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

Optimal. Leaf size=136 \[ \frac {(3 a C+A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}}-\frac {x (-2 x (b B-3 a D)+3 a C+A b)}{8 a b^2 \left (a+b x^2\right )}-\frac {x^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}+\frac {D \log \left (a+b x^2\right )}{2 b^3} \]

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Rubi [A]  time = 0.16, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1804, 635, 205, 260} \begin {gather*} \frac {(3 a C+A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}}-\frac {x (-2 x (b B-3 a D)+3 a C+A b)}{8 a b^2 \left (a+b x^2\right )}-\frac {x^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}+\frac {D \log \left (a+b x^2\right )}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.10, size = 122, normalized size = 0.90 \begin {gather*} \frac {\frac {\sqrt {b} (3 a C+A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}+\frac {-2 a^2 D+2 a b (B+C x)-2 A b^2 x}{\left (a+b x^2\right )^2}+\frac {8 a^2 D-a b (4 B+5 C x)+A b^2 x}{a \left (a+b x^2\right )}+4 D \log \left (a+b x^2\right )}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 447, normalized size = 3.29 \begin {gather*} \left [\frac {12 \, D a^{4} - 4 \, B a^{3} b - 2 \, {\left (5 \, C a^{2} b^{2} - A a b^{3}\right )} x^{3} + 8 \, {\left (2 \, D a^{3} b - B a^{2} b^{2}\right )} x^{2} - {\left ({\left (3 \, C a b^{2} + A b^{3}\right )} x^{4} + 3 \, C a^{3} + A a^{2} b + 2 \, {\left (3 \, C a^{2} b + A 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^{3} b + A a^{2} b^{2}\right )} x + 8 \, {\left (D a^{2} b^{2} x^{4} + 2 \, D a^{3} b x^{2} + D a^{4}\right )} \log \left (b x^{2} + a\right )}{16 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, \frac {6 \, D a^{4} - 2 \, B a^{3} b - {\left (5 \, C a^{2} b^{2} - A a b^{3}\right )} x^{3} + 4 \, {\left (2 \, D a^{3} b - B a^{2} b^{2}\right )} x^{2} + {\left ({\left (3 \, C a b^{2} + A b^{3}\right )} x^{4} + 3 \, C a^{3} + A a^{2} b + 2 \, {\left (3 \, C a^{2} b + A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - {\left (3 \, C a^{3} b + A a^{2} b^{2}\right )} x + 4 \, {\left (D a^{2} b^{2} x^{4} + 2 \, D a^{3} b x^{2} + D a^{4}\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \end {gather*}

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)^3,x, algorithm="fricas")

[Out]

[1/16*(12*D*a^4 - 4*B*a^3*b - 2*(5*C*a^2*b^2 - A*a*b^3)*x^3 + 8*(2*D*a^3*b - B*a^2*b^2)*x^2 - ((3*C*a*b^2 + A*
b^3)*x^4 + 3*C*a^3 + A*a^2*b + 2*(3*C*a^2*b + A*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^3*b + A*a^2*b^2)*x + 8*(D*a^2*b^2*x^4 + 2*D*a^3*b*x^2 + D*a^4)*log(b*x^2 + a))/(a^2*b^5*x^4
+ 2*a^3*b^4*x^2 + a^4*b^3), 1/8*(6*D*a^4 - 2*B*a^3*b - (5*C*a^2*b^2 - A*a*b^3)*x^3 + 4*(2*D*a^3*b - B*a^2*b^2)
*x^2 + ((3*C*a*b^2 + A*b^3)*x^4 + 3*C*a^3 + A*a^2*b + 2*(3*C*a^2*b + A*a*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*
x/a) - (3*C*a^3*b + A*a^2*b^2)*x + 4*(D*a^2*b^2*x^4 + 2*D*a^3*b*x^2 + D*a^4)*log(b*x^2 + a))/(a^2*b^5*x^4 + 2*
a^3*b^4*x^2 + a^4*b^3)]

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giac [A]  time = 0.42, size = 128, normalized size = 0.94 \begin {gather*} \frac {D \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (3 \, C a + A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} - \frac {{\left (5 \, C a b - A b^{2}\right )} x^{3} - 4 \, {\left (2 \, D a^{2} - B a b\right )} x^{2} + {\left (3 \, C a^{2} + A a b\right )} x - \frac {2 \, {\left (3 \, D a^{3} - B a^{2} b\right )}}{b}}{8 \, {\left (b x^{2} + a\right )}^{2} a b^{2}} \end {gather*}

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)^3,x, algorithm="giac")

[Out]

1/2*D*log(b*x^2 + a)/b^3 + 1/8*(3*C*a + A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^2) - 1/8*((5*C*a*b - A*b^2)*
x^3 - 4*(2*D*a^2 - B*a*b)*x^2 + (3*C*a^2 + A*a*b)*x - 2*(3*D*a^3 - B*a^2*b)/b)/((b*x^2 + a)^2*a*b^2)

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maple [A]  time = 0.01, size = 133, normalized size = 0.98 \begin {gather*} \frac {A \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a b}+\frac {3 C \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{2}}+\frac {D \ln \left (b \,x^{2}+a \right )}{2 b^{3}}+\frac {\frac {\left (A b -5 a C \right ) x^{3}}{8 a b}-\frac {\left (b B -2 a D\right ) x^{2}}{2 b^{2}}-\frac {\left (A b +3 a C \right ) x}{8 b^{2}}-\frac {\left (b B -3 a D\right ) a}{4 b^{3}}}{\left (b \,x^{2}+a \right )^{2}} \end {gather*}

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)^3,x)

[Out]

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

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maxima [A]  time = 2.98, size = 146, normalized size = 1.07 \begin {gather*} \frac {6 \, D a^{3} - 2 \, B a^{2} b - {\left (5 \, C a b^{2} - A b^{3}\right )} x^{3} + 4 \, {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2} - {\left (3 \, C a^{2} b + A a b^{2}\right )} x}{8 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}} + \frac {D \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (3 \, C a + A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} \end {gather*}

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)^3,x, algorithm="maxima")

[Out]

1/8*(6*D*a^3 - 2*B*a^2*b - (5*C*a*b^2 - A*b^3)*x^3 + 4*(2*D*a^2*b - B*a*b^2)*x^2 - (3*C*a^2*b + A*a*b^2)*x)/(a
*b^5*x^4 + 2*a^2*b^4*x^2 + a^3*b^3) + 1/2*D*log(b*x^2 + a)/b^3 + 1/8*(3*C*a + A*b)*arctan(b*x/sqrt(a*b))/(sqrt
(a*b)*a*b^2)

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mupad [B]  time = 1.39, size = 195, normalized size = 1.43 \begin {gather*} \frac {\frac {A\,x^3}{8\,a}-\frac {A\,x}{8\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {\frac {B\,x^2}{2\,b}+\frac {B\,a}{4\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {\frac {5\,C\,x^3}{8\,b}+\frac {3\,C\,a\,x}{8\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {D\,\left (\ln \left (b\,x^2+a\right )+\frac {2\,a}{b\,x^2+a}-\frac {a^2}{2\,{\left (b\,x^2+a\right )}^2}\right )}{2\,b^3}+\frac {A\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{3/2}\,b^{3/2}}+\frac {3\,C\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,\sqrt {a}\,b^{5/2}} \end {gather*}

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)^3,x)

[Out]

((A*x^3)/(8*a) - (A*x)/(8*b))/(a^2 + b^2*x^4 + 2*a*b*x^2) - ((B*x^2)/(2*b) + (B*a)/(4*b^2))/(a^2 + b^2*x^4 + 2
*a*b*x^2) - ((5*C*x^3)/(8*b) + (3*C*a*x)/(8*b^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) + (D*(log(a + b*x^2) + (2*a)/(a
+ b*x^2) - a^2/(2*(a + b*x^2)^2)))/(2*b^3) + (A*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(3/2)*b^(3/2)) + (3*C*atan((b^
(1/2)*x)/a^(1/2)))/(8*a^(1/2)*b^(5/2))

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sympy [B]  time = 20.01, size = 304, normalized size = 2.24 \begin {gather*} \left (\frac {D}{2 b^{3}} - \frac {\sqrt {- a^{3} b^{7}} \left (A b + 3 C a\right )}{16 a^{3} b^{6}}\right ) \log {\left (x + \frac {- 8 D a^{2} + 16 a^{2} b^{3} \left (\frac {D}{2 b^{3}} - \frac {\sqrt {- a^{3} b^{7}} \left (A b + 3 C a\right )}{16 a^{3} b^{6}}\right )}{A b^{2} + 3 C a b} \right )} + \left (\frac {D}{2 b^{3}} + \frac {\sqrt {- a^{3} b^{7}} \left (A b + 3 C a\right )}{16 a^{3} b^{6}}\right ) \log {\left (x + \frac {- 8 D a^{2} + 16 a^{2} b^{3} \left (\frac {D}{2 b^{3}} + \frac {\sqrt {- a^{3} b^{7}} \left (A b + 3 C a\right )}{16 a^{3} b^{6}}\right )}{A b^{2} + 3 C a b} \right )} + \frac {- 2 B a^{2} b + 6 D a^{3} + x^{3} \left (A b^{3} - 5 C a b^{2}\right ) + x^{2} \left (- 4 B a b^{2} + 8 D a^{2} b\right ) + x \left (- A a b^{2} - 3 C a^{2} b\right )}{8 a^{3} b^{3} + 16 a^{2} b^{4} x^{2} + 8 a b^{5} x^{4}} \end {gather*}

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)**3,x)

[Out]

(D/(2*b**3) - sqrt(-a**3*b**7)*(A*b + 3*C*a)/(16*a**3*b**6))*log(x + (-8*D*a**2 + 16*a**2*b**3*(D/(2*b**3) - s
qrt(-a**3*b**7)*(A*b + 3*C*a)/(16*a**3*b**6)))/(A*b**2 + 3*C*a*b)) + (D/(2*b**3) + sqrt(-a**3*b**7)*(A*b + 3*C
*a)/(16*a**3*b**6))*log(x + (-8*D*a**2 + 16*a**2*b**3*(D/(2*b**3) + sqrt(-a**3*b**7)*(A*b + 3*C*a)/(16*a**3*b*
*6)))/(A*b**2 + 3*C*a*b)) + (-2*B*a**2*b + 6*D*a**3 + x**3*(A*b**3 - 5*C*a*b**2) + x**2*(-4*B*a*b**2 + 8*D*a**
2*b) + x*(-A*a*b**2 - 3*C*a**2*b))/(8*a**3*b**3 + 16*a**2*b**4*x**2 + 8*a*b**5*x**4)

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