3.5.33 \(\int \frac {(d+e x^r)^{5/2} (a+b \log (c x^n))}{x} \, dx\) [433]

Optimal. Leaf size=327 \[ -\frac {92 b d^2 n \sqrt {d+e x^r}}{15 r^2}-\frac {32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac {92 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 r^2}+\frac {2 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}+\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2}-\frac {2 b d^{5/2} n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2} \]

[Out]

-32/45*b*d*n*(d+e*x^r)^(3/2)/r^2-4/25*b*n*(d+e*x^r)^(5/2)/r^2+92/15*b*d^(5/2)*n*arctanh((d+e*x^r)^(1/2)/d^(1/2
))/r^2+2*b*d^(5/2)*n*arctanh((d+e*x^r)^(1/2)/d^(1/2))^2/r^2-4*b*d^(5/2)*n*arctanh((d+e*x^r)^(1/2)/d^(1/2))*ln(
2*d^(1/2)/(d^(1/2)-(d+e*x^r)^(1/2)))/r^2-2*b*d^(5/2)*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(d+e*x^r)^(1/2)))/r^2-92
/15*b*d^2*n*(d+e*x^r)^(1/2)/r^2+2/15*(a+b*ln(c*x^n))*(5*d*(d+e*x^r)^(3/2)/r+3*(d+e*x^r)^(5/2)/r-15*d^(5/2)*arc
tanh((d+e*x^r)^(1/2)/d^(1/2))/r+15*d^2*(d+e*x^r)^(1/2)/r)

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Rubi [A]
time = 0.34, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {272, 52, 65, 214, 2390, 6131, 6055, 2449, 2352} \begin {gather*} -\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^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}+\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}\right ) \left (a+b \log \left (c x^n\right )\right )+\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 {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} \end {gather*}

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 52

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 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2390

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 2449

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 6055

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 6131

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 {\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) \text {Subst}\left (\int \frac {(d+e x)^{5/2}}{x} \, dx,x,x^r\right )}{5 r^2}-\frac {(2 b d n) \text {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 ) \text {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 ) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [F]
time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (d +e \,x^{r}\right )^{\frac {5}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )}{x}\, dx\]

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

[Out]

1/15*(15*d^(5/2)*log((sqrt(x^r*e + d) - sqrt(d))/(sqrt(x^r*e + d) + sqrt(d)))/r + 2*(3*(x^r*e + d)^(5/2) + 5*(
x^r*e + d)^(3/2)*d + 15*sqrt(x^r*e + d)*d^2)/r)*a + b*integrate((d^2*log(c) + 2*d*e^(r*log(x) + 1)*log(c) + e^
(2*r*log(x) + 2)*log(c) + (d^2 + 2*d*e^(r*log(x) + 1) + e^(2*r*log(x) + 2))*log(x^n))*sqrt(d + e^(r*log(x) + 1
))/x, x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 6189 deep

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x^r\right )}^{5/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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