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

Optimal. Leaf size=201 \[ -\frac{p \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x)}{\sqrt{b} d-\sqrt{-a} e}\right )}{e}-\frac{p \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x)}{\sqrt{-a} e+\sqrt{b} d}\right )}{e}+\frac{\log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e}-\frac{p \log (d+e x) \log \left (\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} d}\right )}{e}-\frac{p \log (d+e x) \log \left (-\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{\sqrt{b} d-\sqrt{-a} e}\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.266316, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2462, 260, 2416, 2394, 2393, 2391} \[ -\frac{p \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x)}{\sqrt{b} d-\sqrt{-a} e}\right )}{e}-\frac{p \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x)}{\sqrt{-a} e+\sqrt{b} d}\right )}{e}+\frac{\log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e}-\frac{p \log (d+e x) \log \left (\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} d}\right )}{e}-\frac{p \log (d+e x) \log \left (-\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{\sqrt{b} d-\sqrt{-a} e}\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/(d + e*x),x]

[Out]

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

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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0806285, size = 201, normalized size = 1. \[ -\frac{p \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x)}{\sqrt{b} d-\sqrt{-a} e}\right )}{e}-\frac{p \text{PolyLog}\left (2,\frac{\sqrt{b} (d+e x)}{\sqrt{-a} e+\sqrt{b} d}\right )}{e}+\frac{\log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e}-\frac{p \log (d+e x) \log \left (\frac{e \left (\sqrt{-a}-\sqrt{b} x\right )}{\sqrt{-a} e+\sqrt{b} d}\right )}{e}-\frac{p \log (d+e x) \log \left (-\frac{e \left (\sqrt{-a}+\sqrt{b} x\right )}{\sqrt{b} d-\sqrt{-a} e}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.511, size = 366, normalized size = 1.8 \begin{align*}{\frac{\ln \left ( ex+d \right ) \ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{e}}-{\frac{p\ln \left ( ex+d \right ) }{e}\ln \left ({ \left ( e\sqrt{-ab}-b \left ( ex+d \right ) +bd \right ) \left ( e\sqrt{-ab}+bd \right ) ^{-1}} \right ) }-{\frac{p\ln \left ( ex+d \right ) }{e}\ln \left ({ \left ( e\sqrt{-ab}+b \left ( ex+d \right ) -bd \right ) \left ( e\sqrt{-ab}-bd \right ) ^{-1}} \right ) }-{\frac{p}{e}{\it dilog} \left ({ \left ( e\sqrt{-ab}-b \left ( ex+d \right ) +bd \right ) \left ( e\sqrt{-ab}+bd \right ) ^{-1}} \right ) }-{\frac{p}{e}{\it dilog} \left ({ \left ( e\sqrt{-ab}+b \left ( ex+d \right ) -bd \right ) \left ( e\sqrt{-ab}-bd \right ) ^{-1}} \right ) }+{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \,{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}}{e}}-{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \,{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) }{e}}-{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \, \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}}{e}}+{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \, \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) }{e}}+{\frac{\ln \left ( ex+d \right ) \ln \left ( c \right ) }{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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