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

Optimal. Leaf size=119 \[ \frac {(3 a D+b B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}}-\frac {2 (a C+A b)-x (b B-5 a D)}{8 a b^2 \left (a+b x^2\right )}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1804, 1814, 12, 205} \[ \frac {(3 a D+b B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}}-\frac {2 (a C+A b)-x (b B-5 a D)}{8 a b^2 \left (a+b x^2\right )}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 99, normalized size = 0.83 \[ \frac {\frac {(3 a D+b B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\sqrt {b} \left (-a^2 (2 C+3 D x)-a b \left (2 A+x \left (B+4 C x+5 D x^2\right )\right )+b^2 B x^3\right )}{a \left (a+b x^2\right )^2}}{8 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 357, normalized size = 3.00 \[ \left [-\frac {8 \, C a^{2} b^{2} x^{2} + 4 \, C a^{3} b + 4 \, A a^{2} b^{2} + 2 \, {\left (5 \, D a^{2} b^{2} - B a b^{3}\right )} x^{3} + {\left ({\left (3 \, D a b^{2} + B b^{3}\right )} x^{4} + 3 \, D a^{3} + B a^{2} b + 2 \, {\left (3 \, D a^{2} b + B a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (3 \, D a^{3} b + B a^{2} b^{2}\right )} x}{16 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, -\frac {4 \, C a^{2} b^{2} x^{2} + 2 \, C a^{3} b + 2 \, A a^{2} b^{2} + {\left (5 \, D a^{2} b^{2} - B a b^{3}\right )} x^{3} - {\left ({\left (3 \, D a b^{2} + B b^{3}\right )} x^{4} + 3 \, D a^{3} + B a^{2} b + 2 \, {\left (3 \, D a^{2} b + B a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (3 \, D a^{3} b + B a^{2} b^{2}\right )} x}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.39, size = 97, normalized size = 0.82 \[ \frac {{\left (3 \, D a + B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} - \frac {5 \, D a b x^{3} - B b^{2} x^{3} + 4 \, C a b x^{2} + 3 \, D a^{2} x + B a b x + 2 \, C a^{2} + 2 \, A a b}{8 \, {\left (b x^{2} + a\right )}^{2} a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 110, normalized size = 0.92 \[ \frac {B \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a b}+\frac {3 D \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{2}}+\frac {-\frac {C \,x^{2}}{2 b}+\frac {\left (b B -5 a D\right ) x^{3}}{8 a b}-\frac {\left (b B +3 a D\right ) x}{8 b^{2}}-\frac {A b +a C}{4 b^{2}}}{\left (b \,x^{2}+a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.99, size = 111, normalized size = 0.93 \[ -\frac {4 \, C a b x^{2} + {\left (5 \, D a b - B b^{2}\right )} x^{3} + 2 \, C a^{2} + 2 \, A a b + {\left (3 \, D a^{2} + B a b\right )} x}{8 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} + \frac {{\left (3 \, D a + B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 16.37, size = 178, normalized size = 1.50 \[ - \frac {\sqrt {- \frac {1}{a^{3} b^{5}}} \left (B b + 3 D a\right ) \log {\left (- a^{2} b^{2} \sqrt {- \frac {1}{a^{3} b^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{3} b^{5}}} \left (B b + 3 D a\right ) \log {\left (a^{2} b^{2} \sqrt {- \frac {1}{a^{3} b^{5}}} + x \right )}}{16} + \frac {- 2 A a b - 2 C a^{2} - 4 C a b x^{2} + x^{3} \left (B b^{2} - 5 D a b\right ) + x \left (- B a b - 3 D a^{2}\right )}{8 a^{3} b^{2} + 16 a^{2} b^{3} x^{2} + 8 a b^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**3*b**5))*(B*b + 3*D*a)*log(-a**2*b**2*sqrt(-1/(a**3*b**5)) + x)/16 + sqrt(-1/(a**3*b**5))*(B*b +
3*D*a)*log(a**2*b**2*sqrt(-1/(a**3*b**5)) + x)/16 + (-2*A*a*b - 2*C*a**2 - 4*C*a*b*x**2 + x**3*(B*b**2 - 5*D*a
*b) + x*(-B*a*b - 3*D*a**2))/(8*a**3*b**2 + 16*a**2*b**3*x**2 + 8*a*b**4*x**4)

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