3.73.55 \(\int \frac {2 e^3 (1+x)^2}{25+25 x} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{4} e^5 \left (4+\frac {4 (1+x)^2}{25 e^2}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 21} \begin {gather*} \frac {e^3 x^2}{25}+\frac {2 e^3 x}{25} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^3*(1 + x)^2)/(25 + 25*x),x]

[Out]

(2*E^3*x)/25 + (E^3*x^2)/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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.76 \begin {gather*} \frac {2}{25} e^3 \left (x+\frac {x^2}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^3*(1 + x)^2)/(25 + 25*x),x]

[Out]

(2*E^3*(x + x^2/2))/25

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fricas [A]  time = 0.67, size = 11, normalized size = 0.52 \begin {gather*} \frac {1}{25} \, {\left (x^{2} + 2 \, x\right )} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(5)*exp(log(x+1)-1)^2/(25*x+25),x, algorithm="fricas")

[Out]

1/25*(x^2 + 2*x)*e^3

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giac [A]  time = 0.16, size = 13, normalized size = 0.62 \begin {gather*} \frac {1}{25} \, x^{2} e^{3} + \frac {2}{25} \, x e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(5)*exp(log(x+1)-1)^2/(25*x+25),x, algorithm="giac")

[Out]

1/25*x^2*e^3 + 2/25*x*e^3

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maple [A]  time = 0.16, size = 14, normalized size = 0.67

method	result	size
default	\(\frac {{\mathrm e}^{5} \left (x +1\right )^{2} {\mathrm e}^{-2}}{25}\)	\(14\)
risch	\(\frac {x^{2} {\mathrm e}^{3}}{25}+\frac {2 x \,{\mathrm e}^{3}}{25}\)	\(14\)
gosper	\(\frac {x \left (2+x \right ) {\mathrm e}^{5} {\mathrm e}^{-2}}{25}\)	\(23\)
norman	\(\left (\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{-1} x}{25}+\frac {{\mathrm e}^{5} {\mathrm e}^{-1} x^{2}}{25}\right ) {\mathrm e}^{-1}\)	\(27\)
meijerg	\(\frac {2 \left (x +1\right )^{-2 \,{\mathrm e}^{-2}+2} {\mathrm e}^{5} \hypergeom \left (\left [1, 1-2 \,{\mathrm e}^{-2}\right ], \relax [2], -x \right ) x \,{\mathrm e}^{-2}}{25}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(5)*exp(ln(x+1)-1)^2/(25*x+25),x,method=_RETURNVERBOSE)

[Out]

1/25*exp(5)*exp(ln(x+1)-1)^2

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maxima [A]  time = 0.35, size = 11, normalized size = 0.52 \begin {gather*} \frac {1}{25} \, {\left (x^{2} + 2 \, x\right )} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(5)*exp(log(x+1)-1)^2/(25*x+25),x, algorithm="maxima")

[Out]

1/25*(x^2 + 2*x)*e^3

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mupad [B]  time = 4.34, size = 8, normalized size = 0.38 \begin {gather*} \frac {x\,{\mathrm {e}}^3\,\left (x+2\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(5)*exp(2*log(x + 1) - 2))/(25*x + 25),x)

[Out]

(x*exp(3)*(x + 2))/25

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sympy [A]  time = 0.08, size = 15, normalized size = 0.71 \begin {gather*} \frac {x^{2} e^{3}}{25} + \frac {2 x e^{3}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(5)*exp(ln(x+1)-1)**2/(25*x+25),x)

[Out]

x**2*exp(3)/25 + 2*x*exp(3)/25

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