3.51.58 \(\int \frac {10 x^3+e^x (e^5 (2-x)-3 x^3-3 x^4)-6 x^3 (i \pi +\log (3))+x^3 (i \pi +\log (3))^2}{x^3} \, dx\)

Optimal. Leaf size=33 \[ x-e^x \left (\frac {e^5}{x^2}+3 x\right )+x (3-i \pi -\log (3))^2 \]

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Rubi [A]  time = 0.18, antiderivative size = 45, normalized size of antiderivative = 1.36, number of steps used = 14, number of rules used = 7, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6, 14, 2199, 2194, 2177, 2178, 2176} \begin {gather*} -\frac {e^{x+5}}{x^2}-3 e^x x+x \left (10-\pi ^2+\log ^2(3)-2 i \pi (3-\log (3))-\log (729)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x^3 + E^x*(E^5*(2 - x) - 3*x^3 - 3*x^4) - 6*x^3*(I*Pi + Log[3]) + x^3*(I*Pi + Log[3])^2)/x^3,x]

[Out]

-(E^(5 + x)/x^2) - 3*E^x*x + x*(10 - Pi^2 - (2*I)*Pi*(3 - Log[3]) + Log[3]^2 - Log[729])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (e^5 (2-x)-3 x^3-3 x^4\right )+x^3 (i \pi +\log (3))^2+x^3 (10-6 (i \pi +\log (3)))}{x^3} \, dx\\ &=\int \frac {e^x \left (e^5 (2-x)-3 x^3-3 x^4\right )+x^3 \left (10-6 (i \pi +\log (3))+(i \pi +\log (3))^2\right )}{x^3} \, dx\\ &=\int \left (-\frac {e^x \left (-2 e^5+e^5 x+3 x^3+3 x^4\right )}{x^3}+10 \left (1+\frac {1}{10} \left (-\pi ^2+2 i \pi (-3+\log (3))+(-6+\log (3)) \log (3)\right )\right )\right ) \, dx\\ &=x \left (10-\pi ^2-2 i \pi (3-\log (3))+\log ^2(3)-\log (729)\right )-\int \frac {e^x \left (-2 e^5+e^5 x+3 x^3+3 x^4\right )}{x^3} \, dx\\ &=x \left (10-\pi ^2-2 i \pi (3-\log (3))+\log ^2(3)-\log (729)\right )-\int \left (3 e^x-\frac {2 e^{5+x}}{x^3}+\frac {e^{5+x}}{x^2}+3 e^x x\right ) \, dx\\ &=x \left (10-\pi ^2-2 i \pi (3-\log (3))+\log ^2(3)-\log (729)\right )+2 \int \frac {e^{5+x}}{x^3} \, dx-3 \int e^x \, dx-3 \int e^x x \, dx-\int \frac {e^{5+x}}{x^2} \, dx\\ &=-3 e^x-\frac {e^{5+x}}{x^2}+\frac {e^{5+x}}{x}-3 e^x x+x \left (10-\pi ^2-2 i \pi (3-\log (3))+\log ^2(3)-\log (729)\right )+3 \int e^x \, dx+\int \frac {e^{5+x}}{x^2} \, dx-\int \frac {e^{5+x}}{x} \, dx\\ &=-\frac {e^{5+x}}{x^2}-3 e^x x-e^5 \text {Ei}(x)+x \left (10-\pi ^2-2 i \pi (3-\log (3))+\log ^2(3)-\log (729)\right )+\int \frac {e^{5+x}}{x} \, dx\\ &=-\frac {e^{5+x}}{x^2}-3 e^x x+x \left (10-\pi ^2-2 i \pi (3-\log (3))+\log ^2(3)-\log (729)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 1.42 \begin {gather*} -\frac {e^{5+x}}{x^2}+10 x-3 e^x x-\pi ^2 x+2 i \pi x (-3+\log (3))-6 x \log (3)+x \log ^2(3) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x^3 + E^x*(E^5*(2 - x) - 3*x^3 - 3*x^4) - 6*x^3*(I*Pi + Log[3]) + x^3*(I*Pi + Log[3])^2)/x^3,x]

[Out]

-(E^(5 + x)/x^2) + 10*x - 3*E^x*x - Pi^2*x + (2*I)*Pi*x*(-3 + Log[3]) - 6*x*Log[3] + x*Log[3]^2

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fricas [A]  time = 0.66, size = 53, normalized size = 1.61 \begin {gather*} -\frac {2 \, {\left (-i \, \pi + 3\right )} x^{3} \log \relax (3) - x^{3} \log \relax (3)^{2} - {\left (-6 i \, \pi - \pi ^{2} + 10\right )} x^{3} + {\left (3 \, x^{3} + e^{5}\right )} e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2-x)*exp(5)-3*x^4-3*x^3)*exp(x)+x^3*(log(3)+I*pi)^2-6*x^3*(log(3)+I*pi)+10*x^3)/x^3,x, algorithm=
"fricas")

[Out]

-(2*(-I*pi + 3)*x^3*log(3) - x^3*log(3)^2 - (-6*I*pi - pi^2 + 10)*x^3 + (3*x^3 + e^5)*e^x)/x^2

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giac [B]  time = 0.15, size = 59, normalized size = 1.79 \begin {gather*} -\frac {\pi ^{2} x^{3} - 2 i \, \pi x^{3} \log \relax (3) - x^{3} \log \relax (3)^{2} + 6 i \, \pi x^{3} + 3 \, x^{3} e^{x} + 6 \, x^{3} \log \relax (3) - 10 \, x^{3} + e^{\left (x + 5\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2-x)*exp(5)-3*x^4-3*x^3)*exp(x)+x^3*(log(3)+I*pi)^2-6*x^3*(log(3)+I*pi)+10*x^3)/x^3,x, algorithm=
"giac")

[Out]

-(pi^2*x^3 - 2*I*pi*x^3*log(3) - x^3*log(3)^2 + 6*I*pi*x^3 + 3*x^3*e^x + 6*x^3*log(3) - 10*x^3 + e^(x + 5))/x^
2

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maple [A]  time = 0.14, size = 48, normalized size = 1.45

method	result	size
norman	\(\frac {\left (2 i \ln \relax (3) \pi -\pi ^{2}-6 i \pi +\ln \relax (3)^{2}-6 \ln \relax (3)+10\right ) x^{3}-{\mathrm e}^{5} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} x^{3}}{x^{2}}\)	\(48\)
risch	\(2 i \ln \relax (3) \pi x -x \,\pi ^{2}-6 i x \pi +x \ln \relax (3)^{2}-6 x \ln \relax (3)+10 x -\frac {\left (3 x^{3}+{\mathrm e}^{5}\right ) {\mathrm e}^{x}}{x^{2}}\)	\(49\)
default	\(10 x +x \ln \relax (3)^{2}-6 i x \pi +2 i \ln \relax (3) \pi x -x \,\pi ^{2}-3 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{2 x^{2}}-\frac {{\mathrm e}^{x}}{2 x}-\frac {\expIntegralEi \left (1, -x \right )}{2}\right )-{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{x}-\expIntegralEi \left (1, -x \right )\right )-6 x \ln \relax (3)\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2-x)*exp(5)-3*x^4-3*x^3)*exp(x)+x^3*(ln(3)+I*Pi)^2-6*x^3*(ln(3)+I*Pi)+10*x^3)/x^3,x,method=_RETURNVERBO
SE)

[Out]

((2*I*ln(3)*Pi-Pi^2-6*I*Pi+ln(3)^2-6*ln(3)+10)*x^3-exp(5)*exp(x)-3*exp(x)*x^3)/x^2

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maxima [C]  time = 0.37, size = 60, normalized size = 1.82 \begin {gather*} -\pi ^{2} x + 2 i \, \pi x \log \relax (3) + x \log \relax (3)^{2} - 6 i \, \pi x - 3 \, {\left (x - 1\right )} e^{x} - e^{5} \Gamma \left (-1, -x\right ) - 2 \, e^{5} \Gamma \left (-2, -x\right ) - 6 \, x \log \relax (3) + 10 \, x - 3 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2-x)*exp(5)-3*x^4-3*x^3)*exp(x)+x^3*(log(3)+I*pi)^2-6*x^3*(log(3)+I*pi)+10*x^3)/x^3,x, algorithm=
"maxima")

[Out]

-pi^2*x + 2*I*pi*x*log(3) + x*log(3)^2 - 6*I*pi*x - 3*(x - 1)*e^x - e^5*gamma(-1, -x) - 2*e^5*gamma(-2, -x) -
6*x*log(3) + 10*x - 3*e^x

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mupad [B]  time = 0.25, size = 40, normalized size = 1.21 \begin {gather*} -x\,\left (\ln \left (729\right )+3\,{\mathrm {e}}^x+\Pi ^2-{\ln \relax (3)}^2-10+\Pi \,6{}\mathrm {i}-\Pi \,\ln \relax (3)\,2{}\mathrm {i}\right )-\frac {{\mathrm {e}}^5\,{\mathrm {e}}^x}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(Pi*1i + log(3))^2 - exp(x)*(exp(5)*(x - 2) + 3*x^3 + 3*x^4) - 6*x^3*(Pi*1i + log(3)) + 10*x^3)/x^3,x
)

[Out]

- x*(Pi*6i + log(729) + 3*exp(x) - Pi*log(3)*2i + Pi^2 - log(3)^2 - 10) - (exp(5)*exp(x))/x^2

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sympy [A]  time = 0.19, size = 46, normalized size = 1.39 \begin {gather*} x \left (- \pi ^{2} - 6 \log {\relax (3 )} + \log {\relax (3 )}^{2} + 10 - 6 i \pi + 2 i \pi \log {\relax (3 )}\right ) + \frac {\left (- 3 x^{3} - e^{5}\right ) e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2-x)*exp(5)-3*x**4-3*x**3)*exp(x)+x**3*(ln(3)+I*pi)**2-6*x**3*(ln(3)+I*pi)+10*x**3)/x**3,x)

[Out]

x*(-pi**2 - 6*log(3) + log(3)**2 + 10 - 6*I*pi + 2*I*pi*log(3)) + (-3*x**3 - exp(5))*exp(x)/x**2

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