3.1.58 \(\int \frac {x^4 (A+B x^2)}{a+b x^2} \, dx\)

Optimal. Leaf size=77 \[ \frac {a^{3/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}-\frac {a x (A b-a B)}{b^3}+\frac {x^3 (A b-a B)}{3 b^2}+\frac {B x^5}{5 b} \]

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {459, 302, 205} \begin {gather*} \frac {a^{3/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {x^3 (A b-a B)}{3 b^2}-\frac {a x (A b-a B)}{b^3}+\frac {B x^5}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x^2))/(a + b*x^2),x]

[Out]

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

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 77, normalized size = 1.00 \begin {gather*} -\frac {a^{3/2} (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {a x (a B-A b)}{b^3}+\frac {x^3 (A b-a B)}{3 b^2}+\frac {B x^5}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^4 \left (A+B x^2\right )}{a+b x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^4*(A + B*x^2))/(a + b*x^2),x]

[Out]

IntegrateAlgebraic[(x^4*(A + B*x^2))/(a + b*x^2), x]

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fricas [A]  time = 0.48, size = 178, normalized size = 2.31 \begin {gather*} \left [\frac {6 \, B b^{2} x^{5} - 10 \, {\left (B a b - A b^{2}\right )} x^{3} - 15 \, {\left (B a^{2} - A 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 ) + 30 \, {\left (B a^{2} - A a b\right )} x}{30 \, b^{3}}, \frac {3 \, B b^{2} x^{5} - 5 \, {\left (B a b - A b^{2}\right )} x^{3} - 15 \, {\left (B a^{2} - A a b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 15 \, {\left (B a^{2} - A a b\right )} x}{15 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(6*B*b^2*x^5 - 10*(B*a*b - A*b^2)*x^3 - 15*(B*a^2 - A*a*b)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)
/(b*x^2 + a)) + 30*(B*a^2 - A*a*b)*x)/b^3, 1/15*(3*B*b^2*x^5 - 5*(B*a*b - A*b^2)*x^3 - 15*(B*a^2 - A*a*b)*sqrt
(a/b)*arctan(b*x*sqrt(a/b)/a) + 15*(B*a^2 - A*a*b)*x)/b^3]

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giac [A]  time = 0.33, size = 85, normalized size = 1.10 \begin {gather*} -\frac {{\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {3 \, B b^{4} x^{5} - 5 \, B a b^{3} x^{3} + 5 \, A b^{4} x^{3} + 15 \, B a^{2} b^{2} x - 15 \, A a b^{3} x}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*a^3 - A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/15*(3*B*b^4*x^5 - 5*B*a*b^3*x^3 + 5*A*b^4*x^3 + 1
5*B*a^2*b^2*x - 15*A*a*b^3*x)/b^5

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maple [A]  time = 0.00, size = 92, normalized size = 1.19 \begin {gather*} \frac {B \,x^{5}}{5 b}+\frac {A \,x^{3}}{3 b}-\frac {B a \,x^{3}}{3 b^{2}}+\frac {A \,a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}-\frac {B \,a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}-\frac {A a x}{b^{2}}+\frac {B \,a^{2} x}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.46, size = 78, normalized size = 1.01 \begin {gather*} -\frac {{\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {3 \, B b^{2} x^{5} - 5 \, {\left (B a b - A b^{2}\right )} x^{3} + 15 \, {\left (B a^{2} - A a b\right )} x}{15 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*a^3 - A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/15*(3*B*b^2*x^5 - 5*(B*a*b - A*b^2)*x^3 + 15*(B*a
^2 - A*a*b)*x)/b^3

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mupad [B]  time = 0.09, size = 96, normalized size = 1.25 \begin {gather*} x^3\,\left (\frac {A}{3\,b}-\frac {B\,a}{3\,b^2}\right )+\frac {B\,x^5}{5\,b}-\frac {a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,x\,\left (A\,b-B\,a\right )}{B\,a^3-A\,a^2\,b}\right )\,\left (A\,b-B\,a\right )}{b^{7/2}}-\frac {a\,x\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x^2))/(a + b*x^2),x)

[Out]

x^3*(A/(3*b) - (B*a)/(3*b^2)) + (B*x^5)/(5*b) - (a^(3/2)*atan((a^(3/2)*b^(1/2)*x*(A*b - B*a))/(B*a^3 - A*a^2*b
))*(A*b - B*a))/b^(7/2) - (a*x*(A/b - (B*a)/b^2))/b

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sympy [B]  time = 0.37, size = 153, normalized size = 1.99 \begin {gather*} \frac {B x^{5}}{5 b} + x^{3} \left (\frac {A}{3 b} - \frac {B a}{3 b^{2}}\right ) + x \left (- \frac {A a}{b^{2}} + \frac {B a^{2}}{b^{3}}\right ) + \frac {\sqrt {- \frac {a^{3}}{b^{7}}} \left (- A b + B a\right ) \log {\left (- \frac {b^{3} \sqrt {- \frac {a^{3}}{b^{7}}} \left (- A b + B a\right )}{- A a b + B a^{2}} + x \right )}}{2} - \frac {\sqrt {- \frac {a^{3}}{b^{7}}} \left (- A b + B a\right ) \log {\left (\frac {b^{3} \sqrt {- \frac {a^{3}}{b^{7}}} \left (- A b + B a\right )}{- A a b + B a^{2}} + x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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