3.98.1 \(\int \frac {1}{4} (109+e^{2 x} x^2 (105+70 x)) \, dx\)

Optimal. Leaf size=19 \[ x \left (1+\frac {35}{4} \left (3+e^{2 x} x^2\right )\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2196, 2176, 2194} \begin {gather*} \frac {35}{4} e^{2 x} x^3+\frac {109 x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(109 + E^(2*x)*x^2*(105 + 70*x))/4,x]

[Out]

(109*x)/4 + (35*E^(2*x)*x^3)/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 2196

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (109+e^{2 x} x^2 (105+70 x)\right ) \, dx\\ &=\frac {109 x}{4}+\frac {1}{4} \int e^{2 x} x^2 (105+70 x) \, dx\\ &=\frac {109 x}{4}+\frac {1}{4} \int \left (105 e^{2 x} x^2+70 e^{2 x} x^3\right ) \, dx\\ &=\frac {109 x}{4}+\frac {35}{2} \int e^{2 x} x^3 \, dx+\frac {105}{4} \int e^{2 x} x^2 \, dx\\ &=\frac {109 x}{4}+\frac {105}{8} e^{2 x} x^2+\frac {35}{4} e^{2 x} x^3-\frac {105}{4} \int e^{2 x} x \, dx-\frac {105}{4} \int e^{2 x} x^2 \, dx\\ &=\frac {109 x}{4}-\frac {105}{8} e^{2 x} x+\frac {35}{4} e^{2 x} x^3+\frac {105}{8} \int e^{2 x} \, dx+\frac {105}{4} \int e^{2 x} x \, dx\\ &=\frac {105 e^{2 x}}{16}+\frac {109 x}{4}+\frac {35}{4} e^{2 x} x^3-\frac {105}{8} \int e^{2 x} \, dx\\ &=\frac {109 x}{4}+\frac {35}{4} e^{2 x} x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.95 \begin {gather*} \frac {1}{4} \left (109 x+35 e^{2 x} x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(109 + E^(2*x)*x^2*(105 + 70*x))/4,x]

[Out]

(109*x + 35*E^(2*x)*x^3)/4

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fricas [A]  time = 0.59, size = 16, normalized size = 0.84 \begin {gather*} \frac {35}{4} \, x e^{\left (2 \, x + \log \left (x^{2}\right )\right )} + \frac {109}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(70*x+105)*exp(log(x^2)+2*x)+109/4,x, algorithm="fricas")

[Out]

35/4*x*e^(2*x + log(x^2)) + 109/4*x

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giac [A]  time = 0.19, size = 13, normalized size = 0.68 \begin {gather*} \frac {35}{4} \, x^{3} e^{\left (2 \, x\right )} + \frac {109}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(70*x+105)*exp(log(x^2)+2*x)+109/4,x, algorithm="giac")

[Out]

35/4*x^3*e^(2*x) + 109/4*x

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maple [A]  time = 0.04, size = 14, normalized size = 0.74

method	result	size
risch	\(\frac {109 x}{4}+\frac {35 \,{\mathrm e}^{2 x} x^{3}}{4}\)	\(14\)
norman	\(\frac {109 x}{4}+\frac {35 \,{\mathrm e}^{\ln \left (x^{2}\right )+2 x} x}{4}\)	\(17\)
default	\(\frac {109 x}{4}+\frac {35 \,{\mathrm e}^{\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x} x^{3}}{4}+\frac {35 \,{\mathrm e}^{\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x} x^{2}}{8}-\frac {35 \,{\mathrm e}^{\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x} x}{4}+\frac {35 \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right ) \left (\frac {\left (\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x \right ) {\mathrm e}^{\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x}}{2}-{\mathrm e}^{\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x}\right )}{8}+\frac {35 \left (\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x \right ) {\mathrm e}^{\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x}}{8}-\frac {35 \,{\mathrm e}^{\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x} \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right )^{2}}{32}-\frac {35 \,{\mathrm e}^{\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x} \left (\ln \left (x^{2}\right )-2 \ln \relax (x )+2 x \right )^{2}}{32}\)	\(193\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(70*x+105)*exp(ln(x^2)+2*x)+109/4,x,method=_RETURNVERBOSE)

[Out]

109/4*x+35/4*exp(2*x)*x^3

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maxima [B]  time = 0.37, size = 41, normalized size = 2.16 \begin {gather*} \frac {35}{16} \, {\left (4 \, x^{3} - 6 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + \frac {105}{16} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + \frac {109}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(70*x+105)*exp(log(x^2)+2*x)+109/4,x, algorithm="maxima")

[Out]

35/16*(4*x^3 - 6*x^2 + 6*x - 3)*e^(2*x) + 105/16*(2*x^2 - 2*x + 1)*e^(2*x) + 109/4*x

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mupad [B]  time = 5.71, size = 13, normalized size = 0.68 \begin {gather*} \frac {109\,x}{4}+\frac {35\,x^3\,{\mathrm {e}}^{2\,x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x + log(x^2))*(70*x + 105))/4 + 109/4,x)

[Out]

(109*x)/4 + (35*x^3*exp(2*x))/4

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sympy [A]  time = 0.09, size = 15, normalized size = 0.79 \begin {gather*} \frac {35 x^{3} e^{2 x}}{4} + \frac {109 x}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(70*x+105)*exp(ln(x**2)+2*x)+109/4,x)

[Out]

35*x**3*exp(2*x)/4 + 109*x/4

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