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

Optimal. Leaf size=247 \[ \frac {3825}{143} \left (x^4+3 x^2+4\right )^{5/2} x+\frac {\left (15365 x^2+53504\right ) \left (x^4+3 x^2+4\right )^{3/2} x}{1001}+\frac {\left (435441 x^2+1653701\right ) \sqrt {x^4+3 x^2+4} x}{5005}+\frac {4525662 \sqrt {x^4+3 x^2+4} x}{5005 \left (x^2+2\right )}+\frac {121826 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{143 \sqrt {x^4+3 x^2+4}}-\frac {4525662 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{5005 \sqrt {x^4+3 x^2+4}}+\frac {125}{13} \left (x^4+3 x^2+4\right )^{5/2} x^3 \]

[Out]

1/1001*x*(15365*x^2+53504)*(x^4+3*x^2+4)^(3/2)+3825/143*x*(x^4+3*x^2+4)^(5/2)+125/13*x^3*(x^4+3*x^2+4)^(5/2)+4
525662/5005*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)+1/5005*x*(435441*x^2+1653701)*(x^4+3*x^2+4)^(1/2)-4525662/5005*(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)+121826/143*(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 = 0.13, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1206, 1679, 1176, 1197, 1103, 1195} \[ \frac {125}{13} \left (x^4+3 x^2+4\right )^{5/2} x^3+\frac {3825}{143} \left (x^4+3 x^2+4\right )^{5/2} x+\frac {\left (15365 x^2+53504\right ) \left (x^4+3 x^2+4\right )^{3/2} x}{1001}+\frac {\left (435441 x^2+1653701\right ) \sqrt {x^4+3 x^2+4} x}{5005}+\frac {4525662 \sqrt {x^4+3 x^2+4} x}{5005 \left (x^2+2\right )}+\frac {121826 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{143 \sqrt {x^4+3 x^2+4}}-\frac {4525662 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{5005 \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4525662*x*Sqrt[4 + 3*x^2 + x^4])/(5005*(2 + x^2)) + (x*(1653701 + 435441*x^2)*Sqrt[4 + 3*x^2 + x^4])/5005 + (
x*(53504 + 15365*x^2)*(4 + 3*x^2 + x^4)^(3/2))/1001 + (3825*x*(4 + 3*x^2 + x^4)^(5/2))/143 + (125*x^3*(4 + 3*x
^2 + x^4)^(5/2))/13 - (4525662*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt
[2]], 1/8])/(5005*Sqrt[4 + 3*x^2 + x^4]) + (121826*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*Ellip
ticF[2*ArcTan[x/Sqrt[2]], 1/8])/(143*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]

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 \left (7+5 x^2\right )^3 \left (4+3 x^2+x^4\right )^{3/2} \, dx &=\frac {125}{13} x^3 \left (4+3 x^2+x^4\right )^{5/2}+\frac {1}{13} \int \left (4+3 x^2+x^4\right )^{3/2} \left (4459+8055 x^2+3825 x^4\right ) \, dx\\ &=\frac {3825}{143} x \left (4+3 x^2+x^4\right )^{5/2}+\frac {125}{13} x^3 \left (4+3 x^2+x^4\right )^{5/2}+\frac {1}{143} \int \left (33749+19755 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2} \, dx\\ &=\frac {x \left (53504+15365 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}}{1001}+\frac {3825}{143} x \left (4+3 x^2+x^4\right )^{5/2}+\frac {125}{13} x^3 \left (4+3 x^2+x^4\right )^{5/2}+\frac {\int \left (2192868+1306323 x^2\right ) \sqrt {4+3 x^2+x^4} \, dx}{3003}\\ &=\frac {x \left (1653701+435441 x^2\right ) \sqrt {4+3 x^2+x^4}}{5005}+\frac {x \left (53504+15365 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}}{1001}+\frac {3825}{143} x \left (4+3 x^2+x^4\right )^{5/2}+\frac {125}{13} x^3 \left (4+3 x^2+x^4\right )^{5/2}+\frac {\int \frac {72038844+40730958 x^2}{\sqrt {4+3 x^2+x^4}} \, dx}{45045}\\ &=\frac {x \left (1653701+435441 x^2\right ) \sqrt {4+3 x^2+x^4}}{5005}+\frac {x \left (53504+15365 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}}{1001}+\frac {3825}{143} x \left (4+3 x^2+x^4\right )^{5/2}+\frac {125}{13} x^3 \left (4+3 x^2+x^4\right )^{5/2}-\frac {9051324 \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx}{5005}+\frac {487304}{143} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {4525662 x \sqrt {4+3 x^2+x^4}}{5005 \left (2+x^2\right )}+\frac {x \left (1653701+435441 x^2\right ) \sqrt {4+3 x^2+x^4}}{5005}+\frac {x \left (53504+15365 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}}{1001}+\frac {3825}{143} x \left (4+3 x^2+x^4\right )^{5/2}+\frac {125}{13} x^3 \left (4+3 x^2+x^4\right )^{5/2}-\frac {4525662 \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 )}{5005 \sqrt {4+3 x^2+x^4}}+\frac {121826 \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 )}{143 \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)^3*(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 \[ {\rm integral}\left ({\left (125 \, x^{10} + 900 \, x^{8} + 2810 \, x^{6} + 4648 \, x^{4} + 3969 \, x^{2} + 1372\right )} \sqrt {x^{4} + 3 \, x^{2} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((125*x^10 + 900*x^8 + 2810*x^6 + 4648*x^4 + 3969*x^2 + 1372)*sqrt(x^4 + 3*x^2 + 4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 309, normalized size = 1.25 \[ \frac {125 \sqrt {x^{4}+3 x^{2}+4}\, x^{11}}{13}+\frac {12075 \sqrt {x^{4}+3 x^{2}+4}\, x^{9}}{143}+\frac {48520 \sqrt {x^{4}+3 x^{2}+4}\, x^{7}}{143}+\frac {71434 \sqrt {x^{4}+3 x^{2}+4}\, x^{5}}{91}+\frac {5528301 \sqrt {x^{4}+3 x^{2}+4}\, x^{3}}{5005}+\frac {4865781 \sqrt {x^{4}+3 x^{2}+4}\, x}{5005}+\frac {32017264 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{5005 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}-\frac {144821184 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )+\EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )\right )}{5005 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4865781/5005*(x^4+3*x^2+4)^(1/2)*x+5528301/5005*(x^4+3*x^2+4)^(1/2)*x^3+48520/143*(x^4+3*x^2+4)^(1/2)*x^7+7143
4/91*(x^4+3*x^2+4)^(1/2)*x^5+125/13*(x^4+3*x^2+4)^(1/2)*x^11+12075/143*(x^4+3*x^2+4)^(1/2)*x^9-144821184/5005/
(-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)))+32017264/5005/(-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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+7)**3*(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)**3, x)

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