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

Optimal. Leaf size=189 \[ -\frac {\left (12 a c+3 b^2+8 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}}-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}-\frac {1}{2} (a-b+c)^{3/2} \tanh ^{-1}\left (\frac {2 a+x (b-2 c)-b}{2 \sqrt {a-b+c} \sqrt {a+b x+c x^2}}\right )+\frac {1}{2} (a+b+c)^{3/2} \tanh ^{-1}\left (\frac {2 a+x (b+2 c)+b}{2 \sqrt {a+b+c} \sqrt {a+b x+c x^2}}\right ) \]

[Out]

-1/2*(a-b+c)^(3/2)*arctanh(1/2*(2*a-b+(b-2*c)*x)/(a-b+c)^(1/2)/(c*x^2+b*x+a)^(1/2))+1/2*(a+b+c)^(3/2)*arctanh(
1/2*(2*a+b+(b+2*c)*x)/(a+b+c)^(1/2)/(c*x^2+b*x+a)^(1/2))-1/8*(12*a*c+3*b^2+8*c^2)*arctanh(1/2*(2*c*x+b)/c^(1/2
)/(c*x^2+b*x+a)^(1/2))/c^(1/2)-1/4*(2*c*x+5*b)*(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {978, 1078, 621, 206, 1033, 724} \[ -\frac {\left (12 a c+3 b^2+8 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}}-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}-\frac {1}{2} (a-b+c)^{3/2} \tanh ^{-1}\left (\frac {2 a+x (b-2 c)-b}{2 \sqrt {a-b+c} \sqrt {a+b x+c x^2}}\right )+\frac {1}{2} (a+b+c)^{3/2} \tanh ^{-1}\left (\frac {2 a+x (b+2 c)+b}{2 \sqrt {a+b+c} \sqrt {a+b x+c x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*b + 2*c*x)*Sqrt[a + b*x + c*x^2])/4 - ((a - b + c)^(3/2)*ArcTanh[(2*a - b + (b - 2*c)*x)/(2*Sqrt[a - b +
c]*Sqrt[a + b*x + c*x^2])])/2 - ((3*b^2 + 12*a*c + 8*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2]
)])/(8*Sqrt[c]) + ((a + b + c)^(3/2)*ArcTanh[(2*a + b + (b + 2*c)*x)/(2*Sqrt[a + b + c]*Sqrt[a + b*x + c*x^2])
])/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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 978

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((b*(3*p + 2*q) +
2*c*(p + q)*x)*(a + b*x + c*x^2)^(p - 1)*(d + f*x^2)^(q + 1))/(2*f*(p + q)*(2*p + 2*q + 1)), x] - Dist[1/(2*f*
(p + q)*(2*p + 2*q + 1)), Int[(a + b*x + c*x^2)^(p - 2)*(d + f*x^2)^q*Simp[b^2*d*(p - 1)*(2*p + q) - (p + q)*(
b^2*d*(1 - p) - 2*a*(c*d - a*f*(2*p + 2*q + 1))) - (2*b*(c*d - a*f)*(1 - p)*(2*p + q) - 2*(p + q)*b*(2*c*d*(2*
p + q) - (c*d + a*f)*(2*p + 2*q + 1)))*x + (b^2*f*p*(1 - p) + 2*c*(p + q)*(c*d*(2*p - 1) - a*f*(4*p + 2*q - 1)
))*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, q}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[p, 1] && NeQ[p + q, 0] && NeQ
[2*p + 2*q + 1, 0] &&  !IGtQ[p, 0] &&  !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 1078

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Sym
bol] :> Dist[C/c, Int[1/Sqrt[d + e*x + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C + B*c*x)/((a + c*x^2)*Sqrt[d
+ e*x + f*x^2]), x], x] /; FreeQ[{a, c, d, e, f, A, B, C}, x] && NeQ[e^2 - 4*d*f, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{3/2}}{1-x^2} \, dx &=-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}+\frac {1}{2} \int \frac {\frac {1}{4} \left (8 a^2+5 b^2+4 a c\right )+4 b (a+c) x+\frac {1}{4} \left (3 b^2+12 a c+8 c^2\right ) x^2}{\left (1-x^2\right ) \sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}-\frac {1}{2} \int \frac {\frac {1}{4} \left (-8 a^2-5 b^2-4 a c\right )+\frac {1}{4} \left (-3 b^2-12 a c-8 c^2\right )-4 b (a+c) x}{\left (1-x^2\right ) \sqrt {a+b x+c x^2}} \, dx+\frac {1}{8} \left (-3 b^2-12 a c-8 c^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}-\frac {1}{2} (a-b+c)^2 \int \frac {1}{(-1-x) \sqrt {a+b x+c x^2}} \, dx+\frac {1}{2} (a+b+c)^2 \int \frac {1}{(1-x) \sqrt {a+b x+c x^2}} \, dx+\frac {1}{4} \left (-3 b^2-12 a c-8 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )\\ &=-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}-\frac {\left (3 b^2+12 a c+8 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}}+(a-b+c)^2 \operatorname {Subst}\left (\int \frac {1}{4 a-4 b+4 c-x^2} \, dx,x,\frac {-2 a+b-(b-2 c) x}{\sqrt {a+b x+c x^2}}\right )-(a+b+c)^2 \operatorname {Subst}\left (\int \frac {1}{4 a+4 b+4 c-x^2} \, dx,x,\frac {-2 a-b-(b+2 c) x}{\sqrt {a+b x+c x^2}}\right )\\ &=-\frac {1}{4} (5 b+2 c x) \sqrt {a+b x+c x^2}-\frac {1}{2} (a-b+c)^{3/2} \tanh ^{-1}\left (\frac {2 a-b+(b-2 c) x}{2 \sqrt {a-b+c} \sqrt {a+b x+c x^2}}\right )-\frac {\left (3 b^2+12 a c+8 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}}+\frac {1}{2} (a+b+c)^{3/2} \tanh ^{-1}\left (\frac {2 a+b+(b+2 c) x}{2 \sqrt {a+b+c} \sqrt {a+b x+c x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.58, size = 181, normalized size = 0.96 \[ \frac {1}{8} \left (-\frac {\left (4 c (3 a+2 c)+3 b^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}-2 (5 b+2 c x) \sqrt {a+x (b+c x)}-4 (a-b+c)^{3/2} \tanh ^{-1}\left (\frac {2 a+b (x-1)-2 c x}{2 \sqrt {a-b+c} \sqrt {a+x (b+c x)}}\right )+4 (a+b+c)^{3/2} \tanh ^{-1}\left (\frac {2 a+b x+b+2 c x}{2 \sqrt {a+b+c} \sqrt {a+x (b+c x)}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*b + 2*c*x)*Sqrt[a + x*(b + c*x)] - 4*(a - b + c)^(3/2)*ArcTanh[(2*a + b*(-1 + x) - 2*c*x)/(2*Sqrt[a - b
+ c]*Sqrt[a + x*(b + c*x)])] - ((3*b^2 + 4*c*(3*a + 2*c))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)
])])/Sqrt[c] + 4*(a + b + c)^(3/2)*ArcTanh[(2*a + b + b*x + 2*c*x)/(2*Sqrt[a + b + c]*Sqrt[a + x*(b + c*x)])])
/8

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.02, size = 1346, normalized size = 7.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^(3/2)*ln((1/2*b-c+c*(x+1))/c^(1/2)+((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2))-1/2*c^(3/2)*ln((1/2*b+c+c*(x-
1))/c^(1/2)+((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2))+1/2*a*((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2)+1/2*c*((x+1)^2*
c+(b-2*c)*(x+1)+a-b+c)^(1/2)-5/8*((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2)*b-1/2*a*((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)
^(1/2)-1/2*c*((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2)-5/8*((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2)*b+3/8*b/c^(1/2)*l
n((1/2*b-c+c*(x+1))/c^(1/2)+((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2))*a-3/8*b/c^(1/2)*ln((1/2*b+c+c*(x-1))/c^(1/2
)+((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2))*a-1/2*a*(a-b+c)^(1/2)*ln((2*a-2*b+2*c+(b-2*c)*(x+1)+2*(a-b+c)^(1/2)*(
(x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2))/(x+1))+3/4*b*ln((1/2*b-c+c*(x+1))/c^(1/2)+((x+1)^2*c+(b-2*c)*(x+1)+a-b+c
)^(1/2))*c^(1/2)+1/2*b*(a-b+c)^(1/2)*ln((2*a-2*b+2*c+(b-2*c)*(x+1)+2*(a-b+c)^(1/2)*((x+1)^2*c+(b-2*c)*(x+1)+a-
b+c)^(1/2))/(x+1))-1/2*c*(a-b+c)^(1/2)*ln((2*a-2*b+2*c+(b-2*c)*(x+1)+2*(a-b+c)^(1/2)*((x+1)^2*c+(b-2*c)*(x+1)+
a-b+c)^(1/2))/(x+1))+1/8*b*((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2)*x+1/16/c*((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2
)*b^2-1/32/c^(3/2)*ln((1/2*b-c+c*(x+1))/c^(1/2)+((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2))*b^3-1/4*c*((x+1)^2*c+(b
-2*c)*(x+1)+a-b+c)^(1/2)*x+1/2*a*(a+b+c)^(1/2)*ln((2*a+2*b+2*c+(b+2*c)*(x-1)+2*(a+b+c)^(1/2)*((x-1)^2*c+(b+2*c
)*(x-1)+a+b+c)^(1/2))/(x-1))-3/4*b*ln((1/2*b+c+c*(x-1))/c^(1/2)+((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2))*c^(1/2)
+1/2*b*(a+b+c)^(1/2)*ln((2*a+2*b+2*c+(b+2*c)*(x-1)+2*(a+b+c)^(1/2)*((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2))/(x-1
))+1/2*c*(a+b+c)^(1/2)*ln((2*a+2*b+2*c+(b+2*c)*(x-1)+2*(a+b+c)^(1/2)*((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2))/(x
-1))-1/8*b*((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2)*x-1/16/c*((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2)*b^2+1/32/c^(3/
2)*ln((1/2*b+c+c*(x-1))/c^(1/2)+((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2))*b^3-1/4*c*((x-1)^2*c+(b+2*c)*(x-1)+a+b+
c)^(1/2)*x-3/16/c^(1/2)*ln((1/2*b-c+c*(x+1))/c^(1/2)+((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2))*b^2-3/4*c^(1/2)*ln
((1/2*b-c+c*(x+1))/c^(1/2)+((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(1/2))*a-3/4*c^(1/2)*ln((1/2*b+c+c*(x-1))/c^(1/2)+(
(x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(1/2))*a-3/16/c^(1/2)*ln((1/2*b+c+c*(x-1))/c^(1/2)+((x-1)^2*c+(b+2*c)*(x-1)+a+b
+c)^(1/2))*b^2+1/6*((x+1)^2*c+(b-2*c)*(x+1)+a-b+c)^(3/2)-1/6*((x-1)^2*c+(b+2*c)*(x-1)+a+b+c)^(3/2)

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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((c*x^2+b*x+a)^(3/2)/(-x^2+1),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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(-x**2+1),x)

[Out]

-Integral(a*sqrt(a + b*x + c*x**2)/(x**2 - 1), x) - Integral(b*x*sqrt(a + b*x + c*x**2)/(x**2 - 1), x) - Integ
ral(c*x**2*sqrt(a + b*x + c*x**2)/(x**2 - 1), x)

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