3.279 \(\int \frac{a+b \log (c x^n)}{x \sqrt{d+e x^2}} \, dx\)

Optimal. Leaf size=166 \[ -\frac{b n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right )}{2 \sqrt{d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt{d}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{2 \sqrt{d}}-\frac{b n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{\sqrt{d}} \]

[Out]

(b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]^2)/(2*Sqrt[d]) - (ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*(a + b*Log[c*x^n]))/S
qrt[d] - (b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/Sqrt[d] - (b*n*Po
lyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*Sqrt[d])

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Rubi [A]  time = 0.262449, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {266, 63, 208, 2348, 12, 5984, 5918, 2402, 2315} \[ -\frac{b n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right )}{2 \sqrt{d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt{d}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{2 \sqrt{d}}-\frac{b n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{\sqrt{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]^2)/(2*Sqrt[d]) - (ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*(a + b*Log[c*x^n]))/S
qrt[d] - (b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/Sqrt[d] - (b*n*Po
lyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*Sqrt[d])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 2348

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.))/(x_), x_Symbol] :> With[{u = IntHi
de[(d + e*x^r)^q/x, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[Dist[1/x, u, x], x], x]] /; FreeQ[{a, b
, c, d, e, n, r}, x] && IntegerQ[q - 1/2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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]

Rubi steps

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

Mathematica [C]  time = 0.198194, size = 162, normalized size = 0.98 \[ \frac{b n \sqrt{\frac{d}{e x^2}+1} \left (-\, _3F_2\left (\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};-\frac{d}{e x^2}\right )-\frac{\sqrt{e} x \log (x) \sinh ^{-1}\left (\frac{\sqrt{d}}{\sqrt{e} x}\right )}{\sqrt{d}}\right )}{\sqrt{d+e x^2}}+\frac{\log \left (\sqrt{d} \sqrt{d+e x^2}+d\right ) \left (-a-b \left (\log \left (c x^n\right )-n \log (x)\right )\right )}{\sqrt{d}}-\frac{\log (x) \left (-a-b \left (\log \left (c x^n\right )-n \log (x)\right )\right )}{\sqrt{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*n*Sqrt[1 + d/(e*x^2)]*(-HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, -(d/(e*x^2))] - (Sqrt[e]*x*ArcSinh[S
qrt[d]/(Sqrt[e]*x)]*Log[x])/Sqrt[d]))/Sqrt[d + e*x^2] - (Log[x]*(-a - b*(-(n*Log[x]) + Log[c*x^n])))/Sqrt[d] +
((-a - b*(-(n*Log[x]) + Log[c*x^n]))*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/Sqrt[d]

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Maple [F]  time = 0.411, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{x}{\frac{1}{\sqrt{e{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/x/(e*x^2+d)^(1/2),x)

[Out]

int((a+b*ln(c*x^n))/x/(e*x^2+d)^(1/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e x^{2} + d} b \log \left (c x^{n}\right ) + \sqrt{e x^{2} + d} a}{e x^{3} + d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x/(e*x**2+d)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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