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

Optimal. Leaf size=130 \[ \frac {a^{3/2} (b B-a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}-\frac {a (A b-a C) \log \left (a+b x^2\right )}{2 b^3}+\frac {x^2 (A b-a C)}{2 b^2}-\frac {a x (b B-a D)}{b^3}+\frac {x^3 (b B-a D)}{3 b^2}+\frac {C x^4}{4 b}+\frac {D x^5}{5 b} \]

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Rubi [A]  time = 0.12, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1802, 635, 205, 260} \begin {gather*} \frac {a^{3/2} (b B-a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {x^2 (A b-a C)}{2 b^2}-\frac {a (A b-a C) \log \left (a+b x^2\right )}{2 b^3}+\frac {x^3 (b B-a D)}{3 b^2}-\frac {a x (b B-a D)}{b^3}+\frac {C x^4}{4 b}+\frac {D x^5}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 114, normalized size = 0.88 \begin {gather*} \frac {x \left (60 a^2 D-10 a b (6 B+x (3 C+2 D x))+b^2 x (30 A+x (20 B+3 x (5 C+4 D x)))\right )+30 a (a C-A b) \log \left (a+b x^2\right )}{60 b^3}-\frac {a^{3/2} (a D-b B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(3/2)*(-(b*B) + a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(7/2)) + (x*(60*a^2*D - 10*a*b*(6*B + x*(3*C + 2*D*x)
) + b^2*x*(30*A + x*(20*B + 3*x*(5*C + 4*D*x)))) + 30*a*(-(A*b) + a*C)*Log[a + b*x^2])/(60*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 )}{a+b x^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),x]

[Out]

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

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fricas [A]  time = 0.94, size = 270, normalized size = 2.08 \begin {gather*} \left [\frac {12 \, D b^{2} x^{5} + 15 \, C b^{2} x^{4} - 20 \, {\left (D a b - B b^{2}\right )} x^{3} - 30 \, {\left (C a b - A b^{2}\right )} x^{2} + 30 \, {\left (D a^{2} - B a b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 60 \, {\left (D a^{2} - B a b\right )} x + 30 \, {\left (C a^{2} - A a b\right )} \log \left (b x^{2} + a\right )}{60 \, b^{3}}, \frac {12 \, D b^{2} x^{5} + 15 \, C b^{2} x^{4} - 20 \, {\left (D a b - B b^{2}\right )} x^{3} - 30 \, {\left (C a b - A b^{2}\right )} x^{2} - 60 \, {\left (D a^{2} - B a b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 60 \, {\left (D a^{2} - B a b\right )} x + 30 \, {\left (C a^{2} - A a b\right )} \log \left (b x^{2} + a\right )}{60 \, b^{3}}\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),x, algorithm="fricas")

[Out]

[1/60*(12*D*b^2*x^5 + 15*C*b^2*x^4 - 20*(D*a*b - B*b^2)*x^3 - 30*(C*a*b - A*b^2)*x^2 + 30*(D*a^2 - B*a*b)*sqrt
(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 60*(D*a^2 - B*a*b)*x + 30*(C*a^2 - A*a*b)*log(b*x^2 +
a))/b^3, 1/60*(12*D*b^2*x^5 + 15*C*b^2*x^4 - 20*(D*a*b - B*b^2)*x^3 - 30*(C*a*b - A*b^2)*x^2 - 60*(D*a^2 - B*
a*b)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) + 60*(D*a^2 - B*a*b)*x + 30*(C*a^2 - A*a*b)*log(b*x^2 + a))/b^3]

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giac [A]  time = 0.35, size = 137, normalized size = 1.05 \begin {gather*} \frac {{\left (C a^{2} - A a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} - \frac {{\left (D a^{3} - B a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {12 \, D b^{4} x^{5} + 15 \, C b^{4} x^{4} - 20 \, D a b^{3} x^{3} + 20 \, B b^{4} x^{3} - 30 \, C a b^{3} x^{2} + 30 \, A b^{4} x^{2} + 60 \, D a^{2} b^{2} x - 60 \, B a b^{3} x}{60 \, b^{5}} \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),x, algorithm="giac")

[Out]

1/2*(C*a^2 - A*a*b)*log(b*x^2 + a)/b^3 - (D*a^3 - B*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/60*(12*D*
b^4*x^5 + 15*C*b^4*x^4 - 20*D*a*b^3*x^3 + 20*B*b^4*x^3 - 30*C*a*b^3*x^2 + 30*A*b^4*x^2 + 60*D*a^2*b^2*x - 60*B
*a*b^3*x)/b^5

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maple [A]  time = 0.01, size = 152, normalized size = 1.17 \begin {gather*} \frac {D x^{5}}{5 b}+\frac {C \,x^{4}}{4 b}+\frac {B \,x^{3}}{3 b}-\frac {D a \,x^{3}}{3 b^{2}}+\frac {A \,x^{2}}{2 b}+\frac {B \,a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}-\frac {C a \,x^{2}}{2 b^{2}}-\frac {D a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}-\frac {A a \ln \left (b \,x^{2}+a \right )}{2 b^{2}}-\frac {B a x}{b^{2}}+\frac {C \,a^{2} \ln \left (b \,x^{2}+a \right )}{2 b^{3}}+\frac {D a^{2} 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),x)

[Out]

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

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maxima [A]  time = 2.93, size = 127, normalized size = 0.98 \begin {gather*} \frac {{\left (C a^{2} - A a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} - \frac {{\left (D a^{3} - B a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {12 \, D b^{2} x^{5} + 15 \, C b^{2} x^{4} - 20 \, {\left (D a b - B b^{2}\right )} x^{3} - 30 \, {\left (C a b - A b^{2}\right )} x^{2} + 60 \, {\left (D a^{2} - B a b\right )} x}{60 \, 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),x, algorithm="maxima")

[Out]

1/2*(C*a^2 - A*a*b)*log(b*x^2 + a)/b^3 - (D*a^3 - B*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/60*(12*D*
b^2*x^5 + 15*C*b^2*x^4 - 20*(D*a*b - B*b^2)*x^3 - 30*(C*a*b - A*b^2)*x^2 + 60*(D*a^2 - B*a*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 )}{b\,x^2+a} \,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),x)

[Out]

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

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sympy [B]  time = 1.35, size = 274, normalized size = 2.11 \begin {gather*} \frac {C x^{4}}{4 b} + \frac {D x^{5}}{5 b} + x^{3} \left (\frac {B}{3 b} - \frac {D a}{3 b^{2}}\right ) + x^{2} \left (\frac {A}{2 b} - \frac {C a}{2 b^{2}}\right ) + x \left (- \frac {B a}{b^{2}} + \frac {D a^{2}}{b^{3}}\right ) + \left (\frac {a \left (- A b + C a\right )}{2 b^{3}} - \frac {\sqrt {- a^{3} b^{7}} \left (- B b + D a\right )}{2 b^{7}}\right ) \log {\left (x + \frac {- A a b + C a^{2} - 2 b^{3} \left (\frac {a \left (- A b + C a\right )}{2 b^{3}} - \frac {\sqrt {- a^{3} b^{7}} \left (- B b + D a\right )}{2 b^{7}}\right )}{- B a b + D a^{2}} \right )} + \left (\frac {a \left (- A b + C a\right )}{2 b^{3}} + \frac {\sqrt {- a^{3} b^{7}} \left (- B b + D a\right )}{2 b^{7}}\right ) \log {\left (x + \frac {- A a b + C a^{2} - 2 b^{3} \left (\frac {a \left (- A b + C a\right )}{2 b^{3}} + \frac {\sqrt {- a^{3} b^{7}} \left (- B b + D a\right )}{2 b^{7}}\right )}{- B a b + D a^{2}} \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),x)

[Out]

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

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