\(\int \frac {(7+5 x^2)^3}{\sqrt {2+3 x^2+x^4}} \, dx\) [300]

Optimal result

Integrand size = 24, antiderivative size = 157 \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\frac {135 x \left (2+x^2\right )}{\sqrt {2+3 x^2+x^4}}+75 x \sqrt {2+3 x^2+x^4}+25 x^3 \sqrt {2+3 x^2+x^4}-\frac {135 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {2+3 x^2+x^4}}+\frac {193 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+3 x^2+x^4}} \]

[Out]

135*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)+193/2*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))
*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)-135*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticE(x/(x^2+1)^(
1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+75*x*(x^4+3*x^2+2)^(1/2)+25*x^3*(x^4+3*x
^2+2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1220, 1693, 1203, 1113, 1149} \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\frac {193 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {135 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+3 x^2+2}}+75 \sqrt {x^4+3 x^2+2} x+\frac {135 \left (x^2+2\right ) x}{\sqrt {x^4+3 x^2+2}}+25 \sqrt {x^4+3 x^2+2} x^3 \]

[In]

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

[Out]

(135*x*(2 + x^2))/Sqrt[2 + 3*x^2 + x^4] + 75*x*Sqrt[2 + 3*x^2 + x^4] + 25*x^3*Sqrt[2 + 3*x^2 + x^4] - (135*Sqr
t[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4] + (193*(1 + x^2)*Sqr
t[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^4])

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 1149

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

Rule 1203

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

Rule 1220

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e^q*x^(2*q - 3)*((
a + b*x^2 + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2*p + 2*q - 1)*e^q*x^(2*q - 2) -
c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && IGtQ[q, 1]

Rule 1693

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[e*x^(2*q - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(2*q + 4*p + 1))), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = 25 x^3 \sqrt {2+3 x^2+x^4}+\frac {1}{5} \int \frac {1715+2925 x^2+1125 x^4}{\sqrt {2+3 x^2+x^4}} \, dx \\ & = 75 x \sqrt {2+3 x^2+x^4}+25 x^3 \sqrt {2+3 x^2+x^4}+\frac {1}{15} \int \frac {2895+2025 x^2}{\sqrt {2+3 x^2+x^4}} \, dx \\ & = 75 x \sqrt {2+3 x^2+x^4}+25 x^3 \sqrt {2+3 x^2+x^4}+135 \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+193 \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx \\ & = \frac {135 x \left (2+x^2\right )}{\sqrt {2+3 x^2+x^4}}+75 x \sqrt {2+3 x^2+x^4}+25 x^3 \sqrt {2+3 x^2+x^4}-\frac {135 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2+3 x^2+x^4}}+\frac {193 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+3 x^2+x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.68 \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\frac {25 x \left (6+11 x^2+6 x^4+x^6\right )-135 i \sqrt {1+x^2} \sqrt {2+x^2} E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-58 i \sqrt {1+x^2} \sqrt {2+x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{\sqrt {2+3 x^2+x^4}} \]

[In]

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

[Out]

(25*x*(6 + 11*x^2 + 6*x^4 + x^6) - (135*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] - (5
8*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2])/Sqrt[2 + 3*x^2 + x^4]

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.61 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.80

method	result	size
risch	\(25 x \left (x^{2}+3\right ) \sqrt {x^{4}+3 x^{2}+2}-\frac {193 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+\frac {135 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}\)	\(126\)
default	\(-\frac {193 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+25 x^{3} \sqrt {x^{4}+3 x^{2}+2}+75 x \sqrt {x^{4}+3 x^{2}+2}+\frac {135 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}\)	\(138\)
elliptic	\(-\frac {193 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+25 x^{3} \sqrt {x^{4}+3 x^{2}+2}+75 x \sqrt {x^{4}+3 x^{2}+2}+\frac {135 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}\)	\(138\)

[In]

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

[Out]

25*x*(x^2+3)*(x^4+3*x^2+2)^(1/2)-193/2*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))+135/2*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))-EllipticE(1/2*I*2^(1/2)*x,2^(1/2)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.34 \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\frac {-135 i \, x E(\arcsin \left (\frac {i}{x}\right )\,|\,2) + 328 i \, x F(\arcsin \left (\frac {i}{x}\right )\,|\,2) + 5 \, {\left (5 \, x^{4} + 15 \, x^{2} + 27\right )} \sqrt {x^{4} + 3 \, x^{2} + 2}}{x} \]

[In]

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

[Out]

(-135*I*x*elliptic_e(arcsin(I/x), 2) + 328*I*x*elliptic_f(arcsin(I/x), 2) + 5*(5*x^4 + 15*x^2 + 27)*sqrt(x^4 +
3*x^2 + 2))/x

Sympy [F]

\[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\int \frac {\left (5 x^{2} + 7\right )^{3}}{\sqrt {\left (x^{2} + 1\right ) \left (x^{2} + 2\right )}}\, dx \]

[In]

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

[Out]

Integral((5*x**2 + 7)**3/sqrt((x**2 + 1)*(x**2 + 2)), x)

Maxima [F]

\[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\int { \frac {{\left (5 \, x^{2} + 7\right )}^{3}}{\sqrt {x^{4} + 3 \, x^{2} + 2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\int { \frac {{\left (5 \, x^{2} + 7\right )}^{3}}{\sqrt {x^{4} + 3 \, x^{2} + 2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\int \frac {{\left (5\,x^2+7\right )}^3}{\sqrt {x^4+3\,x^2+2}} \,d x \]

[In]

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

[Out]

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