3.1.32 \(\int \frac {946+315 x^2}{(7+5 x^2) \sqrt {2+3 x^2+x^4}} \, dx\) [32]

Optimal. Leaf size=106 \[ \frac {631 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {2+3 x^2+x^4}}-\frac {2525 \left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{14 \sqrt {2} \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}} \]

[Out]

-2525/28*(x^2+2)*(1/(x^2+1))^(1/2)*(x^2+1)^(1/2)*EllipticPi(x/(x^2+1)^(1/2),2/7,1/2*2^(1/2))*2^(1/2)/((x^2+2)/
(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+631/4*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2)
)*((x^2+2)/(x^2+1))^(1/2)*2^(1/2)/(x^4+3*x^2+2)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1718, 1113, 1470, 553} \begin {gather*} \frac {631 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{2 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {2525 \left (x^2+2\right ) \Pi \left (\frac {2}{7};\text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{14 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(946 + 315*x^2)/((7 + 5*x^2)*Sqrt[2 + 3*x^2 + x^4]),x]

[Out]

(631*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(2*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) - (2525*
(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(14*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])

Rule 553

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[c*(Sqrt[e +
f*x^2]/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((e + f*x^2)/(e*(c + d*x^2)))]))*EllipticPi[1 - b*(c/(a*d)), Ar
cTan[Rt[d/c, 2]*x], 1 - c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rule 1113

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(2*a + (b + q
)*x^2)*(Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*Elli
pticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1470

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^
(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPar
t[p]), Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p,
q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p]

Rule 1718

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Sqrt[b^2 - 4*a*c]}, Dist[(2*a*B - A*(b + q))/(2*a*e - d*(b + q)), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] -
Dist[(B*d - A*e)/(2*a*e - d*(b + q)), Int[(2*a + (b + q)*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /
; RationalQ[q]] /; FreeQ[{a, b, c, d, e, A, B}, x] && GtQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&
NeQ[c*A^2 - b*A*B + a*B^2, 0]

Rubi steps

\begin {align*} \int \frac {946+315 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx &=\frac {631}{2} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {2525}{8} \int \frac {4+4 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {631 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {2+3 x^2+x^4}}-\frac {\left (2525 \sqrt {\frac {1}{2}+\frac {x^2}{4}} \sqrt {4+4 x^2}\right ) \int \frac {\sqrt {4+4 x^2}}{\sqrt {\frac {1}{2}+\frac {x^2}{4}} \left (7+5 x^2\right )} \, dx}{8 \sqrt {2+3 x^2+x^4}}\\ &=\frac {631 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {2+3 x^2+x^4}}-\frac {2525 \left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{14 \sqrt {2} \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.14, size = 74, normalized size = 0.70 \begin {gather*} -\frac {i \sqrt {1+x^2} \sqrt {2+x^2} \left (441 F\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+505 \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )\right )}{7 \sqrt {2+3 x^2+x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(946 + 315*x^2)/((7 + 5*x^2)*Sqrt[2 + 3*x^2 + x^4]),x]

[Out]

((-1/7*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(441*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] + 505*EllipticPi[10/7, I*ArcSinh
[x/Sqrt[2]], 2]))/Sqrt[2 + 3*x^2 + x^4]

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Maple [C] Result contains complex when optimal does not.
time = 0.14, size = 93, normalized size = 0.88

method	result	size
default	\(-\frac {63 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}-\frac {505 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{7 \sqrt {x^{4}+3 x^{2}+2}}\)	\(93\)
elliptic	\(-\frac {63 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}-\frac {505 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{7 \sqrt {x^{4}+3 x^{2}+2}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((315*x^2+946)/(5*x^2+7)/(x^4+3*x^2+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-63/2*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*2^(1/2)*x,2^(1/2))-505/7*I*2
^(1/2)*(1+1/2*x^2)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticPi(1/2*I*2^(1/2)*x,10/7,2^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((315*x^2+946)/(5*x^2+7)/(x^4+3*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate((315*x^2 + 946)/(sqrt(x^4 + 3*x^2 + 2)*(5*x^2 + 7)), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((315*x^2+946)/(5*x^2+7)/(x^4+3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 3*x^2 + 2)*(315*x^2 + 946)/(5*x^6 + 22*x^4 + 31*x^2 + 14), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {315 x^{2} + 946}{\sqrt {\left (x^{2} + 1\right ) \left (x^{2} + 2\right )} \left (5 x^{2} + 7\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((315*x**2+946)/(5*x**2+7)/(x**4+3*x**2+2)**(1/2),x)

[Out]

Integral((315*x**2 + 946)/(sqrt((x**2 + 1)*(x**2 + 2))*(5*x**2 + 7)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((315*x^2+946)/(5*x^2+7)/(x^4+3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate((315*x^2 + 946)/(sqrt(x^4 + 3*x^2 + 2)*(5*x^2 + 7)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {315\,x^2+946}{\left (5\,x^2+7\right )\,\sqrt {x^4+3\,x^2+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((315*x^2 + 946)/((5*x^2 + 7)*(3*x^2 + x^4 + 2)^(1/2)),x)

[Out]

int((315*x^2 + 946)/((5*x^2 + 7)*(3*x^2 + x^4 + 2)^(1/2)), x)

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