3.3.91 \(\int \frac {a+b \log (c x^n)}{x^3 (d+e x^2)^{3/2}} \, dx\) [291]

Optimal. Leaf size=287 \[ -\frac {b n \sqrt {d+e x^2}}{4 d^2 x^2}-\frac {5 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{5/2}}-\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{5/2}}-\frac {3 e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \sqrt {d+e x^2}}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{5/2}}+\frac {3 b e n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{4 d^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {272, 44, 53, 65, 214, 2392, 457, 79, 6131, 6055, 2449, 2352} \begin {gather*} \frac {3 b e n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{4 d^{5/2}}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}-\frac {3 e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \sqrt {d+e x^2}}-\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{5/2}}-\frac {5 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{5/2}}+\frac {3 b e n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{2 d^{5/2}}-\frac {b n \sqrt {d+e x^2}}{4 d^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(b*n*Sqrt[d + e*x^2])/(d^2*x^2) - (5*b*e*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(4*d^(5/2)) - (3*b*e*n*ArcTa
nh[Sqrt[d + e*x^2]/Sqrt[d]]^2)/(4*d^(5/2)) - (3*e*(a + b*Log[c*x^n]))/(2*d^2*Sqrt[d + e*x^2]) - (a + b*Log[c*x
^n])/(2*d*x^2*Sqrt[d + e*x^2]) + (3*e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*(a + b*Log[c*x^n]))/(2*d^(5/2)) + (3*b*
e*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*d^(5/2)) + (3*b*e*n*Poly
Log[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(4*d^(5/2))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && 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 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 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 {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{3/2}} \, dx &=\frac {a+b \log \left (c x^n\right )}{d x^2 \sqrt {d+e x^2}}-\frac {3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}-(b n) \int \left (-\frac {d+3 e x^2}{2 d^2 x^3 \sqrt {d+e x^2}}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{2 d^{5/2} x}\right ) \, dx\\ &=\frac {a+b \log \left (c x^n\right )}{d x^2 \sqrt {d+e x^2}}-\frac {3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac {(b n) \int \frac {d+3 e x^2}{x^3 \sqrt {d+e x^2}} \, dx}{2 d^2}-\frac {(3 b e n) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{x} \, dx}{2 d^{5/2}}\\ &=\frac {a+b \log \left (c x^n\right )}{d x^2 \sqrt {d+e x^2}}-\frac {3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac {(b n) \text {Subst}\left (\int \frac {d+3 e x}{x^2 \sqrt {d+e x}} \, dx,x,x^2\right )}{4 d^2}-\frac {(3 b e n) \text {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx,x,x^2\right )}{4 d^{5/2}}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d^2 x^2}+\frac {a+b \log \left (c x^n\right )}{d x^2 \sqrt {d+e x^2}}-\frac {3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}-\frac {(3 b e n) \text {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{-d+x^2} \, dx,x,\sqrt {d+e x^2}\right )}{2 d^{5/2}}+\frac {(5 b e n) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{8 d^2}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d^2 x^2}-\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{5/2}}+\frac {a+b \log \left (c x^n\right )}{d x^2 \sqrt {d+e x^2}}-\frac {3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac {(5 b n) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{4 d^2}+\frac {(3 b e n) \text {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{1-\frac {x}{\sqrt {d}}} \, dx,x,\sqrt {d+e x^2}\right )}{2 d^3}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d^2 x^2}-\frac {5 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{5/2}}-\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{5/2}}+\frac {a+b \log \left (c x^n\right )}{d x^2 \sqrt {d+e x^2}}-\frac {3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{5/2}}-\frac {(3 b e n) \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^2}\right )}{2 d^3}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d^2 x^2}-\frac {5 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{5/2}}-\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{5/2}}+\frac {a+b \log \left (c x^n\right )}{d x^2 \sqrt {d+e x^2}}-\frac {3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{5/2}}+\frac {(3 b e n) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {d+e x^2}}{\sqrt {d}}}\right )}{2 d^{5/2}}\\ &=-\frac {b n \sqrt {d+e x^2}}{4 d^2 x^2}-\frac {5 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 d^{5/2}}-\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{5/2}}+\frac {a+b \log \left (c x^n\right )}{d x^2 \sqrt {d+e x^2}}-\frac {3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac {3 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{5/2}}+\frac {3 b e n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x^2}}{\sqrt {d}}}\right )}{4 d^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.21, size = 218, normalized size = 0.76 \begin {gather*} \frac {3 b d^{5/2} n \sqrt {1+\frac {d}{e x^2}} \, _3F_2\left (\frac {5}{2},\frac {5}{2},\frac {5}{2};\frac {7}{2},\frac {7}{2};-\frac {d}{e x^2}\right )-5 b d^{5/2} n \sqrt {1+\frac {d}{e x^2}} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-\frac {d}{e x^2}\right ) (1+2 \log (x))-25 e x^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \left (\sqrt {d} \left (d+3 e x^2\right )+3 e x^2 \sqrt {d+e x^2} \log (x)-3 e x^2 \sqrt {d+e x^2} \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )\right )}{50 d^{5/2} e x^4 \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*d^(5/2)*n*Sqrt[1 + d/(e*x^2)]*HypergeometricPFQ[{5/2, 5/2, 5/2}, {7/2, 7/2}, -(d/(e*x^2))] - 5*b*d^(5/2)*
n*Sqrt[1 + d/(e*x^2)]*Hypergeometric2F1[3/2, 5/2, 7/2, -(d/(e*x^2))]*(1 + 2*Log[x]) - 25*e*x^2*(a - b*n*Log[x]
+ b*Log[c*x^n])*(Sqrt[d]*(d + 3*e*x^2) + 3*e*x^2*Sqrt[d + e*x^2]*Log[x] - 3*e*x^2*Sqrt[d + e*x^2]*Log[d + Sqr
t[d]*Sqrt[d + e*x^2]]))/(50*d^(5/2)*e*x^4*Sqrt[d + e*x^2])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

1/2*a*(3*arcsinh(sqrt(d)*e^(-1/2)/abs(x))*e/d^(5/2) - 3*e/(sqrt(x^2*e + d)*d^2) - 1/(sqrt(x^2*e + d)*d*x^2)) +
b*integrate((log(c) + log(x^n))/((x^5*e + d*x^3)*sqrt(x^2*e + d)), 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))/x^3/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

integral((sqrt(x^2*e + d)*b*log(c*x^n) + sqrt(x^2*e + d)*a)/(x^7*e^2 + 2*d*x^5*e + d^2*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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