3.377 \(\int \frac{1}{(7+5 x^2) (4+3 x^2+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=284 \[ -\frac{\left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ),\frac{1}{8}\right )}{12 \sqrt{2} \sqrt{x^4+3 x^2+4}}+\frac{\sqrt{x^4+3 x^2+4} x}{77 \left (x^2+2\right )}-\frac{\left (4 x^2+13\right ) x}{308 \sqrt{x^4+3 x^2+4}}+\frac{25}{176} \sqrt{\frac{5}{77}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{x^4+3 x^2+4}}\right )-\frac{\sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{77 \sqrt{x^4+3 x^2+4}}+\frac{425 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{3696 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

[Out]

-(x*(13 + 4*x^2))/(308*Sqrt[4 + 3*x^2 + x^4]) + (x*Sqrt[4 + 3*x^2 + x^4])/(77*(2 + x^2)) + (25*Sqrt[5/77]*ArcT
an[(2*Sqrt[11/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/176 - (Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*Elli
pticE[2*ArcTan[x/Sqrt[2]], 1/8])/(77*Sqrt[4 + 3*x^2 + x^4]) - ((2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*E
llipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(12*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) + (425*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)
/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(3696*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.16267, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {1221, 1178, 1197, 1103, 1195, 1216, 1706} \[ \frac{\sqrt{x^4+3 x^2+4} x}{77 \left (x^2+2\right )}-\frac{\left (4 x^2+13\right ) x}{308 \sqrt{x^4+3 x^2+4}}+\frac{25}{176} \sqrt{\frac{5}{77}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{x^4+3 x^2+4}}\right )-\frac{\left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{12 \sqrt{2} \sqrt{x^4+3 x^2+4}}-\frac{\sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{77 \sqrt{x^4+3 x^2+4}}+\frac{425 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{3696 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Int[1/((7 + 5*x^2)*(4 + 3*x^2 + x^4)^(3/2)),x]

[Out]

-(x*(13 + 4*x^2))/(308*Sqrt[4 + 3*x^2 + x^4]) + (x*Sqrt[4 + 3*x^2 + x^4])/(77*(2 + x^2)) + (25*Sqrt[5/77]*ArcT
an[(2*Sqrt[11/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/176 - (Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*Elli
pticE[2*ArcTan[x/Sqrt[2]], 1/8])/(77*Sqrt[4 + 3*x^2 + x^4]) - ((2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*E
llipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(12*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) + (425*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)
/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(3696*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

Rule 1221

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

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*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] && LtQ[p, -1] && IntegerQ[2*p]

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 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 1216

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

Rule 1706

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

Rubi steps

\begin{align*} \int \frac{1}{\left (7+5 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}} \, dx &=\frac{1}{44} \int \frac{-8-5 x^2}{\left (4+3 x^2+x^4\right )^{3/2}} \, dx+\frac{25}{44} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx\\ &=-\frac{x \left (13+4 x^2\right )}{308 \sqrt{4+3 x^2+x^4}}+\frac{\int \frac{-4+16 x^2}{\sqrt{4+3 x^2+x^4}} \, dx}{1232}-\frac{25}{132} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{125}{66} \int \frac{1+\frac{x^2}{2}}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx\\ &=-\frac{x \left (13+4 x^2\right )}{308 \sqrt{4+3 x^2+x^4}}+\frac{25}{176} \sqrt{\frac{5}{77}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{4+3 x^2+x^4}}\right )-\frac{25 \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 )}{264 \sqrt{2} \sqrt{4+3 x^2+x^4}}+\frac{425 \left (2+x^2\right ) \sqrt{\frac{4+3 x^2+x^4}{\left (2+x^2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{3696 \sqrt{2} \sqrt{4+3 x^2+x^4}}+\frac{1}{44} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx-\frac{2}{77} \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx\\ &=-\frac{x \left (13+4 x^2\right )}{308 \sqrt{4+3 x^2+x^4}}+\frac{x \sqrt{4+3 x^2+x^4}}{77 \left (2+x^2\right )}+\frac{25}{176} \sqrt{\frac{5}{77}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{4+3 x^2+x^4}}\right )-\frac{\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 )}{77 \sqrt{4+3 x^2+x^4}}-\frac{\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}}+\frac{425 \left (2+x^2\right ) \sqrt{\frac{4+3 x^2+x^4}{\left (2+x^2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{3696 \sqrt{2} \sqrt{4+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.540215, size = 483, normalized size = 1.7 \[ \frac{\sqrt{2} \left (2 \sqrt{7}+7 i\right ) \sqrt{\frac{-2 i x^2+\sqrt{7}-3 i}{\sqrt{7}-3 i}} \sqrt{\frac{2 i x^2+\sqrt{7}+3 i}{\sqrt{7}+3 i}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{\sqrt{7}-3 i}} x\right ),\frac{-\sqrt{7}+3 i}{\sqrt{7}+3 i}\right )-8 \sqrt{-\frac{i}{\sqrt{7}-3 i}} x^3-2 \sqrt{2} \left (\sqrt{7}+3 i\right ) \sqrt{\frac{-2 i x^2+\sqrt{7}-3 i}{\sqrt{7}-3 i}} \sqrt{\frac{2 i x^2+\sqrt{7}+3 i}{\sqrt{7}+3 i}} E\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )-25 i \sqrt{2} \sqrt{\frac{-2 i x^2+\sqrt{7}-3 i}{\sqrt{7}-3 i}} \sqrt{\frac{2 i x^2+\sqrt{7}+3 i}{\sqrt{7}+3 i}} \Pi \left (\frac{5}{14} \left (3+i \sqrt{7}\right );i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )-26 \sqrt{-\frac{i}{\sqrt{7}-3 i}} x}{616 \sqrt{-\frac{i}{\sqrt{7}-3 i}} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((7 + 5*x^2)*(4 + 3*x^2 + x^4)^(3/2)),x]

[Out]

(-26*Sqrt[(-I)/(-3*I + Sqrt[7])]*x - 8*Sqrt[(-I)/(-3*I + Sqrt[7])]*x^3 - 2*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]*(7*I + 2*Sqrt[7])*Sqrt[(-3*I + S
qrt[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[Sq
rt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] - (25*I)*Sqrt[2]*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])]*EllipticPi[(5*(3 + I*Sqrt[7]))/14, I
*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])])/(616*Sqrt[(-I)/(-3*I + Sqrt[7])]*
Sqrt[4 + 3*x^2 + x^4])

________________________________________________________________________________________

Maple [C]  time = 0.018, size = 409, normalized size = 1.4 \begin{align*} -2\,{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}} \left ({\frac{{x}^{3}}{154}}+{\frac{13\,x}{616}} \right ) }-{\frac{1}{77\,\sqrt{-6+2\,i\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}-{\frac{32}{77\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}+{\frac{32}{77\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticE} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}+{\frac{25}{308\,\sqrt{-3/8+i/8\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticPi} \left ( \sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}x,-{\frac{5}{-{\frac{21}{8}}+{\frac{7\,i}{8}}\sqrt{7}}},{\frac{\sqrt{-{\frac{3}{8}}-{\frac{i}{8}}\sqrt{7}}}{\sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(1/154*x^3+13/616*x)/(x^4+3*x^2+4)^(1/2)-1/77/(-6+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1
+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticF(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2
))^(1/2))-32/77/(-6+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)
/(x^4+3*x^2+4)^(1/2)/(I*7^(1/2)+3)*EllipticF(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))+32/77/(-6
+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2
)/(I*7^(1/2)+3)*EllipticE(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))+25/308/(-3/8+1/8*I*7^(1/2))^
(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticPi((
-3/8+1/8*I*7^(1/2))^(1/2)*x,-5/7/(-3/8+1/8*I*7^(1/2)),(-3/8-1/8*I*7^(1/2))^(1/2)/(-3/8+1/8*I*7^(1/2))^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac{3}{2}}{\left (5 \, x^{2} + 7\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 3 \, x^{2} + 4}}{5 \, x^{10} + 37 \, x^{8} + 127 \, x^{6} + 239 \, x^{4} + 248 \, x^{2} + 112}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 3*x^2 + 4)/(5*x^10 + 37*x^8 + 127*x^6 + 239*x^4 + 248*x^2 + 112), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\left (x^{2} - x + 2\right ) \left (x^{2} + x + 2\right )\right )^{\frac{3}{2}} \left (5 x^{2} + 7\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(((x**2 - x + 2)*(x**2 + x + 2))**(3/2)*(5*x**2 + 7)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac{3}{2}}{\left (5 \, x^{2} + 7\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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