3.2.48 \(\int \frac {a+b \log (c x^n)}{x \sqrt {d+e x}} \, dx\) [148]

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

[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 = 0.14, 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 = {65, 214, 2390, 12, 6131, 6055, 2449, 2352} \begin {gather*} -\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}} \end {gather*}

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 65

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 214

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

Rule 2352

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

Rule 2390

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 2449

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 6055

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 6131

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) \text {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) \text {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) \text {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) \text {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 = 0.07, size = 249, normalized size = 1.64 \begin {gather*} \frac {2 \left (a+b \log \left (c x^n\right )\right ) \log \left (\sqrt {d}-\sqrt {d+e x}\right )-2 \left (a+b \log \left (c x^n\right )\right ) \log \left (\sqrt {d}+\sqrt {d+e x}\right )-b n \left (\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 (1+\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )\right )+2 \text {Li}_2\left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )\right )+b n \left (\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}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )\right )+2 \text {Li}_2\left (\frac {1}{2} \left (1+\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )\right )}{2 \sqrt {d}} \end {gather*}

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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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x \sqrt {e x +d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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

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]

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

[Out]

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

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

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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