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

Optimal. Leaf size=220 \[ \left (\sqrt {d+e x^2}-\sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} b \sqrt {d} n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^2+d}}\right )-b n \sqrt {d+e x^2}+\frac {1}{2} b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2+b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )-b \sqrt {d} n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {266, 50, 63, 208, 2348, 5984, 5918, 2402, 2315} \[ -\frac {1}{2} b \sqrt {d} n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )+\left (\sqrt {d+e x^2}-\sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right ) \left (a+b \log \left (c x^n\right )\right )-b n \sqrt {d+e x^2}+\frac {1}{2} b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2+b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )-b \sqrt {d} n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*n*Sqrt[d + e*x^2]) + b*Sqrt[d]*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]] + (b*Sqrt[d]*n*ArcTanh[Sqrt[d + e*x^2]/S
qrt[d]]^2)/2 + (Sqrt[d + e*x^2] - Sqrt[d]*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])*(a + b*Log[c*x^n]) - b*Sqrt[d]*n*A
rcTanh[Sqrt[d + e*x^2]/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^2])])/2

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]

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

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Mathematica [C]  time = 0.33, size = 203, normalized size = 0.92 \[ \frac {b n \sqrt {d+e x^2} \left (-\, _3F_2\left (-\frac {1}{2},-\frac {1}{2},-\frac {1}{2};\frac {1}{2},\frac {1}{2};-\frac {d}{e x^2}\right )+\log (x) \sqrt {\frac {d}{e x^2}+1}-\frac {\sqrt {d} \log (x) \sinh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {e} x}\right )}{\sqrt {e} x}\right )}{\sqrt {\frac {d}{e x^2}+1}}+\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )-b n \log (x)\right )-\sqrt {d} \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )+\sqrt {d} \log (x) \left (a+b \log \left (c x^n\right )-b n \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} b \log \left (c x^{n}\right ) + \sqrt {e x^{2} + d} a}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) \sqrt {e \,x^{2}+d}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -{\left (\sqrt {d} \operatorname {arsinh}\left (\frac {d}{\sqrt {d e} {\left | x \right |}}\right ) - \sqrt {e x^{2} + d}\right )} a + b \int \frac {\sqrt {e x^{2} + d} {\left (\log \relax (c) + \log \left (x^{n}\right )\right )}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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