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

Optimal. Leaf size=174 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} r}-\frac {2 b n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^r+d}}\right )}{\sqrt {d} r^2}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{\sqrt {d} r^2}-\frac {4 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{\sqrt {d} r^2} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {266, 63, 208, 2348, 12, 5984, 5918, 2402, 2315} \[ -\frac {2 b n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{\sqrt {d} r^2}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} r}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{\sqrt {d} r^2}-\frac {4 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{\sqrt {d} r^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {a+b \log \left (c x^n\right )}{x \sqrt {d+e x^r}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.68, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\sqrt {e \,x^{r}+d}\, x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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