3.30.74 $\int \frac{2048+4096x^2+3072x\log (2)+e^x(100x+75\log (2))}{100x+75\log (2)}dx$

Optimal. Leaf size=21 $$e^x+\frac{512}{25}(x^2+\log (\frac{4x}{3}+\log (2)))$$

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Rubi [A]  time = 0.13, antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 3, integrand size = 35, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.086, Rules used = {6742, 2194, 698} $$\begin{array}{c}\frac{512x^2}{25}+e^x+\frac{512}{25}\log (4x+\log (8))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(2048 + 4096*x^2 + 3072*x*Log[2] + E^x*(100*x + 75*Log[2]))/(100*x + 75*Log[2]),x]

[Out]

E^x + (512*x^2)/25 + (512*Log[4*x + Log[8]])/25

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (e^x+\frac{1024(2+4x^2+x\log (8))}{25(4x+\log (8))})dx\\ & =\frac{1024}{25}\int \frac{2+4x^2+x\log (8)}{4x+\log (8)}dx+\int e^{x}dx\\ & =e^x+\frac{1024}{25}\int (x+\frac{2}{4x+\log (8)})dx\\ & =e^x+\frac{512x^2}{25}+\frac{512}{25}\log (4x+\log (8))\end{array}\end{array}$$

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.43 $$\begin{array}{c}\frac{1}{25}(25e^x+512x^2-32{\log }^2(8)+512\log (4x+\log (8)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(2048 + 4096*x^2 + 3072*x*Log[2] + E^x*(100*x + 75*Log[2]))/(100*x + 75*Log[2]),x]

[Out]

(25*E^x + 512*x^2 - 32*Log[8]^2 + 512*Log[4*x + Log[8]])/25

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fricas [A]  time = 1.11, size = 19, normalized size = 0.90 $$\begin{array}{c}\frac{512}{25}{x}^2+e^x+\frac{512}{25}\log (4x+3\log (2))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((75*log(2)+100*x)*exp(x)+3072*x*log(2)+4096*x^2+2048)/(75*log(2)+100*x),x, algorithm="fricas")

[Out]

512/25*x^2 + e^x + 512/25*log(4*x + 3*log(2))

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giac [A]  time = 0.28, size = 19, normalized size = 0.90 $$\begin{array}{c}\frac{512}{25}{x}^2+e^x+\frac{512}{25}\log (4x+3\log (2))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((75*log(2)+100*x)*exp(x)+3072*x*log(2)+4096*x^2+2048)/(75*log(2)+100*x),x, algorithm="giac")

[Out]

512/25*x^2 + e^x + 512/25*log(4*x + 3*log(2))

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maple [A]  time = 0.44, size = 20, normalized size = 0.95

method	result	size
default	$\frac{512\ln (3\ln (2)+4x)}{25}+\frac{512x^2}{25}+{\mathrm{e}}^x$	$20$
norman	$\frac{512x^2}{25}+{\mathrm{e}}^x+\frac{512\ln (75\ln (2)+100x)}{25}$	$20$
risch	$\frac{512\ln (3\ln (2)+4x)}{25}+\frac{512x^2}{25}+{\mathrm{e}}^x$	$20$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((75*ln(2)+100*x)*exp(x)+3072*x*ln(2)+4096*x^2+2048)/(75*ln(2)+100*x),x,method=_RETURNVERBOSE)

[Out]

512/25*ln(3*ln(2)+4*x)+512/25*x^2+exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\frac{576}{25}\log (2{)}^2\log (4x+3\log (2))-\frac{3}{8}\cdot 2^{\frac{1}{4}}{E}_1(-x-\frac{3}{4}\log (2))\log (2)+\frac{512}{25}{x}^2-\frac{192}{25}(3\log (2)\log (4x+3\log (2))-4x)\log (2)-\frac{768}{25}x\log (2)-12\int \frac{e^x}{16x^2+24x\log (2)+9\log (2{)}^2}dx\log (2)+\frac{4x{e}^x}{4x+3\log (2)}+\frac{512}{25}\log (4x+3\log (2))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((75*log(2)+100*x)*exp(x)+3072*x*log(2)+4096*x^2+2048)/(75*log(2)+100*x),x, algorithm="maxima")

[Out]

576/25*log(2)^2*log(4*x + 3*log(2)) - 3/8*2^(1/4)*exp_integral_e(1, -x - 3/4*log(2))*log(2) + 512/25*x^2 - 192
/25*(3*log(2)*log(4*x + 3*log(2)) - 4*x)*log(2) - 768/25*x*log(2) - 12*integrate(e^x/(16*x^2 + 24*x*log(2) + 9
*log(2)^2), x)*log(2) + 4*x*e^x/(4*x + 3*log(2)) + 512/25*log(4*x + 3*log(2))

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mupad [B]  time = 0.09, size = 17, normalized size = 0.81 $$\begin{array}{c}\frac{512\ln (4x+\ln (8))}{25}+{\mathrm{e}}^x+\frac{512x^2}{25}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(100*x + 75*log(2)) + 3072*x*log(2) + 4096*x^2 + 2048)/(100*x + 75*log(2)),x)

[Out]

(512*log(4*x + log(8)))/25 + exp(x) + (512*x^2)/25

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sympy [A]  time = 0.14, size = 22, normalized size = 1.05 $$\begin{array}{c}\frac{512x^2}{25}+e^x+\frac{512\log (4x+3\log (2))}{25}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((75*ln(2)+100*x)*exp(x)+3072*x*ln(2)+4096*x**2+2048)/(75*ln(2)+100*x),x)

[Out]

512*x**2/25 + exp(x) + 512*log(4*x + 3*log(2))/25

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