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

Optimal. Leaf size=154 \[ \frac {(A b-2 a C) \log \left (a+b x^2\right )}{2 b^3}-\frac {x^2 (A b-2 a C)}{2 a b^2}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}-\frac {\sqrt {a} (3 b B-5 a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}+\frac {x (3 b B-5 a D)}{2 b^3}+\frac {D x^3}{3 b^2} \]

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Rubi [A]  time = 0.24, antiderivative size = 154, 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} \begin {gather*} -\frac {x^2 (A b-2 a C)}{2 a b^2}+\frac {(A b-2 a C) \log \left (a+b x^2\right )}{2 b^3}-\frac {x^3 \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 {x (3 b B-5 a D)}{2 b^3}-\frac {\sqrt {a} (3 b B-5 a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}+\frac {D x^3}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 128, normalized size = 0.83 \begin {gather*} \frac {a (-a (C+D x)+A b+b B x)}{2 b^3 \left (a+b x^2\right )}+\frac {(A b-2 a C) \log \left (a+b x^2\right )}{2 b^3}+\frac {\sqrt {a} (5 a D-3 b B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}+\frac {x (b B-2 a D)}{b^3}+\frac {C x^2}{2 b^2}+\frac {D x^3}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - 2*a*D)*x)/b^3 + (C*x^2)/(2*b^2) + (D*x^3)/(3*b^2) + (a*(A*b + b*B*x - a*(C + D*x)))/(2*b^3*(a + b*x^2)
) + (Sqrt[a]*(-3*b*B + 5*a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(7/2)) + ((A*b - 2*a*C)*Log[a + b*x^2])/(2*b^3
)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 372, normalized size = 2.42 \begin {gather*} \left [\frac {4 \, D b^{2} x^{5} + 6 \, C b^{2} x^{4} + 6 \, C a b x^{2} - 4 \, {\left (5 \, D a b - 3 \, B b^{2}\right )} x^{3} - 6 \, C a^{2} + 6 \, A a b + 3 \, {\left (5 \, D a^{2} - 3 \, B a b + {\left (5 \, D a b - 3 \, B b^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 6 \, {\left (5 \, D a^{2} - 3 \, B a b\right )} x - 6 \, {\left (2 \, C a^{2} - A a b + {\left (2 \, C a b - A b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {2 \, D b^{2} x^{5} + 3 \, C b^{2} x^{4} + 3 \, C a b x^{2} - 2 \, {\left (5 \, D a b - 3 \, B b^{2}\right )} x^{3} - 3 \, C a^{2} + 3 \, A a b + 3 \, {\left (5 \, D a^{2} - 3 \, B a b + {\left (5 \, D a b - 3 \, B b^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 3 \, {\left (5 \, D a^{2} - 3 \, B a b\right )} x - 3 \, {\left (2 \, C a^{2} - A a b + {\left (2 \, C a b - A b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{6 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 131, normalized size = 0.85 \begin {gather*} -\frac {{\left (2 \, C a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (5 \, D a^{2} - 3 \, B a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} - \frac {C a^{2} - A a b + {\left (D a^{2} - B a b\right )} x}{2 \, {\left (b x^{2} + a\right )} b^{3}} + \frac {2 \, D b^{4} x^{3} + 3 \, C b^{4} x^{2} - 12 \, D a b^{3} x + 6 \, B b^{4} x}{6 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 177, normalized size = 1.15 \begin {gather*} \frac {D x^{3}}{3 b^{2}}+\frac {B a x}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {3 B a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{2}}+\frac {C \,x^{2}}{2 b^{2}}-\frac {D a^{2} x}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {5 D a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{3}}+\frac {A a}{2 \left (b \,x^{2}+a \right ) b^{2}}+\frac {A \ln \left (b \,x^{2}+a \right )}{2 b^{2}}+\frac {B x}{b^{2}}-\frac {C \,a^{2}}{2 \left (b \,x^{2}+a \right ) b^{3}}-\frac {C a \ln \left (b \,x^{2}+a \right )}{b^{3}}-\frac {2 D a x}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.00, size = 127, normalized size = 0.82 \begin {gather*} -\frac {C a^{2} - A a b + {\left (D a^{2} - B a b\right )} x}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} - \frac {{\left (2 \, C a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (5 \, D a^{2} - 3 \, B a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {2 \, D b x^{3} + 3 \, C b x^{2} - 6 \, {\left (2 \, D a - B b\right )} x}{6 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 3.87, size = 289, normalized size = 1.88 \begin {gather*} \frac {C x^{2}}{2 b^{2}} + \frac {D x^{3}}{3 b^{2}} + x \left (\frac {B}{b^{2}} - \frac {2 D a}{b^{3}}\right ) + \left (- \frac {- A b + 2 C a}{2 b^{3}} - \frac {\sqrt {- a b^{7}} \left (- 3 B b + 5 D a\right )}{4 b^{7}}\right ) \log {\left (x + \frac {- 2 A b + 4 C a + 4 b^{3} \left (- \frac {- A b + 2 C a}{2 b^{3}} - \frac {\sqrt {- a b^{7}} \left (- 3 B b + 5 D a\right )}{4 b^{7}}\right )}{- 3 B b + 5 D a} \right )} + \left (- \frac {- A b + 2 C a}{2 b^{3}} + \frac {\sqrt {- a b^{7}} \left (- 3 B b + 5 D a\right )}{4 b^{7}}\right ) \log {\left (x + \frac {- 2 A b + 4 C a + 4 b^{3} \left (- \frac {- A b + 2 C a}{2 b^{3}} + \frac {\sqrt {- a b^{7}} \left (- 3 B b + 5 D a\right )}{4 b^{7}}\right )}{- 3 B b + 5 D a} \right )} + \frac {A a b - C a^{2} + x \left (B a b - D a^{2}\right )}{2 a b^{3} + 2 b^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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