3.52.84 \(\int \frac {-3 e^{\frac {x}{5 \log (2)}} x^3+(-150-15 x-180 x^3+90 x^4+495 x^5+420 x^6+75 x^7) \log (2)}{5 x^3 \log (2)} \, dx\)

Optimal. Leaf size=32 \[ 3 \left (-e^{\frac {x}{5 \log (2)}}+(5+x) \left (-\frac {1}{x}+x+x^2\right )^2\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 52, normalized size of antiderivative = 1.62, number of steps used = 6, number of rules used = 4, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {12, 14, 2194, 6742} \begin {gather*} 3 x^5+21 x^4+33 x^3+9 x^2+\frac {15}{x^2}-36 x+\frac {3}{x}-\frac {3 \log (32) e^{\frac {x}{\log (32)}}}{5 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E^(x/(5*Log[2]))*x^3 + (-150 - 15*x - 180*x^3 + 90*x^4 + 495*x^5 + 420*x^6 + 75*x^7)*Log[2])/(5*x^3*Lo
g[2]),x]

[Out]

15/x^2 + 3/x - 36*x + 9*x^2 + 33*x^3 + 21*x^4 + 3*x^5 - (3*E^(x/Log[32])*Log[32])/(5*Log[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 2194

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-3 e^{\frac {x}{5 \log (2)}} x^3+\left (-150-15 x-180 x^3+90 x^4+495 x^5+420 x^6+75 x^7\right ) \log (2)}{x^3} \, dx}{5 \log (2)}\\ &=\frac {\int \left (-3 e^{\frac {x}{\log (32)}}+\frac {15 \left (-1+x^2+x^3\right ) \left (10+x+10 x^2+23 x^3+5 x^4\right ) \log (2)}{x^3}\right ) \, dx}{5 \log (2)}\\ &=3 \int \frac {\left (-1+x^2+x^3\right ) \left (10+x+10 x^2+23 x^3+5 x^4\right )}{x^3} \, dx-\frac {3 \int e^{\frac {x}{\log (32)}} \, dx}{5 \log (2)}\\ &=-\frac {3 e^{\frac {x}{\log (32)}} \log (32)}{5 \log (2)}+3 \int \left (-12-\frac {10}{x^3}-\frac {1}{x^2}+6 x+33 x^2+28 x^3+5 x^4\right ) \, dx\\ &=\frac {15}{x^2}+\frac {3}{x}-36 x+9 x^2+33 x^3+21 x^4+3 x^5-\frac {3 e^{\frac {x}{\log (32)}} \log (32)}{5 \log (2)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 44, normalized size = 1.38 \begin {gather*} -3 e^{\frac {x}{\log (32)}}+\frac {15}{x^2}+\frac {3}{x}-36 x+9 x^2+33 x^3+21 x^4+3 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^(x/(5*Log[2]))*x^3 + (-150 - 15*x - 180*x^3 + 90*x^4 + 495*x^5 + 420*x^6 + 75*x^7)*Log[2])/(5*
x^3*Log[2]),x]

[Out]

-3*E^(x/Log[32]) + 15/x^2 + 3/x - 36*x + 9*x^2 + 33*x^3 + 21*x^4 + 3*x^5

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 44, normalized size = 1.38 \begin {gather*} \frac {3 \, {\left (x^{7} + 7 \, x^{6} + 11 \, x^{5} + 3 \, x^{4} - 12 \, x^{3} - x^{2} e^{\left (\frac {x}{5 \, \log \relax (2)}\right )} + x + 5\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-3*x^3*exp(1/5*x/log(2))+(75*x^7+420*x^6+495*x^5+90*x^4-180*x^3-15*x-150)*log(2))/x^3/log(2),x,
algorithm="fricas")

[Out]

3*(x^7 + 7*x^6 + 11*x^5 + 3*x^4 - 12*x^3 - x^2*e^(1/5*x/log(2)) + x + 5)/x^2

________________________________________________________________________________________

giac [B]  time = 0.14, size = 67, normalized size = 2.09 \begin {gather*} \frac {3 \, {\left (x^{7} \log \relax (2) + 7 \, x^{6} \log \relax (2) + 11 \, x^{5} \log \relax (2) + 3 \, x^{4} \log \relax (2) - 12 \, x^{3} \log \relax (2) - x^{2} e^{\left (\frac {x}{5 \, \log \relax (2)}\right )} \log \relax (2) + x \log \relax (2) + 5 \, \log \relax (2)\right )}}{x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-3*x^3*exp(1/5*x/log(2))+(75*x^7+420*x^6+495*x^5+90*x^4-180*x^3-15*x-150)*log(2))/x^3/log(2),x,
algorithm="giac")

[Out]

3*(x^7*log(2) + 7*x^6*log(2) + 11*x^5*log(2) + 3*x^4*log(2) - 12*x^3*log(2) - x^2*e^(1/5*x/log(2))*log(2) + x*
log(2) + 5*log(2))/(x^2*log(2))

________________________________________________________________________________________

maple [A]  time = 0.12, size = 48, normalized size = 1.50

method	result	size
norman	\(\frac {15+3 x -36 x^{3}+9 x^{4}+33 x^{5}+21 x^{6}+3 x^{7}-3 x^{2} {\mathrm e}^{\frac {x}{5 \ln \relax (2)}}}{x^{2}}\)	\(48\)
risch	\(3 x^{5}+21 x^{4}+33 x^{3}+9 x^{2}-36 x +\frac {15 x \ln \relax (2)+75 \ln \relax (2)}{5 \ln \relax (2) x^{2}}-3 \,{\mathrm e}^{\frac {x}{5 \ln \relax (2)}}\)	\(54\)
derivativedivides	\(\frac {-4500 x \ln \relax (2)^{3}+\frac {1875 \ln \relax (2)^{3}}{x^{2}}+\frac {375 \ln \relax (2)^{3}}{x}+1125 x^{2} \ln \relax (2)^{3}+4125 x^{3} \ln \relax (2)^{3}+2625 x^{4} \ln \relax (2)^{3}+375 x^{5} \ln \relax (2)^{3}-375 \ln \relax (2)^{3} {\mathrm e}^{\frac {x}{5 \ln \relax (2)}}}{125 \ln \relax (2)^{3}}\)	\(83\)
default	\(\frac {-4500 x \ln \relax (2)^{3}+\frac {1875 \ln \relax (2)^{3}}{x^{2}}+\frac {375 \ln \relax (2)^{3}}{x}+1125 x^{2} \ln \relax (2)^{3}+4125 x^{3} \ln \relax (2)^{3}+2625 x^{4} \ln \relax (2)^{3}+375 x^{5} \ln \relax (2)^{3}-375 \ln \relax (2)^{3} {\mathrm e}^{\frac {x}{5 \ln \relax (2)}}}{125 \ln \relax (2)^{3}}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-3*x^3*exp(1/5*x/ln(2))+(75*x^7+420*x^6+495*x^5+90*x^4-180*x^3-15*x-150)*ln(2))/x^3/ln(2),x,method=_R
ETURNVERBOSE)

[Out]

(15+3*x-36*x^3+9*x^4+33*x^5+21*x^6+3*x^7-3*x^2*exp(1/5*x/ln(2)))/x^2

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 64, normalized size = 2.00 \begin {gather*} \frac {3 \, {\left (x^{5} \log \relax (2) + 7 \, x^{4} \log \relax (2) + 11 \, x^{3} \log \relax (2) + 3 \, x^{2} \log \relax (2) - 12 \, x \log \relax (2) - e^{\left (\frac {x}{5 \, \log \relax (2)}\right )} \log \relax (2) + \frac {\log \relax (2)}{x} + \frac {5 \, \log \relax (2)}{x^{2}}\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-3*x^3*exp(1/5*x/log(2))+(75*x^7+420*x^6+495*x^5+90*x^4-180*x^3-15*x-150)*log(2))/x^3/log(2),x,
algorithm="maxima")

[Out]

3*(x^5*log(2) + 7*x^4*log(2) + 11*x^3*log(2) + 3*x^2*log(2) - 12*x*log(2) - e^(1/5*x/log(2))*log(2) + log(2)/x
+ 5*log(2)/x^2)/log(2)

________________________________________________________________________________________

mupad [B]  time = 3.32, size = 43, normalized size = 1.34 \begin {gather*} \frac {3\,x+15}{x^2}-3\,{\mathrm {e}}^{\frac {x}{5\,\ln \relax (2)}}-36\,x+9\,x^2+33\,x^3+21\,x^4+3\,x^5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(2)*(90*x^4 - 180*x^3 - 15*x + 495*x^5 + 420*x^6 + 75*x^7 - 150))/5 - (3*x^3*exp(x/(5*log(2))))/5)/(x
^3*log(2)),x)

[Out]

(3*x + 15)/x^2 - 3*exp(x/(5*log(2))) - 36*x + 9*x^2 + 33*x^3 + 21*x^4 + 3*x^5

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 41, normalized size = 1.28 \begin {gather*} 3 x^{5} + 21 x^{4} + 33 x^{3} + 9 x^{2} - 36 x - 3 e^{\frac {x}{5 \log {\relax (2 )}}} + \frac {3 x + 15}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-3*x**3*exp(1/5*x/ln(2))+(75*x**7+420*x**6+495*x**5+90*x**4-180*x**3-15*x-150)*ln(2))/x**3/ln(2
),x)

[Out]

3*x**5 + 21*x**4 + 33*x**3 + 9*x**2 - 36*x - 3*exp(x/(5*log(2))) + (3*x + 15)/x**2

________________________________________________________________________________________