3.104.32 \(\int (5+10 e^{2 x}+10 e^x \log (7)) \, dx\) [10332]

Optimal. Leaf size=13 \[ 5 \left (-175+x+\left (e^x+\log (7)\right )^2\right ) \]

[Out]

-875+5*(exp(x)+ln(7))^2+5*x

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Rubi [A]
time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2225} \begin {gather*} 5 x+5 e^{2 x}+10 e^x \log (7) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[5 + 10*E^(2*x) + 10*E^x*Log[7],x]

[Out]

5*E^(2*x) + 5*x + 10*E^x*Log[7]

Rule 2225

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} &=5 x+10 \int e^{2 x} \, dx+(10 \log (7)) \int e^x \, dx\\ &=5 e^{2 x}+5 x+10 e^x \log (7)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 15, normalized size = 1.15 \begin {gather*} 5 \left (e^{2 x}+x+e^x \log (49)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[5 + 10*E^(2*x) + 10*E^x*Log[7],x]

[Out]

5*(E^(2*x) + x + E^x*Log[49])

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Maple [A]
time = 0.02, size = 17, normalized size = 1.31

method	result	size
default	\(5 x +5 \,{\mathrm e}^{2 x}+10 \ln \left (7\right ) {\mathrm e}^{x}\)	\(17\)
norman	\(5 x +5 \,{\mathrm e}^{2 x}+10 \ln \left (7\right ) {\mathrm e}^{x}\)	\(17\)
risch	\(5 x +5 \,{\mathrm e}^{2 x}+10 \ln \left (7\right ) {\mathrm e}^{x}\)	\(17\)
derivativedivides	\(5 \,{\mathrm e}^{2 x}+10 \ln \left (7\right ) {\mathrm e}^{x}+5 \ln \left ({\mathrm e}^{x}\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*exp(x)^2+10*ln(7)*exp(x)+5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 16, normalized size = 1.23 \begin {gather*} 10 \, e^{x} \log \left (7\right ) + 5 \, x + 5 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*exp(x)^2+10*log(7)*exp(x)+5,x, algorithm="maxima")

[Out]

10*e^x*log(7) + 5*x + 5*e^(2*x)

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Fricas [A]
time = 0.34, size = 16, normalized size = 1.23 \begin {gather*} 10 \, e^{x} \log \left (7\right ) + 5 \, x + 5 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*exp(x)^2+10*log(7)*exp(x)+5,x, algorithm="fricas")

[Out]

10*e^x*log(7) + 5*x + 5*e^(2*x)

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Sympy [A]
time = 0.04, size = 17, normalized size = 1.31 \begin {gather*} 5 x + 5 e^{2 x} + 10 e^{x} \log {\left (7 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*exp(x)**2+10*ln(7)*exp(x)+5,x)

[Out]

5*x + 5*exp(2*x) + 10*exp(x)*log(7)

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Giac [A]
time = 0.42, size = 16, normalized size = 1.23 \begin {gather*} 10 \, e^{x} \log \left (7\right ) + 5 \, x + 5 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*exp(x)^2+10*log(7)*exp(x)+5,x, algorithm="giac")

[Out]

10*e^x*log(7) + 5*x + 5*e^(2*x)

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Mupad [B]
time = 0.06, size = 16, normalized size = 1.23 \begin {gather*} 5\,x+5\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^x\,\ln \left (7\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*exp(2*x) + 10*exp(x)*log(7) + 5,x)

[Out]

5*x + 5*exp(2*x) + 10*exp(x)*log(7)

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