3.96 \(\int \frac {x^2 (A+B x+C x^2+D x^3)}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=134 \[ \frac {(A b-3 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{5/2}}-\frac {x (A b-3 a C)}{2 a b^2}-\frac {x^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac {(b B-2 a D) \log \left (a+b x^2\right )}{2 b^3}+\frac {D x^2}{2 b^2} \]

[Out]

-1/2*(A*b-3*C*a)*x/a/b^2+1/2*D*x^2/b^2-1/2*x^2*(a*(B-a*D/b)-(A*b-C*a)*x)/a/b/(b*x^2+a)+1/2*(B*b-2*D*a)*ln(b*x^
2+a)/b^3+1/2*(A*b-3*C*a)*arctan(x*b^(1/2)/a^(1/2))/b^(5/2)/a^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1804, 1802, 635, 205, 260} \[ -\frac {x (A b-3 a C)}{2 a b^2}+\frac {(A b-3 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{5/2}}-\frac {x^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac {(b B-2 a D) \log \left (a+b x^2\right )}{2 b^3}+\frac {D x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b - 3*a*C)*x)/(2*a*b^2) + (D*x^2)/(2*b^2) - (x^2*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(2*a*b*(a + b*x^2)) +
((A*b - 3*a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[a]*b^(5/2)) + ((b*B - 2*a*D)*Log[a + b*x^2])/(2*b^3)

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]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1804

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 100, normalized size = 0.75 \[ \frac {\frac {a^2 (-D)+a b (B+C x)-A b^2 x}{a+b x^2}+\frac {\sqrt {b} (A b-3 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a}}+(b B-2 a D) \log \left (a+b x^2\right )+2 b C x+b D x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*C*x + b*D*x^2 + (-(a^2*D) - A*b^2*x + a*b*(B + C*x))/(a + b*x^2) + (Sqrt[b]*(A*b - 3*a*C)*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/Sqrt[a] + (b*B - 2*a*D)*Log[a + b*x^2])/(2*b^3)

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fricas [A]  time = 0.62, size = 357, normalized size = 2.66 \[ \left [\frac {2 \, D a b^{2} x^{4} + 4 \, C a b^{2} x^{3} + 2 \, D a^{2} b x^{2} - 2 \, D a^{3} + 2 \, B a^{2} b + {\left (3 \, C a^{2} - A a b + {\left (3 \, C a b - A b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (3 \, C a^{2} b - A a b^{2}\right )} x - 2 \, {\left (2 \, D a^{3} - B a^{2} b + {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}, \frac {D a b^{2} x^{4} + 2 \, C a b^{2} x^{3} + D a^{2} b x^{2} - D a^{3} + B a^{2} b - {\left (3 \, C a^{2} - A a b + {\left (3 \, C a b - A b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (3 \, C a^{2} b - A a b^{2}\right )} x - {\left (2 \, D a^{3} - B a^{2} b + {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.39, size = 111, normalized size = 0.83 \[ -\frac {{\left (3 \, C a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{2}} - \frac {{\left (2 \, D a - B b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {D b^{2} x^{2} + 2 \, C b^{2} x}{2 \, b^{4}} - \frac {D a^{2} - B a b - {\left (C a b - A b^{2}\right )} x}{2 \, {\left (b x^{2} + a\right )} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*C*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2) - 1/2*(2*D*a - B*b)*log(b*x^2 + a)/b^3 + 1/2*(D*b^2*x
^2 + 2*C*b^2*x)/b^4 - 1/2*(D*a^2 - B*a*b - (C*a*b - A*b^2)*x)/((b*x^2 + a)*b^3)

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maple [A]  time = 0.01, size = 154, normalized size = 1.15 \[ -\frac {A x}{2 \left (b \,x^{2}+a \right ) b}+\frac {A \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b}+\frac {C a x}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {3 C a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{2}}+\frac {D x^{2}}{2 b^{2}}+\frac {B a}{2 \left (b \,x^{2}+a \right ) b^{2}}+\frac {B \ln \left (b \,x^{2}+a \right )}{2 b^{2}}+\frac {C x}{b^{2}}-\frac {D a^{2}}{2 \left (b \,x^{2}+a \right ) b^{3}}-\frac {D a \ln \left (b \,x^{2}+a \right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*D*x^2/b^2+1/b^2*C*x-1/2/b/(b*x^2+a)*A*x+1/2/b^2/(b*x^2+a)*a*C*x+1/2/b^2/(b*x^2+a)*B*a-1/2/b^3/(b*x^2+a)*a^
2*D+1/2/b^2*ln(b*x^2+a)*B-1/b^3*ln(b*x^2+a)*a*D+1/2/b/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*A-3/2/b^2/(a*b)^(1
/2)*arctan(1/(a*b)^(1/2)*b*x)*a*C

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maxima [A]  time = 2.96, size = 108, normalized size = 0.81 \[ -\frac {D a^{2} - B a b - {\left (C a b - A b^{2}\right )} x}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} - \frac {{\left (3 \, C a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{2}} + \frac {D x^{2} + 2 \, C x}{2 \, b^{2}} - \frac {{\left (2 \, D a - B b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(D*a^2 - B*a*b - (C*a*b - A*b^2)*x)/(b^4*x^2 + a*b^3) - 1/2*(3*C*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b
)*b^2) + 1/2*(D*x^2 + 2*C*x)/b^2 - 1/2*(2*D*a - B*b)*log(b*x^2 + a)/b^3

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mupad [B]  time = 1.29, size = 152, normalized size = 1.13 \[ \frac {B\,\ln \left (b\,x^2+a\right )}{2\,b^2}+\frac {x^2\,D}{2\,b^2}+\frac {C\,x}{b^2}-\frac {a^2\,D}{2\,b^3\,\left (b\,x^2+a\right )}+\frac {B\,a}{2\,b^2\,\left (b\,x^2+a\right )}-\frac {A\,x}{2\,b\,\left (b\,x^2+a\right )}+\frac {C\,a\,x}{2\,\left (b^3\,x^2+a\,b^2\right )}+\frac {A\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,\sqrt {a}\,b^{3/2}}-\frac {3\,C\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,b^{5/2}}-\frac {a\,\ln \left (b\,x^2+a\right )\,D}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*log(a + b*x^2))/(2*b^2) + (x^2*D)/(2*b^2) + (C*x)/b^2 - (a^2*D)/(2*b^3*(a + b*x^2)) + (B*a)/(2*b^2*(a + b*x
^2)) - (A*x)/(2*b*(a + b*x^2)) + (C*a*x)/(2*(a*b^2 + b^3*x^2)) + (A*atan((b^(1/2)*x)/a^(1/2)))/(2*a^(1/2)*b^(3
/2)) - (3*C*a^(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(2*b^(5/2)) - (a*log(a + b*x^2)*D)/b^3

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sympy [B]  time = 4.61, size = 284, normalized size = 2.12 \[ \frac {C x}{b^{2}} + \frac {D x^{2}}{2 b^{2}} + \left (- \frac {- B b + 2 D a}{2 b^{3}} - \frac {\sqrt {- a b^{7}} \left (- A b + 3 C a\right )}{4 a b^{6}}\right ) \log {\left (x + \frac {2 B a b - 4 D a^{2} - 4 a b^{3} \left (- \frac {- B b + 2 D a}{2 b^{3}} - \frac {\sqrt {- a b^{7}} \left (- A b + 3 C a\right )}{4 a b^{6}}\right )}{- A b^{2} + 3 C a b} \right )} + \left (- \frac {- B b + 2 D a}{2 b^{3}} + \frac {\sqrt {- a b^{7}} \left (- A b + 3 C a\right )}{4 a b^{6}}\right ) \log {\left (x + \frac {2 B a b - 4 D a^{2} - 4 a b^{3} \left (- \frac {- B b + 2 D a}{2 b^{3}} + \frac {\sqrt {- a b^{7}} \left (- A b + 3 C a\right )}{4 a b^{6}}\right )}{- A b^{2} + 3 C a b} \right )} + \frac {B a b - D a^{2} + x \left (- A b^{2} + C a b\right )}{2 a b^{3} + 2 b^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C*x/b**2 + D*x**2/(2*b**2) + (-(-B*b + 2*D*a)/(2*b**3) - sqrt(-a*b**7)*(-A*b + 3*C*a)/(4*a*b**6))*log(x + (2*B
*a*b - 4*D*a**2 - 4*a*b**3*(-(-B*b + 2*D*a)/(2*b**3) - sqrt(-a*b**7)*(-A*b + 3*C*a)/(4*a*b**6)))/(-A*b**2 + 3*
C*a*b)) + (-(-B*b + 2*D*a)/(2*b**3) + sqrt(-a*b**7)*(-A*b + 3*C*a)/(4*a*b**6))*log(x + (2*B*a*b - 4*D*a**2 - 4
*a*b**3*(-(-B*b + 2*D*a)/(2*b**3) + sqrt(-a*b**7)*(-A*b + 3*C*a)/(4*a*b**6)))/(-A*b**2 + 3*C*a*b)) + (B*a*b -
D*a**2 + x*(-A*b**2 + C*a*b))/(2*a*b**3 + 2*b**4*x**2)

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