3.53.29 \(\int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x (48-16 x+8 x^5+16 x^6+4 x^7)}{x^4} \, dx\)

Optimal. Leaf size=21 \[ \left (-\frac {4}{x^4}+x+x^2\right ) \left (x+4 x \left (e^x+x\right )\right ) \]

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Rubi [B]  time = 0.19, antiderivative size = 48, normalized size of antiderivative = 2.29, number of steps used = 22, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {14, 2199, 2177, 2178, 2176, 2194} \begin {gather*} 4 x^4+4 e^x x^3+5 x^3-\frac {16 e^x}{x^3}-\frac {4}{x^3}+4 e^x x^2+x^2-\frac {16}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + 32*x + 2*x^5 + 15*x^6 + 16*x^7 + E^x*(48 - 16*x + 8*x^5 + 16*x^6 + 4*x^7))/x^4,x]

[Out]

-4/x^3 - (16*E^x)/x^3 - 16/x^2 + x^2 + 4*E^x*x^2 + 5*x^3 + 4*E^x*x^3 + 4*x^4

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 \left (\frac {4 e^x \left (12-4 x+2 x^5+4 x^6+x^7\right )}{x^4}+\frac {12+32 x+2 x^5+15 x^6+16 x^7}{x^4}\right ) \, dx\\ &=4 \int \frac {e^x \left (12-4 x+2 x^5+4 x^6+x^7\right )}{x^4} \, dx+\int \frac {12+32 x+2 x^5+15 x^6+16 x^7}{x^4} \, dx\\ &=4 \int \left (\frac {12 e^x}{x^4}-\frac {4 e^x}{x^3}+2 e^x x+4 e^x x^2+e^x x^3\right ) \, dx+\int \left (\frac {12}{x^4}+\frac {32}{x^3}+2 x+15 x^2+16 x^3\right ) \, dx\\ &=-\frac {4}{x^3}-\frac {16}{x^2}+x^2+5 x^3+4 x^4+4 \int e^x x^3 \, dx+8 \int e^x x \, dx-16 \int \frac {e^x}{x^3} \, dx+16 \int e^x x^2 \, dx+48 \int \frac {e^x}{x^4} \, dx\\ &=-\frac {4}{x^3}-\frac {16 e^x}{x^3}-\frac {16}{x^2}+\frac {8 e^x}{x^2}+8 e^x x+x^2+16 e^x x^2+5 x^3+4 e^x x^3+4 x^4-8 \int e^x \, dx-8 \int \frac {e^x}{x^2} \, dx-12 \int e^x x^2 \, dx+16 \int \frac {e^x}{x^3} \, dx-32 \int e^x x \, dx\\ &=-8 e^x-\frac {4}{x^3}-\frac {16 e^x}{x^3}-\frac {16}{x^2}+\frac {8 e^x}{x}-24 e^x x+x^2+4 e^x x^2+5 x^3+4 e^x x^3+4 x^4+8 \int \frac {e^x}{x^2} \, dx-8 \int \frac {e^x}{x} \, dx+24 \int e^x x \, dx+32 \int e^x \, dx\\ &=24 e^x-\frac {4}{x^3}-\frac {16 e^x}{x^3}-\frac {16}{x^2}+x^2+4 e^x x^2+5 x^3+4 e^x x^3+4 x^4-8 \text {Ei}(x)+8 \int \frac {e^x}{x} \, dx-24 \int e^x \, dx\\ &=-\frac {4}{x^3}-\frac {16 e^x}{x^3}-\frac {16}{x^2}+x^2+4 e^x x^2+5 x^3+4 e^x x^3+4 x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 22, normalized size = 1.05 \begin {gather*} \frac {\left (1+4 e^x+4 x\right ) \left (-4+x^5+x^6\right )}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + 32*x + 2*x^5 + 15*x^6 + 16*x^7 + E^x*(48 - 16*x + 8*x^5 + 16*x^6 + 4*x^7))/x^4,x]

[Out]

((1 + 4*E^x + 4*x)*(-4 + x^5 + x^6))/x^3

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fricas [A]  time = 0.55, size = 34, normalized size = 1.62 \begin {gather*} \frac {4 \, x^{7} + 5 \, x^{6} + x^{5} + 4 \, {\left (x^{6} + x^{5} - 4\right )} e^{x} - 16 \, x - 4}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^7+16*x^6+8*x^5-16*x+48)*exp(x)+16*x^7+15*x^6+2*x^5+32*x+12)/x^4,x, algorithm="fricas")

[Out]

(4*x^7 + 5*x^6 + x^5 + 4*(x^6 + x^5 - 4)*e^x - 16*x - 4)/x^3

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giac [A]  time = 0.21, size = 40, normalized size = 1.90 \begin {gather*} \frac {4 \, x^{7} + 4 \, x^{6} e^{x} + 5 \, x^{6} + 4 \, x^{5} e^{x} + x^{5} - 16 \, x - 16 \, e^{x} - 4}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^7+16*x^6+8*x^5-16*x+48)*exp(x)+16*x^7+15*x^6+2*x^5+32*x+12)/x^4,x, algorithm="giac")

[Out]

(4*x^7 + 4*x^6*e^x + 5*x^6 + 4*x^5*e^x + x^5 - 16*x - 16*e^x - 4)/x^3

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maple [A]  time = 0.04, size = 39, normalized size = 1.86

method	result	size
risch	\(4 x^{4}+5 x^{3}+x^{2}+\frac {-16 x -4}{x^{3}}+\frac {4 \left (x^{6}+x^{5}-4\right ) {\mathrm e}^{x}}{x^{3}}\)	\(39\)
norman	\(\frac {-4+x^{5}-16 x +5 x^{6}+4 x^{7}+4 x^{5} {\mathrm e}^{x}+4 x^{6} {\mathrm e}^{x}-16 \,{\mathrm e}^{x}}{x^{3}}\)	\(41\)
default	\(x^{2}-\frac {4}{x^{3}}-\frac {16}{x^{2}}+5 x^{3}+4 x^{4}-\frac {16 \,{\mathrm e}^{x}}{x^{3}}+4 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^7+16*x^6+8*x^5-16*x+48)*exp(x)+16*x^7+15*x^6+2*x^5+32*x+12)/x^4,x,method=_RETURNVERBOSE)

[Out]

4*x^4+5*x^3+x^2+(-16*x-4)/x^3+4*(x^6+x^5-4)/x^3*exp(x)

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maxima [C]  time = 0.38, size = 74, normalized size = 3.52 \begin {gather*} 4 \, x^{4} + 5 \, x^{3} + x^{2} + 4 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} + 16 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 8 \, {\left (x - 1\right )} e^{x} - \frac {16}{x^{2}} - \frac {4}{x^{3}} + 16 \, \Gamma \left (-2, -x\right ) + 48 \, \Gamma \left (-3, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^7+16*x^6+8*x^5-16*x+48)*exp(x)+16*x^7+15*x^6+2*x^5+32*x+12)/x^4,x, algorithm="maxima")

[Out]

4*x^4 + 5*x^3 + x^2 + 4*(x^3 - 3*x^2 + 6*x - 6)*e^x + 16*(x^2 - 2*x + 2)*e^x + 8*(x - 1)*e^x - 16/x^2 - 4/x^3
+ 16*gamma(-2, -x) + 48*gamma(-3, -x)

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mupad [B]  time = 3.48, size = 21, normalized size = 1.00 \begin {gather*} \frac {\left (x^6+x^5-4\right )\,\left (4\,x+4\,{\mathrm {e}}^x+1\right )}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + exp(x)*(8*x^5 - 16*x + 16*x^6 + 4*x^7 + 48) + 2*x^5 + 15*x^6 + 16*x^7 + 12)/x^4,x)

[Out]

((x^5 + x^6 - 4)*(4*x + 4*exp(x) + 1))/x^3

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sympy [B]  time = 0.12, size = 41, normalized size = 1.95 \begin {gather*} 4 x^{4} + 5 x^{3} + x^{2} + \frac {- 16 x - 4}{x^{3}} + \frac {\left (4 x^{6} + 4 x^{5} - 16\right ) e^{x}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**7+16*x**6+8*x**5-16*x+48)*exp(x)+16*x**7+15*x**6+2*x**5+32*x+12)/x**4,x)

[Out]

4*x**4 + 5*x**3 + x**2 + (-16*x - 4)/x**3 + (4*x**6 + 4*x**5 - 16)*exp(x)/x**3

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