3.76.99 \(\int \frac {-12-403 x-2438 x^2-535 x^3+2360 x^4-672 x^5-128 x^6+48 x^7+e^2 (4 x^4+x^5)+e (-8 x^2-2 x^3-158 x^4+24 x^5+16 x^6)+(-16 x-204 x^2-48 x^3+32 x^4+8 x^5) \log (4+x)+(-4 x^2-x^3) \log ^2(4+x)}{4 x^4+x^5} \, dx\)

Optimal. Leaf size=24 \[ \frac {\left ((5-2 x)^2+\frac {1}{x}+e x+\log (4+x)\right )^2}{x} \]

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Rubi [B]  time = 0.97, antiderivative size = 114, normalized size of antiderivative = 4.75, number of steps used = 38, number of rules used = 17, integrand size = 135, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.126, Rules used = {1593, 6742, 44, 36, 29, 31, 43, 1620, 2418, 2389, 2295, 2395, 2392, 2391, 2390, 2301, 2397} \begin {gather*} 16 x^3+\frac {1}{x^3}+8 e x^2-160 x^2+\frac {50}{x^2}+\frac {2 \log (x+4)}{x^2}+e^2 x-40 e x+600 x+\frac {2 e}{x}+\frac {585}{x}-\frac {1}{4} \log ^2(x+4)+\frac {(x+4) \log ^2(x+4)}{4 x}+8 (x+4) \log (x+4)+2 e \log (x+4)-72 \log (x+4)+\frac {50 \log (x+4)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12 - 403*x - 2438*x^2 - 535*x^3 + 2360*x^4 - 672*x^5 - 128*x^6 + 48*x^7 + E^2*(4*x^4 + x^5) + E*(-8*x^2
- 2*x^3 - 158*x^4 + 24*x^5 + 16*x^6) + (-16*x - 204*x^2 - 48*x^3 + 32*x^4 + 8*x^5)*Log[4 + x] + (-4*x^2 - x^3)
*Log[4 + x]^2)/(4*x^4 + x^5),x]

[Out]

x^(-3) + 50/x^2 + 585/x + (2*E)/x + 600*x - 40*E*x + E^2*x - 160*x^2 + 8*E*x^2 + 16*x^3 - 72*Log[4 + x] + 2*E*
Log[4 + x] + (2*Log[4 + x])/x^2 + (50*Log[4 + x])/x + 8*(4 + x)*Log[4 + x] - Log[4 + x]^2/4 + ((4 + x)*Log[4 +
x]^2)/(4*x)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2390

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

Rule 2391

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

Rule 2392

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + (e*x)/d]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rule 2395

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

Rule 2397

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

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-12-403 x-2438 x^2-535 x^3+2360 x^4-672 x^5-128 x^6+48 x^7+e^2 \left (4 x^4+x^5\right )+e \left (-8 x^2-2 x^3-158 x^4+24 x^5+16 x^6\right )+\left (-16 x-204 x^2-48 x^3+32 x^4+8 x^5\right ) \log (4+x)+\left (-4 x^2-x^3\right ) \log ^2(4+x)}{x^4 (4+x)} \, dx\\ &=\int \left (e^2+\frac {2360}{4+x}-\frac {12}{x^4 (4+x)}-\frac {403}{x^3 (4+x)}-\frac {2438}{x^2 (4+x)}-\frac {535}{x (4+x)}-\frac {672 x}{4+x}-\frac {128 x^2}{4+x}+\frac {48 x^3}{4+x}+\frac {2 e \left (-4-x-79 x^2+12 x^3+8 x^4\right )}{x^2 (4+x)}+\frac {4 \left (-4-51 x-12 x^2+8 x^3+2 x^4\right ) \log (4+x)}{x^3 (4+x)}-\frac {\log ^2(4+x)}{x^2}\right ) \, dx\\ &=e^2 x+2360 \log (4+x)+4 \int \frac {\left (-4-51 x-12 x^2+8 x^3+2 x^4\right ) \log (4+x)}{x^3 (4+x)} \, dx-12 \int \frac {1}{x^4 (4+x)} \, dx+48 \int \frac {x^3}{4+x} \, dx-128 \int \frac {x^2}{4+x} \, dx-403 \int \frac {1}{x^3 (4+x)} \, dx-535 \int \frac {1}{x (4+x)} \, dx-672 \int \frac {x}{4+x} \, dx-2438 \int \frac {1}{x^2 (4+x)} \, dx+(2 e) \int \frac {-4-x-79 x^2+12 x^3+8 x^4}{x^2 (4+x)} \, dx-\int \frac {\log ^2(4+x)}{x^2} \, dx\\ &=e^2 x+2360 \log (4+x)+\frac {(4+x) \log ^2(4+x)}{4 x}-\frac {1}{2} \int \frac {\log (4+x)}{x} \, dx+4 \int \left (2 \log (4+x)-\frac {\log (4+x)}{x^3}-\frac {25 \log (4+x)}{2 x^2}+\frac {\log (4+x)}{8 x}-\frac {\log (4+x)}{8 (4+x)}\right ) \, dx-12 \int \left (\frac {1}{4 x^4}-\frac {1}{16 x^3}+\frac {1}{64 x^2}-\frac {1}{256 x}+\frac {1}{256 (4+x)}\right ) \, dx+48 \int \left (16-4 x+x^2-\frac {64}{4+x}\right ) \, dx-128 \int \left (-4+x+\frac {16}{4+x}\right ) \, dx-\frac {535}{4} \int \frac {1}{x} \, dx+\frac {535}{4} \int \frac {1}{4+x} \, dx-403 \int \left (\frac {1}{4 x^3}-\frac {1}{16 x^2}+\frac {1}{64 x}-\frac {1}{64 (4+x)}\right ) \, dx-672 \int \left (1-\frac {4}{4+x}\right ) \, dx-2438 \int \left (\frac {1}{4 x^2}-\frac {1}{16 x}+\frac {1}{16 (4+x)}\right ) \, dx+(2 e) \int \left (-20-\frac {1}{x^2}+8 x+\frac {1}{4+x}\right ) \, dx\\ &=\frac {1}{x^3}+\frac {50}{x^2}+\frac {1169}{2 x}+\frac {2 e}{x}+608 x-40 e x+e^2 x-160 x^2+8 e x^2+16 x^3+\frac {99 \log (x)}{8}-\frac {1}{2} \log (4) \log (x)-\frac {675}{8} \log (4+x)+2 e \log (4+x)+\frac {(4+x) \log ^2(4+x)}{4 x}-\frac {1}{2} \int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx+\frac {1}{2} \int \frac {\log (4+x)}{x} \, dx-\frac {1}{2} \int \frac {\log (4+x)}{4+x} \, dx-4 \int \frac {\log (4+x)}{x^3} \, dx+8 \int \log (4+x) \, dx-50 \int \frac {\log (4+x)}{x^2} \, dx\\ &=\frac {1}{x^3}+\frac {50}{x^2}+\frac {1169}{2 x}+\frac {2 e}{x}+608 x-40 e x+e^2 x-160 x^2+8 e x^2+16 x^3+\frac {99 \log (x)}{8}-\frac {675}{8} \log (4+x)+2 e \log (4+x)+\frac {2 \log (4+x)}{x^2}+\frac {50 \log (4+x)}{x}+\frac {(4+x) \log ^2(4+x)}{4 x}+\frac {\text {Li}_2\left (-\frac {x}{4}\right )}{2}+\frac {1}{2} \int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,4+x\right )-2 \int \frac {1}{x^2 (4+x)} \, dx+8 \operatorname {Subst}(\int \log (x) \, dx,x,4+x)-50 \int \frac {1}{x (4+x)} \, dx\\ &=\frac {1}{x^3}+\frac {50}{x^2}+\frac {1169}{2 x}+\frac {2 e}{x}+600 x-40 e x+e^2 x-160 x^2+8 e x^2+16 x^3+\frac {99 \log (x)}{8}-\frac {675}{8} \log (4+x)+2 e \log (4+x)+\frac {2 \log (4+x)}{x^2}+\frac {50 \log (4+x)}{x}+8 (4+x) \log (4+x)-\frac {1}{4} \log ^2(4+x)+\frac {(4+x) \log ^2(4+x)}{4 x}-2 \int \left (\frac {1}{4 x^2}-\frac {1}{16 x}+\frac {1}{16 (4+x)}\right ) \, dx-\frac {25}{2} \int \frac {1}{x} \, dx+\frac {25}{2} \int \frac {1}{4+x} \, dx\\ &=\frac {1}{x^3}+\frac {50}{x^2}+\frac {585}{x}+\frac {2 e}{x}+600 x-40 e x+e^2 x-160 x^2+8 e x^2+16 x^3-72 \log (4+x)+2 e \log (4+x)+\frac {2 \log (4+x)}{x^2}+\frac {50 \log (4+x)}{x}+8 (4+x) \log (4+x)-\frac {1}{4} \log ^2(4+x)+\frac {(4+x) \log ^2(4+x)}{4 x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 79, normalized size = 3.29 \begin {gather*} \frac {1}{x^3}+\frac {50}{x^2}+\frac {585+2 e}{x}+\left (600-40 e+e^2\right ) x+8 (-20+e) x^2+16 x^3+2 (-20+e) \log (4+x)+\frac {2 \left (1+25 x+4 x^3\right ) \log (4+x)}{x^2}+\frac {\log ^2(4+x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 - 403*x - 2438*x^2 - 535*x^3 + 2360*x^4 - 672*x^5 - 128*x^6 + 48*x^7 + E^2*(4*x^4 + x^5) + E*(-
8*x^2 - 2*x^3 - 158*x^4 + 24*x^5 + 16*x^6) + (-16*x - 204*x^2 - 48*x^3 + 32*x^4 + 8*x^5)*Log[4 + x] + (-4*x^2
- x^3)*Log[4 + x]^2)/(4*x^4 + x^5),x]

[Out]

x^(-3) + 50/x^2 + (585 + 2*E)/x + (600 - 40*E + E^2)*x + 8*(-20 + E)*x^2 + 16*x^3 + 2*(-20 + E)*Log[4 + x] + (
2*(1 + 25*x + 4*x^3)*Log[4 + x])/x^2 + Log[4 + x]^2/x

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fricas [B]  time = 1.52, size = 92, normalized size = 3.83 \begin {gather*} \frac {16 \, x^{6} - 160 \, x^{5} + x^{4} e^{2} + 600 \, x^{4} + x^{2} \log \left (x + 4\right )^{2} + 585 \, x^{2} + 2 \, {\left (4 \, x^{5} - 20 \, x^{4} + x^{2}\right )} e + 2 \, {\left (4 \, x^{4} + x^{3} e - 20 \, x^{3} + 25 \, x^{2} + x\right )} \log \left (x + 4\right ) + 50 \, x + 1}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-4*x^2)*log(4+x)^2+(8*x^5+32*x^4-48*x^3-204*x^2-16*x)*log(4+x)+(x^5+4*x^4)*exp(1)^2+(16*x^6+24
*x^5-158*x^4-2*x^3-8*x^2)*exp(1)+48*x^7-128*x^6-672*x^5+2360*x^4-535*x^3-2438*x^2-403*x-12)/(x^5+4*x^4),x, alg
orithm="fricas")

[Out]

(16*x^6 - 160*x^5 + x^4*e^2 + 600*x^4 + x^2*log(x + 4)^2 + 585*x^2 + 2*(4*x^5 - 20*x^4 + x^2)*e + 2*(4*x^4 + x
^3*e - 20*x^3 + 25*x^2 + x)*log(x + 4) + 50*x + 1)/x^3

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giac [B]  time = 0.16, size = 111, normalized size = 4.62 \begin {gather*} \frac {16 \, x^{6} + 8 \, x^{5} e - 160 \, x^{5} + x^{4} e^{2} - 40 \, x^{4} e + 8 \, x^{4} \log \left (x + 4\right ) + 2 \, x^{3} e \log \left (x + 4\right ) + 600 \, x^{4} - 40 \, x^{3} \log \left (x + 4\right ) + x^{2} \log \left (x + 4\right )^{2} + 2 \, x^{2} e + 50 \, x^{2} \log \left (x + 4\right ) + 585 \, x^{2} + 2 \, x \log \left (x + 4\right ) + 50 \, x + 1}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-4*x^2)*log(4+x)^2+(8*x^5+32*x^4-48*x^3-204*x^2-16*x)*log(4+x)+(x^5+4*x^4)*exp(1)^2+(16*x^6+24
*x^5-158*x^4-2*x^3-8*x^2)*exp(1)+48*x^7-128*x^6-672*x^5+2360*x^4-535*x^3-2438*x^2-403*x-12)/(x^5+4*x^4),x, alg
orithm="giac")

[Out]

(16*x^6 + 8*x^5*e - 160*x^5 + x^4*e^2 - 40*x^4*e + 8*x^4*log(x + 4) + 2*x^3*e*log(x + 4) + 600*x^4 - 40*x^3*lo
g(x + 4) + x^2*log(x + 4)^2 + 2*x^2*e + 50*x^2*log(x + 4) + 585*x^2 + 2*x*log(x + 4) + 50*x + 1)/x^3

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maple [B]  time = 0.23, size = 98, normalized size = 4.08

method	result	size
norman	\(\frac {1+x^{2} \ln \left (4+x \right )^{2}+\left (-160+8 \,{\mathrm e}\right ) x^{5}+\left (585+2 \,{\mathrm e}\right ) x^{2}+\left (600-40 \,{\mathrm e}+{\mathrm e}^{2}\right ) x^{4}+\left (-40+2 \,{\mathrm e}\right ) x^{3} \ln \left (4+x \right )+50 x +16 x^{6}+50 x^{2} \ln \left (4+x \right )+2 \ln \left (4+x \right ) x +8 \ln \left (4+x \right ) x^{4}}{x^{3}}\)	\(98\)
risch	\(\frac {\ln \left (4+x \right )^{2}}{x}+\frac {2 \left (4 x^{3}+25 x +1\right ) \ln \left (4+x \right )}{x^{2}}+\frac {x^{4} {\mathrm e}^{2}+8 x^{5} {\mathrm e}+16 x^{6}+2 \ln \left (-x -4\right ) x^{3} {\mathrm e}-40 x^{4} {\mathrm e}-160 x^{5}-40 \ln \left (-x -4\right ) x^{3}+600 x^{4}+2 x^{2} {\mathrm e}+585 x^{2}+50 x +1}{x^{3}}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3-4*x^2)*ln(4+x)^2+(8*x^5+32*x^4-48*x^3-204*x^2-16*x)*ln(4+x)+(x^5+4*x^4)*exp(1)^2+(16*x^6+24*x^5-158
*x^4-2*x^3-8*x^2)*exp(1)+48*x^7-128*x^6-672*x^5+2360*x^4-535*x^3-2438*x^2-403*x-12)/(x^5+4*x^4),x,method=_RETU
RNVERBOSE)

[Out]

(1+x^2*ln(4+x)^2+(-160+8*exp(1))*x^5+(585+2*exp(1))*x^2+(600-40*exp(1)+exp(1)^2)*x^4+(-40+2*exp(1))*x^3*ln(4+x
)+50*x+16*x^6+50*x^2*ln(4+x)+2*ln(4+x)*x+8*ln(4+x)*x^4)/x^3

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maxima [B]  time = 0.41, size = 179, normalized size = 7.46 \begin {gather*} 16 \, x^{3} - 160 \, x^{2} + {\left (x - 4 \, \log \left (x + 4\right )\right )} e^{2} + 8 \, {\left (x^{2} - 8 \, x + 32 \, \log \left (x + 4\right )\right )} e + 24 \, {\left (x - 4 \, \log \left (x + 4\right )\right )} e + \frac {1}{2} \, {\left (\frac {4}{x} - \log \left (x + 4\right ) + \log \relax (x)\right )} e + \frac {1}{2} \, {\left (\log \left (x + 4\right ) - \log \relax (x)\right )} e + 4 \, e^{2} \log \left (x + 4\right ) - 158 \, e \log \left (x + 4\right ) + 608 \, x - \frac {64 \, x^{3} - 8 \, x \log \left (x + 4\right )^{2} - {\left (64 \, x^{3} + 355 \, x^{2} + 400 \, x + 16\right )} \log \left (x + 4\right ) - 4 \, x}{8 \, x^{2}} - \frac {403 \, {\left (x - 2\right )}}{16 \, x^{2}} + \frac {1219}{2 \, x} + \frac {3 \, x^{2} - 6 \, x + 16}{16 \, x^{3}} - \frac {675}{8} \, \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-4*x^2)*log(4+x)^2+(8*x^5+32*x^4-48*x^3-204*x^2-16*x)*log(4+x)+(x^5+4*x^4)*exp(1)^2+(16*x^6+24
*x^5-158*x^4-2*x^3-8*x^2)*exp(1)+48*x^7-128*x^6-672*x^5+2360*x^4-535*x^3-2438*x^2-403*x-12)/(x^5+4*x^4),x, alg
orithm="maxima")

[Out]

16*x^3 - 160*x^2 + (x - 4*log(x + 4))*e^2 + 8*(x^2 - 8*x + 32*log(x + 4))*e + 24*(x - 4*log(x + 4))*e + 1/2*(4
/x - log(x + 4) + log(x))*e + 1/2*(log(x + 4) - log(x))*e + 4*e^2*log(x + 4) - 158*e*log(x + 4) + 608*x - 1/8*
(64*x^3 - 8*x*log(x + 4)^2 - (64*x^3 + 355*x^2 + 400*x + 16)*log(x + 4) - 4*x)/x^2 - 403/16*(x - 2)/x^2 + 1219
/2/x + 1/16*(3*x^2 - 6*x + 16)/x^3 - 675/8*log(x + 4)

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mupad [B]  time = 6.15, size = 88, normalized size = 3.67 \begin {gather*} 600\,x-40\,\ln \left (x+4\right )+2\,\ln \left (x+4\right )\,\mathrm {e}+8\,x\,\ln \left (x+4\right )+\frac {2\,\ln \left (x+4\right )+50}{x^2}-40\,x\,\mathrm {e}+x\,{\mathrm {e}}^2+8\,x^2\,\mathrm {e}+\frac {{\ln \left (x+4\right )}^2+50\,\ln \left (x+4\right )+2\,\mathrm {e}+585}{x}-160\,x^2+\frac {1}{x^3}+16\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(403*x - exp(2)*(4*x^4 + x^5) + exp(1)*(8*x^2 + 2*x^3 + 158*x^4 - 24*x^5 - 16*x^6) + log(x + 4)^2*(4*x^2
+ x^3) + log(x + 4)*(16*x + 204*x^2 + 48*x^3 - 32*x^4 - 8*x^5) + 2438*x^2 + 535*x^3 - 2360*x^4 + 672*x^5 + 128
*x^6 - 48*x^7 + 12)/(4*x^4 + x^5),x)

[Out]

600*x - 40*log(x + 4) + 2*log(x + 4)*exp(1) + 8*x*log(x + 4) + (2*log(x + 4) + 50)/x^2 - 40*x*exp(1) + x*exp(2
) + 8*x^2*exp(1) + (50*log(x + 4) + 2*exp(1) + log(x + 4)^2 + 585)/x - 160*x^2 + 1/x^3 + 16*x^3

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sympy [B]  time = 0.80, size = 83, normalized size = 3.46 \begin {gather*} 16 x^{3} + x^{2} \left (-160 + 8 e\right ) + x \left (- 40 e + e^{2} + 600\right ) + 2 \left (-20 + e\right ) \log {\left (x + 4 \right )} + \frac {\log {\left (x + 4 \right )}^{2}}{x} + \frac {\left (8 x^{3} + 50 x + 2\right ) \log {\left (x + 4 \right )}}{x^{2}} + \frac {x^{2} \left (2 e + 585\right ) + 50 x + 1}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3-4*x**2)*ln(4+x)**2+(8*x**5+32*x**4-48*x**3-204*x**2-16*x)*ln(4+x)+(x**5+4*x**4)*exp(1)**2+(1
6*x**6+24*x**5-158*x**4-2*x**3-8*x**2)*exp(1)+48*x**7-128*x**6-672*x**5+2360*x**4-535*x**3-2438*x**2-403*x-12)
/(x**5+4*x**4),x)

[Out]

16*x**3 + x**2*(-160 + 8*E) + x*(-40*E + exp(2) + 600) + 2*(-20 + E)*log(x + 4) + log(x + 4)**2/x + (8*x**3 +
50*x + 2)*log(x + 4)/x**2 + (x**2*(2*E + 585) + 50*x + 1)/x**3

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