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

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

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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} $$\begin{array}{c}\frac{e^3{x}^2}{25}+\frac{2e^{3}x}{25}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\left(2e^3\right)\int \frac{(1+x{)}^2}{25+25x}dx\\ & =\frac{1}{25}\left(2e^3\right)\int (1+x)dx\\ & =\frac{2e^{3}x}{25}+\frac{e^3{x}^2}{25}\end{array}\end{array}$$

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.76 $$\begin{array}{c}\frac{2}{25}{e}^3(x+\frac{x^2}{2})\end{array}$$

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 $$\begin{array}{c}\frac{1}{25}(x^2+2x)e^3\end{array}$$

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 $$\begin{array}{c}\frac{1}{25}{x}^2{e}^3+\frac{2}{25}x{e}^3\end{array}$$

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	$\frac{{\mathrm{e}}^5{(x+1)}^2{\mathrm{e}}^{-2}}{25}$	$14$
risch	$\frac{x^2{\mathrm{e}}^3}{25}+\frac{2x{\mathrm{e}}^3}{25}$	$14$
gosper	$\frac{x(2+x){\mathrm{e}}^5{\mathrm{e}}^{-2}}{25}$	$23$
norman	$(\frac{2{\mathrm{e}}^5{\mathrm{e}}^{-1}x}{25}+\frac{{\mathrm{e}}^5{\mathrm{e}}^{-1}{x}^2}{25}){\mathrm{e}}^{-1}$	$27$
meijerg	$\frac{2{(x+1)}^{-2{\mathrm{e}}^{-2}+2}{\mathrm{e}}^5\mathrm{hypergeom}([1,1-2{\mathrm{e}}^{-2}],[2],-x)x{\mathrm{e}}^{-2}}{25}$	$32$

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 $$\begin{array}{c}\frac{1}{25}(x^2+2x)e^3\end{array}$$

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 $$\begin{array}{c}\frac{x{\mathrm{e}}^3(x+2)}{25}\end{array}$$

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 $$\begin{array}{c}\frac{x^2{e}^3}{25}+\frac{2x{e}^3}{25}\end{array}$$

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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