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

Optimal. Leaf size=211 \[ -2 b \sqrt{d} n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right )+2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )-2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )-4 b n \sqrt{d+e x}+2 b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )^2+4 b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )-4 b \sqrt{d} n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.330594, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {2346, 63, 208, 2348, 12, 5984, 5918, 2402, 2315, 2319, 50} \[ -2 b \sqrt{d} n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right )+2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )-2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )-4 b n \sqrt{d+e x}+2 b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )^2+4 b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )-4 b \sqrt{d} n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2346

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

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]

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rubi steps

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

Mathematica [A]  time = 0.205856, size = 331, normalized size = 1.57 \[ -\frac{1}{2} b \sqrt{d} n \left (2 \text{PolyLog}\left (2,\frac{1}{2}-\frac{\sqrt{d+e x}}{2 \sqrt{d}}\right )+\log \left (\sqrt{d}-\sqrt{d+e x}\right ) \left (\log \left (\sqrt{d}-\sqrt{d+e x}\right )+2 \log \left (\frac{1}{2} \left (\frac{\sqrt{d+e x}}{\sqrt{d}}+1\right )\right )\right )\right )+\frac{1}{2} b \sqrt{d} n \left (2 \text{PolyLog}\left (2,\frac{1}{2} \left (\frac{\sqrt{d+e x}}{\sqrt{d}}+1\right )\right )+\log \left (\sqrt{d+e x}+\sqrt{d}\right ) \left (\log \left (\sqrt{d+e x}+\sqrt{d}\right )+2 \log \left (\frac{1}{2}-\frac{\sqrt{d+e x}}{2 \sqrt{d}}\right )\right )\right )+\sqrt{d} \log \left (\sqrt{d}-\sqrt{d+e x}\right ) \left (a+b \log \left (c x^n\right )\right )-\sqrt{d} \log \left (\sqrt{d+e x}+\sqrt{d}\right ) \left (a+b \log \left (c x^n\right )\right )+2 a \sqrt{d+e x}+2 b \sqrt{d+e x} \log \left (c x^n\right )-4 b n \left (\sqrt{d+e x}-\sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.53, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{x}\sqrt{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*ln(c*x^n))*(e*x+d)^(1/2)/x,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))*(e*x+d)^(1/2)/x,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 + d} b \log \left (c x^{n}\right ) + \sqrt{e x + d} a}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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