3.372 \(\int \frac {(7+5 x^2)^4}{(4+3 x^2+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=200 \[ \frac {14523 \sqrt {x^4+3 x^2+4} x}{28 \left (x^2+2\right )}+\frac {625}{3} \sqrt {x^4+3 x^2+4} x+\frac {\left (2719-4023 x^2\right ) x}{28 \sqrt {x^4+3 x^2+4}}+\frac {4243 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{12 \sqrt {2} \sqrt {x^4+3 x^2+4}}-\frac {14523 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{14 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

[Out]

1/28*x*(-4023*x^2+2719)/(x^4+3*x^2+4)^(1/2)+625/3*x*(x^4+3*x^2+4)^(1/2)+14523/28*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)
-14523/28*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2)))*EllipticE(sin(2*arctan(1
/2*x*2^(1/2))),1/4*2^(1/2))*2^(1/2)*((x^4+3*x^2+4)/(x^2+2)^2)^(1/2)/(x^4+3*x^2+4)^(1/2)+4243/24*(x^2+2)*(cos(2
*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2)))*EllipticF(sin(2*arctan(1/2*x*2^(1/2))),1/4*2^(1/
2))*((x^4+3*x^2+4)/(x^2+2)^2)^(1/2)*2^(1/2)/(x^4+3*x^2+4)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1205, 1679, 1197, 1103, 1195} \[ \frac {14523 \sqrt {x^4+3 x^2+4} x}{28 \left (x^2+2\right )}+\frac {625}{3} \sqrt {x^4+3 x^2+4} x+\frac {\left (2719-4023 x^2\right ) x}{28 \sqrt {x^4+3 x^2+4}}+\frac {4243 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{12 \sqrt {2} \sqrt {x^4+3 x^2+4}}-\frac {14523 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{14 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2719 - 4023*x^2))/(28*Sqrt[4 + 3*x^2 + x^4]) + (625*x*Sqrt[4 + 3*x^2 + x^4])/3 + (14523*x*Sqrt[4 + 3*x^2 +
x^4])/(28*(2 + x^2)) - (14523*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/
8])/(14*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) + (4243*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTa
n[x/Sqrt[2]], 1/8])/(12*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

Rule 1103

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

Rule 1195

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

Rule 1197

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

Rule 1205

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coeff[Polynom
ialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x
^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2))/(
2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToS
um[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c
*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*
a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]

Rule 1679

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*} \int \frac {\left (7+5 x^2\right )^4}{\left (4+3 x^2+x^4\right )^{3/2}} \, dx &=\frac {x \left (2719-4023 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+\frac {1}{28} \int \frac {14088+49523 x^2+17500 x^4}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {x \left (2719-4023 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+\frac {625}{3} x \sqrt {4+3 x^2+x^4}+\frac {1}{84} \int \frac {-27736+43569 x^2}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {x \left (2719-4023 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+\frac {625}{3} x \sqrt {4+3 x^2+x^4}+\frac {4243}{6} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx-\frac {14523}{14} \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {x \left (2719-4023 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+\frac {625}{3} x \sqrt {4+3 x^2+x^4}+\frac {14523 x \sqrt {4+3 x^2+x^4}}{28 \left (2+x^2\right )}-\frac {14523 \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{14 \sqrt {2} \sqrt {4+3 x^2+x^4}}+\frac {4243 \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{12 \sqrt {2} \sqrt {4+3 x^2+x^4}}\\ \end {align*}

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Mathematica [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (625 \, x^{8} + 3500 \, x^{6} + 7350 \, x^{4} + 6860 \, x^{2} + 2401\right )} \sqrt {x^{4} + 3 \, x^{2} + 4}}{x^{8} + 6 \, x^{6} + 17 \, x^{4} + 24 \, x^{2} + 16}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((625*x^8 + 3500*x^6 + 7350*x^4 + 6860*x^2 + 2401)*sqrt(x^4 + 3*x^2 + 4)/(x^8 + 6*x^6 + 17*x^4 + 24*x^
2 + 16), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{4}}{{\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 339, normalized size = 1.70 \[ \frac {625 \sqrt {x^{4}+3 x^{2}+4}\, x}{3}-\frac {27736 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{21 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}-\frac {1250 \left (-\frac {9}{14} x^{3}+\frac {2}{7} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}-\frac {116184 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )+\EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )\right )}{7 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )}-\frac {7000 \left (-\frac {1}{14} x^{3}-\frac {6}{7} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}-\frac {14700 \left (\frac {3}{14} x^{3}+\frac {4}{7} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}-\frac {13720 \left (-\frac {1}{7} x^{3}-\frac {3}{14} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}-\frac {4802 \left (\frac {3}{56} x^{3}+\frac {1}{56} x \right )}{\sqrt {x^{4}+3 x^{2}+4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1250*(-9/14*x^3+2/7*x)/(x^4+3*x^2+4)^(1/2)+625/3*(x^4+3*x^2+4)^(1/2)*x-27736/21/(-6+2*I*7^(1/2))^(1/2)*(-(-3/
8+1/8*I*7^(1/2))*x^2+1)^(1/2)*(-(-3/8-1/8*I*7^(1/2))*x^2+1)^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticF(1/4*(-6+2*I*7^
(1/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1/2))-116184/7/(-6+2*I*7^(1/2))^(1/2)*(-(-3/8+1/8*I*7^(1/2))*x^2+1)^(1/2)*
(-(-3/8-1/8*I*7^(1/2))*x^2+1)^(1/2)/(x^4+3*x^2+4)^(1/2)/(I*7^(1/2)+3)*(EllipticF(1/4*(-6+2*I*7^(1/2))^(1/2)*x,
1/4*(2+6*I*7^(1/2))^(1/2))-EllipticE(1/4*(-6+2*I*7^(1/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1/2)))-7000*(-1/14*x^3-
6/7*x)/(x^4+3*x^2+4)^(1/2)-14700*(3/14*x^3+4/7*x)/(x^4+3*x^2+4)^(1/2)-13720*(-1/7*x^3-3/14*x)/(x^4+3*x^2+4)^(1
/2)-4802*(3/56*x^3+1/56*x)/(x^4+3*x^2+4)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{4}}{{\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (5\,x^2+7\right )}^4}{{\left (x^4+3\,x^2+4\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (5 x^{2} + 7\right )^{4}}{\left (\left (x^{2} - x + 2\right ) \left (x^{2} + x + 2\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((5*x**2 + 7)**4/((x**2 - x + 2)*(x**2 + x + 2))**(3/2), x)

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