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

Optimal. Leaf size=259 \[ -\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}+3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-3 b \sqrt {d} e n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )-4 b e n \sqrt {d+e x}-\frac {b d n \sqrt {d+e x}}{x}+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )-6 b \sqrt {d} e n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \]

[Out]

-(e*x+d)^(3/2)*(a+b*ln(c*x^n))/x+3*b*e*n*arctanh((e*x+d)^(1/2)/d^(1/2))*d^(1/2)+3*b*e*n*arctanh((e*x+d)^(1/2)/
d^(1/2))^2*d^(1/2)-3*e*arctanh((e*x+d)^(1/2)/d^(1/2))*(a+b*ln(c*x^n))*d^(1/2)-6*b*e*n*arctanh((e*x+d)^(1/2)/d^
(1/2))*ln(2*d^(1/2)/(d^(1/2)-(e*x+d)^(1/2)))*d^(1/2)-3*b*e*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(e*x+d)^(1/2)))*d^
(1/2)-4*b*e*n*(e*x+d)^(1/2)-b*d*n*(e*x+d)^(1/2)/x+3*e*(a+b*ln(c*x^n))*(e*x+d)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {47, 50, 63, 208, 2350, 14, 5984, 5918, 2402, 2315} \[ -3 b \sqrt {d} e n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}+3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-4 b e n \sqrt {d+e x}-\frac {b d n \sqrt {d+e x}}{x}+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )-6 b \sqrt {d} e n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-4*b*e*n*Sqrt[d + e*x] - (b*d*n*Sqrt[d + e*x])/x + 3*b*Sqrt[d]*e*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 3*b*Sqrt[d
]*e*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]^2 + 3*e*Sqrt[d + e*x]*(a + b*Log[c*x^n]) - ((d + e*x)^(3/2)*(a + b*Log[c*
x^n]))/x - 3*Sqrt[d]*e*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*(a + b*Log[c*x^n]) - 6*b*Sqrt[d]*e*n*ArcTanh[Sqrt[d + e*
x]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])] - 3*b*Sqrt[d]*e*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] -
Sqrt[d + e*x])]

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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 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 {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx &=3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {-(d-2 e x) \sqrt {d+e x}-3 \sqrt {d} e x \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x^2} \, dx\\ &=3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (-\frac {d \sqrt {d+e x}}{x^2}+\frac {2 e \sqrt {d+e x}}{x}-\frac {3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x}\right ) \, dx\\ &=3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )+(b d n) \int \frac {\sqrt {d+e x}}{x^2} \, dx-(2 b e n) \int \frac {\sqrt {d+e x}}{x} \, dx+\left (3 b \sqrt {d} e n\right ) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx\\ &=-4 b e n \sqrt {d+e x}-\frac {b d n \sqrt {d+e x}}{x}+3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )+\left (6 b \sqrt {d} e 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}\right )+\frac {1}{2} (b d e n) \int \frac {1}{x \sqrt {d+e x}} \, dx-(2 b d e n) \int \frac {1}{x \sqrt {d+e x}} \, dx\\ &=-4 b e n \sqrt {d+e x}-\frac {b d n \sqrt {d+e x}}{x}+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2+3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )+(b d n) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )-(4 b d n) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )-(6 b e 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}\right )\\ &=-4 b e n \sqrt {d+e x}-\frac {b d n \sqrt {d+e x}}{x}+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2+3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-6 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )+(6 b e 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}\right )\\ &=-4 b e n \sqrt {d+e x}-\frac {b d n \sqrt {d+e x}}{x}+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2+3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-6 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )-\left (6 b \sqrt {d} e n\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {d+e x}}{\sqrt {d}}}\right )\\ &=-4 b e n \sqrt {d+e x}-\frac {b d n \sqrt {d+e x}}{x}+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+3 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2+3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x}-3 \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-6 b \sqrt {d} e n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )-3 b \sqrt {d} e n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x}}{\sqrt {d}}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 480, normalized size = 1.85 \[ \frac {-4 a d \sqrt {d+e x}+8 a e x \sqrt {d+e x}+6 a \sqrt {d} e x \log \left (\sqrt {d}-\sqrt {d+e x}\right )-6 a \sqrt {d} e x \log \left (\sqrt {d+e x}+\sqrt {d}\right )+6 b \sqrt {d} e x \log \left (c x^n\right ) \log \left (\sqrt {d}-\sqrt {d+e x}\right )-4 b d \sqrt {d+e x} \log \left (c x^n\right )+8 b e x \sqrt {d+e x} \log \left (c x^n\right )-6 b \sqrt {d} e x \log \left (c x^n\right ) \log \left (\sqrt {d+e x}+\sqrt {d}\right )-6 b \sqrt {d} e n x \text {Li}_2\left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )+6 b \sqrt {d} e n x \text {Li}_2\left (\frac {1}{2} \left (\frac {\sqrt {d+e x}}{\sqrt {d}}+1\right )\right )-4 b d n \sqrt {d+e x}-16 b e n x \sqrt {d+e x}-3 b \sqrt {d} e n x \log ^2\left (\sqrt {d}-\sqrt {d+e x}\right )+3 b \sqrt {d} e n x \log ^2\left (\sqrt {d+e x}+\sqrt {d}\right )-6 b \sqrt {d} e n x \log \left (\frac {1}{2} \left (\frac {\sqrt {d+e x}}{\sqrt {d}}+1\right )\right ) \log \left (\sqrt {d}-\sqrt {d+e x}\right )+6 b \sqrt {d} e n x \log \left (\sqrt {d+e x}+\sqrt {d}\right ) \log \left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )+12 b \sqrt {d} e n x \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a*d*Sqrt[d + e*x] - 4*b*d*n*Sqrt[d + e*x] + 8*a*e*x*Sqrt[d + e*x] - 16*b*e*n*x*Sqrt[d + e*x] + 12*b*Sqrt[d
]*e*n*x*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] - 4*b*d*Sqrt[d + e*x]*Log[c*x^n] + 8*b*e*x*Sqrt[d + e*x]*Log[c*x^n] + 6
*a*Sqrt[d]*e*x*Log[Sqrt[d] - Sqrt[d + e*x]] + 6*b*Sqrt[d]*e*x*Log[c*x^n]*Log[Sqrt[d] - Sqrt[d + e*x]] - 3*b*Sq
rt[d]*e*n*x*Log[Sqrt[d] - Sqrt[d + e*x]]^2 - 6*a*Sqrt[d]*e*x*Log[Sqrt[d] + Sqrt[d + e*x]] - 6*b*Sqrt[d]*e*x*Lo
g[c*x^n]*Log[Sqrt[d] + Sqrt[d + e*x]] + 3*b*Sqrt[d]*e*n*x*Log[Sqrt[d] + Sqrt[d + e*x]]^2 + 6*b*Sqrt[d]*e*n*x*L
og[Sqrt[d] + Sqrt[d + e*x]]*Log[1/2 - Sqrt[d + e*x]/(2*Sqrt[d])] - 6*b*Sqrt[d]*e*n*x*Log[Sqrt[d] - Sqrt[d + e*
x]]*Log[(1 + Sqrt[d + e*x]/Sqrt[d])/2] - 6*b*Sqrt[d]*e*n*x*PolyLog[2, 1/2 - Sqrt[d + e*x]/(2*Sqrt[d])] + 6*b*S
qrt[d]*e*n*x*PolyLog[2, (1 + Sqrt[d + e*x]/Sqrt[d])/2])/(4*x)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b e x + b d\right )} \sqrt {e x + d} \log \left (c x^{n}\right ) + {\left (a e x + a d\right )} \sqrt {e x + d}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{\frac {3}{2}} \left (b \ln \left (c \,x^{n}\right )+a \right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (3 \, \sqrt {d} e \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right ) + 4 \, \sqrt {e x + d} e - \frac {2 \, \sqrt {e x + d} d}{x}\right )} a + b \int \frac {{\left (e x \log \relax (c) + d \log \relax (c) + {\left (e x + d\right )} \log \left (x^{n}\right )\right )} \sqrt {e x + d}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*sqrt(d)*e*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d))) + 4*sqrt(e*x + d)*e - 2*sqrt(e*x + d
)*d/x)*a + b*integrate((e*x*log(c) + d*log(c) + (e*x + d)*log(x^n))*sqrt(e*x + d)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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