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

Optimal. Leaf size=328 \[ \frac {\sqrt {d} \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^{3/2} \sqrt {d+e x^2}}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d+e x^2}}-\frac {b \sqrt {d} n \sqrt {\frac {e x^2}{d}+1} \text {Li}_2\left (e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{3/2} \sqrt {d+e x^2}}+\frac {b \sqrt {d} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 e^{3/2} \sqrt {d+e x^2}}+\frac {b \sqrt {d} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2} \sqrt {d+e x^2}}-\frac {b \sqrt {d} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{e^{3/2} \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2341, 288, 215, 2350, 14, 21, 5659, 3716, 2190, 2279, 2391} \[ -\frac {b \sqrt {d} n \sqrt {\frac {e x^2}{d}+1} \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{3/2} \sqrt {d+e x^2}}+\frac {\sqrt {d} \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^{3/2} \sqrt {d+e x^2}}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d+e x^2}}+\frac {b \sqrt {d} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 e^{3/2} \sqrt {d+e x^2}}+\frac {b \sqrt {d} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2} \sqrt {d+e x^2}}-\frac {b \sqrt {d} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{e^{3/2} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*n*Sqrt[1 + (e*x^2)/d]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/(e^(3/2)*Sqrt[d + e*x^2]) + (b*Sqrt[d]*n*Sqrt[1
+ (e*x^2)/d]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]^2)/(2*e^(3/2)*Sqrt[d + e*x^2]) - (b*Sqrt[d]*n*Sqrt[1 + (e*x^2)/d]*A
rcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[1 - E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/(e^(3/2)*Sqrt[d + e*x^2]) - (x*(a + b
*Log[c*x^n]))/(e*Sqrt[d + e*x^2]) + (Sqrt[d]*Sqrt[1 + (e*x^2)/d]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n
]))/(e^(3/2)*Sqrt[d + e*x^2]) - (b*Sqrt[d]*n*Sqrt[1 + (e*x^2)/d]*PolyLog[2, E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])
])/(2*e^(3/2)*Sqrt[d + e*x^2])

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 215

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 288

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2341

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*x^2)/d)^FracPart[q], Int[x^m*(1 + (e*x^2)/d)^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 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 2391

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 3716

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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5659

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

Rubi steps

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

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Mathematica [C]  time = 0.48, size = 217, normalized size = 0.66 \[ -\frac {b n \sqrt {\frac {e x^2}{d}+1} \left (e^{3/2} x^3 \left (d+e x^2\right ) \, _3F_2\left (\frac {3}{2},\frac {3}{2},\frac {3}{2};\frac {5}{2},\frac {5}{2};-\frac {e x^2}{d}\right )-9 d^{3/2} \log (x) \left (d+e x^2\right ) \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+9 d^2 \sqrt {e} x \log (x) \sqrt {\frac {e x^2}{d}+1}\right )}{9 d e^{3/2} \left (d+e x^2\right )^{3/2}}+\frac {\log \left (\sqrt {e} \sqrt {d+e x^2}+e x\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{e^{3/2}}-\frac {x \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{e \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} b x^{2} \log \left (c x^{n}\right ) + \sqrt {e x^{2} + d} a x^{2}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((sqrt(e*x^2 + d)*b*x^2*log(c*x^n) + sqrt(e*x^2 + d)*a*x^2)/(e^2*x^4 + 2*d*e*x^2 + d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{2}}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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