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

Optimal. Leaf size=219 \[ -\frac {220779 \sqrt {x^4+3 x^2+4} x}{28 \left (x^2+2\right )}+\frac {5000}{3} \sqrt {x^4+3 x^2+4} x+\frac {\left (45779 x^2+99493\right ) x}{28 \sqrt {x^4+3 x^2+4}}-\frac {130729 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{12 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {220779 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{14 \sqrt {2} \sqrt {x^4+3 x^2+4}}+625 \sqrt {x^4+3 x^2+4} x^3 \]

[Out]

1/28*x*(45779*x^2+99493)/(x^4+3*x^2+4)^(1/2)+5000/3*x*(x^4+3*x^2+4)^(1/2)+625*x^3*(x^4+3*x^2+4)^(1/2)-220779/2
8*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)+220779/28*(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)-130729/24*(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 = 0.12, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ 625 \sqrt {x^4+3 x^2+4} x^3-\frac {220779 \sqrt {x^4+3 x^2+4} x}{28 \left (x^2+2\right )}+\frac {5000}{3} \sqrt {x^4+3 x^2+4} x+\frac {\left (45779 x^2+99493\right ) x}{28 \sqrt {x^4+3 x^2+4}}-\frac {130729 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{12 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {220779 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{14 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(99493 + 45779*x^2))/(28*Sqrt[4 + 3*x^2 + x^4]) + (5000*x*Sqrt[4 + 3*x^2 + x^4])/3 + 625*x^3*Sqrt[4 + 3*x^2
+ x^4] - (220779*x*Sqrt[4 + 3*x^2 + x^4])/(28*(2 + x^2)) + (220779*(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]) - (130729*(2 + x^2)*Sqrt[(4 + 3*x^
2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[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 )^5}{\left (4+3 x^2+x^4\right )^{3/2}} \, dx &=\frac {x \left (99493+45779 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+\frac {1}{28} \int \frac {18156+269221 x^2+350000 x^4+87500 x^6}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {x \left (99493+45779 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+625 x^3 \sqrt {4+3 x^2+x^4}+\frac {1}{140} \int \frac {90780+296105 x^2+700000 x^4}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {x \left (99493+45779 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+\frac {5000}{3} x \sqrt {4+3 x^2+x^4}+625 x^3 \sqrt {4+3 x^2+x^4}+\frac {1}{420} \int \frac {-2527660-3311685 x^2}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {x \left (99493+45779 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+\frac {5000}{3} x \sqrt {4+3 x^2+x^4}+625 x^3 \sqrt {4+3 x^2+x^4}+\frac {220779}{14} \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx-\frac {130729}{6} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {x \left (99493+45779 x^2\right )}{28 \sqrt {4+3 x^2+x^4}}+\frac {5000}{3} x \sqrt {4+3 x^2+x^4}+625 x^3 \sqrt {4+3 x^2+x^4}-\frac {220779 x \sqrt {4+3 x^2+x^4}}{28 \left (2+x^2\right )}+\frac {220779 \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 {130729 \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 [C]  time = 0.52, size = 339, normalized size = 1.55 \[ \frac {-\sqrt {2} \left (662337 \sqrt {7}+975947 i\right ) \sqrt {\frac {-2 i x^2+\sqrt {7}-3 i}{\sqrt {7}-3 i}} \sqrt {\frac {2 i x^2+\sqrt {7}+3 i}{\sqrt {7}+3 i}} F\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )+662337 \sqrt {2} \left (\sqrt {7}+3 i\right ) \sqrt {\frac {-2 i x^2+\sqrt {7}-3 i}{\sqrt {7}-3 i}} \sqrt {\frac {2 i x^2+\sqrt {7}+3 i}{\sqrt {7}+3 i}} E\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )+4 \sqrt {-\frac {i}{\sqrt {7}-3 i}} x \left (52500 x^6+297500 x^4+767337 x^2+858479\right )}{336 \sqrt {-\frac {i}{\sqrt {7}-3 i}} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[(-I)/(-3*I + Sqrt[7])]*x*(858479 + 767337*x^2 + 297500*x^4 + 52500*x^6) + 662337*Sqrt[2]*(3*I + Sqrt[7
])*Sqrt[(-3*I + Sqrt[7] - (2*I)*x^2)/(-3*I + Sqrt[7])]*Sqrt[(3*I + Sqrt[7] + (2*I)*x^2)/(3*I + Sqrt[7])]*Ellip
ticE[I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] - Sqrt[2]*(975947*I + 662337
*Sqrt[7])*Sqrt[(-3*I + Sqrt[7] - (2*I)*x^2)/(-3*I + Sqrt[7])]*Sqrt[(3*I + Sqrt[7] + (2*I)*x^2)/(3*I + Sqrt[7])
]*EllipticF[I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])])/(336*Sqrt[(-I)/(-3*I
+ Sqrt[7])]*Sqrt[4 + 3*x^2 + x^4])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3125 \, x^{10} + 21875 \, x^{8} + 61250 \, x^{6} + 85750 \, x^{4} + 60025 \, x^{2} + 16807\right )} \sqrt {x^{4} + 3 \, x^{2} + 4}}{x^{8} + 6 \, x^{6} + 17 \, x^{4} + 24 \, x^{2} + 16}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3125*x^10 + 21875*x^8 + 61250*x^6 + 85750*x^4 + 60025*x^2 + 16807)*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 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{5}}{{\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.05, size = 379, normalized size = 1.73 \[ 625 \sqrt {x^{4}+3 x^{2}+4}\, x^{3}+\frac {5000 \sqrt {x^{4}+3 x^{2}+4}\, x}{3}-\frac {505532 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{21 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}-\frac {6250 \left (\frac {31}{14} x^{3}+\frac {18}{7} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}+\frac {1766232 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )+\EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )\right )}{7 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )}-\frac {43750 \left (-\frac {9}{14} x^{3}+\frac {2}{7} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}-\frac {122500 \left (-\frac {1}{14} x^{3}-\frac {6}{7} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}-\frac {171500 \left (\frac {3}{14} x^{3}+\frac {4}{7} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}-\frac {120050 \left (-\frac {1}{7} x^{3}-\frac {3}{14} x \right )}{\sqrt {x^{4}+3 x^{2}+4}}-\frac {33614 \left (\frac {3}{56} x^{3}+\frac {1}{56} x \right )}{\sqrt {x^{4}+3 x^{2}+4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6250*(31/14*x^3+18/7*x)/(x^4+3*x^2+4)^(1/2)+625*(x^4+3*x^2+4)^(1/2)*x^3+5000/3*(x^4+3*x^2+4)^(1/2)*x-505532/2
1/(-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))+1766232/7/(-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)*(Ellipti
cF(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)))-43750*(-9/14*x^3+2/7*x)/(x^4+3*x^2+4)^(1/2)-122500*(-1/14*x^3-6/7*x)/(x^4+3*x^2+4)^(1/2)-17150
0*(3/14*x^3+4/7*x)/(x^4+3*x^2+4)^(1/2)-120050*(-1/7*x^3-3/14*x)/(x^4+3*x^2+4)^(1/2)-33614*(1/56*x+3/56*x^3)/(x
^4+3*x^2+4)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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