3.1.68 \(\int \frac {A+B x^2}{x^6 (a+b x^2)} \, dx\)

Optimal. Leaf size=80 \[ -\frac {b^{3/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2}}-\frac {b (A b-a B)}{a^3 x}+\frac {A b-a B}{3 a^2 x^3}-\frac {A}{5 a x^5} \]

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Rubi [A]  time = 0.05, antiderivative size = 80, 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 = {453, 325, 205} \begin {gather*} -\frac {b^{3/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2}}+\frac {A b-a B}{3 a^2 x^3}-\frac {b (A b-a B)}{a^3 x}-\frac {A}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^6*(a + b*x^2)),x]

[Out]

-A/(5*a*x^5) + (A*b - a*B)/(3*a^2*x^3) - (b*(A*b - a*B))/(a^3*x) - (b^(3/2)*(A*b - a*B)*ArcTan[(Sqrt[b]*x)/Sqr
t[a]])/a^(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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 453

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

Rubi steps

\begin {align*} \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )} \, dx &=-\frac {A}{5 a x^5}-\frac {(5 A b-5 a B) \int \frac {1}{x^4 \left (a+b x^2\right )} \, dx}{5 a}\\ &=-\frac {A}{5 a x^5}+\frac {A b-a B}{3 a^2 x^3}+\frac {(b (A b-a B)) \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx}{a^2}\\ &=-\frac {A}{5 a x^5}+\frac {A b-a B}{3 a^2 x^3}-\frac {b (A b-a B)}{a^3 x}-\frac {\left (b^2 (A b-a B)\right ) \int \frac {1}{a+b x^2} \, dx}{a^3}\\ &=-\frac {A}{5 a x^5}+\frac {A b-a B}{3 a^2 x^3}-\frac {b (A b-a B)}{a^3 x}-\frac {b^{3/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 78, normalized size = 0.98 \begin {gather*} \frac {b^{3/2} (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2}}+\frac {b (a B-A b)}{a^3 x}+\frac {A b-a B}{3 a^2 x^3}-\frac {A}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^6*(a + b*x^2)),x]

[Out]

-1/5*A/(a*x^5) + (A*b - a*B)/(3*a^2*x^3) + (b*(-(A*b) + a*B))/(a^3*x) + (b^(3/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b
]*x)/Sqrt[a]])/a^(7/2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x^2)/(x^6*(a + b*x^2)),x]

[Out]

IntegrateAlgebraic[(A + B*x^2)/(x^6*(a + b*x^2)), x]

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fricas [A]  time = 0.48, size = 184, normalized size = 2.30 \begin {gather*} \left [-\frac {15 \, {\left (B a b - A b^{2}\right )} x^{5} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 30 \, {\left (B a b - A b^{2}\right )} x^{4} + 6 \, A a^{2} + 10 \, {\left (B a^{2} - A a b\right )} x^{2}}{30 \, a^{3} x^{5}}, \frac {15 \, {\left (B a b - A b^{2}\right )} x^{5} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 15 \, {\left (B a b - A b^{2}\right )} x^{4} - 3 \, A a^{2} - 5 \, {\left (B a^{2} - A a b\right )} x^{2}}{15 \, a^{3} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a),x, algorithm="fricas")

[Out]

[-1/30*(15*(B*a*b - A*b^2)*x^5*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 30*(B*a*b - A*b^2)
*x^4 + 6*A*a^2 + 10*(B*a^2 - A*a*b)*x^2)/(a^3*x^5), 1/15*(15*(B*a*b - A*b^2)*x^5*sqrt(b/a)*arctan(x*sqrt(b/a))
+ 15*(B*a*b - A*b^2)*x^4 - 3*A*a^2 - 5*(B*a^2 - A*a*b)*x^2)/(a^3*x^5)]

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giac [A]  time = 0.30, size = 81, normalized size = 1.01 \begin {gather*} \frac {{\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {15 \, B a b x^{4} - 15 \, A b^{2} x^{4} - 5 \, B a^{2} x^{2} + 5 \, A a b x^{2} - 3 \, A a^{2}}{15 \, a^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 96, normalized size = 1.20 \begin {gather*} -\frac {A \,b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{3}}+\frac {B \,b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}-\frac {A \,b^{2}}{a^{3} x}+\frac {B b}{a^{2} x}+\frac {A b}{3 a^{2} x^{3}}-\frac {B}{3 a \,x^{3}}-\frac {A}{5 a \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^6/(b*x^2+a),x)

[Out]

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

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maxima [A]  time = 2.46, size = 79, normalized size = 0.99 \begin {gather*} \frac {{\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {15 \, {\left (B a b - A b^{2}\right )} x^{4} - 3 \, A a^{2} - 5 \, {\left (B a^{2} - A a b\right )} x^{2}}{15 \, a^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.10, size = 70, normalized size = 0.88 \begin {gather*} -\frac {\frac {A}{5\,a}-\frac {x^2\,\left (A\,b-B\,a\right )}{3\,a^2}+\frac {b\,x^4\,\left (A\,b-B\,a\right )}{a^3}}{x^5}-\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^6*(a + b*x^2)),x)

[Out]

- (A/(5*a) - (x^2*(A*b - B*a))/(3*a^2) + (b*x^4*(A*b - B*a))/a^3)/x^5 - (b^(3/2)*atan((b^(1/2)*x)/a^(1/2))*(A*
b - B*a))/a^(7/2)

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sympy [B]  time = 0.49, size = 163, normalized size = 2.04 \begin {gather*} - \frac {\sqrt {- \frac {b^{3}}{a^{7}}} \left (- A b + B a\right ) \log {\left (- \frac {a^{4} \sqrt {- \frac {b^{3}}{a^{7}}} \left (- A b + B a\right )}{- A b^{3} + B a b^{2}} + x \right )}}{2} + \frac {\sqrt {- \frac {b^{3}}{a^{7}}} \left (- A b + B a\right ) \log {\left (\frac {a^{4} \sqrt {- \frac {b^{3}}{a^{7}}} \left (- A b + B a\right )}{- A b^{3} + B a b^{2}} + x \right )}}{2} + \frac {- 3 A a^{2} + x^{4} \left (- 15 A b^{2} + 15 B a b\right ) + x^{2} \left (5 A a b - 5 B a^{2}\right )}{15 a^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**6/(b*x**2+a),x)

[Out]

-sqrt(-b**3/a**7)*(-A*b + B*a)*log(-a**4*sqrt(-b**3/a**7)*(-A*b + B*a)/(-A*b**3 + B*a*b**2) + x)/2 + sqrt(-b**
3/a**7)*(-A*b + B*a)*log(a**4*sqrt(-b**3/a**7)*(-A*b + B*a)/(-A*b**3 + B*a*b**2) + x)/2 + (-3*A*a**2 + x**4*(-
15*A*b**2 + 15*B*a*b) + x**2*(5*A*a*b - 5*B*a**2))/(15*a**3*x**5)

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