3.301 \(\int \frac{a+b \log (c x^n)}{x (d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=251 \[ -\frac{b n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right )}{2 d^{5/2}}+\frac{1}{3} \left (\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{1}{d \left (d+e x^2\right )^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{b n}{3 d^2 \sqrt{d+e x^2}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{2 d^{5/2}}+\frac{4 b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{b n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}} \]

[Out]

-(b*n)/(3*d^2*Sqrt[d + e*x^2]) + (4*b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(3*d^(5/2)) + (b*n*ArcTanh[Sqrt[d +
e*x^2]/Sqrt[d]]^2)/(2*d^(5/2)) + ((1/(d*(d + e*x^2)^(3/2)) + 3/(d^2*Sqrt[d + e*x^2]) - (3*ArcTanh[Sqrt[d + e*x
^2]/Sqrt[d]])/d^(5/2))*(a + b*Log[c*x^n]))/3 - (b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d]
- Sqrt[d + e*x^2])])/d^(5/2) - (b*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*d^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.403881, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {266, 51, 63, 208, 2348, 5984, 5918, 2402, 2315} \[ -\frac{b n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right )}{2 d^{5/2}}+\frac{1}{3} \left (\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{1}{d \left (d+e x^2\right )^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{b n}{3 d^2 \sqrt{d+e x^2}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{2 d^{5/2}}+\frac{4 b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{b n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x*(d + e*x^2)^(5/2)),x]

[Out]

-(b*n)/(3*d^2*Sqrt[d + e*x^2]) + (4*b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(3*d^(5/2)) + (b*n*ArcTanh[Sqrt[d +
e*x^2]/Sqrt[d]]^2)/(2*d^(5/2)) + ((1/(d*(d + e*x^2)^(3/2)) + 3/(d^2*Sqrt[d + e*x^2]) - (3*ArcTanh[Sqrt[d + e*x
^2]/Sqrt[d]])/d^(5/2))*(a + b*Log[c*x^n]))/3 - (b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d]
- Sqrt[d + e*x^2])])/d^(5/2) - (b*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*d^(5/2))

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^{5/2}} \, dx &=\frac{1}{3} \left (\frac{1}{d \left (d+e x^2\right )^{3/2}}+\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac{1}{3 d x \left (d+e x^2\right )^{3/2}}+\frac{1}{d^2 x \sqrt{d+e x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2} x}\right ) \, dx\\ &=\frac{1}{3} \left (\frac{1}{d \left (d+e x^2\right )^{3/2}}+\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}\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^{5/2}}-\frac{(b n) \int \frac{1}{x \sqrt{d+e x^2}} \, dx}{d^2}-\frac{(b n) \int \frac{1}{x \left (d+e x^2\right )^{3/2}} \, dx}{3 d}\\ &=\frac{1}{3} \left (\frac{1}{d \left (d+e x^2\right )^{3/2}}+\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}\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^{5/2}}-\frac{(b n) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{2 d^2}-\frac{(b n) \operatorname{Subst}\left (\int \frac{1}{x (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 d}\\ &=-\frac{b n}{3 d^2 \sqrt{d+e x^2}}+\frac{1}{3} \left (\frac{1}{d \left (d+e x^2\right )^{3/2}}+\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}\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^{5/2}}-\frac{(b n) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{6 d^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^2 e}\\ &=-\frac{b n}{3 d^2 \sqrt{d+e x^2}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{2 d^{5/2}}+\frac{1}{3} \left (\frac{1}{d \left (d+e x^2\right )^{3/2}}+\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}\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^3}-\frac{(b n) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{3 d^2 e}\\ &=-\frac{b n}{3 d^2 \sqrt{d+e x^2}}+\frac{4 b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{2 d^{5/2}}+\frac{1}{3} \left (\frac{1}{d \left (d+e x^2\right )^{3/2}}+\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}\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^{5/2}}+\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^3}\\ &=-\frac{b n}{3 d^2 \sqrt{d+e x^2}}+\frac{4 b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{2 d^{5/2}}+\frac{1}{3} \left (\frac{1}{d \left (d+e x^2\right )^{3/2}}+\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}\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^{5/2}}-\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^{5/2}}\\ &=-\frac{b n}{3 d^2 \sqrt{d+e x^2}}+\frac{4 b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{2 d^{5/2}}+\frac{1}{3} \left (\frac{1}{d \left (d+e x^2\right )^{3/2}}+\frac{3}{d^2 \sqrt{d+e x^2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{d^{5/2}}\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^{5/2}}-\frac{b n \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{d+e x^2}}{\sqrt{d}}}\right )}{2 d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.447829, size = 273, normalized size = 1.09 \[ \frac{b n \sqrt{\frac{d}{e x^2}+1} \left (-3 d^{5/2} \left (d+e x^2\right )^2 \, _3F_2\left (\frac{5}{2},\frac{5}{2},\frac{5}{2};\frac{7}{2},\frac{7}{2};-\frac{d}{e x^2}\right )+25 \sqrt{d} e^3 x^6 \log (x) \sqrt{\frac{d}{e x^2}+1} \left (4 d+3 e x^2\right )-75 e^{5/2} x^5 \log (x) \left (d+e x^2\right )^2 \sinh ^{-1}\left (\frac{\sqrt{d}}{\sqrt{e} x}\right )\right )}{75 d^{5/2} e^2 x^4 \left (d+e x^2\right )^{5/2}}+\frac{\left (4 d+3 e x^2\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{\log \left (\sqrt{d} \sqrt{d+e x^2}+d\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d^{5/2}}+\frac{\log (x) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x*(d + e*x^2)^(5/2)),x]

[Out]

(b*n*Sqrt[1 + d/(e*x^2)]*(-3*d^(5/2)*(d + e*x^2)^2*HypergeometricPFQ[{5/2, 5/2, 5/2}, {7/2, 7/2}, -(d/(e*x^2))
] + 25*Sqrt[d]*e^3*Sqrt[1 + d/(e*x^2)]*x^6*(4*d + 3*e*x^2)*Log[x] - 75*e^(5/2)*x^5*(d + e*x^2)^2*ArcSinh[Sqrt[
d]/(Sqrt[e]*x)]*Log[x]))/(75*d^(5/2)*e^2*x^4*(d + e*x^2)^(5/2)) + ((4*d + 3*e*x^2)*(a - b*n*Log[x] + b*Log[c*x
^n]))/(3*d^2*(d + e*x^2)^(3/2)) + (Log[x]*(a - b*n*Log[x] + b*Log[c*x^n]))/d^(5/2) - ((a - b*n*Log[x] + b*Log[
c*x^n])*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/d^(5/2)

________________________________________________________________________________________

Maple [F]  time = 0.413, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{x} \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/x/(e*x^2+d)^(5/2),x)

[Out]

int((a+b*ln(c*x^n))/x/(e*x^2+d)^(5/2),x)

________________________________________________________________________________________

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))/x/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e x^{2} + d} b \log \left (c x^{n}\right ) + \sqrt{e x^{2} + d} a}{e^{3} x^{7} + 3 \, d e^{2} x^{5} + 3 \, d^{2} e x^{3} + d^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral((sqrt(e*x^2 + d)*b*log(c*x^n) + sqrt(e*x^2 + d)*a)/(e^3*x^7 + 3*d*e^2*x^5 + 3*d^2*e*x^3 + d^3*x), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x/(e*x**2+d)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac{5}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

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