3.3.27 \(\int \frac {x^2 \log (c (a+b x^2)^p)}{d+e x} \, dx\) [227]

Optimal. Leaf size=313 \[ \frac {2 d p x}{e^2}-\frac {p x^2}{2 e}-\frac {2 \sqrt {a} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} e^2}-\frac {d^2 p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{e^3}-\frac {d^2 p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{e^3}-\frac {d x \log \left (c \left (a+b x^2\right )^p\right )}{e^2}+\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 b e}+\frac {d^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e^3}-\frac {d^2 p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e^3}-\frac {d^2 p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {2516, 2498, 327, 211, 2504, 2436, 2332, 2512, 266, 2463, 2441, 2440, 2438} \begin {gather*} -\frac {d^2 p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e^3}-\frac {d^2 p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {-a} e+\sqrt {b} d}\right )}{e^3}-\frac {2 \sqrt {a} d p \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} e^2}+\frac {d^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e^3}-\frac {d x \log \left (c \left (a+b x^2\right )^p\right )}{e^2}+\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 b e}-\frac {d^2 p \log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )}{e^3}-\frac {d^2 p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )}{e^3}+\frac {2 d p x}{e^2}-\frac {p x^2}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 327

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 2332

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

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 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 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 2498

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

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Mathematica [A]
time = 0.11, size = 271, normalized size = 0.87 \begin {gather*} \frac {-e^2 p x^2+4 d e p \left (x-\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}\right )-2 d e x \log \left (c \left (a+b x^2\right )^p\right )+\frac {e^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}+2 d^2 \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )-2 d^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 )}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.48, size = 825, normalized size = 2.64

method	result	size
risch	\(\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} d^{2} \ln \left (e x +d \right )}{2 e^{3}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) d x}{2 e^{2}}+\frac {2 d p x}{e^{2}}-\frac {p \,d^{2} \dilog \left (\frac {e \sqrt {-b a}-\left (e x +d \right ) b +b d}{e \sqrt {-b a}+b d}\right )}{e^{3}}-\frac {p \,d^{2} \dilog \left (\frac {e \sqrt {-b a}+\left (e x +d \right ) b -b d}{e \sqrt {-b a}-b d}\right )}{e^{3}}-\frac {p \,d^{2} \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-b a}-\left (e x +d \right ) b +b d}{e \sqrt {-b a}+b d}\right )}{e^{3}}-\frac {p \,d^{2} \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-b a}+\left (e x +d \right ) b -b d}{e \sqrt {-b a}-b d}\right )}{e^{3}}-\frac {\ln \left (c \right ) d x}{e^{2}}+\frac {\ln \left (c \right ) d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {2 p a d \arctan \left (\frac {2 \left (e x +d \right ) b -2 b d}{2 e \sqrt {b a}}\right )}{e^{2} \sqrt {b a}}-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right ) d x}{e^{2}}+\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right ) d^{2} \ln \left (e x +d \right )}{e^{3}}+\frac {\ln \left (c \right ) x^{2}}{2 e}+\frac {p a \ln \left (\left (e x +d \right )^{2} b -2 d \left (e x +d \right ) b +a \,e^{2}+b \,d^{2}\right )}{2 b e}+\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 ) \mathrm {csgn}\left (i c \right ) d x}{2 e^{2}}-\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 ) \mathrm {csgn}\left (i c \right ) d^{2} \ln \left (e x +d \right )}{2 e^{3}}+\frac {5 p \,d^{2}}{2 e^{3}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3} x^{2}}{4 e}-\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3} d^{2} \ln \left (e x +d \right )}{2 e^{3}}-\frac {p \,x^{2}}{2 e}-\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 ) \mathrm {csgn}\left (i c \right ) x^{2}}{4 e}+\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right ) x^{2}}{2 e}+\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) x^{2}}{4 e}+\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} x^{2}}{4 e}-\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} d x}{2 e^{2}}+\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) d^{2} \ln \left (e x +d \right )}{2 e^{3}}+\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3} d x}{2 e^{2}}\)	\(825\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*p*x/e^2-p/e^3*d^2*ln(e*x+d)*ln((e*(-b*a)^(1/2)-(e*x+d)*b+b*d)/(e*(-b*a)^(1/2)+b*d))-p/e^3*d^2*ln(e*x+d)*ln
((e*(-b*a)^(1/2)+(e*x+d)*b-b*d)/(e*(-b*a)^(1/2)-b*d))-ln(c)/e^2*d*x+ln(c)*d^2/e^3*ln(e*x+d)-1/4*I*Pi*csgn(I*(b
*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)/e*x^2-2*p/e^2*a*d/(b*a)^(1/2)*arctan(1/2*(2*(e*x+d)*b-2*b*d)/e/(b*a
)^(1/2))-ln((b*x^2+a)^p)/e^2*d*x+ln((b*x^2+a)^p)*d^2/e^3*ln(e*x+d)+1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)^3/e^2*d*x+1/
2*ln(c)/e*x^2-1/4*I*Pi*csgn(I*c*(b*x^2+a)^p)^3/e*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^2*d*x+1/2/b*p/e*a*ln((e*x+d)^2*b-2*d*(e*x+d)*b+a*e^2+b*d^2)-1/2*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2
+a)^p)*csgn(I*c)*d^2/e^3*ln(e*x+d)+5/2*p/e^3*d^2-p/e^3*d^2*dilog((e*(-b*a)^(1/2)-(e*x+d)*b+b*d)/(e*(-b*a)^(1/2
)+b*d))-p/e^3*d^2*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)^3*
d^2/e^3*ln(e*x+d)-1/2*p*x^2/e+1/2*ln((b*x^2+a)^p)/e*x^2-1/2*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2/e
^2*d*x+1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)*d^2/e^3*ln(e*x+d)+1/4*I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)
/e*x^2+1/4*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2/e*x^2+1/2*I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2
+a)^p)^2*d^2/e^3*ln(e*x+d)-1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)/e^2*d*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(x^2*log(c*(b*x^2+a)^p)/(e*x+d),x, algorithm="maxima")

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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