3.290 \(\int \frac {a+b \log (c x^n)}{x (d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=209 \[ \left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^2+d}}\right )}{2 d^{3/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{3/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}} \]

[Out]

b*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(3/2)+1/2*b*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))^2/d^(3/2)-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^(3/2)-1/2*b*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-
(e*x^2+d)^(1/2)))/d^(3/2)+(a+b*ln(c*x^n))*(-arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(3/2)+1/d/(e*x^2+d)^(1/2))

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Rubi [A]  time = 0.33, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {266, 51, 63, 208, 2348, 5984, 5918, 2402, 2315} \[ -\frac {b n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{3/2}}+\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{3/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x*(d + e*x^2)^(3/2)),x]

[Out]

(b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/d^(3/2) + (b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]^2)/(2*d^(3/2)) + (1/(d*
Sqrt[d + e*x^2]) - ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]/d^(3/2))*(a + b*Log[c*x^n]) - (b*n*ArcTanh[Sqrt[d + e*x^2]
/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/d^(3/2) - (b*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - S
qrt[d + e*x^2])])/(2*d^(3/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^{3/2}} \, dx &=\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {1}{d x \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2} x}\right ) \, dx\\ &=\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b n) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{x} \, dx}{d^{3/2}}-\frac {(b n) \int \frac {1}{x \sqrt {d+e x^2}} \, dx}{d}\\ &=\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b n) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx,x,x^2\right )}{2 d^{3/2}}-\frac {(b n) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{2 d}\\ &=\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(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^2}\right )}{d^{3/2}}-\frac {(b n) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{d e}\\ &=\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{3/2}}+\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(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 )}{d^2}\\ &=\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{3/2}}+\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b 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 )}{d^{3/2}}+\frac {(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 )}{d^2}\\ &=\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{3/2}}+\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b 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 )}{d^{3/2}}-\frac {(b n) \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 )}{d^{3/2}}\\ &=\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{3/2}}+\left (\frac {1}{d \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b 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 )}{d^{3/2}}-\frac {b n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x^2}}{\sqrt {d}}}\right )}{2 d^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.39, size = 241, normalized size = 1.15 \[ \frac {9 e x^2 \left (\log (x) \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )+b n \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )\right )+\left (\sqrt {d}-\sqrt {d+e x^2} \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )\right ) \left (a+b \log \left (c x^n\right )\right )-b n \log ^2(x) \sqrt {d+e x^2}-b \sqrt {e} n x \log (x) \sqrt {\frac {d}{e x^2}+1} \sinh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {e} x}\right )\right )-b d^{3/2} n \sqrt {\frac {d}{e x^2}+1} \, _3F_2\left (\frac {3}{2},\frac {3}{2},\frac {3}{2};\frac {5}{2},\frac {5}{2};-\frac {d}{e x^2}\right )}{9 d^{3/2} e x^2 \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x*(d + e*x^2)^(3/2)),x]

[Out]

(-(b*d^(3/2)*n*Sqrt[1 + d/(e*x^2)]*HypergeometricPFQ[{3/2, 3/2, 3/2}, {5/2, 5/2}, -(d/(e*x^2))]) + 9*e*x^2*(-(
b*Sqrt[e]*n*Sqrt[1 + d/(e*x^2)]*x*ArcSinh[Sqrt[d]/(Sqrt[e]*x)]*Log[x]) - b*n*Sqrt[d + e*x^2]*Log[x]^2 + Sqrt[d
+ e*x^2]*Log[x]*(a + b*Log[c*x^n] + b*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]]) + (a + b*Log[c*x^n])*(Sqrt[d] - Sqr
t[d + e*x^2]*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])))/(9*d^(3/2)*e*x^2*Sqrt[d + e*x^2])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} b \log \left (c x^{n}\right ) + \sqrt {e x^{2} + d} a}{e^{2} x^{5} + 2 \, d e x^{3} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

integral((sqrt(e*x^2 + d)*b*log(c*x^n) + sqrt(e*x^2 + d)*a)/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((e*x^2 + d)^(3/2)*x), x)

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\left (e \,x^{2}+d \right )^{\frac {3}{2}} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*x^n)+a)/x/(e*x^2+d)^(3/2),x)

[Out]

int((b*ln(c*x^n)+a)/x/(e*x^2+d)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a {\left (\frac {\operatorname {arsinh}\left (\frac {d}{\sqrt {d e} {\left | x \right |}}\right )}{d^{\frac {3}{2}}} - \frac {1}{\sqrt {e x^{2} + d} d}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{{\left (e x^{3} + d x\right )} \sqrt {e x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

-a*(arcsinh(d/(sqrt(d*e)*abs(x)))/d^(3/2) - 1/(sqrt(e*x^2 + d)*d)) + b*integrate((log(c) + log(x^n))/((e*x^3 +
d*x)*sqrt(e*x^2 + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x^n))/(x*(d + e*x^2)^(3/2)),x)

[Out]

int((a + b*log(c*x^n))/(x*(d + e*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \log {\left (c x^{n} \right )}}{x \left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x/(e*x**2+d)**(3/2),x)

[Out]

Integral((a + b*log(c*x**n))/(x*(d + e*x**2)**(3/2)), x)

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