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

Optimal. Leaf size=187 \[ -\frac {15 \sqrt {x^4+3 x^2+4} x}{x^2+2}+75 \sqrt {x^4+3 x^2+4} x+\frac {13 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{2 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {15 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{\sqrt {x^4+3 x^2+4}}+25 \sqrt {x^4+3 x^2+4} x^3 \]

[Out]

75*x*(x^4+3*x^2+4)^(1/2)+25*x^3*(x^4+3*x^2+4)^(1/2)-15*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)+13/4*(x^2+2)*(cos(2*arcta
n(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)+15*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/co
s(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)

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Rubi [A]  time = 0.09, antiderivative size = 187, 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 = {1206, 1679, 1197, 1103, 1195} \[ 25 \sqrt {x^4+3 x^2+4} x^3-\frac {15 \sqrt {x^4+3 x^2+4} x}{x^2+2}+75 \sqrt {x^4+3 x^2+4} x+\frac {13 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{2 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {15 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{\sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

75*x*Sqrt[4 + 3*x^2 + x^4] + 25*x^3*Sqrt[4 + 3*x^2 + x^4] - (15*x*Sqrt[4 + 3*x^2 + x^4])/(2 + x^2) + (15*Sqrt[
2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/Sqrt[4 + 3*x^2 + x^4] +
(13*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(2*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 1206

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 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 )^3}{\sqrt {4+3 x^2+x^4}} \, dx &=25 x^3 \sqrt {4+3 x^2+x^4}+\frac {1}{5} \int \frac {1715+2175 x^2+1125 x^4}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=75 x \sqrt {4+3 x^2+x^4}+25 x^3 \sqrt {4+3 x^2+x^4}+\frac {1}{15} \int \frac {645-225 x^2}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=75 x \sqrt {4+3 x^2+x^4}+25 x^3 \sqrt {4+3 x^2+x^4}+13 \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx+30 \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=75 x \sqrt {4+3 x^2+x^4}+25 x^3 \sqrt {4+3 x^2+x^4}-\frac {15 x \sqrt {4+3 x^2+x^4}}{2+x^2}+\frac {15 \sqrt {2} \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 )}{\sqrt {4+3 x^2+x^4}}+\frac {13 \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 )}{2 \sqrt {2} \sqrt {4+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.48, size = 337, normalized size = 1.80 \[ \frac {-\sqrt {2} \left (15 \sqrt {7}+131 i\right ) \sqrt {\frac {-2 i x^2+\sqrt {7}-3 i}{\sqrt {7}-3 i}} \sqrt {\frac {2 i x^2+\sqrt {7}+3 i}{\sqrt {7}+3 i}} F\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )+15 \sqrt {2} \left (\sqrt {7}+3 i\right ) \sqrt {\frac {-2 i x^2+\sqrt {7}-3 i}{\sqrt {7}-3 i}} \sqrt {\frac {2 i x^2+\sqrt {7}+3 i}{\sqrt {7}+3 i}} E\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )+100 \sqrt {-\frac {i}{\sqrt {7}-3 i}} x \left (x^6+6 x^4+13 x^2+12\right )}{4 \sqrt {-\frac {i}{\sqrt {7}-3 i}} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(100*Sqrt[(-I)/(-3*I + Sqrt[7])]*x*(12 + 13*x^2 + 6*x^4 + x^6) + 15*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])]*EllipticE[I*ArcSinh[Sqrt[(
-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] - Sqrt[2]*(131*I + 15*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])])/(4*Sqrt[(-I)/(-3*I + Sqrt[7])]*Sqrt[4 + 3*x^2 +
x^4])

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {125 \, x^{6} + 525 \, x^{4} + 735 \, x^{2} + 343}{\sqrt {x^{4} + 3 \, x^{2} + 4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((125*x^6 + 525*x^4 + 735*x^2 + 343)/sqrt(x^4 + 3*x^2 + 4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.03, size = 241, normalized size = 1.29 \[ 25 \sqrt {x^{4}+3 x^{2}+4}\, x^{3}+75 \sqrt {x^{4}+3 x^{2}+4}\, x +\frac {172 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{\sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}+\frac {480 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )+\EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )\right )}{\sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*(x^4+3*x^2+4)^(1/2)*x^3+75*(x^4+3*x^2+4)^(1/2)*x+172/(-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))+480/(-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))-Ell
ipticE(1/4*(-6+2*I*7^(1/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (5\,x^2+7\right )}^3}{\sqrt {x^4+3\,x^2+4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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