3.71 \(\int \frac{x^3}{(a+c x^2)^{3/2} (d+e x+f x^2)} \, dx\)

Optimal. Leaf size=499 \[ \frac{c e x \left (a \left (e^2-2 d f\right )+c d^2\right )+a f \left (a \left (e^2-d f\right )+c d^2\right )}{a f^2 \sqrt{a+c x^2} \left ((c d-a f)^2+a c e^2\right )}-\frac{\left (2 a d e f-\left (e-\sqrt{e^2-4 d f}\right ) \left (a \left (e^2-d f\right )+c d^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (2 a d e f-\left (\sqrt{e^2-4 d f}+e\right ) \left (a \left (e^2-d f\right )+c d^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{e x}{a f^2 \sqrt{a+c x^2}}-\frac{1}{c f \sqrt{a+c x^2}} \]

[Out]

-(1/(c*f*Sqrt[a + c*x^2])) - (e*x)/(a*f^2*Sqrt[a + c*x^2]) + (a*f*(c*d^2 + a*(e^2 - d*f)) + c*e*(c*d^2 + a*(e^
2 - 2*d*f))*x)/(a*f^2*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[a + c*x^2]) - ((2*a*d*e*f - (e - Sqrt[e^2 - 4*d*f])*(c*d^
2 + a*(e^2 - d*f)))*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*S
qrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^2 + c*(e
^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + ((2*a*d*e*f - (e + Sqrt[e^2 - 4*d*f])*(c*d^2 + a*(e^2 - d*f)))*ArcTanh[(
2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c
*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d
*f])])

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Rubi [A]  time = 2.1068, antiderivative size = 499, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {6728, 191, 261, 1017, 1034, 725, 206} \[ \frac{c e x \left (a \left (e^2-2 d f\right )+c d^2\right )+a f \left (a \left (e^2-d f\right )+c d^2\right )}{a f^2 \sqrt{a+c x^2} \left ((c d-a f)^2+a c e^2\right )}-\frac{\left (2 a d e f-\left (e-\sqrt{e^2-4 d f}\right ) \left (a \left (e^2-d f\right )+c d^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (2 a d e f-\left (\sqrt{e^2-4 d f}+e\right ) \left (a \left (e^2-d f\right )+c d^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{e x}{a f^2 \sqrt{a+c x^2}}-\frac{1}{c f \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(c*f*Sqrt[a + c*x^2])) - (e*x)/(a*f^2*Sqrt[a + c*x^2]) + (a*f*(c*d^2 + a*(e^2 - d*f)) + c*e*(c*d^2 + a*(e^
2 - 2*d*f))*x)/(a*f^2*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[a + c*x^2]) - ((2*a*d*e*f - (e - Sqrt[e^2 - 4*d*f])*(c*d^
2 + a*(e^2 - d*f)))*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*S
qrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^2 + c*(e
^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + ((2*a*d*e*f - (e + Sqrt[e^2 - 4*d*f])*(c*d^2 + a*(e^2 - d*f)))*ArcTanh[(
2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c
*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d
*f])])

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1017

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[((a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1)*(g*c*(2*a*c*e) + (-(a*h))*(2*c^2*d - c*(2*a*f)) + c*(g*(2*c^2*
d - c*(2*a*f)) - h*(-2*a*c*e))*x))/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)), x] + Dist[1/((-4*a*c)*(a*c*e^
2 + (c*d - a*f)^2)*(p + 1)), Int[(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Simp[(-2*g*c)*((c*d - a*f)^2 - (-(a*e
))*(c*e))*(p + 1) + (2*(g*c*(c*d - a*f) - a*(-(h*c*e))))*(a*f*(p + 1) - c*d*(p + 2)) - e*((g*c)*(2*a*c*e) + (-
(a*h))*(2*c^2*d - c*((Plus[2])*a*f)))*(p + q + 2) - (2*f*((g*c)*(2*a*c*e) + (-(a*h))*(2*c^2*d - c*((Plus[2])*a
*f)))*(p + q + 2) - (2*(g*c*(c*d - a*f) - a*(-(h*c*e))))*(-(c*e*(2*p + q + 4))))*x - c*f*(2*(g*c*(c*d - a*f) -
a*(-(h*c*e))))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, g, h, q}, x] && NeQ[e^2 - 4*d*f, 0] &
& LtQ[p, -1] && NeQ[a*c*e^2 + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1])

Rule 1034

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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])

Rubi steps

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

Mathematica [A]  time = 2.90962, size = 577, normalized size = 1.16 \[ \frac{\left (-\frac{e \left (e^2-3 d f\right )}{\sqrt{e^2-4 d f}}-d f+e^2\right ) \left (2 a f+c x \left (e-\sqrt{e^2-4 d f}\right )\right )}{a f^2 \sqrt{a+c x^2} \left (4 a f^2+c \left (e-\sqrt{e^2-4 d f}\right )^2\right )}+\frac{\left (\frac{e \left (e^2-3 d f\right )}{\sqrt{e^2-4 d f}}-d f+e^2\right ) \left (2 a f+c x \left (\sqrt{e^2-4 d f}+e\right )\right )}{a f^2 \sqrt{a+c x^2} \left (4 a f^2+c \left (\sqrt{e^2-4 d f}+e\right )^2\right )}+\frac{\sqrt{2} \left (-e^2 \sqrt{e^2-4 d f}+d f \sqrt{e^2-4 d f}-3 d e f+e^3\right ) \tanh ^{-1}\left (\frac{2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt{e^2-4 d f} \left (2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}-\frac{\sqrt{2} \left (e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}-3 d e f+e^3\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{e^2-4 d f} \left (2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}-\frac{e x}{a f^2 \sqrt{a+c x^2}}-\frac{1}{c f \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(c*f*Sqrt[a + c*x^2])) - (e*x)/(a*f^2*Sqrt[a + c*x^2]) + ((e^2 - d*f - (e*(e^2 - 3*d*f))/Sqrt[e^2 - 4*d*f]
)*(2*a*f + c*(e - Sqrt[e^2 - 4*d*f])*x))/(a*f^2*(4*a*f^2 + c*(e - Sqrt[e^2 - 4*d*f])^2)*Sqrt[a + c*x^2]) + ((e
^2 - d*f + (e*(e^2 - 3*d*f))/Sqrt[e^2 - 4*d*f])*(2*a*f + c*(e + Sqrt[e^2 - 4*d*f])*x))/(a*f^2*(4*a*f^2 + c*(e
+ Sqrt[e^2 - 4*d*f])^2)*Sqrt[a + c*x^2]) + (Sqrt[2]*(e^3 - 3*d*e*f - e^2*Sqrt[e^2 - 4*d*f] + d*f*Sqrt[e^2 - 4*
d*f])*ArcTanh[(2*a*f + c*(-e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])]
*Sqrt[a + c*x^2])])/(Sqrt[e^2 - 4*d*f]*(2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]))^(3/2)) - (Sqrt[2]*(e^
3 - 3*d*e*f + e^2*Sqrt[e^2 - 4*d*f] - d*f*Sqrt[e^2 - 4*d*f])*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sq
rt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[e^2 - 4*d*f]*(2*a*f^2 + c*(e^2
- 2*d*f + e*Sqrt[e^2 - 4*d*f]))^(3/2))

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Maple [B]  time = 0.276, size = 6124, normalized size = 12.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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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(x^3/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a + c x^{2}\right )^{\frac{3}{2}} \left (d + e x + f x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/((a + c*x**2)**(3/2)*(d + e*x + f*x**2)), 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(x^3/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="giac")

[Out]

Exception raised: TypeError