3.16.72 \(\int \frac {-160-832 x-1602 x^2+801 x^3+(32+136 x) \log (x^2)}{64 x^3} \, dx\)

Optimal. Leaf size=27 \[ 4 \left (2+\left (1+\frac {4+x}{16 x}\right )^2\right ) \left (4+x-\log \left (x^2\right )\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps used = 9, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {12, 14, 37, 2334, 43} \begin {gather*} \frac {1}{x^2}-\frac {(17 x+4)^2 \log \left (x^2\right )}{64 x^2}+\frac {801 x}{64}+\frac {35}{4 x}-16 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-160 - 832*x - 1602*x^2 + 801*x^3 + (32 + 136*x)*Log[x^2])/(64*x^3),x]

[Out]

x^(-2) + 35/(4*x) + (801*x)/64 - 16*Log[x] - ((4 + 17*x)^2*Log[x^2])/(64*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 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 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])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{64} \int \frac {-160-832 x-1602 x^2+801 x^3+(32+136 x) \log \left (x^2\right )}{x^3} \, dx\\ &=\frac {1}{64} \int \left (\frac {-160-832 x-1602 x^2+801 x^3}{x^3}+\frac {8 (4+17 x) \log \left (x^2\right )}{x^3}\right ) \, dx\\ &=\frac {1}{64} \int \frac {-160-832 x-1602 x^2+801 x^3}{x^3} \, dx+\frac {1}{8} \int \frac {(4+17 x) \log \left (x^2\right )}{x^3} \, dx\\ &=-\frac {(4+17 x)^2 \log \left (x^2\right )}{64 x^2}+\frac {1}{64} \int \left (801-\frac {160}{x^3}-\frac {832}{x^2}-\frac {1602}{x}\right ) \, dx-\frac {1}{4} \int -\frac {(4+17 x)^2}{8 x^3} \, dx\\ &=\frac {5}{4 x^2}+\frac {13}{x}+\frac {801 x}{64}-\frac {801 \log (x)}{32}-\frac {(4+17 x)^2 \log \left (x^2\right )}{64 x^2}+\frac {1}{32} \int \frac {(4+17 x)^2}{x^3} \, dx\\ &=\frac {5}{4 x^2}+\frac {13}{x}+\frac {801 x}{64}-\frac {801 \log (x)}{32}-\frac {(4+17 x)^2 \log \left (x^2\right )}{64 x^2}+\frac {1}{32} \int \left (\frac {16}{x^3}+\frac {136}{x^2}+\frac {289}{x}\right ) \, dx\\ &=\frac {1}{x^2}+\frac {35}{4 x}+\frac {801 x}{64}-16 \log (x)-\frac {(4+17 x)^2 \log \left (x^2\right )}{64 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 44, normalized size = 1.63 \begin {gather*} \frac {1}{x^2}+\frac {35}{4 x}+\frac {801 x}{64}-\frac {801 \log (x)}{32}-\frac {\log \left (x^2\right )}{4 x^2}-\frac {17 \log \left (x^2\right )}{8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-160 - 832*x - 1602*x^2 + 801*x^3 + (32 + 136*x)*Log[x^2])/(64*x^3),x]

[Out]

x^(-2) + 35/(4*x) + (801*x)/64 - (801*Log[x])/32 - Log[x^2]/(4*x^2) - (17*Log[x^2])/(8*x)

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fricas [A]  time = 0.53, size = 31, normalized size = 1.15 \begin {gather*} \frac {801 \, x^{3} - {\left (801 \, x^{2} + 136 \, x + 16\right )} \log \left (x^{2}\right ) + 560 \, x + 64}{64 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((136*x+32)*log(x^2)+801*x^3-1602*x^2-832*x-160)/x^3,x, algorithm="fricas")

[Out]

1/64*(801*x^3 - (801*x^2 + 136*x + 16)*log(x^2) + 560*x + 64)/x^2

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giac [A]  time = 0.31, size = 32, normalized size = 1.19 \begin {gather*} \frac {801}{64} \, x - \frac {{\left (17 \, x + 2\right )} \log \left (x^{2}\right )}{8 \, x^{2}} + \frac {35 \, x + 4}{4 \, x^{2}} - \frac {801}{32} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((136*x+32)*log(x^2)+801*x^3-1602*x^2-832*x-160)/x^3,x, algorithm="giac")

[Out]

801/64*x - 1/8*(17*x + 2)*log(x^2)/x^2 + 1/4*(35*x + 4)/x^2 - 801/32*log(x)

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maple [A]  time = 0.03, size = 33, normalized size = 1.22

method	result	size
norman	\(\frac {1+\frac {35 x}{4}+\frac {801 x^{3}}{64}-\frac {17 x \ln \left (x^{2}\right )}{8}-\frac {\ln \left (x^{2}\right )}{4}}{x^{2}}-\frac {801 \ln \relax (x )}{32}\)	\(33\)
default	\(\frac {801 x}{64}+\frac {35}{4 x}+\frac {1}{x^{2}}-\frac {801 \ln \relax (x )}{32}-\frac {17 \ln \left (x^{2}\right )}{8 x}-\frac {\ln \left (x^{2}\right )}{4 x^{2}}\)	\(35\)
risch	\(-\frac {\left (17 x +2\right ) \ln \left (x^{2}\right )}{8 x^{2}}-\frac {1602 x^{2} \ln \relax (x )-801 x^{3}-560 x -64}{64 x^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/64*((136*x+32)*ln(x^2)+801*x^3-1602*x^2-832*x-160)/x^3,x,method=_RETURNVERBOSE)

[Out]

(1+35/4*x+801/64*x^3-17/8*x*ln(x^2)-1/4*ln(x^2))/x^2-801/32*ln(x)

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maxima [A]  time = 0.69, size = 34, normalized size = 1.26 \begin {gather*} \frac {801}{64} \, x - \frac {17 \, \log \left (x^{2}\right )}{8 \, x} + \frac {35}{4 \, x} - \frac {\log \left (x^{2}\right )}{4 \, x^{2}} + \frac {1}{x^{2}} - \frac {801}{32} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((136*x+32)*log(x^2)+801*x^3-1602*x^2-832*x-160)/x^3,x, algorithm="maxima")

[Out]

801/64*x - 17/8*log(x^2)/x + 35/4/x - 1/4*log(x^2)/x^2 + 1/x^2 - 801/32*log(x)

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mupad [B]  time = 1.02, size = 38, normalized size = 1.41 \begin {gather*} \frac {801\,x}{64}-\frac {801\,\ln \left (x^2\right )}{64}-\frac {x^2\,\left (\frac {17\,\ln \left (x^2\right )}{8}-\frac {35}{4}\right )+x\,\left (\frac {\ln \left (x^2\right )}{4}-1\right )}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(13*x + (801*x^2)/32 - (801*x^3)/64 - (log(x^2)*(136*x + 32))/64 + 5/2)/x^3,x)

[Out]

(801*x)/64 - (801*log(x^2))/64 - (x^2*((17*log(x^2))/8 - 35/4) + x*(log(x^2)/4 - 1))/x^3

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sympy [A]  time = 0.16, size = 37, normalized size = 1.37 \begin {gather*} \frac {801 x}{64} - \frac {801 \log {\relax (x )}}{32} + \frac {\left (- 17 x - 2\right ) \log {\left (x^{2} \right )}}{8 x^{2}} + \frac {560 x + 64}{64 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((136*x+32)*ln(x**2)+801*x**3-1602*x**2-832*x-160)/x**3,x)

[Out]

801*x/64 - 801*log(x)/32 + (-17*x - 2)*log(x**2)/(8*x**2) + (560*x + 64)/(64*x**2)

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