3.81 \(\int \frac {x^6 (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=92 \[ \frac {(3 A c+b B) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{5/2} c^{3/2}}+\frac {x (3 A c+b B)}{8 b^2 c \left (b+c x^2\right )}-\frac {x (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

[Out]

-1/4*(-A*c+B*b)*x/b/c/(c*x^2+b)^2+1/8*(3*A*c+B*b)*x/b^2/c/(c*x^2+b)+1/8*(3*A*c+B*b)*arctan(x*c^(1/2)/b^(1/2))/
b^(5/2)/c^(3/2)

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Rubi [A]  time = 0.05, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1584, 385, 199, 205} \[ \frac {(3 A c+b B) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{5/2} c^{3/2}}+\frac {x (3 A c+b B)}{8 b^2 c \left (b+c x^2\right )}-\frac {x (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

-((b*B - A*c)*x)/(4*b*c*(b + c*x^2)^2) + ((b*B + 3*A*c)*x)/(8*b^2*c*(b + c*x^2)) + ((b*B + 3*A*c)*ArcTan[(Sqrt
[c]*x)/Sqrt[b]])/(8*b^(5/2)*c^(3/2))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 385

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

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 {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {A+B x^2}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac {(b B-A c) x}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+3 A c) \int \frac {1}{\left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {(b B-A c) x}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+3 A c) x}{8 b^2 c \left (b+c x^2\right )}+\frac {(b B+3 A c) \int \frac {1}{b+c x^2} \, dx}{8 b^2 c}\\ &=-\frac {(b B-A c) x}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+3 A c) x}{8 b^2 c \left (b+c x^2\right )}+\frac {(b B+3 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{5/2} c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 84, normalized size = 0.91 \[ \frac {(3 A c+b B) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{5/2} c^{3/2}}+\frac {x \left (b c \left (5 A+B x^2\right )+3 A c^2 x^2+b^2 (-B)\right )}{8 b^2 c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(x*(-(b^2*B) + 3*A*c^2*x^2 + b*c*(5*A + B*x^2)))/(8*b^2*c*(b + c*x^2)^2) + ((b*B + 3*A*c)*ArcTan[(Sqrt[c]*x)/S
qrt[b]])/(8*b^(5/2)*c^(3/2))

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fricas [A]  time = 1.06, size = 300, normalized size = 3.26 \[ \left [\frac {2 \, {\left (B b^{2} c^{2} + 3 \, A b c^{3}\right )} x^{3} - {\left ({\left (B b c^{2} + 3 \, A c^{3}\right )} x^{4} + B b^{3} + 3 \, A b^{2} c + 2 \, {\left (B b^{2} c + 3 \, A b c^{2}\right )} x^{2}\right )} \sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right ) - 2 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x}{16 \, {\left (b^{3} c^{4} x^{4} + 2 \, b^{4} c^{3} x^{2} + b^{5} c^{2}\right )}}, \frac {{\left (B b^{2} c^{2} + 3 \, A b c^{3}\right )} x^{3} + {\left ({\left (B b c^{2} + 3 \, A c^{3}\right )} x^{4} + B b^{3} + 3 \, A b^{2} c + 2 \, {\left (B b^{2} c + 3 \, A b c^{2}\right )} x^{2}\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right ) - {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x}{8 \, {\left (b^{3} c^{4} x^{4} + 2 \, b^{4} c^{3} x^{2} + b^{5} c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

[1/16*(2*(B*b^2*c^2 + 3*A*b*c^3)*x^3 - ((B*b*c^2 + 3*A*c^3)*x^4 + B*b^3 + 3*A*b^2*c + 2*(B*b^2*c + 3*A*b*c^2)*
x^2)*sqrt(-b*c)*log((c*x^2 - 2*sqrt(-b*c)*x - b)/(c*x^2 + b)) - 2*(B*b^3*c - 5*A*b^2*c^2)*x)/(b^3*c^4*x^4 + 2*
b^4*c^3*x^2 + b^5*c^2), 1/8*((B*b^2*c^2 + 3*A*b*c^3)*x^3 + ((B*b*c^2 + 3*A*c^3)*x^4 + B*b^3 + 3*A*b^2*c + 2*(B
*b^2*c + 3*A*b*c^2)*x^2)*sqrt(b*c)*arctan(sqrt(b*c)*x/b) - (B*b^3*c - 5*A*b^2*c^2)*x)/(b^3*c^4*x^4 + 2*b^4*c^3
*x^2 + b^5*c^2)]

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giac [A]  time = 0.16, size = 78, normalized size = 0.85 \[ \frac {{\left (B b + 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{2} c} + \frac {B b c x^{3} + 3 \, A c^{2} x^{3} - B b^{2} x + 5 \, A b c x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

1/8*(B*b + 3*A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2*c) + 1/8*(B*b*c*x^3 + 3*A*c^2*x^3 - B*b^2*x + 5*A*b*c*x
)/((c*x^2 + b)^2*b^2*c)

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maple [A]  time = 0.06, size = 90, normalized size = 0.98 \[ \frac {3 A \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b^{2}}+\frac {B \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b c}+\frac {\frac {\left (3 A c +b B \right ) x^{3}}{8 b^{2}}+\frac {\left (5 A c -b B \right ) x}{8 b c}}{\left (c \,x^{2}+b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

(1/8*(3*A*c+B*b)/b^2*x^3+1/8*(5*A*c-B*b)/b/c*x)/(c*x^2+b)^2+3/8/b^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*A+1/
8/b/c/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*B

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maxima [A]  time = 2.95, size = 92, normalized size = 1.00 \[ \frac {{\left (B b c + 3 \, A c^{2}\right )} x^{3} - {\left (B b^{2} - 5 \, A b c\right )} x}{8 \, {\left (b^{2} c^{3} x^{4} + 2 \, b^{3} c^{2} x^{2} + b^{4} c\right )}} + \frac {{\left (B b + 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/8*((B*b*c + 3*A*c^2)*x^3 - (B*b^2 - 5*A*b*c)*x)/(b^2*c^3*x^4 + 2*b^3*c^2*x^2 + b^4*c) + 1/8*(B*b + 3*A*c)*ar
ctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2*c)

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mupad [B]  time = 0.14, size = 82, normalized size = 0.89 \[ \frac {\frac {x^3\,\left (3\,A\,c+B\,b\right )}{8\,b^2}+\frac {x\,\left (5\,A\,c-B\,b\right )}{8\,b\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (3\,A\,c+B\,b\right )}{8\,b^{5/2}\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

((x^3*(3*A*c + B*b))/(8*b^2) + (x*(5*A*c - B*b))/(8*b*c))/(b^2 + c^2*x^4 + 2*b*c*x^2) + (atan((c^(1/2)*x)/b^(1
/2))*(3*A*c + B*b))/(8*b^(5/2)*c^(3/2))

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sympy [A]  time = 0.60, size = 150, normalized size = 1.63 \[ - \frac {\sqrt {- \frac {1}{b^{5} c^{3}}} \left (3 A c + B b\right ) \log {\left (- b^{3} c \sqrt {- \frac {1}{b^{5} c^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{b^{5} c^{3}}} \left (3 A c + B b\right ) \log {\left (b^{3} c \sqrt {- \frac {1}{b^{5} c^{3}}} + x \right )}}{16} + \frac {x^{3} \left (3 A c^{2} + B b c\right ) + x \left (5 A b c - B b^{2}\right )}{8 b^{4} c + 16 b^{3} c^{2} x^{2} + 8 b^{2} c^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

-sqrt(-1/(b**5*c**3))*(3*A*c + B*b)*log(-b**3*c*sqrt(-1/(b**5*c**3)) + x)/16 + sqrt(-1/(b**5*c**3))*(3*A*c + B
*b)*log(b**3*c*sqrt(-1/(b**5*c**3)) + x)/16 + (x**3*(3*A*c**2 + B*b*c) + x*(5*A*b*c - B*b**2))/(8*b**4*c + 16*
b**3*c**2*x**2 + 8*b**2*c**3*x**4)

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