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]

1/2*f*arctanh(1/2*(b*d^(1/2)-2*a*f^(1/2)+x*(2*c*d^(1/2)-b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f-b*d^(1/2)*f^(
1/2))^(1/2))/d^(1/2)/(c*d+a*f-b*d^(1/2)*f^(1/2))^(3/2)+1/2*f*arctanh(1/2*(b*d^(1/2)+2*a*f^(1/2)+x*(2*c*d^(1/2)
+b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(1/2))/d^(1/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(3/2)-
2*(b*(b^2*f-c*(3*a*f+c*d))-c*(2*a*c*f-b^2*f+2*c^2*d)*x)/(-4*a*c+b^2)/(b^2*d*f-(a*f+c*d)^2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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 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 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)]

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*}

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Mathematica [A]  time = 0.40, size = 360, normalized size = 1.16 \[ \frac {2 \left (\frac {f \left (b^2-4 a c\right ) \left (a f+b \sqrt {d} \sqrt {f}+c d\right ) \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+b \left (\sqrt {d}-\sqrt {f} x\right )+2 c \sqrt {d} x}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{4 \sqrt {d} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}+\frac {f \left (4 a c-b^2\right ) \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right ) \tanh ^{-1}\left (\frac {-2 \left (a \sqrt {f}+c \sqrt {d} x\right )-b \left (\sqrt {d}+\sqrt {f} x\right )}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{4 \sqrt {d} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}+\frac {-b c (3 a f+c d)-2 c^2 x (a f+c d)+b^3 f+b^2 c f x}{\sqrt {a+x (b+c x)}}\right )}{\left (b^2-4 a c\right ) \left ((a f+c d)^2-b^2 d f\right )} \]

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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valueint()  Bad Argument Typ
e

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maple [B]  time = 0.02, size = 1376, normalized size = 4.44 \[ \text {result too large to display} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((c*sqrt(4*d*f))/(2*f^2)>0)',
see `assume?` for more details)Is ((c*sqrt(4*d*f))/(2*f^2)    +b/(2*f))    ^2    -(c*((b*sqrt(4*d*f))
/(2*f)                  +(c*d)/f+a))     /f^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\left (d-f\,x^2\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{- a d \sqrt {a + b x + c x^{2}} + a f x^{2} \sqrt {a + b x + c x^{2}} - b d x \sqrt {a + b x + c x^{2}} + b f x^{3} \sqrt {a + b x + c x^{2}} - c d x^{2} \sqrt {a + b x + c x^{2}} + c f x^{4} \sqrt {a + b x + c x^{2}}}\, dx \]

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]

-Integral(1/(-a*d*sqrt(a + b*x + c*x**2) + a*f*x**2*sqrt(a + b*x + c*x**2) - b*d*x*sqrt(a + b*x + c*x**2) + b*
f*x**3*sqrt(a + b*x + c*x**2) - c*d*x**2*sqrt(a + b*x + c*x**2) + c*f*x**4*sqrt(a + b*x + c*x**2)), x)

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