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

Optimal. Leaf size=116 \[ \frac {(a C+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}}-\frac {4 a^2 D-b x (a C+3 A b)}{8 a^2 b^2 \left (a+b x^2\right )}+\frac {x (A b-a C)-a \left (B-\frac {a D}{b}\right )}{4 a b \left (a+b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1814, 639, 205} \[ -\frac {4 a^2 D-b x (a C+3 A b)}{8 a^2 b^2 \left (a+b x^2\right )}+\frac {(a C+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

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

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Mathematica [A]  time = 0.11, size = 104, normalized size = 0.90 \[ \frac {\frac {\sqrt {a} \left (-2 a^3 D-a^2 b (2 B+x (C+4 D x))+a b^2 x \left (5 A+C x^2\right )+3 A b^3 x^3\right )}{\left (a+b x^2\right )^2}+\sqrt {b} (a C+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 346, normalized size = 2.98 \[ \left [-\frac {8 \, D a^{3} b x^{2} + 4 \, D a^{4} + 4 \, B a^{3} b - 2 \, {\left (C a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} + {\left ({\left (C a b^{2} + 3 \, A b^{3}\right )} x^{4} + C a^{3} + 3 \, A a^{2} b + 2 \, {\left (C a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (C a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{16 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}, -\frac {4 \, D a^{3} b x^{2} + 2 \, D a^{4} + 2 \, B a^{3} b - {\left (C a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} - {\left ({\left (C a b^{2} + 3 \, A b^{3}\right )} x^{4} + C a^{3} + 3 \, A a^{2} b + 2 \, {\left (C a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (C a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{8 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.39, size = 106, normalized size = 0.91 \[ \frac {{\left (C a + 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} + \frac {C a b^{2} x^{3} + 3 \, A b^{3} x^{3} - 4 \, D a^{2} b x^{2} - C a^{2} b x + 5 \, A a b^{2} x - 2 \, D a^{3} - 2 \, B a^{2} b}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 111, normalized size = 0.96 \[ \frac {3 A \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{2}}+\frac {C \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a b}+\frac {-\frac {D x^{2}}{2 b}+\frac {\left (3 A b +a C \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A b -a C \right ) x}{8 a b}-\frac {b B +a D}{4 b^{2}}}{\left (b \,x^{2}+a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.00, size = 122, normalized size = 1.05 \[ -\frac {4 \, D a^{2} b x^{2} + 2 \, D a^{3} + 2 \, B a^{2} b - {\left (C a b^{2} + 3 \, A b^{3}\right )} x^{3} + {\left (C a^{2} b - 5 \, A a b^{2}\right )} x}{8 \, {\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} + \frac {{\left (C a + 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.33, size = 163, normalized size = 1.41 \[ \frac {\frac {C\,x^3}{8\,a}-\frac {C\,x}{8\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\frac {5\,A\,x}{8\,a}+\frac {3\,A\,b\,x^3}{8\,a^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {B}{4\,b\,\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}-\frac {\left (2\,b\,x^2+a\right )\,D}{4\,b^2\,{\left (b\,x^2+a\right )}^2}+\frac {3\,A\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{5/2}\,\sqrt {b}}+\frac {C\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{3/2}\,b^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 11.27, size = 184, normalized size = 1.59 \[ - \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \left (3 A b + C a\right ) \log {\left (- a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \left (3 A b + C a\right ) \log {\left (a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{16} + \frac {- 2 B a^{2} b - 2 D a^{3} - 4 D a^{2} b x^{2} + x^{3} \left (3 A b^{3} + C a b^{2}\right ) + x \left (5 A a b^{2} - C a^{2} b\right )}{8 a^{4} b^{2} + 16 a^{3} b^{3} x^{2} + 8 a^{2} b^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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