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

Optimal. Leaf size=207 \[ \frac {1}{63} x \left (35 x^2+108\right ) \left (x^4+3 x^2+4\right )^{3/2}+\frac {1}{105} x \left (289 x^2+1029\right ) \sqrt {x^4+3 x^2+4}+\frac {2798 x \sqrt {x^4+3 x^2+4}}{105 \left (x^2+2\right )}+\frac {74 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{3 \sqrt {x^4+3 x^2+4}}-\frac {2798 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{105 \sqrt {x^4+3 x^2+4}} \]

[Out]

1/63*x*(35*x^2+108)*(x^4+3*x^2+4)^(3/2)+2798/105*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)+1/105*x*(289*x^2+1029)*(x^4+3*x
^2+4)^(1/2)-2798/105*(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)+74/3*(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 = 0.07, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1176, 1197, 1103, 1195} \[ \frac {1}{63} x \left (35 x^2+108\right ) \left (x^4+3 x^2+4\right )^{3/2}+\frac {1}{105} x \left (289 x^2+1029\right ) \sqrt {x^4+3 x^2+4}+\frac {2798 x \sqrt {x^4+3 x^2+4}}{105 \left (x^2+2\right )}+\frac {74 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{3 \sqrt {x^4+3 x^2+4}}-\frac {2798 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{105 \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2798*x*Sqrt[4 + 3*x^2 + x^4])/(105*(2 + x^2)) + (x*(1029 + 289*x^2)*Sqrt[4 + 3*x^2 + x^4])/105 + (x*(108 + 35
*x^2)*(4 + 3*x^2 + x^4)^(3/2))/63 - (2798*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*Ar
cTan[x/Sqrt[2]], 1/8])/(105*Sqrt[4 + 3*x^2 + x^4]) + (74*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]
*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(3*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 1176

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(2*b*e*p + c*d*(4*p
+ 3) + c*e*(4*p + 1)*x^2)*(a + b*x^2 + c*x^4)^p)/(c*(4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/(c*(4*p + 1)*(4*p +
3)), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*x^2, x]*(a +
b*x^2 + c*x^4)^(p - 1), 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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

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]

Rubi steps

\begin {align*} \int \left (7+5 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2} \, dx &=\frac {1}{63} x \left (108+35 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}+\frac {1}{21} \int \left (444+289 x^2\right ) \sqrt {4+3 x^2+x^4} \, dx\\ &=\frac {1}{105} x \left (1029+289 x^2\right ) \sqrt {4+3 x^2+x^4}+\frac {1}{63} x \left (108+35 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}+\frac {1}{315} \int \frac {14292+8394 x^2}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {1}{105} x \left (1029+289 x^2\right ) \sqrt {4+3 x^2+x^4}+\frac {1}{63} x \left (108+35 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}-\frac {5596}{105} \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx+\frac {296}{3} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {2798 x \sqrt {4+3 x^2+x^4}}{105 \left (2+x^2\right )}+\frac {1}{105} x \left (1029+289 x^2\right ) \sqrt {4+3 x^2+x^4}+\frac {1}{63} x \left (108+35 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}-\frac {2798 \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 )}{105 \sqrt {4+3 x^2+x^4}}+\frac {74 \sqrt {2} \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 )}{3 \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 + 3*x^2 + x^4)^(3/2),x]

[Out]

$Aborted

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (5 \, x^{6} + 22 \, x^{4} + 41 \, x^{2} + 28\right )} \sqrt {x^{4} + 3 \, x^{2} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((5*x^6 + 22*x^4 + 41*x^2 + 28)*sqrt(x^4 + 3*x^2 + 4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 275, normalized size = 1.33 \[ \frac {5 \sqrt {x^{4}+3 x^{2}+4}\, x^{7}}{9}+\frac {71 \sqrt {x^{4}+3 x^{2}+4}\, x^{5}}{21}+\frac {3187 \sqrt {x^{4}+3 x^{2}+4}\, x^{3}}{315}+\frac {583 \sqrt {x^{4}+3 x^{2}+4}\, x}{35}+\frac {6352 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{35 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}-\frac {89536 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )+\EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )\right )}{105 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/9*(x^4+3*x^2+4)^(1/2)*x^7+71/21*(x^4+3*x^2+4)^(1/2)*x^5+3187/315*(x^4+3*x^2+4)^(1/2)*x^3+583/35*(x^4+3*x^2+4
)^(1/2)*x+6352/35/(-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))-89536/105/(-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)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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