x+(-b*a)^(1/2))/(-b*a)^(1/2))+p*e^2/d^3*ln(e*x+d)*ln((e*(-b*a)^(1/2)-(e*x+d)*b+b*d)/(e*(-b*a)^(1/2)+b*
d))-1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)*e^2/d^3*ln(e*x+d)-1/4*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+
a)^p)^2/d/x^2-1/4*I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)/d/x^2-2*p*b/d^2*e/(b*a)^(1/2)*arctan(b*x/(b*a)^(1/2))
+1/2*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2*e/d^2/x

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^3/(e*x+d),x, algorithm="maxima")

[Out]

integrate(log((b*x^2 + a)^p*c)/((x*e + d)*x^3), x)

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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(log(c*(b*x^2+a)^p)/x^3/(e*x+d),x, algorithm="fricas")

[Out]

integral(log((b*x^2 + a)^p*c)/(x^4*e + d*x^3), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**2+a)**p)/x**3/(e*x+d),x)

[Out]

Timed out

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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(log(c*(b*x^2+a)^p)/x^3/(e*x+d),x, algorithm="giac")

[Out]

integrate(log((b*x^2 + a)^p*c)/((x*e + d)*x^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{x^3\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^2)^p)/(x^3*(d + e*x)),x)

[Out]

int(log(c*(a + b*x^2)^p)/(x^3*(d + e*x)), x)

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