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

Optimal. Leaf size=155 \[ -\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}+\frac {3 (b B-5 a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}-\frac {3 x (b B-5 a D)}{8 a b^3}+\frac {C \log \left (a+b x^2\right )}{2 b^3}-\frac {x^2 (4 a C-x (3 b B-7 a D))}{8 a b^2 \left (a+b x^2\right )} \]

[Out]

-3/8*(B*b-5*D*a)*x/a/b^3-1/4*x^3*(a*(B-a*D/b)-(A*b-C*a)*x)/a/b/(b*x^2+a)^2-1/8*x^2*(4*a*C-(3*B*b-7*D*a)*x)/a/b
^2/(b*x^2+a)+1/2*C*ln(b*x^2+a)/b^3+3/8*(B*b-5*D*a)*arctan(x*b^(1/2)/a^(1/2))/b^(7/2)/a^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 155, 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, 774, 635, 205, 260} \[ -\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}-\frac {x^2 (4 a C-x (3 b B-7 a D))}{8 a b^2 \left (a+b x^2\right )}-\frac {3 x (b B-5 a D)}{8 a b^3}+\frac {3 (b B-5 a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {C \log \left (a+b x^2\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*B - 5*a*D)*x)/(8*a*b^3) - (x^3*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(4*a*b*(a + b*x^2)^2) - (x^2*(4*a*C -
(3*b*B - 7*a*D)*x))/(8*a*b^2*(a + b*x^2)) + (3*(b*B - 5*a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2))
+ (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 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

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 )^3} \, dx &=-\frac {x^3 \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^2 \left (-3 a \left (B-\frac {a D}{b}\right )-4 a C x-4 a D x^2\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=-\frac {x^3 \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^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {\int \frac {x \left (8 a^2 C-3 a (b B-5 a D) x\right )}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=-\frac {3 (b B-5 a D) x}{8 a b^3}-\frac {x^3 \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^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {\int \frac {3 a^2 (b B-5 a D)+8 a^2 b C x}{a+b x^2} \, dx}{8 a^2 b^3}\\ &=-\frac {3 (b B-5 a D) x}{8 a b^3}-\frac {x^3 \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^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {C \int \frac {x}{a+b x^2} \, dx}{b^2}+\frac {(3 (b B-5 a D)) \int \frac {1}{a+b x^2} \, dx}{8 b^3}\\ &=-\frac {3 (b B-5 a D) x}{8 a b^3}-\frac {x^3 \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^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {3 (b B-5 a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {C \log \left (a+b x^2\right )}{2 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 126, normalized size = 0.81 \[ \frac {a (-a (C+D x)+A b+b B x)}{4 b^3 \left (a+b x^2\right )^2}+\frac {8 a C+9 a D x-4 A b-5 b B x}{8 b^3 \left (a+b x^2\right )}+\frac {3 (b B-5 a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {C \log \left (a+b x^2\right )}{2 b^3}+\frac {D x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(D*x)/b^3 + (-4*A*b + 8*a*C - 5*b*B*x + 9*a*D*x)/(8*b^3*(a + b*x^2)) + (a*(A*b + b*B*x - a*(C + D*x)))/(4*b^3*
(a + b*x^2)^2) + (3*(b*B - 5*a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2)) + (C*Log[a + b*x^2])/(2*b^3
)

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fricas [A]  time = 0.74, size = 480, normalized size = 3.10 \[ \left [\frac {16 \, D a b^{3} x^{5} + 12 \, C a^{3} b - 4 \, A a^{2} b^{2} + 10 \, {\left (5 \, D a^{2} b^{2} - B a b^{3}\right )} x^{3} + 8 \, {\left (2 \, C a^{2} b^{2} - A a b^{3}\right )} x^{2} - 3 \, {\left ({\left (5 \, D a b^{2} - B b^{3}\right )} x^{4} + 5 \, D a^{3} - B a^{2} b + 2 \, {\left (5 \, D a^{2} b - B a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 6 \, {\left (5 \, D a^{3} b - B a^{2} b^{2}\right )} x + 8 \, {\left (C a b^{3} x^{4} + 2 \, C a^{2} b^{2} x^{2} + C a^{3} b\right )} \log \left (b x^{2} + a\right )}{16 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {8 \, D a b^{3} x^{5} + 6 \, C a^{3} b - 2 \, A a^{2} b^{2} + 5 \, {\left (5 \, D a^{2} b^{2} - B a b^{3}\right )} x^{3} + 4 \, {\left (2 \, C a^{2} b^{2} - A a b^{3}\right )} x^{2} - 3 \, {\left ({\left (5 \, D a b^{2} - B b^{3}\right )} x^{4} + 5 \, D a^{3} - B a^{2} b + 2 \, {\left (5 \, D a^{2} b - B a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, {\left (5 \, D a^{3} b - B a^{2} b^{2}\right )} x + 4 \, {\left (C a b^{3} x^{4} + 2 \, C a^{2} b^{2} x^{2} + C a^{3} b\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \]

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)^3,x, algorithm="fricas")

[Out]

[1/16*(16*D*a*b^3*x^5 + 12*C*a^3*b - 4*A*a^2*b^2 + 10*(5*D*a^2*b^2 - B*a*b^3)*x^3 + 8*(2*C*a^2*b^2 - A*a*b^3)*
x^2 - 3*((5*D*a*b^2 - B*b^3)*x^4 + 5*D*a^3 - B*a^2*b + 2*(5*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)) + 6*(5*D*a^3*b - B*a^2*b^2)*x + 8*(C*a*b^3*x^4 + 2*C*a^2*b^2*x^2 + C*a^3*b)*log
(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4), 1/8*(8*D*a*b^3*x^5 + 6*C*a^3*b - 2*A*a^2*b^2 + 5*(5*D*a^2*
b^2 - B*a*b^3)*x^3 + 4*(2*C*a^2*b^2 - A*a*b^3)*x^2 - 3*((5*D*a*b^2 - B*b^3)*x^4 + 5*D*a^3 - B*a^2*b + 2*(5*D*a
^2*b - B*a*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + 3*(5*D*a^3*b - B*a^2*b^2)*x + 4*(C*a*b^3*x^4 + 2*C*a^2*
b^2*x^2 + C*a^3*b)*log(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4)]

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giac [A]  time = 0.38, size = 122, normalized size = 0.79 \[ \frac {D x}{b^{3}} + \frac {C \log \left (b x^{2} + a\right )}{2 \, b^{3}} - \frac {3 \, {\left (5 \, D a - B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {{\left (9 \, D a b - 5 \, B b^{2}\right )} x^{3} + 6 \, C a^{2} - 2 \, A a b + 4 \, {\left (2 \, C a b - A b^{2}\right )} x^{2} + {\left (7 \, D a^{2} - 3 \, B a b\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{3}} \]

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)^3,x, algorithm="giac")

[Out]

D*x/b^3 + 1/2*C*log(b*x^2 + a)/b^3 - 3/8*(5*D*a - B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/8*((9*D*a*b -
5*B*b^2)*x^3 + 6*C*a^2 - 2*A*a*b + 4*(2*C*a*b - A*b^2)*x^2 + (7*D*a^2 - 3*B*a*b)*x)/((b*x^2 + a)^2*b^3)

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maple [A]  time = 0.01, size = 206, normalized size = 1.33 \[ -\frac {5 B \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b}+\frac {9 D a \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}-\frac {A \,x^{2}}{2 \left (b \,x^{2}+a \right )^{2} b}+\frac {C a \,x^{2}}{\left (b \,x^{2}+a \right )^{2} b^{2}}-\frac {3 B a x}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}+\frac {7 D a^{2} x}{8 \left (b \,x^{2}+a \right )^{2} b^{3}}-\frac {A a}{4 \left (b \,x^{2}+a \right )^{2} b^{2}}+\frac {3 B \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{2}}+\frac {3 C \,a^{2}}{4 \left (b \,x^{2}+a \right )^{2} b^{3}}-\frac {15 D a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{3}}+\frac {C \ln \left (b \,x^{2}+a \right )}{2 b^{3}}+\frac {D x}{b^{3}} \]

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)^3,x)

[Out]

D/b^3*x-5/8/b/(b*x^2+a)^2*B*x^3+9/8/b^2/(b*x^2+a)^2*D*x^3*a-1/2/b/(b*x^2+a)^2*A*x^2+1/b^2/(b*x^2+a)^2*C*x^2*a-
3/8/b^2/(b*x^2+a)^2*B*x*a+7/8/b^3/(b*x^2+a)^2*a^2*D*x-1/4/(b*x^2+a)^2*A*a/b^2+3/4/b^3/(b*x^2+a)^2*a^2*C+1/2*C*
ln(b*x^2+a)/b^3+3/8/(a*b)^(1/2)*B/b^2*arctan(1/(a*b)^(1/2)*b*x)-15/8/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)
*a*D

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maxima [A]  time = 2.98, size = 136, normalized size = 0.88 \[ \frac {{\left (9 \, D a b - 5 \, B b^{2}\right )} x^{3} + 6 \, C a^{2} - 2 \, A a b + 4 \, {\left (2 \, C a b - A b^{2}\right )} x^{2} + {\left (7 \, D a^{2} - 3 \, B a b\right )} x}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {D x}{b^{3}} + \frac {C \log \left (b x^{2} + a\right )}{2 \, b^{3}} - \frac {3 \, {\left (5 \, D a - B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} \]

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)^3,x, algorithm="maxima")

[Out]

1/8*((9*D*a*b - 5*B*b^2)*x^3 + 6*C*a^2 - 2*A*a*b + 4*(2*C*a*b - A*b^2)*x^2 + (7*D*a^2 - 3*B*a*b)*x)/(b^5*x^4 +
2*a*b^4*x^2 + a^2*b^3) + D*x/b^3 + 1/2*C*log(b*x^2 + a)/b^3 - 3/8*(5*D*a - B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a
*b)*b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 29.73, size = 282, normalized size = 1.82 \[ \frac {D x}{b^{3}} + \left (\frac {C}{2 b^{3}} - \frac {3 \sqrt {- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right ) \log {\left (x + \frac {8 C a - 16 a b^{3} \left (\frac {C}{2 b^{3}} - \frac {3 \sqrt {- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right )}{- 3 B b + 15 D a} \right )} + \left (\frac {C}{2 b^{3}} + \frac {3 \sqrt {- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right ) \log {\left (x + \frac {8 C a - 16 a b^{3} \left (\frac {C}{2 b^{3}} + \frac {3 \sqrt {- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right )}{- 3 B b + 15 D a} \right )} + \frac {- 2 A a b + 6 C a^{2} + x^{3} \left (- 5 B b^{2} + 9 D a b\right ) + x^{2} \left (- 4 A b^{2} + 8 C a b\right ) + x \left (- 3 B a b + 7 D a^{2}\right )}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} \]

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)**3,x)

[Out]

D*x/b**3 + (C/(2*b**3) - 3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b**7))*log(x + (8*C*a - 16*a*b**3*(C/(2*b**3) -
3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b**7)))/(-3*B*b + 15*D*a)) + (C/(2*b**3) + 3*sqrt(-a*b**7)*(-B*b + 5*D*a)
/(16*a*b**7))*log(x + (8*C*a - 16*a*b**3*(C/(2*b**3) + 3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b**7)))/(-3*B*b +
15*D*a)) + (-2*A*a*b + 6*C*a**2 + x**3*(-5*B*b**2 + 9*D*a*b) + x**2*(-4*A*b**2 + 8*C*a*b) + x*(-3*B*a*b + 7*D*
a**2))/(8*a**2*b**3 + 16*a*b**4*x**2 + 8*b**5*x**4)

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