3.397 \(\int \frac{1}{(d+e x^2)^{3/2} (a+b x^2+c x^4)} \, dx\)
Optimal. Leaf size=341 \[ -\frac{c \left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac{c \left (\frac{2 c d-b e}{\sqrt{b^2-4 a c}}+e\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}+\frac{e^2 x}{d \sqrt{d+e x^2} \left (a e^2-b d e+c d^2\right )} \]
[Out]
(e^2*x)/(d*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x^2]) - (c*(e - (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*
c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b - Sqrt[b^2 - 4*a*c
]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)) - (c*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])
*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b +
Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2))
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Rubi [A] time = 0.773841, antiderivative size = 341, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used =
{1172, 191, 1692, 377, 205} \[ -\frac{c \left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac{c \left (\frac{2 c d-b e}{\sqrt{b^2-4 a c}}+e\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}+\frac{e^2 x}{d \sqrt{d+e x^2} \left (a e^2-b d e+c d^2\right )} \]
Antiderivative was successfully verified.
[In]
Int[1/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]
[Out]
(e^2*x)/(d*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x^2]) - (c*(e - (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*
c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b - Sqrt[b^2 - 4*a*c
]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)) - (c*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])
*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b +
Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2))
Rule 1172
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*
e^2), Int[(d + e*x^2)^q, x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d + e*x^2)^(q + 1)*(c*d - b*e - c*e*x^
2))/(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] && !IntegerQ[q] && LtQ[q, -1]
Rule 191
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]
Rule 1692
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]
Rule 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx &=\frac{\int \frac{c d-b e-c e x^2}{\sqrt{d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{c d^2-b d e+a e^2}+\frac{e^2 \int \frac{1}{\left (d+e x^2\right )^{3/2}} \, dx}{c d^2-b d e+a e^2}\\ &=\frac{e^2 x}{d \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}+\frac{\int \left (\frac{-c e-\frac{c (-2 c d+b e)}{\sqrt{b^2-4 a c}}}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}}+\frac{-c e+\frac{c (-2 c d+b e)}{\sqrt{b^2-4 a c}}}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}}\right ) \, dx}{c d^2-b d e+a e^2}\\ &=\frac{e^2 x}{d \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}-\frac{\left (c \left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{c d^2-b d e+a e^2}-\frac{\left (c \left (e+\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{c d^2-b d e+a e^2}\\ &=\frac{e^2 x}{d \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}-\frac{\left (c \left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b-\sqrt{b^2-4 a c}-\left (-2 c d+\left (b-\sqrt{b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{c d^2-b d e+a e^2}-\frac{\left (c \left (e+\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b+\sqrt{b^2-4 a c}-\left (-2 c d+\left (b+\sqrt{b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{c d^2-b d e+a e^2}\\ &=\frac{e^2 x}{d \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x^2}}-\frac{c \left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} x}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )}-\frac{c \left (e+\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} x}{\sqrt{b+\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b+\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )}\\ \end{align*}
Mathematica [C] time = 10.0012, size = 2112, normalized size = 6.19 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]
[Out]
(2*c*x*(45*Sqrt[-(((-b + Sqrt[b^2 - 4*a*c])*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2*(d + e*x^2))/(d^2*(b - Sq
rt[b^2 - 4*a*c] + 2*c*x^2)^2))] + (30*e*x^2*Sqrt[-(((-b + Sqrt[b^2 - 4*a*c])*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])
*e)*x^2*(d + e*x^2))/(d^2*(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)^2))])/d - 45*ArcSin[Sqrt[-(((2*c*d + (-b + Sqrt[b^
2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))]] - (30*e*x^2*ArcSin[Sqrt[-(((2*c*d + (-b + Sqrt[b
^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))]])/d - (45*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x
^2*ArcSin[Sqrt[-(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))]])/(d*(-b
+ Sqrt[b^2 - 4*a*c] - 2*c*x^2)) - (30*e*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^4*ArcSin[Sqrt[-(((2*c*d + (-b +
Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))]])/(d^2*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)
) + 4*(-(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))))^(5/2)*Sqrt[((-b +
Sqrt[b^2 - 4*a*c])*(d + e*x^2))/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))]*Hypergeometric2F1[2, 2, 7/2, -(((2*c*
d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))] + (4*e*x^2*(-(((2*c*d + (-b + Sq
rt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))))^(5/2)*Sqrt[((-b + Sqrt[b^2 - 4*a*c])*(d + e*
x^2))/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))]*Hypergeometric2F1[2, 2, 7/2, -(((2*c*d + (-b + Sqrt[b^2 - 4*a*c]
)*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))])/d))/(15*Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*(-(((2
*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))))^(3/2)*(1 - (2*c*x^2)/(-b + Sq
rt[b^2 - 4*a*c]))*Sqrt[d + e*x^2]*Sqrt[((-b + Sqrt[b^2 - 4*a*c])*(d + e*x^2))/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c
*x^2))]) - (2*c*x*(45*Sqrt[-(((b + Sqrt[b^2 - 4*a*c])*(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*x^2*(d + e*x^2))/(d
^2*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)^2))] + (30*e*x^2*Sqrt[-(((b + Sqrt[b^2 - 4*a*c])*(-2*c*d + (b + Sqrt[b^2
- 4*a*c])*e)*x^2*(d + e*x^2))/(d^2*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)^2))])/d - 45*ArcSin[Sqrt[((2*c*d - (b + S
qrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]] - (30*e*x^2*ArcSin[Sqrt[((2*c*d - (b + Sqrt[
b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]])/d + (45*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2
*ArcSin[Sqrt[((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]])/(d*(b + Sqrt[b
^2 - 4*a*c] + 2*c*x^2)) - (30*e*(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*x^4*ArcSin[Sqrt[((2*c*d - (b + Sqrt[b^2 -
4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]])/(d^2*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)) + 4*(((2*c*d
- (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)))^(5/2)*Sqrt[((b + Sqrt[b^2 - 4*a*c])*
(d + e*x^2))/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]*Hypergeometric2F1[2, 2, 7/2, ((2*c*d - (b + Sqrt[b^2 - 4*a
*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))] + (4*e*x^2*(((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(
b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)))^(5/2)*Sqrt[((b + Sqrt[b^2 - 4*a*c])*(d + e*x^2))/(d*(b + Sqrt[b^2 - 4*a*c]
+ 2*c*x^2))]*Hypergeometric2F1[2, 2, 7/2, ((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c]
+ 2*c*x^2))])/d))/(15*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*(((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(
d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)))^(3/2)*(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))*Sqrt[d + e*x^2]*Sqrt[((b +
Sqrt[b^2 - 4*a*c])*(d + e*x^2))/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))])
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Maple [C] time = 0.02, size = 246, normalized size = 0.7 \begin{align*} 32\,{\frac{{e}^{3/2}}{ \left ( 16\,a{e}^{2}-16\,deb+16\,c{d}^{2} \right ) \left ( 2\,e{x}^{2}-2\,\sqrt{e}\sqrt{e{x}^{2}+d}x+2\,d \right ) }}+8\,{\frac{{e}^{3/2}}{16\,a{e}^{2}-16\,deb+16\,c{d}^{2}}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{4}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{3}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{2}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){\it \_Z}+c{d}^{4} \right ) }{\frac{ \left ( c{{\it \_R}}^{2}+2\, \left ( 2\,be-3\,cd \right ){\it \_R}+c{d}^{2} \right ) \ln \left ( \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{2}-{\it \_R} \right ) }{{{\it \_R}}^{3}c+3\,{{\it \_R}}^{2}be-3\,{{\it \_R}}^{2}cd+8\,{\it \_R}\,a{e}^{2}-4\,{\it \_R}\,bde+3\,{\it \_R}\,c{d}^{2}+b{d}^{2}e-c{d}^{3}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x)
[Out]
32*e^(3/2)/(16*a*e^2-16*b*d*e+16*c*d^2)/(2*e*x^2-2*e^(1/2)*(e*x^2+d)^(1/2)*x+2*d)+8*e^(3/2)/(16*a*e^2-16*b*d*e
+16*c*d^2)*sum((c*_R^2+2*(2*b*e-3*c*d)*_R+c*d^2)/(_R^3*c+3*_R^2*b*e-3*_R^2*c*d+8*_R*a*e^2-4*_R*b*d*e+3*_R*c*d^
2+b*d^2*e-c*d^3)*ln(((e*x^2+d)^(1/2)-e^(1/2)*x)^2-_R),_R=RootOf(c*_Z^4+(4*b*e-4*c*d)*_Z^3+(16*a*e^2-8*b*d*e+6*
c*d^2)*_Z^2+(4*b*d^2*e-4*c*d^3)*_Z+c*d^4))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2} + a\right )}{\left (e x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")
[Out]
integrate(1/((c*x^4 + b*x^2 + a)*(e*x^2 + d)^(3/2)), x)
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x^{2}\right )^{\frac{3}{2}} \left (a + b x^{2} + c x^{4}\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x**2+d)**(3/2)/(c*x**4+b*x**2+a),x)
[Out]
Integral(1/((d + e*x**2)**(3/2)*(a + b*x**2 + c*x**4)), x)
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="giac")
[Out]
Exception raised: TypeError