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

Optimal. Leaf size=310 \[ -\frac{2 \left (b \left (b^2 f-c (3 a f+c d)\right )-c x \left (2 a c f+b^2 (-f)+2 c^2 d\right )\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac{f \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 \sqrt{d} \left (a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d\right )^{3/2}}+\frac{f \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 \sqrt{d} \left (a f+b \sqrt{d} \sqrt{f}+c d\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.41061, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {975, 1033, 724, 206} \[ -\frac{2 \left (b \left (b^2 f-c (3 a f+c d)\right )-c x \left (2 a c f+b^2 (-f)+2 c^2 d\right )\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac{f \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 \sqrt{d} \left (a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d\right )^{3/2}}+\frac{f \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 \sqrt{d} \left (a f+b \sqrt{d} \sqrt{f}+c d\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 975

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((b^3*f + b*c*(c*d
- 3*a*f) + c*(2*c^2*d + b^2*f - c*(2*a*f))*x)*(a + b*x + c*x^2)^(p + 1)*(d + f*x^2)^(q + 1))/((b^2 - 4*a*c)*(
b^2*d*f + (c*d - a*f)^2)*(p + 1)), x] - Dist[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)), Int[(a + b*x
+ c*x^2)^(p + 1)*(d + f*x^2)^q*Simp[2*c*(b^2*d*f + (c*d - a*f)^2)*(p + 1) - (2*c^2*d + b^2*f - c*(2*a*f))*(a*
f*(p + 1) - c*d*(p + 2)) + (2*f*(b^3*f + b*c*(c*d - 3*a*f))*(p + q + 2) - (2*c^2*d + b^2*f - c*(2*a*f))*(b*f*(
p + 1)))*x + c*f*(2*c^2*d + b^2*f - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &
&  !IGtQ[q, 0]

Rule 1033

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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])

Rubi steps

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

Mathematica [A]  time = 0.446705, size = 360, normalized size = 1.16 \[ \frac{2 \left (\frac{-b c (3 a f+c d)-2 c^2 x (a f+c d)+b^2 c f x+b^3 f}{\sqrt{a+x (b+c x)}}+\frac{f \left (b^2-4 a c\right ) \left (a f+b \sqrt{d} \sqrt{f}+c d\right ) \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+b \left (\sqrt{d}-\sqrt{f} x\right )+2 c \sqrt{d} x}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{4 \sqrt{d} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}+\frac{f \left (4 a c-b^2\right ) \left (a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d\right ) \tanh ^{-1}\left (\frac{-2 \left (a \sqrt{f}+c \sqrt{d} x\right )-b \left (\sqrt{d}+\sqrt{f} x\right )}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{4 \sqrt{d} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{\left (b^2-4 a c\right ) \left ((a f+c d)^2-b^2 d f\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.292, size = 1376, normalized size = 4.4 \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+b*x+a)^(3/2)/(-f*x^2+d),x)

[Out]

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

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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(1/(c*x^2+b*x+a)^(3/2)/(-f*x^2+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(1/(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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+b*x+a)**(3/2)/(-f*x**2+d),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x, algorithm="giac")

[Out]

sage2