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

Optimal. Leaf size=152 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {2 b n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {4 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}} \]

[Out]

2*b*n*arctanh((e*x+d)^(1/2)/d^(1/2))^2/d^(1/2)-2*arctanh((e*x+d)^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/d^(1/2)-4*b*n*
arctanh((e*x+d)^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)-(e*x+d)^(1/2)))/d^(1/2)-2*b*n*polylog(2,1-2*d^(1/2)/(d^(1
/2)-(e*x+d)^(1/2)))/d^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {63, 208, 2348, 12, 5984, 5918, 2402, 2315} \[ -\frac {2 b n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {4 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]^2)/Sqrt[d] - (2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*(a + b*Log[c*x^n]))/Sqrt[
d] - (4*b*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/Sqrt[d] - (2*b*n*PolyLo
g[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/Sqrt[d]

Rule 12

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

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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{x \sqrt {d+e x}} \, dx &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-(b n) \int -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d} x} \, dx\\ &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}+\frac {(2 b n) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx}{\sqrt {d}}\\ &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}+\frac {(4 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}\right )}{\sqrt {d}}\\ &=\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {(4 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}\right )}{d}\\ &=\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}+\frac {(4 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}\right )}{d}\\ &=\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {(4 b n) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {d+e x}}{\sqrt {d}}}\right )}{\sqrt {d}}\\ &=\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {2 b n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x}}{\sqrt {d}}}\right )}{\sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 249, normalized size = 1.64 \[ \frac {2 \log \left (\sqrt {d}-\sqrt {d+e x}\right ) \left (a+b \log \left (c x^n\right )\right )-2 \log \left (\sqrt {d+e x}+\sqrt {d}\right ) \left (a+b \log \left (c x^n\right )\right )-b n \left (2 \text {Li}_2\left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )+\log \left (\sqrt {d}-\sqrt {d+e x}\right ) \left (\log \left (\sqrt {d}-\sqrt {d+e x}\right )+2 \log \left (\frac {1}{2} \left (\frac {\sqrt {d+e x}}{\sqrt {d}}+1\right )\right )\right )\right )+b n \left (2 \text {Li}_2\left (\frac {1}{2} \left (\frac {\sqrt {d+e x}}{\sqrt {d}}+1\right )\right )+\log \left (\sqrt {d+e x}+\sqrt {d}\right ) \left (\log \left (\sqrt {d+e x}+\sqrt {d}\right )+2 \log \left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )\right )\right )}{2 \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*Log[c*x^n])*Log[Sqrt[d] - Sqrt[d + e*x]] - 2*(a + b*Log[c*x^n])*Log[Sqrt[d] + Sqrt[d + e*x]] - b*n*(
Log[Sqrt[d] - Sqrt[d + e*x]]*(Log[Sqrt[d] - Sqrt[d + e*x]] + 2*Log[(1 + Sqrt[d + e*x]/Sqrt[d])/2]) + 2*PolyLog
[2, 1/2 - Sqrt[d + e*x]/(2*Sqrt[d])]) + b*n*(Log[Sqrt[d] + Sqrt[d + e*x]]*(Log[Sqrt[d] + Sqrt[d + e*x]] + 2*Lo
g[1/2 - Sqrt[d + e*x]/(2*Sqrt[d])]) + 2*PolyLog[2, (1 + Sqrt[d + e*x]/Sqrt[d])/2]))/(2*Sqrt[d])

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x + d} b \log \left (c x^{n}\right ) + \sqrt {e x + d} a}{e x^{2} + d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\sqrt {e x +d}\, x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,\sqrt {d+e\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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