3.60 \(\int \frac{A+B x+C x^2}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=98 \[ \frac{(a C+3 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} c^{3/2}}+\frac{x (a C+3 A c)}{8 a^2 c \left (a+c x^2\right )}-\frac{a B-x (A c-a C)}{4 a c \left (a+c x^2\right )^2} \]

[Out]

-(a*B - (A*c - a*C)*x)/(4*a*c*(a + c*x^2)^2) + ((3*A*c + a*C)*x)/(8*a^2*c*(a + c*x^2)) + ((3*A*c + a*C)*ArcTan
[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))

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Rubi [A]  time = 0.0630782, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1814, 12, 199, 205} \[ \frac{(a C+3 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} c^{3/2}}+\frac{x (a C+3 A c)}{8 a^2 c \left (a+c x^2\right )}-\frac{a B-x (A c-a C)}{4 a c \left (a+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*B - (A*c - a*C)*x)/(4*a*c*(a + c*x^2)^2) + ((3*A*c + a*C)*x)/(8*a^2*c*(a + c*x^2)) + ((3*A*c + a*C)*ArcTan
[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && 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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx &=-\frac{a B-(A c-a C) x}{4 a c \left (a+c x^2\right )^2}-\frac{\int \frac{-3 A-\frac{a C}{c}}{\left (a+c x^2\right )^2} \, dx}{4 a}\\ &=-\frac{a B-(A c-a C) x}{4 a c \left (a+c x^2\right )^2}+\frac{(3 A c+a C) \int \frac{1}{\left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac{a B-(A c-a C) x}{4 a c \left (a+c x^2\right )^2}+\frac{(3 A c+a C) x}{8 a^2 c \left (a+c x^2\right )}+\frac{(3 A c+a C) \int \frac{1}{a+c x^2} \, dx}{8 a^2 c}\\ &=-\frac{a B-(A c-a C) x}{4 a c \left (a+c x^2\right )^2}+\frac{(3 A c+a C) x}{8 a^2 c \left (a+c x^2\right )}+\frac{(3 A c+a C) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0742712, size = 90, normalized size = 0.92 \[ \frac{-a^2 (2 B+C x)+a c x \left (5 A+C x^2\right )+3 A c^2 x^3}{8 a^2 c \left (a+c x^2\right )^2}+\frac{(a C+3 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*c^2*x^3 - a^2*(2*B + C*x) + a*c*x*(5*A + C*x^2))/(8*a^2*c*(a + c*x^2)^2) + ((3*A*c + a*C)*ArcTan[(Sqrt[c]
*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))

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Maple [A]  time = 0.049, size = 96, normalized size = 1. \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( 3\,Ac+aC \right ){x}^{3}}{8\,{a}^{2}}}+{\frac{ \left ( 5\,Ac-aC \right ) x}{8\,ac}}-{\frac{B}{4\,c}} \right ) }+{\frac{3\,A}{8\,{a}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{C}{8\,ac}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/8*(3*A*c+C*a)/a^2*x^3+1/8*(5*A*c-C*a)/a/c*x-1/4*B/c)/(c*x^2+a)^2+3/8/a^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2)
)*A+1/8/a/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*C

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76369, size = 656, normalized size = 6.69 \begin{align*} \left [-\frac{4 \, B a^{3} c - 2 \,{\left (C a^{2} c^{2} + 3 \, A a c^{3}\right )} x^{3} +{\left ({\left (C a c^{2} + 3 \, A c^{3}\right )} x^{4} + C a^{3} + 3 \, A a^{2} c + 2 \,{\left (C a^{2} c + 3 \, A a c^{2}\right )} x^{2}\right )} \sqrt{-a c} \log \left (\frac{c x^{2} - 2 \, \sqrt{-a c} x - a}{c x^{2} + a}\right ) + 2 \,{\left (C a^{3} c - 5 \, A a^{2} c^{2}\right )} x}{16 \,{\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, -\frac{2 \, B a^{3} c -{\left (C a^{2} c^{2} + 3 \, A a c^{3}\right )} x^{3} -{\left ({\left (C a c^{2} + 3 \, A c^{3}\right )} x^{4} + C a^{3} + 3 \, A a^{2} c + 2 \,{\left (C a^{2} c + 3 \, A a c^{2}\right )} x^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) +{\left (C a^{3} c - 5 \, A a^{2} c^{2}\right )} x}{8 \,{\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(4*B*a^3*c - 2*(C*a^2*c^2 + 3*A*a*c^3)*x^3 + ((C*a*c^2 + 3*A*c^3)*x^4 + C*a^3 + 3*A*a^2*c + 2*(C*a^2*c
+ 3*A*a*c^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(C*a^3*c - 5*A*a^2*c^2)*x)/(a^3
*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), -1/8*(2*B*a^3*c - (C*a^2*c^2 + 3*A*a*c^3)*x^3 - ((C*a*c^2 + 3*A*c^3)*x^4
+ C*a^3 + 3*A*a^2*c + 2*(C*a^2*c + 3*A*a*c^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + (C*a^3*c - 5*A*a^2*c^2)*x
)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2)]

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Sympy [A]  time = 1.56703, size = 156, normalized size = 1.59 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} c^{3}}} \left (3 A c + C a\right ) \log{\left (- a^{3} c \sqrt{- \frac{1}{a^{5} c^{3}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{5} c^{3}}} \left (3 A c + C a\right ) \log{\left (a^{3} c \sqrt{- \frac{1}{a^{5} c^{3}}} + x \right )}}{16} + \frac{- 2 B a^{2} + x^{3} \left (3 A c^{2} + C a c\right ) + x \left (5 A a c - C a^{2}\right )}{8 a^{4} c + 16 a^{3} c^{2} x^{2} + 8 a^{2} c^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**5*c**3))*(3*A*c + C*a)*log(-a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + sqrt(-1/(a**5*c**3))*(3*A*c + C
*a)*log(a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + (-2*B*a**2 + x**3*(3*A*c**2 + C*a*c) + x*(5*A*a*c - C*a**2))/(8*
a**4*c + 16*a**3*c**2*x**2 + 8*a**2*c**3*x**4)

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Giac [A]  time = 1.15251, size = 113, normalized size = 1.15 \begin{align*} \frac{{\left (C a + 3 \, A c\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{8 \, \sqrt{a c} a^{2} c} + \frac{C a c x^{3} + 3 \, A c^{2} x^{3} - C a^{2} x + 5 \, A a c x - 2 \, B a^{2}}{8 \,{\left (c x^{2} + a\right )}^{2} a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(C*a + 3*A*c)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c) + 1/8*(C*a*c*x^3 + 3*A*c^2*x^3 - C*a^2*x + 5*A*a*c*x
- 2*B*a^2)/((c*x^2 + a)^2*a^2*c)