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

Optimal. Leaf size=258 \[ \frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {b e n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^2+d}}\right )}{4 d^{3/2}}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{3/2}}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{3/2}}+\frac {b e n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{2 d^{3/2}}-\frac {b n \sqrt {d+e x^2}}{4 d x^2} \]

[Out]

-1/4*b*e*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(3/2)-1/4*b*e*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))^2/d^(3/2)+1/2*e
*arctanh((e*x^2+d)^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/d^(3/2)+1/2*b*e*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)+1/4*b*e*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(e*x^2+d)^(1/2)))/d^(3/2)-1/4*
b*n*(e*x^2+d)^(1/2)/d/x^2-1/2*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/d/x^2

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Rubi [A]  time = 0.37, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {266, 51, 63, 208, 2350, 12, 14, 47, 5984, 5918, 2402, 2315} \[ \frac {b e n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{4 d^{3/2}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{3/2}}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{3/2}}+\frac {b e n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{2 d^{3/2}}-\frac {b n \sqrt {d+e x^2}}{4 d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 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 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 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 {a+b \log \left (c x^n\right )}{x^3 \sqrt {d+e x^2}} \, dx &=-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}-(b n) \int \frac {-\frac {\sqrt {d+e x^2}}{d}+\frac {e x^2 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}}{2 x^3} \, dx\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}-\frac {1}{2} (b n) \int \frac {-\frac {\sqrt {d+e x^2}}{d}+\frac {e x^2 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}}{x^3} \, dx\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}-\frac {1}{2} (b n) \int \left (-\frac {\sqrt {d+e x^2}}{d x^3}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2} x}\right ) \, dx\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}+\frac {(b n) \int \frac {\sqrt {d+e x^2}}{x^3} \, dx}{2 d}-\frac {(b e n) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{x} \, dx}{2 d^{3/2}}\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}+\frac {(b n) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x}}{x^2} \, dx,x,x^2\right )}{4 d}-\frac {(b e n) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx,x,x^2\right )}{4 d^{3/2}}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}-\frac {(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^2}\right )}{2 d^{3/2}}+\frac {(b e n) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{8 d}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d x^2}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{3/2}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 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 )}{4 d}+\frac {(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^2}\right )}{2 d^2}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d x^2}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{3/2}}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{3/2}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}+\frac {b e 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 )}{2 d^{3/2}}-\frac {(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^2}\right )}{2 d^2}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d x^2}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{3/2}}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{3/2}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}+\frac {b e 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 )}{2 d^{3/2}}+\frac {(b e 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 )}{2 d^{3/2}}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d x^2}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{3/2}}-\frac {b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{3/2}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{3/2}}+\frac {b e 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 )}{2 d^{3/2}}+\frac {b e n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x^2}}{\sqrt {d}}}\right )}{4 d^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 1.10, size = 229, normalized size = 0.89 \[ \frac {\frac {b n \sqrt {\frac {d}{e x^2}+1} \left (2 d^{3/2} \, _3F_2\left (\frac {3}{2},\frac {3}{2},\frac {3}{2};\frac {5}{2},\frac {5}{2};-\frac {d}{e x^2}\right )+9 e x^2 (2 \log (x)+1) \left (\sqrt {e} x \sinh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {e} x}\right )-\sqrt {d} \sqrt {\frac {d}{e x^2}+1}\right )\right )}{x^2 \sqrt {d+e x^2}}-\frac {18 \sqrt {d} \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{x^2}+18 e \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )-18 e \log (x) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{36 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*n*Sqrt[1 + d/(e*x^2)]*(2*d^(3/2)*HypergeometricPFQ[{3/2, 3/2, 3/2}, {5/2, 5/2}, -(d/(e*x^2))] + 9*e*x^2*(-
(Sqrt[d]*Sqrt[1 + d/(e*x^2)]) + Sqrt[e]*x*ArcSinh[Sqrt[d]/(Sqrt[e]*x)])*(1 + 2*Log[x])))/(x^2*Sqrt[d + e*x^2])
- (18*Sqrt[d]*Sqrt[d + e*x^2]*(a - b*n*Log[x] + b*Log[c*x^n]))/x^2 - 18*e*Log[x]*(a - b*n*Log[x] + b*Log[c*x^
n]) + 18*e*(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(36*d^(3/2))

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} b \log \left (c x^{n}\right ) + \sqrt {e x^{2} + d} a}{e x^{5} + d x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^3/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^3/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\sqrt {e \,x^{2}+d}\, x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^3/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x^n))/(x^3*(d + e*x^2)^(1/2)),x)

[Out]

int((a + b*log(c*x^n))/(x^3*(d + e*x^2)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x**3/(e*x**2+d)**(1/2),x)

[Out]

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

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