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

Optimal. Leaf size=256 \[ -\frac {d \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e^2}+\frac {x \log \left (c \left (a+b x^2\right )^p\right )}{e}+\frac {d p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e^2}+\frac {d p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e^2}+\frac {d p \log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )}{e^2}+\frac {d p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )}{e^2}+\frac {2 \sqrt {a} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} e}-\frac {2 p x}{e} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2466, 2448, 321, 205, 2462, 260, 2416, 2394, 2393, 2391} \[ \frac {d p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e^2}+\frac {d p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {-a} e+\sqrt {b} d}\right )}{e^2}-\frac {d \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e^2}+\frac {x \log \left (c \left (a+b x^2\right )^p\right )}{e}+\frac {d p \log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )}{e^2}+\frac {d p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )}{e^2}+\frac {2 \sqrt {a} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} e}-\frac {2 p x}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 2448

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

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

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Mathematica [A]  time = 0.12, size = 225, normalized size = 0.88 \[ \frac {-d \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )+e x \log \left (c \left (a+b x^2\right )^p\right )+d p \left (\text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )+\text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )+\log (d+e x) \left (\log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )+\log \left (\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {-a} e-\sqrt {b} d}\right )\right )\right )-2 e p \left (x-\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}\right )}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x \log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.31, size = 576, normalized size = 2.25 \[ \frac {i \pi d \,\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 e^{2}}-\frac {i \pi d \,\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 e^{2}}-\frac {i \pi d \,\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 e^{2}}+\frac {i \pi d \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3} \ln \left (e x +d \right )}{2 e^{2}}-\frac {i \pi x \,\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 )}{2 e}+\frac {i \pi x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{2 e}+\frac {i \pi x \,\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}}{2 e}-\frac {i \pi x \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}}{2 e}+\frac {2 a p \arctan \left (\frac {-2 b d +2 \left (e x +d \right ) b}{2 \sqrt {a b}\, e}\right )}{\sqrt {a b}\, e}+\frac {d 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 )}{e^{2}}+\frac {d 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 )}{e^{2}}+\frac {d p \dilog \left (\frac {b d -\left (e x +d \right ) b +\sqrt {-a b}\, e}{b d +\sqrt {-a b}\, e}\right )}{e^{2}}+\frac {d p \dilog \left (\frac {-b d +\left (e x +d \right ) b +\sqrt {-a b}\, e}{-b d +\sqrt {-a b}\, e}\right )}{e^{2}}-\frac {d \ln \relax (c ) \ln \left (e x +d \right )}{e^{2}}-\frac {d \ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \ln \left (e x +d \right )}{e^{2}}-\frac {2 p x}{e}+\frac {x \ln \relax (c )}{e}+\frac {x \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{e}-\frac {2 d p}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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