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

Optimal. Leaf size=327 \[ -\frac{2 b d^{5/2} n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^r}}\right )}{r^2}+\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{92 b d^2 n \sqrt{d+e x^r}}{15 r^2}+\frac{2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )^2}{r^2}+\frac{92 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{15 r^2}-\frac{4 b d^{5/2} n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^r}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r^2}-\frac{32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac{4 b n \left (d+e x^r\right )^{5/2}}{25 r^2} \]

[Out]

(-92*b*d^2*n*Sqrt[d + e*x^r])/(15*r^2) - (32*b*d*n*(d + e*x^r)^(3/2))/(45*r^2) - (4*b*n*(d + e*x^r)^(5/2))/(25
*r^2) + (92*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]])/(15*r^2) + (2*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^r]/Sq
rt[d]]^2)/r^2 + (2*((15*d^2*Sqrt[d + e*x^r])/r + (5*d*(d + e*x^r)^(3/2))/r + (3*(d + e*x^r)^(5/2))/r - (15*d^(
5/2)*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]])/r)*(a + b*Log[c*x^n]))/15 - (4*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^r]/Sqrt
[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^r])])/r^2 - (2*b*d^(5/2)*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] -
Sqrt[d + e*x^r])])/r^2

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Rubi [A]  time = 0.480287, antiderivative size = 327, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {266, 50, 63, 208, 2348, 5984, 5918, 2402, 2315} \[ -\frac{2 b d^{5/2} n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^r}}\right )}{r^2}+\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{92 b d^2 n \sqrt{d+e x^r}}{15 r^2}+\frac{2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )^2}{r^2}+\frac{92 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{15 r^2}-\frac{4 b d^{5/2} n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^r}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r^2}-\frac{32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac{4 b n \left (d+e x^r\right )^{5/2}}{25 r^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-92*b*d^2*n*Sqrt[d + e*x^r])/(15*r^2) - (32*b*d*n*(d + e*x^r)^(3/2))/(45*r^2) - (4*b*n*(d + e*x^r)^(5/2))/(25
*r^2) + (92*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]])/(15*r^2) + (2*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^r]/Sq
rt[d]]^2)/r^2 + (2*((15*d^2*Sqrt[d + e*x^r])/r + (5*d*(d + e*x^r)^(3/2))/r + (3*(d + e*x^r)^(5/2))/r - (15*d^(
5/2)*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]])/r)*(a + b*Log[c*x^n]))/15 - (4*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^r]/Sqrt
[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^r])])/r^2 - (2*b*d^(5/2)*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] -
Sqrt[d + e*x^r])])/r^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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx &=\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac{2 d^2 \sqrt{d+e x^r}}{r x}+\frac{2 d \left (d+e x^r\right )^{3/2}}{3 r x}+\frac{2 \left (d+e x^r\right )^{5/2}}{5 r x}-\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r x}\right ) \, dx\\ &=\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{(2 b n) \int \frac{\left (d+e x^r\right )^{5/2}}{x} \, dx}{5 r}-\frac{(2 b d n) \int \frac{\left (d+e x^r\right )^{3/2}}{x} \, dx}{3 r}-\frac{\left (2 b d^2 n\right ) \int \frac{\sqrt{d+e x^r}}{x} \, dx}{r}+\frac{\left (2 b d^{5/2} n\right ) \int \frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{x} \, dx}{r}\\ &=\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{(2 b n) \operatorname{Subst}\left (\int \frac{(d+e x)^{5/2}}{x} \, dx,x,x^r\right )}{5 r^2}-\frac{(2 b d n) \operatorname{Subst}\left (\int \frac{(d+e x)^{3/2}}{x} \, dx,x,x^r\right )}{3 r^2}-\frac{\left (2 b d^2 n\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x}}{x} \, dx,x,x^r\right )}{r^2}+\frac{\left (2 b d^{5/2} n\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{x} \, dx,x,x^r\right )}{r^2}\\ &=-\frac{4 b d^2 n \sqrt{d+e x^r}}{r^2}-\frac{4 b d n \left (d+e x^r\right )^{3/2}}{9 r^2}-\frac{4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{(2 b d n) \operatorname{Subst}\left (\int \frac{(d+e x)^{3/2}}{x} \, dx,x,x^r\right )}{5 r^2}-\frac{\left (2 b d^2 n\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x}}{x} \, dx,x,x^r\right )}{3 r^2}+\frac{\left (4 b d^{5/2} n\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}\left (\frac{x}{\sqrt{d}}\right )}{-d+x^2} \, dx,x,\sqrt{d+e x^r}\right )}{r^2}-\frac{\left (2 b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^r\right )}{r^2}\\ &=-\frac{16 b d^2 n \sqrt{d+e x^r}}{3 r^2}-\frac{32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac{4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac{2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )^2}{r^2}+\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{\left (2 b d^2 n\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x}}{x} \, dx,x,x^r\right )}{5 r^2}-\frac{\left (4 b d^2 n\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{x}{\sqrt{d}}\right )}{1-\frac{x}{\sqrt{d}}} \, dx,x,\sqrt{d+e x^r}\right )}{r^2}-\frac{\left (2 b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^r\right )}{3 r^2}-\frac{\left (4 b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^r}\right )}{e r^2}\\ &=-\frac{92 b d^2 n \sqrt{d+e x^r}}{15 r^2}-\frac{32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac{4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac{4 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r^2}+\frac{2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )^2}{r^2}+\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{4 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right ) \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^r}}\right )}{r^2}+\frac{\left (4 b d^2 n\right ) \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^r}\right )}{r^2}-\frac{\left (2 b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^r\right )}{5 r^2}-\frac{\left (4 b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^r}\right )}{3 e r^2}\\ &=-\frac{92 b d^2 n \sqrt{d+e x^r}}{15 r^2}-\frac{32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac{4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac{16 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{3 r^2}+\frac{2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )^2}{r^2}+\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{4 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right ) \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^r}}\right )}{r^2}-\frac{\left (4 b d^{5/2} n\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-\frac{\sqrt{d+e x^r}}{\sqrt{d}}}\right )}{r^2}-\frac{\left (4 b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^r}\right )}{5 e r^2}\\ &=-\frac{92 b d^2 n \sqrt{d+e x^r}}{15 r^2}-\frac{32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac{4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac{92 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{15 r^2}+\frac{2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )^2}{r^2}+\frac{2}{15} \left (\frac{15 d^2 \sqrt{d+e x^r}}{r}+\frac{5 d \left (d+e x^r\right )^{3/2}}{r}+\frac{3 \left (d+e x^r\right )^{5/2}}{r}-\frac{15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{4 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^r}}{\sqrt{d}}\right ) \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^r}}\right )}{r^2}-\frac{2 b d^{5/2} n \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{d+e x^r}}{\sqrt{d}}}\right )}{r^2}\\ \end{align*}

Mathematica [F]  time = 0.53774, size = 0, normalized size = 0. \[ \int \frac{\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.769, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{x} \left ( d+e{x}^{r} \right ) ^{{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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