3.68 \(\int \frac{A+B x+C x^2}{(a+c x^2)^4} \, dx\)

Optimal. Leaf size=126 \[ \frac{(a C+5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{3/2}}+\frac{x (a C+5 A c)}{16 a^3 c \left (a+c x^2\right )}+\frac{x (a C+5 A c)}{24 a^2 c \left (a+c x^2\right )^2}-\frac{a B-x (A c-a C)}{6 a c \left (a+c x^2\right )^3} \]

[Out]

-(a*B - (A*c - a*C)*x)/(6*a*c*(a + c*x^2)^3) + ((5*A*c + a*C)*x)/(24*a^2*c*(a + c*x^2)^2) + ((5*A*c + a*C)*x)/
(16*a^3*c*(a + c*x^2)) + ((5*A*c + a*C)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*c^(3/2))

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Rubi [A]  time = 0.078232, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1814, 12, 199, 205} \[ \frac{(a C+5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{3/2}}+\frac{x (a C+5 A c)}{16 a^3 c \left (a+c x^2\right )}+\frac{x (a C+5 A c)}{24 a^2 c \left (a+c x^2\right )^2}-\frac{a B-x (A c-a C)}{6 a c \left (a+c x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/(a + c*x^2)^4,x]

[Out]

-(a*B - (A*c - a*C)*x)/(6*a*c*(a + c*x^2)^3) + ((5*A*c + a*C)*x)/(24*a^2*c*(a + c*x^2)^2) + ((5*A*c + a*C)*x)/
(16*a^3*c*(a + c*x^2)) + ((5*A*c + a*C)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*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 )^4} \, dx &=-\frac{a B-(A c-a C) x}{6 a c \left (a+c x^2\right )^3}-\frac{\int \frac{-5 A-\frac{a C}{c}}{\left (a+c x^2\right )^3} \, dx}{6 a}\\ &=-\frac{a B-(A c-a C) x}{6 a c \left (a+c x^2\right )^3}+\frac{(5 A c+a C) \int \frac{1}{\left (a+c x^2\right )^3} \, dx}{6 a c}\\ &=-\frac{a B-(A c-a C) x}{6 a c \left (a+c x^2\right )^3}+\frac{(5 A c+a C) x}{24 a^2 c \left (a+c x^2\right )^2}+\frac{(5 A c+a C) \int \frac{1}{\left (a+c x^2\right )^2} \, dx}{8 a^2 c}\\ &=-\frac{a B-(A c-a C) x}{6 a c \left (a+c x^2\right )^3}+\frac{(5 A c+a C) x}{24 a^2 c \left (a+c x^2\right )^2}+\frac{(5 A c+a C) x}{16 a^3 c \left (a+c x^2\right )}+\frac{(5 A c+a C) \int \frac{1}{a+c x^2} \, dx}{16 a^3 c}\\ &=-\frac{a B-(A c-a C) x}{6 a c \left (a+c x^2\right )^3}+\frac{(5 A c+a C) x}{24 a^2 c \left (a+c x^2\right )^2}+\frac{(5 A c+a C) x}{16 a^3 c \left (a+c x^2\right )}+\frac{(5 A c+a C) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0885953, size = 112, normalized size = 0.89 \[ \frac{a^2 c x \left (33 A+8 C x^2\right )-a^3 (8 B+3 C x)+a c^2 x^3 \left (40 A+3 C x^2\right )+15 A c^3 x^5}{48 a^3 c \left (a+c x^2\right )^3}+\frac{(a C+5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/(a + c*x^2)^4,x]

[Out]

(15*A*c^3*x^5 - a^3*(8*B + 3*C*x) + a*c^2*x^3*(40*A + 3*C*x^2) + a^2*c*x*(33*A + 8*C*x^2))/(48*a^3*c*(a + c*x^
2)^3) + ((5*A*c + a*C)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*c^(3/2))

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Maple [A]  time = 0.05, size = 113, normalized size = 0.9 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{3}} \left ({\frac{ \left ( 5\,Ac+aC \right ) c{x}^{5}}{16\,{a}^{3}}}+{\frac{ \left ( 5\,Ac+aC \right ){x}^{3}}{6\,{a}^{2}}}+{\frac{ \left ( 11\,Ac-aC \right ) x}{16\,ac}}-{\frac{B}{6\,c}} \right ) }+{\frac{5\,A}{16\,{a}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{C}{16\,{a}^{2}c}\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)^4,x)

[Out]

(1/16*(5*A*c+C*a)/a^3*c*x^5+1/6/a^2*(5*A*c+C*a)*x^3+1/16*(11*A*c-C*a)/a/c*x-1/6*B/c)/(c*x^2+a)^3+5/16/a^3/(a*c
)^(1/2)*arctan(x*c/(a*c)^(1/2))*A+1/16/a^2/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)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.00482, size = 900, normalized size = 7.14 \begin{align*} \left [-\frac{16 \, B a^{4} c - 6 \,{\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} x^{5} - 16 \,{\left (C a^{3} c^{2} + 5 \, A a^{2} c^{3}\right )} x^{3} + 3 \,{\left ({\left (C a c^{3} + 5 \, A c^{4}\right )} x^{6} + C a^{4} + 5 \, A a^{3} c + 3 \,{\left (C a^{2} c^{2} + 5 \, A a c^{3}\right )} x^{4} + 3 \,{\left (C a^{3} c + 5 \, A a^{2} 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 ) + 6 \,{\left (C a^{4} c - 11 \, A a^{3} c^{2}\right )} x}{96 \,{\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}, -\frac{8 \, B a^{4} c - 3 \,{\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} x^{5} - 8 \,{\left (C a^{3} c^{2} + 5 \, A a^{2} c^{3}\right )} x^{3} - 3 \,{\left ({\left (C a c^{3} + 5 \, A c^{4}\right )} x^{6} + C a^{4} + 5 \, A a^{3} c + 3 \,{\left (C a^{2} c^{2} + 5 \, A a c^{3}\right )} x^{4} + 3 \,{\left (C a^{3} c + 5 \, A a^{2} c^{2}\right )} x^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) + 3 \,{\left (C a^{4} c - 11 \, A a^{3} c^{2}\right )} x}{48 \,{\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} 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)^4,x, algorithm="fricas")

[Out]

[-1/96*(16*B*a^4*c - 6*(C*a^2*c^3 + 5*A*a*c^4)*x^5 - 16*(C*a^3*c^2 + 5*A*a^2*c^3)*x^3 + 3*((C*a*c^3 + 5*A*c^4)
*x^6 + C*a^4 + 5*A*a^3*c + 3*(C*a^2*c^2 + 5*A*a*c^3)*x^4 + 3*(C*a^3*c + 5*A*a^2*c^2)*x^2)*sqrt(-a*c)*log((c*x^
2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 6*(C*a^4*c - 11*A*a^3*c^2)*x)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*
x^2 + a^7*c^2), -1/48*(8*B*a^4*c - 3*(C*a^2*c^3 + 5*A*a*c^4)*x^5 - 8*(C*a^3*c^2 + 5*A*a^2*c^3)*x^3 - 3*((C*a*c
^3 + 5*A*c^4)*x^6 + C*a^4 + 5*A*a^3*c + 3*(C*a^2*c^2 + 5*A*a*c^3)*x^4 + 3*(C*a^3*c + 5*A*a^2*c^2)*x^2)*sqrt(a*
c)*arctan(sqrt(a*c)*x/a) + 3*(C*a^4*c - 11*A*a^3*c^2)*x)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^
2)]

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Sympy [A]  time = 2.22933, size = 196, normalized size = 1.56 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{7} c^{3}}} \left (5 A c + C a\right ) \log{\left (- a^{4} c \sqrt{- \frac{1}{a^{7} c^{3}}} + x \right )}}{32} + \frac{\sqrt{- \frac{1}{a^{7} c^{3}}} \left (5 A c + C a\right ) \log{\left (a^{4} c \sqrt{- \frac{1}{a^{7} c^{3}}} + x \right )}}{32} + \frac{- 8 B a^{3} + x^{5} \left (15 A c^{3} + 3 C a c^{2}\right ) + x^{3} \left (40 A a c^{2} + 8 C a^{2} c\right ) + x \left (33 A a^{2} c - 3 C a^{3}\right )}{48 a^{6} c + 144 a^{5} c^{2} x^{2} + 144 a^{4} c^{3} x^{4} + 48 a^{3} c^{4} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/(c*x**2+a)**4,x)

[Out]

-sqrt(-1/(a**7*c**3))*(5*A*c + C*a)*log(-a**4*c*sqrt(-1/(a**7*c**3)) + x)/32 + sqrt(-1/(a**7*c**3))*(5*A*c + C
*a)*log(a**4*c*sqrt(-1/(a**7*c**3)) + x)/32 + (-8*B*a**3 + x**5*(15*A*c**3 + 3*C*a*c**2) + x**3*(40*A*a*c**2 +
8*C*a**2*c) + x*(33*A*a**2*c - 3*C*a**3))/(48*a**6*c + 144*a**5*c**2*x**2 + 144*a**4*c**3*x**4 + 48*a**3*c**4
*x**6)

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Giac [A]  time = 1.15811, size = 147, normalized size = 1.17 \begin{align*} \frac{{\left (C a + 5 \, A c\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} a^{3} c} + \frac{3 \, C a c^{2} x^{5} + 15 \, A c^{3} x^{5} + 8 \, C a^{2} c x^{3} + 40 \, A a c^{2} x^{3} - 3 \, C a^{3} x + 33 \, A a^{2} c x - 8 \, B a^{3}}{48 \,{\left (c x^{2} + a\right )}^{3} a^{3} 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)^4,x, algorithm="giac")

[Out]

1/16*(C*a + 5*A*c)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c) + 1/48*(3*C*a*c^2*x^5 + 15*A*c^3*x^5 + 8*C*a^2*c*x^
3 + 40*A*a*c^2*x^3 - 3*C*a^3*x + 33*A*a^2*c*x - 8*B*a^3)/((c*x^2 + a)^3*a^3*c)