3.18.14 $\int \frac{e^{\frac{21-4x\log (5)\log (2+x^2)}{4x}}(-42-21x^2-8x^3\log (5))}{8x^2+4x^4}dx$

Optimal. Leaf size=20 $$e^{\frac{21}{4x}-\log (5)\log (2+x^2)}$$

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Rubi [A]  time = 0.55, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 50, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.040, Rules used = {1593, 6706} $$\begin{array}{c}{e}^{\frac{21}{4}/x}{5}^{-\log (x^2+2)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^((21 - 4*x*Log[5]*Log[2 + x^2])/(4*x))*(-42 - 21*x^2 - 8*x^3*Log[5]))/(8*x^2 + 4*x^4),x]

[Out]

E^(21/(4*x))/5^Log[2 + x^2]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{\frac{21-4x\log (5)\log (2+x^2)}{4x}}(-42-21x^2-8x^3\log (5))}{x^2(8+4x^2)}dx\\ & =5^{-\log (2+x^2)}{e}^{\frac{21}{4}/x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.22, size = 20, normalized size = 1.00 $$\begin{array}{c}{e}^{\frac{21}{4x}-\log (5)\log (2+x^2)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^((21 - 4*x*Log[5]*Log[2 + x^2])/(4*x))*(-42 - 21*x^2 - 8*x^3*Log[5]))/(8*x^2 + 4*x^4),x]

[Out]

E^(21/(4*x) - Log[5]*Log[2 + x^2])

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fricas [A]  time = 0.56, size = 19, normalized size = 0.95 $$\begin{array}{c}{e}^{(-\frac{4x\log (5)\log (x^2+2)-21}{4x})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^3*log(5)-21*x^2-42)*exp(1/4*(-4*x*log(5)*log(x^2+2)+21)/x)/(4*x^4+8*x^2),x, algorithm="fricas"
)

[Out]

e^(-1/4*(4*x*log(5)*log(x^2 + 2) - 21)/x)

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giac [A]  time = 0.20, size = 17, normalized size = 0.85 $$\begin{array}{c}{e}^{(-\log (5)\log (x^2+2)+\frac{21}{4x})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^3*log(5)-21*x^2-42)*exp(1/4*(-4*x*log(5)*log(x^2+2)+21)/x)/(4*x^4+8*x^2),x, algorithm="giac")

[Out]

e^(-log(5)*log(x^2 + 2) + 21/4/x)

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maple [A]  time = 0.27, size = 18, normalized size = 0.90

method	result	size
risch	${(x^2+2)}^{-\ln (5)}{\mathrm{e}}^{\frac{21}{4x}}$	$18$
gosper	${\mathrm{e}}^{-\frac{4x\ln (5)\ln (x^2+2)-21}{4x}}$	$20$
norman	${\mathrm{e}}^{\frac{-4x\ln (5)\ln (x^2+2)+21}{4x}}$	$20$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x^3*ln(5)-21*x^2-42)*exp(1/4*(-4*x*ln(5)*ln(x^2+2)+21)/x)/(4*x^4+8*x^2),x,method=_RETURNVERBOSE)

[Out]

(x^2+2)^(-ln(5))*exp(21/4/x)

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maxima [A]  time = 0.65, size = 17, normalized size = 0.85 $$\begin{array}{c}{e}^{(-\log (5)\log (x^2+2)+\frac{21}{4x})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^3*log(5)-21*x^2-42)*exp(1/4*(-4*x*log(5)*log(x^2+2)+21)/x)/(4*x^4+8*x^2),x, algorithm="maxima"
)

[Out]

e^(-log(5)*log(x^2 + 2) + 21/4/x)

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mupad [B]  time = 1.43, size = 17, normalized size = 0.85 $$\begin{array}{c}\frac{{\mathrm{e}}^{\frac{21}{4x}}}{{(x^2+2)}^{\ln (5)}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x*log(5)*log(x^2 + 2) - 21/4)/x)*(8*x^3*log(5) + 21*x^2 + 42))/(8*x^2 + 4*x^4),x)

[Out]

exp(21/(4*x))/(x^2 + 2)^log(5)

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sympy [A]  time = 0.57, size = 17, normalized size = 0.85 $$\begin{array}{c}{e}^{\frac{-x\log (5)\log (x^2+2)+\frac{21}{4}}{x}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x**3*ln(5)-21*x**2-42)*exp(1/4*(-4*x*ln(5)*ln(x**2+2)+21)/x)/(4*x**4+8*x**2),x)

[Out]

exp((-x*log(5)*log(x**2 + 2) + 21/4)/x)

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