3.230 \(\int \frac {\log (c (a+b x^2)^p)}{x (d+e x)} \, dx\)

Optimal. Leaf size=247 \[ -\frac {\log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d}+\frac {\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d}+\frac {p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{d}+\frac {p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{d}+\frac {p \log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )}{d}+\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )}{d}+\frac {p \text {Li}_2\left (\frac {b x^2}{a}+1\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {2466, 2454, 2394, 2315, 2462, 260, 2416, 2393, 2391} \[ \frac {p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{d}+\frac {p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {-a} e+\sqrt {b} d}\right )}{d}+\frac {p \text {PolyLog}\left (2,\frac {b x^2}{a}+1\right )}{2 d}-\frac {\log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d}+\frac {\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d}+\frac {p \log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )}{d}+\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 260

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 2315

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

Rule 2391

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 2393

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 2394

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 2416

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 2454

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 2462

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 2466

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

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Mathematica [A]  time = 0.07, size = 232, normalized size = 0.94 \[ -\frac {\log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d}+\frac {\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+p \text {Li}_2\left (\frac {b x^2+a}{a}\right )}{2 d}+\frac {p \left (\text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )+\text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )+\log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )+\log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Log[d + e*x]*Log[c*(a + b*x^2)^p])/d) + (p*(Log[(e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*d + Sqrt[-a]*e)]*Log[d
+ e*x] + 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)]))/d + (Log[-((b*x^2)
/a)]*Log[c*(a + b*x^2)^p] + p*PolyLog[2, (a + b*x^2)/a])/(2*d)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{e x^{2} + d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{{\left (e x + d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.25, size = 624, normalized size = 2.53 \[ -\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \ln \relax (x )}{2 d}+\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \ln \left (e x +d \right )}{2 d}+\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \ln \relax (x )}{2 d}-\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 d}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \ln \relax (x )}{2 d}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 d}-\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3} \ln \relax (x )}{2 d}+\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3} \ln \left (e x +d \right )}{2 d}-\frac {p \ln \relax (x ) \ln \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{d}-\frac {p \ln \relax (x ) \ln \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{d}+\frac {p \ln \left (\frac {b d -\left (e x +d \right ) b +\sqrt {-a b}\, e}{b d +\sqrt {-a b}\, e}\right ) \ln \left (e x +d \right )}{d}+\frac {p \ln \left (\frac {-b d +\left (e x +d \right ) b +\sqrt {-a b}\, e}{-b d +\sqrt {-a b}\, e}\right ) \ln \left (e x +d \right )}{d}-\frac {p \dilog \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{d}-\frac {p \dilog \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{d}+\frac {p \dilog \left (\frac {b d -\left (e x +d \right ) b +\sqrt {-a b}\, e}{b d +\sqrt {-a b}\, e}\right )}{d}+\frac {p \dilog \left (\frac {-b d +\left (e x +d \right ) b +\sqrt {-a b}\, e}{-b d +\sqrt {-a b}\, e}\right )}{d}+\frac {\ln \relax (c ) \ln \relax (x )}{d}-\frac {\ln \relax (c ) \ln \left (e x +d \right )}{d}+\frac {\ln \relax (x ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{d}-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \ln \left (e x +d \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{{\left (e x + d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{x \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(a + b*x**2)**p)/(x*(d + e*x)), x)

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