3.74 \(\int \frac{1}{(a+c x^2)^{3/2} (d+e x+f x^2)} \, dx\)

Optimal. Leaf size=416 \[ -\frac{f \left (2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{c (x (c d-a f)+a e)}{a \sqrt{a+c x^2} \left ((c d-a f)^2+a c e^2\right )} \]

[Out]

(c*(a*e + (c*d - a*f)*x))/(a*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[a + c*x^2]) - (f*(2*a*f^2 + c*(e^2 - 2*d*f + e*Sqr
t[e^2 - 4*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])]) + (f*(2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*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])])/(Sqr
t[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])])

________________________________________________________________________________________

Rubi [A]  time = 0.61705, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {976, 1034, 725, 206} \[ -\frac{f \left (2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{c (x (c d-a f)+a e)}{a \sqrt{a+c x^2} \left ((c d-a f)^2+a c e^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]

[Out]

(c*(a*e + (c*d - a*f)*x))/(a*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[a + c*x^2]) - (f*(2*a*f^2 + c*(e^2 - 2*d*f + e*Sqr
t[e^2 - 4*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])]) + (f*(2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*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])])/(Sqr
t[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 976

Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((2*a*c^2*e + c*(2
*c^2*d - c*(2*a*f))*x)*(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1))/((-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*Si
mp[2*c*((c*d - a*f)^2 - (-(a*e))*(c*e))*(p + 1) - (2*c^2*d - c*(2*a*f))*(a*f*(p + 1) - c*d*(p + 2)) - e*(-2*a*
c^2*e)*(p + q + 2) + (2*f*(2*a*c^2*e)*(p + q + 2) - (2*c^2*d - c*(2*a*f))*(-(c*e*(2*p + q + 4))))*x + c*f*(2*c
^2*d - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, q}, x] && NeQ[e^2 - 4*d*f, 0] && Lt
Q[p, -1] && NeQ[a*c*e^2 + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]

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{1}{\left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx &=\frac{c (a e+(c d-a f) x)}{a \left (a c e^2+(c d-a f)^2\right ) \sqrt{a+c x^2}}-\frac{\int \frac{-2 a c \left (a f^2+c \left (e^2-d f\right )\right )-2 a c^2 e f x}{\sqrt{a+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 a c \left (a c e^2+(c d-a f)^2\right )}\\ &=\frac{c (a e+(c d-a f) x)}{a \left (a c e^2+(c d-a f)^2\right ) \sqrt{a+c x^2}}-\frac{\left (f \left (2 a f^2+c \left (e^2-2 d f-e \sqrt{e^2-4 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 (f \left (2 a f^2+c \left (e^2-2 d f+e \sqrt{e^2-4 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{c (a e+(c d-a f) x)}{a \left (a c e^2+(c d-a f)^2\right ) \sqrt{a+c x^2}}+\frac{\left (f \left (2 a f^2+c \left (e^2-2 d f-e \sqrt{e^2-4 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 (f \left (2 a f^2+c \left (e^2-2 d f+e \sqrt{e^2-4 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{c (a e+(c d-a f) x)}{a \left (a c e^2+(c d-a f)^2\right ) \sqrt{a+c x^2}}-\frac{f \left (2 a f^2+c \left (e^2-2 d f+e \sqrt{e^2-4 d f}\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{f \left (2 a f^2+c \left (e^2-2 d f-e \sqrt{e^2-4 d f}\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 = 2.33532, size = 320, normalized size = 0.77 \[ \frac{c (a (e-f x)+c d x)}{a \sqrt{a+c x^2} \left (a^2 f^2+a c \left (e^2-2 d f\right )+c^2 d^2\right )}-\frac{2 \sqrt{2} f^3 \tanh ^{-1}\left (\frac{2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt{e^2-4 d f} \left (2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac{2 \sqrt{2} f^3 \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{e^2-4 d f} \left (2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.304, size = 1713, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x)

[Out]

2/(-4*d*f+e^2)^(1/2)/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)*f^2/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*
c-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*
e^2)/f^2)^(1/2)-4*c^2*f/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-1/f^2*(-4
*d*f+e^2)*c^2)/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2)
)/f)+1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*x+4/(-4*d*f+e^2)^(1/2)*c^2*f/(-(-4*d*f+e^2
)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-1/f^2*(-4*d*f+e^2)*c^2)/((x-1/2*(-e+(-4*d*f+e^
2)^(1/2))/f)^2*c-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a
*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*x*e-2/(-4*d*f+e^2)^(1/2)/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)*f^2*2^
(1/2)/((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f
+c*e^2)/f^2-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*((-(-4*d*f+e^2)^(1/2)*c*e
+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-4*c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/
2*(-e+(-4*d*f+e^2)^(1/2))/f)+2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2))/(x-1/2*(-e+(-4*d*f+
e^2)^(1/2))/f))-2/(-4*d*f+e^2)^(1/2)/((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)*f^2/((x+1/2*(e+(-4*d*f+e^2
)^(1/2))/f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^
2-2*c*d*f+c*e^2)/f^2)^(1/2)-4*c^2*f/((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/(4*a*c-4*c^2/f*d+c^2/f^2*e^
2-1/f^2*(-4*d*f+e^2)*c^2)/((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e
^2)^(1/2))/f)+1/2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*x-4/(-4*d*f+e^2)^(1/2)*c^2*f/((-4*
d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-1/f^2*(-4*d*f+e^2)*c^2)/((x+1/2*(e+(-4*
d*f+e^2)^(1/2))/f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*((-4*d*f+e^2)^(1/2)*c*e
+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*x*e+2/(-4*d*f+e^2)^(1/2)/((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)*f^2
*2^(1/2)/(((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*ln((((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*
f+c*e^2)/f^2-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*(((-4*d*f+e^2)^(1/2)*c*e+
2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-4*c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*
(e+(-4*d*f+e^2)^(1/2))/f)+2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2))/(x+1/2*(e+(-4*d*f+e^2)^
(1/2))/f))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\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(1/(c*x**2+a)**(3/2)/(f*x**2+e*x+d),x)

[Out]

Integral(1/((a + c*x**2)**(3/2)*(d + e*x + f*x**2)), x)

________________________________________________________________________________________

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

[Out]

Exception raised: TypeError