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

Optimal. Leaf size=341 \[ \frac {b d^4 n \log \left (d+\frac {e}{\sqrt {x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )}{e^4}-\frac {4 b d^3 n \left (d+\frac {e}{\sqrt {x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )}{e^4}+\frac {3 b d^2 n \left (d+\frac {e}{\sqrt {x}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )}{e^4}-\frac {4 b d n \left (d+\frac {e}{\sqrt {x}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )}{3 e^4}+\frac {b n \left (d+\frac {e}{\sqrt {x}}\right )^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )}{4 e^4}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}-\frac {b^2 d^4 n^2 \log ^2\left (d+\frac {e}{\sqrt {x}}\right )}{2 e^4}+\frac {4 b^2 d^3 n^2}{e^3 \sqrt {x}}-\frac {3 b^2 d^2 n^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{2 e^4}+\frac {4 b^2 d n^2 \left (d+\frac {e}{\sqrt {x}}\right )^3}{9 e^4}-\frac {b^2 n^2 \left (d+\frac {e}{\sqrt {x}}\right )^4}{16 e^4} \]

[Out]

-1/2*b^2*d^4*n^2*ln(d+e/x^(1/2))^2/e^4+b*d^4*n*ln(d+e/x^(1/2))*(a+b*ln(c*(d+e/x^(1/2))^n))/e^4-1/2*(a+b*ln(c*(
d+e/x^(1/2))^n))^2/x^2-4*b*d^3*n*(a+b*ln(c*(d+e/x^(1/2))^n))*(d+e/x^(1/2))/e^4-3/2*b^2*d^2*n^2*(d+e/x^(1/2))^2
/e^4+3*b*d^2*n*(a+b*ln(c*(d+e/x^(1/2))^n))*(d+e/x^(1/2))^2/e^4+4/9*b^2*d*n^2*(d+e/x^(1/2))^3/e^4-4/3*b*d*n*(a+
b*ln(c*(d+e/x^(1/2))^n))*(d+e/x^(1/2))^3/e^4-1/16*b^2*n^2*(d+e/x^(1/2))^4/e^4+1/4*b*n*(a+b*ln(c*(d+e/x^(1/2))^
n))*(d+e/x^(1/2))^4/e^4+4*b^2*d^3*n^2/e^3/x^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 263, normalized size of antiderivative = 0.77, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2454, 2398, 2411, 43, 2334, 12, 14, 2301} \[ -\frac {1}{12} b n \left (\frac {48 d^3 \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}-\frac {36 d^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{e^4}-\frac {12 d^4 \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}+\frac {16 d \left (d+\frac {e}{\sqrt {x}}\right )^3}{e^4}-\frac {3 \left (d+\frac {e}{\sqrt {x}}\right )^4}{e^4}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}+\frac {4 b^2 d^3 n^2}{e^3 \sqrt {x}}-\frac {3 b^2 d^2 n^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{2 e^4}-\frac {b^2 d^4 n^2 \log ^2\left (d+\frac {e}{\sqrt {x}}\right )}{2 e^4}+\frac {4 b^2 d n^2 \left (d+\frac {e}{\sqrt {x}}\right )^3}{9 e^4}-\frac {b^2 n^2 \left (d+\frac {e}{\sqrt {x}}\right )^4}{16 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^2*d^2*n^2*(d + e/Sqrt[x])^2)/(2*e^4) + (4*b^2*d*n^2*(d + e/Sqrt[x])^3)/(9*e^4) - (b^2*n^2*(d + e/Sqrt[x]
)^4)/(16*e^4) + (4*b^2*d^3*n^2)/(e^3*Sqrt[x]) - (b^2*d^4*n^2*Log[d + e/Sqrt[x]]^2)/(2*e^4) - (b*n*((48*d^3*(d
+ e/Sqrt[x]))/e^4 - (36*d^2*(d + e/Sqrt[x])^2)/e^4 + (16*d*(d + e/Sqrt[x])^3)/e^4 - (3*(d + e/Sqrt[x])^4)/e^4
- (12*d^4*Log[d + e/Sqrt[x]])/e^4)*(a + b*Log[c*(d + e/Sqrt[x])^n]))/12 - (a + b*Log[c*(d + e/Sqrt[x])^n])^2/(
2*x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2398

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
f + g*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(g*(q + 1)), x] - Dist[(b*e*n*p)/(g*(q + 1)), Int[((f + g*x)^(q
+ 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{x^3} \, dx &=-\left (2 \operatorname {Subst}\left (\int x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}+(b e n) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{d+e x} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}+(b n) \operatorname {Subst}\left (\int \frac {\left (-\frac {d}{e}+\frac {x}{e}\right )^4 \left (a+b \log \left (c x^n\right )\right )}{x} \, dx,x,d+\frac {e}{\sqrt {x}}\right )\\ &=-\frac {1}{12} b n \left (\frac {48 d^3 \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}-\frac {36 d^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{e^4}+\frac {16 d \left (d+\frac {e}{\sqrt {x}}\right )^3}{e^4}-\frac {3 \left (d+\frac {e}{\sqrt {x}}\right )^4}{e^4}-\frac {12 d^4 \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}-\left (b^2 n^2\right ) \operatorname {Subst}\left (\int \frac {x \left (-48 d^3+36 d^2 x-16 d x^2+3 x^3\right )+12 d^4 \log (x)}{12 e^4 x} \, dx,x,d+\frac {e}{\sqrt {x}}\right )\\ &=-\frac {1}{12} b n \left (\frac {48 d^3 \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}-\frac {36 d^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{e^4}+\frac {16 d \left (d+\frac {e}{\sqrt {x}}\right )^3}{e^4}-\frac {3 \left (d+\frac {e}{\sqrt {x}}\right )^4}{e^4}-\frac {12 d^4 \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}-\frac {\left (b^2 n^2\right ) \operatorname {Subst}\left (\int \frac {x \left (-48 d^3+36 d^2 x-16 d x^2+3 x^3\right )+12 d^4 \log (x)}{x} \, dx,x,d+\frac {e}{\sqrt {x}}\right )}{12 e^4}\\ &=-\frac {1}{12} b n \left (\frac {48 d^3 \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}-\frac {36 d^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{e^4}+\frac {16 d \left (d+\frac {e}{\sqrt {x}}\right )^3}{e^4}-\frac {3 \left (d+\frac {e}{\sqrt {x}}\right )^4}{e^4}-\frac {12 d^4 \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}-\frac {\left (b^2 n^2\right ) \operatorname {Subst}\left (\int \left (-48 d^3+36 d^2 x-16 d x^2+3 x^3+\frac {12 d^4 \log (x)}{x}\right ) \, dx,x,d+\frac {e}{\sqrt {x}}\right )}{12 e^4}\\ &=-\frac {3 b^2 d^2 n^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{2 e^4}+\frac {4 b^2 d n^2 \left (d+\frac {e}{\sqrt {x}}\right )^3}{9 e^4}-\frac {b^2 n^2 \left (d+\frac {e}{\sqrt {x}}\right )^4}{16 e^4}+\frac {4 b^2 d^3 n^2}{e^3 \sqrt {x}}-\frac {1}{12} b n \left (\frac {48 d^3 \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}-\frac {36 d^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{e^4}+\frac {16 d \left (d+\frac {e}{\sqrt {x}}\right )^3}{e^4}-\frac {3 \left (d+\frac {e}{\sqrt {x}}\right )^4}{e^4}-\frac {12 d^4 \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}-\frac {\left (b^2 d^4 n^2\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,d+\frac {e}{\sqrt {x}}\right )}{e^4}\\ &=-\frac {3 b^2 d^2 n^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{2 e^4}+\frac {4 b^2 d n^2 \left (d+\frac {e}{\sqrt {x}}\right )^3}{9 e^4}-\frac {b^2 n^2 \left (d+\frac {e}{\sqrt {x}}\right )^4}{16 e^4}+\frac {4 b^2 d^3 n^2}{e^3 \sqrt {x}}-\frac {b^2 d^4 n^2 \log ^2\left (d+\frac {e}{\sqrt {x}}\right )}{2 e^4}-\frac {1}{12} b n \left (\frac {48 d^3 \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}-\frac {36 d^2 \left (d+\frac {e}{\sqrt {x}}\right )^2}{e^4}+\frac {16 d \left (d+\frac {e}{\sqrt {x}}\right )^3}{e^4}-\frac {3 \left (d+\frac {e}{\sqrt {x}}\right )^4}{e^4}-\frac {12 d^4 \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^4}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{2 x^2}\\ \end {align*}

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Mathematica [C]  time = 0.39, size = 473, normalized size = 1.39 \[ -\frac {b n \left (-144 a d^4 x^2 \log \left (d \sqrt {x}+e\right )-144 a d^4 x^2 \log \left (-\frac {e}{d \sqrt {x}}\right )+144 a d^3 e x^{3/2}-72 a d^2 e^2 x+48 a d e^3 \sqrt {x}-36 a e^4+144 b d^4 x^2 \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )-144 b d^4 x^2 \log \left (d \sqrt {x}+e\right ) \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )-144 b d^4 x^2 \log \left (-\frac {e}{d \sqrt {x}}\right ) \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )+144 b d^3 e x^{3/2} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )-72 b d^2 e^2 x \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )-36 b e^4 \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )+48 b d e^3 \sqrt {x} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )-144 b d^4 n x^2 \text {Li}_2\left (\frac {e}{d \sqrt {x}}+1\right )-144 b d^4 n x^2 \text {Li}_2\left (\frac {\sqrt {x} d}{e}+1\right )+72 b d^4 n x^2 \log ^2\left (d \sqrt {x}+e\right )+156 b d^4 n x^2 \log \left (d+\frac {e}{\sqrt {x}}\right )-144 b d^4 n x^2 \log \left (d \sqrt {x}+e\right ) \log \left (-\frac {d \sqrt {x}}{e}\right )-300 b d^3 e n x^{3/2}+78 b d^2 e^2 n x-28 b d e^3 n \sqrt {x}+9 b e^4 n\right )+72 e^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )^2}{144 e^4 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/144*(72*e^4*(a + b*Log[c*(d + e/Sqrt[x])^n])^2 + b*n*(-36*a*e^4 + 9*b*e^4*n + 48*a*d*e^3*Sqrt[x] - 28*b*d*e
^3*n*Sqrt[x] - 72*a*d^2*e^2*x + 78*b*d^2*e^2*n*x + 144*a*d^3*e*x^(3/2) - 300*b*d^3*e*n*x^(3/2) + 156*b*d^4*n*x
^2*Log[d + e/Sqrt[x]] - 36*b*e^4*Log[c*(d + e/Sqrt[x])^n] + 48*b*d*e^3*Sqrt[x]*Log[c*(d + e/Sqrt[x])^n] - 72*b
*d^2*e^2*x*Log[c*(d + e/Sqrt[x])^n] + 144*b*d^3*e*x^(3/2)*Log[c*(d + e/Sqrt[x])^n] + 144*b*d^4*x^2*Log[c*(d +
e/Sqrt[x])^n] - 144*a*d^4*x^2*Log[e + d*Sqrt[x]] - 144*b*d^4*x^2*Log[c*(d + e/Sqrt[x])^n]*Log[e + d*Sqrt[x]] +
72*b*d^4*n*x^2*Log[e + d*Sqrt[x]]^2 - 144*a*d^4*x^2*Log[-(e/(d*Sqrt[x]))] - 144*b*d^4*x^2*Log[c*(d + e/Sqrt[x
])^n]*Log[-(e/(d*Sqrt[x]))] - 144*b*d^4*n*x^2*Log[e + d*Sqrt[x]]*Log[-((d*Sqrt[x])/e)] - 144*b*d^4*n*x^2*PolyL
og[2, 1 + e/(d*Sqrt[x])] - 144*b*d^4*n*x^2*PolyLog[2, 1 + (d*Sqrt[x])/e]))/(e^4*x^2)

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fricas [A]  time = 0.44, size = 361, normalized size = 1.06 \[ -\frac {9 \, b^{2} e^{4} n^{2} + 72 \, b^{2} e^{4} \log \relax (c)^{2} - 36 \, a b e^{4} n + 72 \, a^{2} e^{4} - 72 \, {\left (b^{2} d^{4} n^{2} x^{2} - b^{2} e^{4} n^{2}\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right )^{2} + 6 \, {\left (13 \, b^{2} d^{2} e^{2} n^{2} - 12 \, a b d^{2} e^{2} n\right )} x - 36 \, {\left (2 \, b^{2} d^{2} e^{2} n x + b^{2} e^{4} n - 4 \, a b e^{4}\right )} \log \relax (c) - 12 \, {\left (6 \, b^{2} d^{2} e^{2} n^{2} x + 3 \, b^{2} e^{4} n^{2} - 12 \, a b e^{4} n - {\left (25 \, b^{2} d^{4} n^{2} - 12 \, a b d^{4} n\right )} x^{2} + 12 \, {\left (b^{2} d^{4} n x^{2} - b^{2} e^{4} n\right )} \log \relax (c) - 4 \, {\left (3 \, b^{2} d^{3} e n^{2} x + b^{2} d e^{3} n^{2}\right )} \sqrt {x}\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right ) - 4 \, {\left (7 \, b^{2} d e^{3} n^{2} - 12 \, a b d e^{3} n + 3 \, {\left (25 \, b^{2} d^{3} e n^{2} - 12 \, a b d^{3} e n\right )} x - 12 \, {\left (3 \, b^{2} d^{3} e n x + b^{2} d e^{3} n\right )} \log \relax (c)\right )} \sqrt {x}}{144 \, e^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/144*(9*b^2*e^4*n^2 + 72*b^2*e^4*log(c)^2 - 36*a*b*e^4*n + 72*a^2*e^4 - 72*(b^2*d^4*n^2*x^2 - b^2*e^4*n^2)*l
og((d*x + e*sqrt(x))/x)^2 + 6*(13*b^2*d^2*e^2*n^2 - 12*a*b*d^2*e^2*n)*x - 36*(2*b^2*d^2*e^2*n*x + b^2*e^4*n -
4*a*b*e^4)*log(c) - 12*(6*b^2*d^2*e^2*n^2*x + 3*b^2*e^4*n^2 - 12*a*b*e^4*n - (25*b^2*d^4*n^2 - 12*a*b*d^4*n)*x
^2 + 12*(b^2*d^4*n*x^2 - b^2*e^4*n)*log(c) - 4*(3*b^2*d^3*e*n^2*x + b^2*d*e^3*n^2)*sqrt(x))*log((d*x + e*sqrt(
x))/x) - 4*(7*b^2*d*e^3*n^2 - 12*a*b*d*e^3*n + 3*(25*b^2*d^3*e*n^2 - 12*a*b*d^3*e*n)*x - 12*(3*b^2*d^3*e*n*x +
b^2*d*e^3*n)*log(c))*sqrt(x))/(e^4*x^2)

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giac [B]  time = 0.44, size = 1071, normalized size = 3.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(1/2))^n))^2/x^3,x, algorithm="giac")

[Out]

1/144*(288*(d*sqrt(x) + e)*b^2*d^3*n^2*log((d*sqrt(x) + e)/sqrt(x))^2/sqrt(x) - 432*(d*sqrt(x) + e)^2*b^2*d^2*
n^2*log((d*sqrt(x) + e)/sqrt(x))^2/x - 576*(d*sqrt(x) + e)*b^2*d^3*n^2*log((d*sqrt(x) + e)/sqrt(x))/sqrt(x) +
576*(d*sqrt(x) + e)*b^2*d^3*n*log(c)*log((d*sqrt(x) + e)/sqrt(x))/sqrt(x) + 288*(d*sqrt(x) + e)^3*b^2*d*n^2*lo
g((d*sqrt(x) + e)/sqrt(x))^2/x^(3/2) + 432*(d*sqrt(x) + e)^2*b^2*d^2*n^2*log((d*sqrt(x) + e)/sqrt(x))/x - 864*
(d*sqrt(x) + e)^2*b^2*d^2*n*log(c)*log((d*sqrt(x) + e)/sqrt(x))/x - 72*(d*sqrt(x) + e)^4*b^2*n^2*log((d*sqrt(x
) + e)/sqrt(x))^2/x^2 + 576*(d*sqrt(x) + e)*b^2*d^3*n^2/sqrt(x) - 576*(d*sqrt(x) + e)*b^2*d^3*n*log(c)/sqrt(x)
+ 288*(d*sqrt(x) + e)*b^2*d^3*log(c)^2/sqrt(x) - 192*(d*sqrt(x) + e)^3*b^2*d*n^2*log((d*sqrt(x) + e)/sqrt(x))
/x^(3/2) + 576*(d*sqrt(x) + e)*a*b*d^3*n*log((d*sqrt(x) + e)/sqrt(x))/sqrt(x) + 576*(d*sqrt(x) + e)^3*b^2*d*n*
log(c)*log((d*sqrt(x) + e)/sqrt(x))/x^(3/2) - 216*(d*sqrt(x) + e)^2*b^2*d^2*n^2/x + 432*(d*sqrt(x) + e)^2*b^2*
d^2*n*log(c)/x - 432*(d*sqrt(x) + e)^2*b^2*d^2*log(c)^2/x + 36*(d*sqrt(x) + e)^4*b^2*n^2*log((d*sqrt(x) + e)/s
qrt(x))/x^2 - 864*(d*sqrt(x) + e)^2*a*b*d^2*n*log((d*sqrt(x) + e)/sqrt(x))/x - 144*(d*sqrt(x) + e)^4*b^2*n*log
(c)*log((d*sqrt(x) + e)/sqrt(x))/x^2 + 64*(d*sqrt(x) + e)^3*b^2*d*n^2/x^(3/2) - 576*(d*sqrt(x) + e)*a*b*d^3*n/
sqrt(x) - 192*(d*sqrt(x) + e)^3*b^2*d*n*log(c)/x^(3/2) + 576*(d*sqrt(x) + e)*a*b*d^3*log(c)/sqrt(x) + 288*(d*s
qrt(x) + e)^3*b^2*d*log(c)^2/x^(3/2) + 576*(d*sqrt(x) + e)^3*a*b*d*n*log((d*sqrt(x) + e)/sqrt(x))/x^(3/2) - 9*
(d*sqrt(x) + e)^4*b^2*n^2/x^2 + 432*(d*sqrt(x) + e)^2*a*b*d^2*n/x + 36*(d*sqrt(x) + e)^4*b^2*n*log(c)/x^2 - 86
4*(d*sqrt(x) + e)^2*a*b*d^2*log(c)/x - 72*(d*sqrt(x) + e)^4*b^2*log(c)^2/x^2 - 144*(d*sqrt(x) + e)^4*a*b*n*log
((d*sqrt(x) + e)/sqrt(x))/x^2 - 192*(d*sqrt(x) + e)^3*a*b*d*n/x^(3/2) + 288*(d*sqrt(x) + e)*a^2*d^3/sqrt(x) +
576*(d*sqrt(x) + e)^3*a*b*d*log(c)/x^(3/2) + 36*(d*sqrt(x) + e)^4*a*b*n/x^2 - 432*(d*sqrt(x) + e)^2*a^2*d^2/x
- 144*(d*sqrt(x) + e)^4*a*b*log(c)/x^2 + 288*(d*sqrt(x) + e)^3*a^2*d/x^(3/2) - 72*(d*sqrt(x) + e)^4*a^2/x^2)*e
^(-4)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \left (d +\frac {e}{\sqrt {x}}\right )^{n}\right )+a \right )^{2}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*(d+e/x^(1/2))^n)+a)^2/x^3,x)

[Out]

int((b*ln(c*(d+e/x^(1/2))^n)+a)^2/x^3,x)

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maxima [A]  time = 0.88, size = 321, normalized size = 0.94 \[ \frac {1}{12} \, a b e n {\left (\frac {12 \, d^{4} \log \left (d \sqrt {x} + e\right )}{e^{5}} - \frac {6 \, d^{4} \log \relax (x)}{e^{5}} - \frac {12 \, d^{3} x^{\frac {3}{2}} - 6 \, d^{2} e x + 4 \, d e^{2} \sqrt {x} - 3 \, e^{3}}{e^{4} x^{2}}\right )} + \frac {1}{144} \, {\left (12 \, e n {\left (\frac {12 \, d^{4} \log \left (d \sqrt {x} + e\right )}{e^{5}} - \frac {6 \, d^{4} \log \relax (x)}{e^{5}} - \frac {12 \, d^{3} x^{\frac {3}{2}} - 6 \, d^{2} e x + 4 \, d e^{2} \sqrt {x} - 3 \, e^{3}}{e^{4} x^{2}}\right )} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right ) - \frac {{\left (72 \, d^{4} x^{2} \log \left (d \sqrt {x} + e\right )^{2} + 18 \, d^{4} x^{2} \log \relax (x)^{2} - 150 \, d^{4} x^{2} \log \relax (x) - 300 \, d^{3} e x^{\frac {3}{2}} + 78 \, d^{2} e^{2} x - 28 \, d e^{3} \sqrt {x} + 9 \, e^{4} - 12 \, {\left (6 \, d^{4} x^{2} \log \relax (x) - 25 \, d^{4} x^{2}\right )} \log \left (d \sqrt {x} + e\right )\right )} n^{2}}{e^{4} x^{2}}\right )} b^{2} - \frac {b^{2} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right )^{2}}{2 \, x^{2}} - \frac {a b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right )}{x^{2}} - \frac {a^{2}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(1/2))^n))^2/x^3,x, algorithm="maxima")

[Out]

1/12*a*b*e*n*(12*d^4*log(d*sqrt(x) + e)/e^5 - 6*d^4*log(x)/e^5 - (12*d^3*x^(3/2) - 6*d^2*e*x + 4*d*e^2*sqrt(x)
- 3*e^3)/(e^4*x^2)) + 1/144*(12*e*n*(12*d^4*log(d*sqrt(x) + e)/e^5 - 6*d^4*log(x)/e^5 - (12*d^3*x^(3/2) - 6*d
^2*e*x + 4*d*e^2*sqrt(x) - 3*e^3)/(e^4*x^2))*log(c*(d + e/sqrt(x))^n) - (72*d^4*x^2*log(d*sqrt(x) + e)^2 + 18*
d^4*x^2*log(x)^2 - 150*d^4*x^2*log(x) - 300*d^3*e*x^(3/2) + 78*d^2*e^2*x - 28*d*e^3*sqrt(x) + 9*e^4 - 12*(6*d^
4*x^2*log(x) - 25*d^4*x^2)*log(d*sqrt(x) + e))*n^2/(e^4*x^2))*b^2 - 1/2*b^2*log(c*(d + e/sqrt(x))^n)^2/x^2 - a
*b*log(c*(d + e/sqrt(x))^n)/x^2 - 1/2*a^2/x^2

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mupad [B]  time = 0.56, size = 424, normalized size = 1.24 \[ \ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )\,\left (\frac {\frac {b\,d\,\left (4\,a-b\,n\right )}{3\,e}-\frac {4\,a\,b\,d}{3\,e}}{x^{3/2}}-\frac {b\,\left (4\,a-b\,n\right )}{4\,x^2}-\frac {d\,\left (\frac {b\,d\,\left (4\,a-b\,n\right )}{e}-\frac {4\,a\,b\,d}{e}\right )}{2\,e\,x}+\frac {d^2\,\left (\frac {b\,d\,\left (4\,a-b\,n\right )}{e}-\frac {4\,a\,b\,d}{e}\right )}{e^2\,\sqrt {x}}\right )+\frac {\frac {d\,\left (2\,a^2-a\,b\,n+\frac {b^2\,n^2}{4}\right )}{3\,e}-\frac {d\,\left (6\,a^2-b^2\,n^2\right )}{9\,e}}{x^{3/2}}-{\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )}^2\,\left (\frac {b^2}{2\,x^2}-\frac {b^2\,d^4}{2\,e^4}\right )-\frac {\frac {a^2}{2}-\frac {a\,b\,n}{4}+\frac {b^2\,n^2}{16}}{x^2}-\frac {\frac {d\,\left (\frac {d\,\left (2\,a^2-a\,b\,n+\frac {b^2\,n^2}{4}\right )}{e}-\frac {d\,\left (6\,a^2-b^2\,n^2\right )}{3\,e}\right )}{2\,e}+\frac {b^2\,d^2\,n^2}{4\,e^2}}{x}+\frac {\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (2\,a^2-a\,b\,n+\frac {b^2\,n^2}{4}\right )}{e}-\frac {d\,\left (6\,a^2-b^2\,n^2\right )}{3\,e}\right )}{e}+\frac {b^2\,d^2\,n^2}{2\,e^2}\right )}{e}+\frac {b^2\,d^3\,n^2}{e^3}}{\sqrt {x}}-\frac {\ln \left (d+\frac {e}{\sqrt {x}}\right )\,\left (25\,b^2\,d^4\,n^2-12\,a\,b\,d^4\,n\right )}{12\,e^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d + e/x^(1/2))^n))^2/x^3,x)

[Out]

log(c*(d + e/x^(1/2))^n)*(((b*d*(4*a - b*n))/(3*e) - (4*a*b*d)/(3*e))/x^(3/2) - (b*(4*a - b*n))/(4*x^2) - (d*(
(b*d*(4*a - b*n))/e - (4*a*b*d)/e))/(2*e*x) + (d^2*((b*d*(4*a - b*n))/e - (4*a*b*d)/e))/(e^2*x^(1/2))) + ((d*(
2*a^2 + (b^2*n^2)/4 - a*b*n))/(3*e) - (d*(6*a^2 - b^2*n^2))/(9*e))/x^(3/2) - log(c*(d + e/x^(1/2))^n)^2*(b^2/(
2*x^2) - (b^2*d^4)/(2*e^4)) - (a^2/2 + (b^2*n^2)/16 - (a*b*n)/4)/x^2 - ((d*((d*(2*a^2 + (b^2*n^2)/4 - a*b*n))/
e - (d*(6*a^2 - b^2*n^2))/(3*e)))/(2*e) + (b^2*d^2*n^2)/(4*e^2))/x + ((d*((d*((d*(2*a^2 + (b^2*n^2)/4 - a*b*n)
)/e - (d*(6*a^2 - b^2*n^2))/(3*e)))/e + (b^2*d^2*n^2)/(2*e^2)))/e + (b^2*d^3*n^2)/e^3)/x^(1/2) - (log(d + e/x^
(1/2))*(25*b^2*d^4*n^2 - 12*a*b*d^4*n))/(12*e^4)

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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*(d+e/x**(1/2))**n))**2/x**3,x)

[Out]

Timed out

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