3.182 \(\int \frac{A+C x^2}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=98 \[ \frac{C \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2}}-\frac{2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0770029, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1660, 12, 621, 206} \[ \frac{C \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2}}-\frac{2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.763087, size = 104, normalized size = 1.06 \[ \frac{\frac{2 \sqrt{c} \left (a C (b-2 c x)+A c (b+2 c x)+b^2 C x\right )}{\sqrt{a+x (b+c x)}}-C \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{c^{3/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.052, size = 169, normalized size = 1.7 \begin{align*} -{\frac{Cx}{c}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{Cb}{2\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{C{b}^{2}x}{c \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{C{b}^{3}}{2\,{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{C\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+2\,{\frac{A \left ( 2\,cx+b \right ) }{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+A)/(c*x^2+b*x+a)^(3/2),x)

[Out]

-C*x/c/(c*x^2+b*x+a)^(1/2)+1/2*C*b/c^2/(c*x^2+b*x+a)^(1/2)+C*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+1/2*C*b^3
/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+C/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2*A*(2*c*x+b)/(4*a*
c-b^2)/(c*x^2+b*x+a)^(1/2)

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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((C*x^2+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.14125, size = 892, normalized size = 9.1 \begin{align*} \left [\frac{{\left (C a b^{2} - 4 \, C a^{2} c +{\left (C b^{2} c - 4 \, C a c^{2}\right )} x^{2} +{\left (C b^{3} - 4 \, C a b c\right )} x\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) - 4 \,{\left (C a b c + A b c^{2} +{\left (C b^{2} c - 2 \, C a c^{2} + 2 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{2 \,{\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} +{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} +{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )}}, -\frac{{\left (C a b^{2} - 4 \, C a^{2} c +{\left (C b^{2} c - 4 \, C a c^{2}\right )} x^{2} +{\left (C b^{3} - 4 \, C a b c\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \,{\left (C a b c + A b c^{2} +{\left (C b^{2} c - 2 \, C a c^{2} + 2 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{a b^{2} c^{2} - 4 \, a^{2} c^{3} +{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} +{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[1/2*((C*a*b^2 - 4*C*a^2*c + (C*b^2*c - 4*C*a*c^2)*x^2 + (C*b^3 - 4*C*a*b*c)*x)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c
*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(C*a*b*c + A*b*c^2 + (C*b^2*c - 2*C*a*c^2
+ 2*A*c^3)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*
x), -((C*a*b^2 - 4*C*a^2*c + (C*b^2*c - 4*C*a*c^2)*x^2 + (C*b^3 - 4*C*a*b*c)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2
+ b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(C*a*b*c + A*b*c^2 + (C*b^2*c - 2*C*a*c^2 + 2*A*
c^3)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+A)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((A + C*x**2)/(a + b*x + c*x**2)**(3/2), x)

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Giac [A]  time = 1.33972, size = 149, normalized size = 1.52 \begin{align*} -\frac{2 \,{\left (\frac{{\left (C b^{2} - 2 \, C a c + 2 \, A c^{2}\right )} x}{b^{2} c - 4 \, a c^{2}} + \frac{C a b + A b c}{b^{2} c - 4 \, a c^{2}}\right )}}{\sqrt{c x^{2} + b x + a}} - \frac{C \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

-2*((C*b^2 - 2*C*a*c + 2*A*c^2)*x/(b^2*c - 4*a*c^2) + (C*a*b + A*b*c)/(b^2*c - 4*a*c^2))/sqrt(c*x^2 + b*x + a)
- C*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(3/2)