3.94 \(\int \frac {x^4 (A+B x+C x^2+D x^3)}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=176 \[ -\frac {\sqrt {a} (3 A b-5 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}+\frac {x (3 A b-5 a C)}{2 b^3}-\frac {x^3 (3 A b-5 a C)}{6 a b^2}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}-\frac {a (2 b B-3 a D) \log \left (a+b x^2\right )}{2 b^4}+\frac {x^2 (2 b B-3 a D)}{2 b^3}+\frac {D x^4}{4 b^2} \]

[Out]

1/2*(3*A*b-5*C*a)*x/b^3+1/2*(2*B*b-3*D*a)*x^2/b^3-1/6*(3*A*b-5*C*a)*x^3/a/b^2+1/4*D*x^4/b^2-1/2*x^4*(a*(B-a*D/
b)-(A*b-C*a)*x)/a/b/(b*x^2+a)-1/2*a*(2*B*b-3*D*a)*ln(b*x^2+a)/b^4-1/2*(3*A*b-5*C*a)*arctan(x*b^(1/2)/a^(1/2))*
a^(1/2)/b^(7/2)

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Rubi [A]  time = 0.27, antiderivative size = 176, 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} \[ -\frac {x^3 (3 A b-5 a C)}{6 a b^2}+\frac {x (3 A b-5 a C)}{2 b^3}-\frac {\sqrt {a} (3 A b-5 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac {x^2 (2 b B-3 a D)}{2 b^3}-\frac {a (2 b B-3 a D) \log \left (a+b x^2\right )}{2 b^4}+\frac {D x^4}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^2,x]

[Out]

((3*A*b - 5*a*C)*x)/(2*b^3) + ((2*b*B - 3*a*D)*x^2)/(2*b^3) - ((3*A*b - 5*a*C)*x^3)/(6*a*b^2) + (D*x^4)/(4*b^2
) - (x^4*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(2*a*b*(a + b*x^2)) - (Sqrt[a]*(3*A*b - 5*a*C)*ArcTan[(Sqrt[b]*x)/
Sqrt[a]])/(2*b^(7/2)) - (a*(2*b*B - 3*a*D)*Log[a + b*x^2])/(2*b^4)

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^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx &=-\frac {x^4 \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^3 \left (-4 a \left (B-\frac {a D}{b}\right )+(3 A b-5 a C) x-2 a D x^2\right )}{a+b x^2} \, dx}{2 a b}\\ &=-\frac {x^4 \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 A b-5 a C)}{b^2}-\frac {2 a (2 b B-3 a D) x}{b^2}+\frac {(3 A b-5 a C) x^2}{b}-\frac {2 a D x^3}{b}+\frac {a^2 (3 A b-5 a C)+2 a^2 (2 b B-3 a D) x}{b^2 \left (a+b x^2\right )}\right ) \, dx}{2 a b}\\ &=\frac {(3 A b-5 a C) x}{2 b^3}+\frac {(2 b B-3 a D) x^2}{2 b^3}-\frac {(3 A b-5 a C) x^3}{6 a b^2}+\frac {D x^4}{4 b^2}-\frac {x^4 \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 A b-5 a C)+2 a^2 (2 b B-3 a D) x}{a+b x^2} \, dx}{2 a b^3}\\ &=\frac {(3 A b-5 a C) x}{2 b^3}+\frac {(2 b B-3 a D) x^2}{2 b^3}-\frac {(3 A b-5 a C) x^3}{6 a b^2}+\frac {D x^4}{4 b^2}-\frac {x^4 \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 (3 A b-5 a C)) \int \frac {1}{a+b x^2} \, dx}{2 b^3}-\frac {(a (2 b B-3 a D)) \int \frac {x}{a+b x^2} \, dx}{b^3}\\ &=\frac {(3 A b-5 a C) x}{2 b^3}+\frac {(2 b B-3 a D) x^2}{2 b^3}-\frac {(3 A b-5 a C) x^3}{6 a b^2}+\frac {D x^4}{4 b^2}-\frac {x^4 \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 A b-5 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}-\frac {a (2 b B-3 a D) \log \left (a+b x^2\right )}{2 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 139, normalized size = 0.79 \[ \frac {\frac {6 a \left (a^2 D-a b (B+C x)+A b^2 x\right )}{a+b x^2}+12 b x (A b-2 a C)+6 \sqrt {a} \sqrt {b} (5 a C-3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+6 b x^2 (b B-2 a D)+6 a (3 a D-2 b B) \log \left (a+b x^2\right )+4 b^2 C x^3+3 b^2 D x^4}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^2,x]

[Out]

(12*b*(A*b - 2*a*C)*x + 6*b*(b*B - 2*a*D)*x^2 + 4*b^2*C*x^3 + 3*b^2*D*x^4 + (6*a*(a^2*D + A*b^2*x - a*b*(B + C
*x)))/(a + b*x^2) + 6*Sqrt[a]*Sqrt[b]*(-3*A*b + 5*a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]] + 6*a*(-2*b*B + 3*a*D)*Log[
a + b*x^2])/(12*b^4)

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fricas [A]  time = 0.58, size = 468, normalized size = 2.66 \[ \left [\frac {3 \, D b^{3} x^{6} + 4 \, C b^{3} x^{5} - 3 \, {\left (3 \, D a b^{2} - 2 \, B b^{3}\right )} x^{4} + 6 \, D a^{3} - 6 \, B a^{2} b - 4 \, {\left (5 \, C a b^{2} - 3 \, A b^{3}\right )} x^{3} - 6 \, {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2} - 3 \, {\left (5 \, C a^{2} b - 3 \, A a b^{2} + {\left (5 \, C a b^{2} - 3 \, A b^{3}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 6 \, {\left (5 \, C a^{2} b - 3 \, A a b^{2}\right )} x + 6 \, {\left (3 \, D a^{3} - 2 \, B a^{2} b + {\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{5} x^{2} + a b^{4}\right )}}, \frac {3 \, D b^{3} x^{6} + 4 \, C b^{3} x^{5} - 3 \, {\left (3 \, D a b^{2} - 2 \, B b^{3}\right )} x^{4} + 6 \, D a^{3} - 6 \, B a^{2} b - 4 \, {\left (5 \, C a b^{2} - 3 \, A b^{3}\right )} x^{3} - 6 \, {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2} + 6 \, {\left (5 \, C a^{2} b - 3 \, A a b^{2} + {\left (5 \, C a b^{2} - 3 \, A b^{3}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 6 \, {\left (5 \, C a^{2} b - 3 \, A a b^{2}\right )} x + 6 \, {\left (3 \, D a^{3} - 2 \, B a^{2} b + {\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.43, size = 159, normalized size = 0.90 \[ \frac {{\left (5 \, C a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {{\left (3 \, D a^{2} - 2 \, B a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac {D a^{3} - B a^{2} b - {\left (C a^{2} b - A a b^{2}\right )} x}{2 \, {\left (b x^{2} + a\right )} b^{4}} + \frac {3 \, D b^{6} x^{4} + 4 \, C b^{6} x^{3} - 12 \, D a b^{5} x^{2} + 6 \, B b^{6} x^{2} - 24 \, C a b^{5} x + 12 \, A b^{6} x}{12 \, b^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

1/2*(5*C*a^2 - 3*A*a*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/2*(3*D*a^2 - 2*B*a*b)*log(b*x^2 + a)/b^4 + 1
/2*(D*a^3 - B*a^2*b - (C*a^2*b - A*a*b^2)*x)/((b*x^2 + a)*b^4) + 1/12*(3*D*b^6*x^4 + 4*C*b^6*x^3 - 12*D*a*b^5*
x^2 + 6*B*b^6*x^2 - 24*C*a*b^5*x + 12*A*b^6*x)/b^8

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maple [A]  time = 0.01, size = 201, normalized size = 1.14 \[ \frac {D x^{4}}{4 b^{2}}+\frac {C \,x^{3}}{3 b^{2}}+\frac {A a x}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {3 A a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{2}}+\frac {B \,x^{2}}{2 b^{2}}-\frac {C \,a^{2} x}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {5 C \,a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{3}}-\frac {D a \,x^{2}}{b^{3}}+\frac {A x}{b^{2}}-\frac {B \,a^{2}}{2 \left (b \,x^{2}+a \right ) b^{3}}-\frac {B a \ln \left (b \,x^{2}+a \right )}{b^{3}}-\frac {2 C a x}{b^{3}}+\frac {D a^{3}}{2 \left (b \,x^{2}+a \right ) b^{4}}+\frac {3 D a^{2} \ln \left (b \,x^{2}+a \right )}{2 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^2,x)

[Out]

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

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maxima [A]  time = 2.83, size = 150, normalized size = 0.85 \[ \frac {D a^{3} - B a^{2} b - {\left (C a^{2} b - A a b^{2}\right )} x}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} + \frac {{\left (5 \, C a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {3 \, D b x^{4} + 4 \, C b x^{3} - 6 \, {\left (2 \, D a - B b\right )} x^{2} - 12 \, {\left (2 \, C a - A b\right )} x}{12 \, b^{3}} + \frac {{\left (3 \, D a^{2} - 2 \, B a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^2,x)

[Out]

int((x^4*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^2, x)

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sympy [B]  time = 4.77, size = 335, normalized size = 1.90 \[ \frac {C x^{3}}{3 b^{2}} + \frac {D x^{4}}{4 b^{2}} + x^{2} \left (\frac {B}{2 b^{2}} - \frac {D a}{b^{3}}\right ) + x \left (\frac {A}{b^{2}} - \frac {2 C a}{b^{3}}\right ) + \left (\frac {a \left (- 2 B b + 3 D a\right )}{2 b^{4}} - \frac {\sqrt {- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right ) \log {\left (x + \frac {4 B a b - 6 D a^{2} + 4 b^{4} \left (\frac {a \left (- 2 B b + 3 D a\right )}{2 b^{4}} - \frac {\sqrt {- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right )}{- 3 A b^{2} + 5 C a b} \right )} + \left (\frac {a \left (- 2 B b + 3 D a\right )}{2 b^{4}} + \frac {\sqrt {- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right ) \log {\left (x + \frac {4 B a b - 6 D a^{2} + 4 b^{4} \left (\frac {a \left (- 2 B b + 3 D a\right )}{2 b^{4}} + \frac {\sqrt {- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right )}{- 3 A b^{2} + 5 C a b} \right )} + \frac {- B a^{2} b + D a^{3} + x \left (A a b^{2} - C a^{2} b\right )}{2 a b^{4} + 2 b^{5} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**2,x)

[Out]

C*x**3/(3*b**2) + D*x**4/(4*b**2) + x**2*(B/(2*b**2) - D*a/b**3) + x*(A/b**2 - 2*C*a/b**3) + (a*(-2*B*b + 3*D*
a)/(2*b**4) - sqrt(-a*b**9)*(-3*A*b + 5*C*a)/(4*b**8))*log(x + (4*B*a*b - 6*D*a**2 + 4*b**4*(a*(-2*B*b + 3*D*a
)/(2*b**4) - sqrt(-a*b**9)*(-3*A*b + 5*C*a)/(4*b**8)))/(-3*A*b**2 + 5*C*a*b)) + (a*(-2*B*b + 3*D*a)/(2*b**4) +
sqrt(-a*b**9)*(-3*A*b + 5*C*a)/(4*b**8))*log(x + (4*B*a*b - 6*D*a**2 + 4*b**4*(a*(-2*B*b + 3*D*a)/(2*b**4) +
sqrt(-a*b**9)*(-3*A*b + 5*C*a)/(4*b**8)))/(-3*A*b**2 + 5*C*a*b)) + (-B*a**2*b + D*a**3 + x*(A*a*b**2 - C*a**2*
b))/(2*a*b**4 + 2*b**5*x**2)

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