3.57.70 \(\int \frac {40 x+e^x (20+20 x)+(-3-7 x^6) \log (25)}{\log (25)} \, dx\)

Optimal. Leaf size=22 \[ x+x \left (-4-x^6+\frac {20 \left (e^x+x\right )}{\log (25)}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.77, number of steps used = 5, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 2176, 2194} \begin {gather*} -x^7+\frac {20 x^2}{\log (25)}-3 x-\frac {20 e^x}{\log (25)}+\frac {20 e^x (x+1)}{\log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40*x + E^x*(20 + 20*x) + (-3 - 7*x^6)*Log[25])/Log[25],x]

[Out]

-3*x - x^7 - (20*E^x)/Log[25] + (20*x^2)/Log[25] + (20*E^x*(1 + x))/Log[25]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (40 x+e^x (20+20 x)+\left (-3-7 x^6\right ) \log (25)\right ) \, dx}{\log (25)}\\ &=\frac {20 x^2}{\log (25)}+\frac {\int e^x (20+20 x) \, dx}{\log (25)}+\int \left (-3-7 x^6\right ) \, dx\\ &=-3 x-x^7+\frac {20 x^2}{\log (25)}+\frac {20 e^x (1+x)}{\log (25)}-\frac {20 \int e^x \, dx}{\log (25)}\\ &=-3 x-x^7-\frac {20 e^x}{\log (25)}+\frac {20 x^2}{\log (25)}+\frac {20 e^x (1+x)}{\log (25)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 29, normalized size = 1.32 \begin {gather*} \frac {20 e^x x+20 x^2-3 x \log (25)-x^7 \log (25)}{\log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40*x + E^x*(20 + 20*x) + (-3 - 7*x^6)*Log[25])/Log[25],x]

[Out]

(20*E^x*x + 20*x^2 - 3*x*Log[25] - x^7*Log[25])/Log[25]

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fricas [A]  time = 0.62, size = 27, normalized size = 1.23 \begin {gather*} \frac {10 \, x^{2} + 10 \, x e^{x} - {\left (x^{7} + 3 \, x\right )} \log \relax (5)}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((20*x+20)*exp(x)+2*(-7*x^6-3)*log(5)+40*x)/log(5),x, algorithm="fricas")

[Out]

(10*x^2 + 10*x*e^x - (x^7 + 3*x)*log(5))/log(5)

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giac [A]  time = 0.17, size = 27, normalized size = 1.23 \begin {gather*} \frac {10 \, x^{2} + 10 \, x e^{x} - {\left (x^{7} + 3 \, x\right )} \log \relax (5)}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((20*x+20)*exp(x)+2*(-7*x^6-3)*log(5)+40*x)/log(5),x, algorithm="giac")

[Out]

(10*x^2 + 10*x*e^x - (x^7 + 3*x)*log(5))/log(5)

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maple [A]  time = 0.06, size = 28, normalized size = 1.27

method	result	size
norman	\(-3 x -x^{7}+\frac {10 x^{2}}{\ln \relax (5)}+\frac {10 \,{\mathrm e}^{x} x}{\ln \relax (5)}\)	\(28\)
risch	\(-3 x -x^{7}+\frac {10 x^{2}}{\ln \relax (5)}+\frac {10 \,{\mathrm e}^{x} x}{\ln \relax (5)}\)	\(28\)
default	\(\frac {20 \,{\mathrm e}^{x} x +20 x^{2}-2 x^{7} \ln \relax (5)-6 x \ln \relax (5)}{2 \ln \relax (5)}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((20*x+20)*exp(x)+2*(-7*x^6-3)*ln(5)+40*x)/ln(5),x,method=_RETURNVERBOSE)

[Out]

-3*x-x^7+10*x^2/ln(5)+10*exp(x)/ln(5)*x

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maxima [A]  time = 0.36, size = 27, normalized size = 1.23 \begin {gather*} \frac {10 \, x^{2} + 10 \, x e^{x} - {\left (x^{7} + 3 \, x\right )} \log \relax (5)}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((20*x+20)*exp(x)+2*(-7*x^6-3)*log(5)+40*x)/log(5),x, algorithm="maxima")

[Out]

(10*x^2 + 10*x*e^x - (x^7 + 3*x)*log(5))/log(5)

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mupad [B]  time = 0.06, size = 25, normalized size = 1.14 \begin {gather*} \frac {x\,\left (10\,x-3\,\ln \relax (5)+10\,{\mathrm {e}}^x-x^6\,\ln \relax (5)\right )}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x - log(5)*(7*x^6 + 3) + (exp(x)*(20*x + 20))/2)/log(5),x)

[Out]

(x*(10*x - 3*log(5) + 10*exp(x) - x^6*log(5)))/log(5)

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sympy [A]  time = 0.11, size = 24, normalized size = 1.09 \begin {gather*} - x^{7} + \frac {10 x^{2}}{\log {\relax (5 )}} + \frac {10 x e^{x}}{\log {\relax (5 )}} - 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((20*x+20)*exp(x)+2*(-7*x**6-3)*ln(5)+40*x)/ln(5),x)

[Out]

-x**7 + 10*x**2/log(5) + 10*x*exp(x)/log(5) - 3*x

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