3.373 \(\int \frac{(d+e x^2)^{3/2}}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=487 \[ \frac{\left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c \sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (3 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}{2 c \sqrt{b^2-4 a c}}-\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (3 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}{2 c \sqrt{b^2-4 a c}} \]

[Out]

((2*c^2*d^2 + b*(b - Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*(b*d - Sqrt[b^2 - 4*a*c]*d + a*e))*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])])/(c*Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt
[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]) - ((2*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*
(b*d + Sqrt[b^2 - 4*a*c]*d + a*e))*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])])/(c*Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*
e]) + (Sqrt[e]*(3*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c*Sqrt[b^2 - 4*a*c
]) - (Sqrt[e]*(3*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c*Sqrt[b^2 - 4*a*c]
)

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Rubi [A]  time = 1.57217, antiderivative size = 487, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {1174, 416, 523, 217, 206, 377, 205} \[ \frac{\left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c \sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (3 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}{2 c \sqrt{b^2-4 a c}}-\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (3 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}{2 c \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^(3/2)/(a + b*x^2 + c*x^4),x]

[Out]

((2*c^2*d^2 + b*(b - Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*(b*d - Sqrt[b^2 - 4*a*c]*d + a*e))*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])])/(c*Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt
[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]) - ((2*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*
(b*d + Sqrt[b^2 - 4*a*c]*d + a*e))*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])])/(c*Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*
e]) + (Sqrt[e]*(3*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c*Sqrt[b^2 - 4*a*c
]) - (Sqrt[e]*(3*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c*Sqrt[b^2 - 4*a*c]
)

Rule 1174

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]
}, Dist[(2*c)/r, Int[(d + e*x^2)^q/(b - r + 2*c*x^2), x], x] - Dist[(2*c)/r, Int[(d + e*x^2)^q/(b + r + 2*c*x^
2), x], x]] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !Integ
erQ[q]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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{\left (d+e x^2\right )^{3/2}}{a+b x^2+c x^4} \, dx &=\frac{(2 c) \int \frac{\left (d+e x^2\right )^{3/2}}{b-\sqrt{b^2-4 a c}+2 c x^2} \, dx}{\sqrt{b^2-4 a c}}-\frac{(2 c) \int \frac{\left (d+e x^2\right )^{3/2}}{b+\sqrt{b^2-4 a c}+2 c x^2} \, dx}{\sqrt{b^2-4 a c}}\\ &=\frac{\int \frac{d \left (4 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right )+2 e \left (3 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) x^2}{\left (b-\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{2 \sqrt{b^2-4 a c}}-\frac{\int \frac{d \left (4 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right )+2 e \left (3 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) x^2}{\left (b+\sqrt{b^2-4 a c}+2 c x^2\right ) \sqrt{d+e x^2}} \, dx}{2 \sqrt{b^2-4 a c}}\\ &=\frac{\left (e \left (3 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right )\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{2 c \sqrt{b^2-4 a c}}-\frac{\left (e \left (3 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right )\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{2 c \sqrt{b^2-4 a c}}+\frac{\left (2 c^2 d^2+b \left (b-\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt{b^2-4 a c} d+a e\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 \sqrt{b^2-4 a c}}-\frac{\left (2 c^2 d^2+b \left (b+\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt{b^2-4 a c} d+a e\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 \sqrt{b^2-4 a c}}\\ &=\frac{\left (e \left (3 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{2 c \sqrt{b^2-4 a c}}-\frac{\left (e \left (3 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{2 c \sqrt{b^2-4 a c}}+\frac{\left (2 c^2 d^2+b \left (b-\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt{b^2-4 a c} d+a e\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 \sqrt{b^2-4 a c}}-\frac{\left (2 c^2 d^2+b \left (b+\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt{b^2-4 a c} d+a e\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 \sqrt{b^2-4 a c}}\\ &=\frac{\left (2 c^2 d^2+b \left (b-\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt{b^2-4 a c} d+a e\right )\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 )}{c \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}-\frac{\left (2 c^2 d^2+b \left (b+\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt{b^2-4 a c} d+a e\right )\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 )}{c \sqrt{b^2-4 a c} \sqrt{b+\sqrt{b^2-4 a c}} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}+\frac{\sqrt{e} \left (3 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c \sqrt{b^2-4 a c}}-\frac{\sqrt{e} \left (3 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c \sqrt{b^2-4 a c}}\\ \end{align*}

Mathematica [B]  time = 6.16337, size = 9290, normalized size = 19.08 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d + e*x^2)^(3/2)/(a + b*x^2 + c*x^4),x]

[Out]

Result too large to show

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Maple [C]  time = 0.021, size = 217, normalized size = 0.5 \begin{align*} -{\frac{1}{c}{e}^{{\frac{3}{2}}}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) }+{\frac{1}{2\,c}{e}^{{\frac{3}{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 ( be-2\,cd \right ){{\it \_R}}^{2}+2\,e \left ( 2\,ae-bd \right ){\it \_R}+b{d}^{2}e-2\,c{d}^{3}}{{{\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}}\ln \left ( \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{2}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x)

[Out]

-e^(3/2)/c*ln((e*x^2+d)^(1/2)-e^(1/2)*x)+1/2*e^(3/2)/c*sum(((b*e-2*c*d)*_R^2+2*e*(2*a*e-b*d)*_R+b*d^2*e-2*c*d^
3)/(_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{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^(3/2)/(c*x^4 + b*x^2 + a), 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((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{\left (d + e x^{2}\right )^{\frac{3}{2}}}{a + b x^{2} + c x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**(3/2)/(c*x**4+b*x**2+a),x)

[Out]

Integral((d + e*x**2)**(3/2)/(a + b*x**2 + c*x**4), x)

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Giac [A]  time = 1.35158, size = 36, normalized size = 0.07 \begin{align*} -\frac{e^{\frac{3}{2}} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

-1/2*e^(3/2)*log((x*e^(1/2) - sqrt(x^2*e + d))^2)/c