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

Optimal. Leaf size=221 \[ -\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {b e n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {b n \sqrt {d+e x}}{x}+\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {2 b e n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {47, 63, 208, 2350, 14, 5984, 5918, 2402, 2315} \[ -\frac {b e n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {b n \sqrt {d+e x}}{x}+\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {2 b e n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 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 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 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 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} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx &=-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \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 {-\sqrt {d+e x}-\frac {e x \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}}{x^2} \, dx\\ &=-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \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 \left (-\frac {\sqrt {d+e x}}{x^2}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d} x}\right ) \, dx\\ &=-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \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 {\sqrt {d+e x}}{x^2} \, dx+\frac {(b e n) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx}{\sqrt {d}}\\ &=-\frac {b n \sqrt {d+e x}}{x}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}+\frac {1}{2} (b e n) \int \frac {1}{x \sqrt {d+e x}} \, dx+\frac {(2 b e 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 {b n \sqrt {d+e x}}{x}+\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \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) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )-\frac {(2 b e 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 {b n \sqrt {d+e x}}{x}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \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 e 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 e 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 {b n \sqrt {d+e x}}{x}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \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 e 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 e 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 {b n \sqrt {d+e x}}{x}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \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 e 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 {b e 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 = 0.34, size = 392, normalized size = 1.77 \[ -\frac {4 a \sqrt {d} \sqrt {d+e x}-2 a e x \log \left (\sqrt {d}-\sqrt {d+e x}\right )+2 a e x \log \left (\sqrt {d+e x}+\sqrt {d}\right )-2 b e x \log \left (c x^n\right ) \log \left (\sqrt {d}-\sqrt {d+e x}\right )+4 b \sqrt {d} \sqrt {d+e x} \log \left (c x^n\right )+2 b e x \log \left (c x^n\right ) \log \left (\sqrt {d+e x}+\sqrt {d}\right )+2 b e n x \text {Li}_2\left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )-2 b e n x \text {Li}_2\left (\frac {1}{2} \left (\frac {\sqrt {d+e x}}{\sqrt {d}}+1\right )\right )+4 b \sqrt {d} n \sqrt {d+e x}+b e n x \log ^2\left (\sqrt {d}-\sqrt {d+e x}\right )-b e n x \log ^2\left (\sqrt {d+e x}+\sqrt {d}\right )+2 b e n x \log \left (\frac {1}{2} \left (\frac {\sqrt {d+e x}}{\sqrt {d}}+1\right )\right ) \log \left (\sqrt {d}-\sqrt {d+e x}\right )-2 b e n x \log \left (\sqrt {d+e x}+\sqrt {d}\right ) \log \left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )+4 b e n x \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{4 \sqrt {d} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x + d} b \log \left (c x^{n}\right ) + \sqrt {e x + d} a}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) \sqrt {e x +d}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (\frac {e \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{\sqrt {d}} - \frac {2 \, \sqrt {e x + d}}{x}\right )} a + b \int \frac {\sqrt {e x + d} {\left (\log \relax (c) + \log \left (x^{n}\right )\right )}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/sqrt(d) - 2*sqrt(e*x + d)/x)*a + b*integrate(s
qrt(e*x + d)*(log(c) + log(x^n))/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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