3.103.79 \(\int \frac {-5+e^{4 x^2+4 x^3} (14 x^2+112 x^4+168 x^5)}{x^2} \, dx\) [10279]

Optimal. Leaf size=20 \[ 5+\frac {5}{x}+14 e^{4 x \left (x+x^2\right )} x \]

[Out]

5+5/x+14*exp(4*x*(x^2+x))*x

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(41\) vs. \(2(20)=40\).
time = 0.05, antiderivative size = 41, normalized size of antiderivative = 2.05, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14, 2326} \begin {gather*} \frac {14 e^{4 x^2 (x+1)} \left (3 x^3+2 x^2\right )}{x^2+2 (x+1) x}+\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + E^(4*x^2 + 4*x^3)*(14*x^2 + 112*x^4 + 168*x^5))/x^2,x]

[Out]

5/x + (14*E^(4*x^2*(1 + x))*(2*x^2 + 3*x^3))/(x^2 + 2*x*(1 + 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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {5}{x^2}+14 e^{4 x^2 (1+x)} \left (1+8 x^2+12 x^3\right )\right ) \, dx\\ &=\frac {5}{x}+14 \int e^{4 x^2 (1+x)} \left (1+8 x^2+12 x^3\right ) \, dx\\ &=\frac {5}{x}+\frac {14 e^{4 x^2 (1+x)} \left (2 x^2+3 x^3\right )}{x^2+2 x (1+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.07, size = 19, normalized size = 0.95 \begin {gather*} \frac {5}{x}+14 e^{4 x^2 (1+x)} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + E^(4*x^2 + 4*x^3)*(14*x^2 + 112*x^4 + 168*x^5))/x^2,x]

[Out]

5/x + 14*E^(4*x^2*(1 + x))*x

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Maple [A]
time = 0.03, size = 19, normalized size = 0.95

method	result	size
risch	\(\frac {5}{x}+14 x \,{\mathrm e}^{4 \left (x +1\right ) x^{2}}\)	\(19\)
norman	\(\frac {5+14 \,{\mathrm e}^{4 x^{3}+4 x^{2}} x^{2}}{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((168*x^5+112*x^4+14*x^2)*exp(4*x^3+4*x^2)-5)/x^2,x,method=_RETURNVERBOSE)

[Out]

5/x+14*x*exp(4*(x+1)*x^2)

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Maxima [A]
time = 0.31, size = 21, normalized size = 1.05 \begin {gather*} 14 \, x e^{\left (4 \, x^{3} + 4 \, x^{2}\right )} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((168*x^5+112*x^4+14*x^2)*exp(4*x^3+4*x^2)-5)/x^2,x, algorithm="maxima")

[Out]

14*x*e^(4*x^3 + 4*x^2) + 5/x

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Fricas [A]
time = 0.35, size = 23, normalized size = 1.15 \begin {gather*} \frac {14 \, x^{2} e^{\left (4 \, x^{3} + 4 \, x^{2}\right )} + 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((168*x^5+112*x^4+14*x^2)*exp(4*x^3+4*x^2)-5)/x^2,x, algorithm="fricas")

[Out]

(14*x^2*e^(4*x^3 + 4*x^2) + 5)/x

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Sympy [A]
time = 0.05, size = 17, normalized size = 0.85 \begin {gather*} 14 x e^{4 x^{3} + 4 x^{2}} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((168*x**5+112*x**4+14*x**2)*exp(4*x**3+4*x**2)-5)/x**2,x)

[Out]

14*x*exp(4*x**3 + 4*x**2) + 5/x

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Giac [A]
time = 0.41, size = 23, normalized size = 1.15 \begin {gather*} \frac {14 \, x^{2} e^{\left (4 \, x^{3} + 4 \, x^{2}\right )} + 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((168*x^5+112*x^4+14*x^2)*exp(4*x^3+4*x^2)-5)/x^2,x, algorithm="giac")

[Out]

(14*x^2*e^(4*x^3 + 4*x^2) + 5)/x

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Mupad [B]
time = 6.81, size = 21, normalized size = 1.05 \begin {gather*} 14\,x\,{\mathrm {e}}^{4\,x^3+4\,x^2}+\frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x^2 + 4*x^3)*(14*x^2 + 112*x^4 + 168*x^5) - 5)/x^2,x)

[Out]

14*x*exp(4*x^2 + 4*x^3) + 5/x

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