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

Optimal. Leaf size=78 \[ \frac {c^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}}+\frac {c (b B-A c)}{b^3 x}-\frac {b B-A c}{3 b^2 x^3}-\frac {A}{5 b x^5} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1584, 453, 325, 205} \[ \frac {c^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}}-\frac {b B-A c}{3 b^2 x^3}+\frac {c (b B-A c)}{b^3 x}-\frac {A}{5 b x^5} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^4*(b*x^2 + c*x^4)),x]

[Out]

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

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 78, normalized size = 1.00 \[ \frac {c^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}}+\frac {c (b B-A c)}{b^3 x}+\frac {A c-b B}{3 b^2 x^3}-\frac {A}{5 b x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.13, size = 184, normalized size = 2.36 \[ \left [-\frac {15 \, {\left (B b c - A c^{2}\right )} x^{5} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) - 30 \, {\left (B b c - A c^{2}\right )} x^{4} + 6 \, A b^{2} + 10 \, {\left (B b^{2} - A b c\right )} x^{2}}{30 \, b^{3} x^{5}}, \frac {15 \, {\left (B b c - A c^{2}\right )} x^{5} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) + 15 \, {\left (B b c - A c^{2}\right )} x^{4} - 3 \, A b^{2} - 5 \, {\left (B b^{2} - A b c\right )} x^{2}}{15 \, b^{3} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 81, normalized size = 1.04 \[ \frac {{\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} + \frac {15 \, B b c x^{4} - 15 \, A c^{2} x^{4} - 5 \, B b^{2} x^{2} + 5 \, A b c x^{2} - 3 \, A b^{2}}{15 \, b^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 96, normalized size = 1.23 \[ -\frac {A \,c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{3}}+\frac {B \,c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{2}}-\frac {A \,c^{2}}{b^{3} x}+\frac {B c}{b^{2} x}+\frac {A c}{3 b^{2} x^{3}}-\frac {B}{3 b \,x^{3}}-\frac {A}{5 b \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.05, size = 79, normalized size = 1.01 \[ \frac {{\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} + \frac {15 \, {\left (B b c - A c^{2}\right )} x^{4} - 3 \, A b^{2} - 5 \, {\left (B b^{2} - A b c\right )} x^{2}}{15 \, b^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 70, normalized size = 0.90 \[ -\frac {\frac {A}{5\,b}-\frac {x^2\,\left (A\,c-B\,b\right )}{3\,b^2}+\frac {c\,x^4\,\left (A\,c-B\,b\right )}{b^3}}{x^5}-\frac {c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.58, size = 163, normalized size = 2.09 \[ - \frac {\sqrt {- \frac {c^{3}}{b^{7}}} \left (- A c + B b\right ) \log {\left (- \frac {b^{4} \sqrt {- \frac {c^{3}}{b^{7}}} \left (- A c + B b\right )}{- A c^{3} + B b c^{2}} + x \right )}}{2} + \frac {\sqrt {- \frac {c^{3}}{b^{7}}} \left (- A c + B b\right ) \log {\left (\frac {b^{4} \sqrt {- \frac {c^{3}}{b^{7}}} \left (- A c + B b\right )}{- A c^{3} + B b c^{2}} + x \right )}}{2} + \frac {- 3 A b^{2} + x^{4} \left (- 15 A c^{2} + 15 B b c\right ) + x^{2} \left (5 A b c - 5 B b^{2}\right )}{15 b^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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