3.3.59 \(\int \frac {\sqrt {d+e x^2} (a+b \log (c x^n))}{x^2} \, dx\) [259]

Optimal. Leaf size=345 \[ -\frac {b n \sqrt {d+e x^2}}{x}+\frac {b \sqrt {e} n \sqrt {d+e x^2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {1+\frac {e x^2}{d}}}+\frac {b \sqrt {e} n \sqrt {d+e x^2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 \sqrt {d} \sqrt {1+\frac {e x^2}{d}}}-\frac {b \sqrt {e} n \sqrt {d+e x^2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{\sqrt {d} \sqrt {1+\frac {e x^2}{d}}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {\sqrt {e} \sqrt {d+e x^2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} \sqrt {1+\frac {e x^2}{d}}}-\frac {b \sqrt {e} n \sqrt {d+e x^2} \text {Li}_2\left (e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 \sqrt {d} \sqrt {1+\frac {e x^2}{d}}} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2386, 283, 221, 2392, 14, 5775, 3797, 2221, 2317, 2438} \begin {gather*} -\frac {b \sqrt {e} n \sqrt {d+e x^2} \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 \sqrt {d} \sqrt {\frac {e x^2}{d}+1}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {\sqrt {e} \sqrt {d+e x^2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} \sqrt {\frac {e x^2}{d}+1}}-\frac {b n \sqrt {d+e x^2}}{x}+\frac {b \sqrt {e} n \sqrt {d+e x^2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 \sqrt {d} \sqrt {\frac {e x^2}{d}+1}}+\frac {b \sqrt {e} n \sqrt {d+e x^2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {\frac {e x^2}{d}+1}}-\frac {b \sqrt {e} n \sqrt {d+e x^2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{\sqrt {d} \sqrt {\frac {e x^2}{d}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*n*Sqrt[d + e*x^2])/x) + (b*Sqrt[e]*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*Sqrt[1 + (e*x
^2)/d]) + (b*Sqrt[e]*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]^2)/(2*Sqrt[d]*Sqrt[1 + (e*x^2)/d]) - (b*Sq
rt[e]*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[1 - E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/(Sqrt[d]*Sqr
t[1 + (e*x^2)/d]) - (Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/x + (Sqrt[e]*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt
[d]]*(a + b*Log[c*x^n]))/(Sqrt[d]*Sqrt[1 + (e*x^2)/d]) - (b*Sqrt[e]*n*Sqrt[d + e*x^2]*PolyLog[2, E^(2*ArcSinh[
(Sqrt[e]*x)/Sqrt[d]])])/(2*Sqrt[d]*Sqrt[1 + (e*x^2)/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 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2386

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^IntPart[q
]*((d + e*x^2)^FracPart[q]/(1 + (e/d)*x^2)^FracPart[q]), Int[x^m*(1 + (e/d)*x^2)^q*(a + b*Log[c*x^n]), x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[m/2] && IntegerQ[q - 1/2] &&  !(LtQ[m + 2*q, -2] || GtQ[d, 0])

Rule 2392

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 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.38, size = 183, normalized size = 0.53 \begin {gather*} \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 {e x^2}{d}\right )-\sqrt {1+\frac {e x^2}{d}} \log (x)+\frac {\sqrt {e} x \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)}{\sqrt {d}}\right )}{x \sqrt {1+\frac {e x^2}{d}}}-\frac {\sqrt {d+e x^2} \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{x}+\sqrt {e} \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^2,x, algorithm="fricas")

[Out]

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

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

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**2,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^2,x, algorithm="giac")

[Out]

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

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

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^2,x)

[Out]

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

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