3.439 \(\int \frac {a+b \log (c x^n)}{x (d+e x^r)^{7/2}} \, dx\)

Optimal. Leaf size=314 \[ \frac {2}{15} \left (-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}+\frac {15}{d^3 r \sqrt {d+e x^r}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {3}{d r \left (d+e x^r\right )^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {2 b n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^r+d}}\right )}{d^{7/2} r^2}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{d^{7/2} r^2}+\frac {92 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 d^{7/2} r^2}-\frac {4 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r^2}-\frac {32 b n}{15 d^3 r^2 \sqrt {d+e x^r}}-\frac {4 b n}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}} \]

[Out]

-4/15*b*n/d^2/r^2/(d+e*x^r)^(3/2)+92/15*b*n*arctanh((d+e*x^r)^(1/2)/d^(1/2))/d^(7/2)/r^2+2*b*n*arctanh((d+e*x^
r)^(1/2)/d^(1/2))^2/d^(7/2)/r^2-4*b*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)))
/d^(7/2)/r^2-2*b*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(d+e*x^r)^(1/2)))/d^(7/2)/r^2+2/15*(a+b*ln(c*x^n))*(3/d/r/(d
+e*x^r)^(5/2)+5/d^2/r/(d+e*x^r)^(3/2)-15*arctanh((d+e*x^r)^(1/2)/d^(1/2))/d^(7/2)/r+15/d^3/r/(d+e*x^r)^(1/2))-
32/15*b*n/d^3/r^2/(d+e*x^r)^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {266, 51, 63, 208, 2348, 5984, 5918, 2402, 2315} \[ -\frac {2 b n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{d^{7/2} r^2}+\frac {2}{15} \left (\frac {15}{d^3 r \sqrt {d+e x^r}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}+\frac {3}{d r \left (d+e x^r\right )^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {32 b n}{15 d^3 r^2 \sqrt {d+e x^r}}-\frac {4 b n}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{d^{7/2} r^2}+\frac {92 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 d^{7/2} r^2}-\frac {4 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x*(d + e*x^r)^(7/2)),x]

[Out]

(-4*b*n)/(15*d^2*r^2*(d + e*x^r)^(3/2)) - (32*b*n)/(15*d^3*r^2*Sqrt[d + e*x^r]) + (92*b*n*ArcTanh[Sqrt[d + e*x
^r]/Sqrt[d]])/(15*d^(7/2)*r^2) + (2*b*n*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]]^2)/(d^(7/2)*r^2) + (2*(3/(d*r*(d + e*
x^r)^(5/2)) + 5/(d^2*r*(d + e*x^r)^(3/2)) + 15/(d^3*r*Sqrt[d + e*x^r]) - (15*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]])
/(d^(7/2)*r))*(a + b*Log[c*x^n]))/15 - (4*b*n*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt
[d + e*x^r])])/(d^(7/2)*r^2) - (2*b*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^r])])/(d^(7/2)*r^2)

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 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 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 \left (d+e x^r\right )^{7/2}} \, dx &=\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {2}{5 d r x \left (d+e x^r\right )^{5/2}}+\frac {2}{3 d^2 r x \left (d+e x^r\right )^{3/2}}+\frac {2}{d^3 r x \sqrt {d+e x^r}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r x}\right ) \, dx\\ &=\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(2 b n) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{x} \, dx}{d^{7/2} r}-\frac {(2 b n) \int \frac {1}{x \sqrt {d+e x^r}} \, dx}{d^3 r}-\frac {(2 b n) \int \frac {1}{x \left (d+e x^r\right )^{3/2}} \, dx}{3 d^2 r}-\frac {(2 b n) \int \frac {1}{x \left (d+e x^r\right )^{5/2}} \, dx}{5 d r}\\ &=\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(2 b n) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx,x,x^r\right )}{d^{7/2} r^2}-\frac {(2 b n) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^r\right )}{d^3 r^2}-\frac {(2 b n) \operatorname {Subst}\left (\int \frac {1}{x (d+e x)^{3/2}} \, dx,x,x^r\right )}{3 d^2 r^2}-\frac {(2 b n) \operatorname {Subst}\left (\int \frac {1}{x (d+e x)^{5/2}} \, dx,x,x^r\right )}{5 d r^2}\\ &=-\frac {4 b n}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}}-\frac {4 b n}{3 d^3 r^2 \sqrt {d+e x^r}}+\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(4 b n) \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 )}{d^{7/2} r^2}-\frac {(2 b n) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^r\right )}{3 d^3 r^2}-\frac {(2 b n) \operatorname {Subst}\left (\int \frac {1}{x (d+e x)^{3/2}} \, dx,x,x^r\right )}{5 d^2 r^2}-\frac {(4 b n) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^r}\right )}{d^3 e r^2}\\ &=-\frac {4 b n}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}}-\frac {32 b n}{15 d^3 r^2 \sqrt {d+e x^r}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r^2}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{d^{7/2} r^2}+\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(4 b 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^r}\right )}{d^4 r^2}-\frac {(2 b n) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^r\right )}{5 d^3 r^2}-\frac {(4 b n) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^r}\right )}{3 d^3 e r^2}\\ &=-\frac {4 b n}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}}-\frac {32 b n}{15 d^3 r^2 \sqrt {d+e x^r}}+\frac {16 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{3 d^{7/2} r^2}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{d^{7/2} r^2}+\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b 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 )}{d^{7/2} r^2}+\frac {(4 b 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^r}\right )}{d^4 r^2}-\frac {(4 b n) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^r}\right )}{5 d^3 e r^2}\\ &=-\frac {4 b n}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}}-\frac {32 b n}{15 d^3 r^2 \sqrt {d+e x^r}}+\frac {92 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 d^{7/2} r^2}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{d^{7/2} r^2}+\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b 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 )}{d^{7/2} r^2}-\frac {(4 b n) \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 )}{d^{7/2} r^2}\\ &=-\frac {4 b n}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}}-\frac {32 b n}{15 d^3 r^2 \sqrt {d+e x^r}}+\frac {92 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 d^{7/2} r^2}+\frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{d^{7/2} r^2}+\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b 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 )}{d^{7/2} r^2}-\frac {2 b n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x^r}}{\sqrt {d}}}\right )}{d^{7/2} r^2}\\ \end {align*}

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Mathematica [F]  time = 0.42, size = 0, normalized size = 0.00 \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Log[c*x^n])/(x*(d + e*x^r)^(7/2)),x]

[Out]

Integrate[(a + b*Log[c*x^n])/(x*(d + e*x^r)^(7/2)), x]

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(d+e*x^r)^(7/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{r} + d\right )}^{\frac {7}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(d+e*x^r)^(7/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((e*x^r + d)^(7/2)*x), x)

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maple [F]  time = 0.42, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\left (e \,x^{r}+d \right )^{\frac {7}{2}} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*x^n)+a)/x/(e*x^r+d)^(7/2),x)

[Out]

int((b*ln(c*x^n)+a)/x/(e*x^r+d)^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{15} \, a {\left (\frac {15 \, \log \left (\frac {\sqrt {e x^{r} + d} - \sqrt {d}}{\sqrt {e x^{r} + d} + \sqrt {d}}\right )}{d^{\frac {7}{2}} r} + \frac {2 \, {\left (15 \, {\left (e x^{r} + d\right )}^{2} + 5 \, {\left (e x^{r} + d\right )} d + 3 \, d^{2}\right )}}{{\left (e x^{r} + d\right )}^{\frac {5}{2}} d^{3} r}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{{\left (e^{3} x x^{3 \, r} + 3 \, d e^{2} x x^{2 \, r} + 3 \, d^{2} e x x^{r} + d^{3} x\right )} \sqrt {e x^{r} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(d+e*x^r)^(7/2),x, algorithm="maxima")

[Out]

1/15*a*(15*log((sqrt(e*x^r + d) - sqrt(d))/(sqrt(e*x^r + d) + sqrt(d)))/(d^(7/2)*r) + 2*(15*(e*x^r + d)^2 + 5*
(e*x^r + d)*d + 3*d^2)/((e*x^r + d)^(5/2)*d^3*r)) + b*integrate((log(c) + log(x^n))/((e^3*x*x^(3*r) + 3*d*e^2*
x*x^(2*r) + 3*d^2*e*x*x^r + d^3*x)*sqrt(e*x^r + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,{\left (d+e\,x^r\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x^n))/(x*(d + e*x^r)^(7/2)),x)

[Out]

int((a + b*log(c*x^n))/(x*(d + e*x^r)^(7/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x/(d+e*x**r)**(7/2),x)

[Out]

Timed out

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