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

Optimal. Leaf size=96 \[ \frac {3 (b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2} \sqrt {c}}+\frac {x (3 b B-7 A c)}{8 b^3 \left (b+c x^2\right )}-\frac {A}{b^3 x}+\frac {x (b B-A c)}{4 b^2 \left (b+c x^2\right )^2} \]

[Out]

-A/b^3/x+1/4*(-A*c+B*b)*x/b^2/(c*x^2+b)^2+1/8*(-7*A*c+3*B*b)*x/b^3/(c*x^2+b)+3/8*(-5*A*c+B*b)*arctan(x*c^(1/2)
/b^(1/2))/b^(7/2)/c^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 96, 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, 456, 453, 205} \[ \frac {x (3 b B-7 A c)}{8 b^3 \left (b+c x^2\right )}+\frac {x (b B-A c)}{4 b^2 \left (b+c x^2\right )^2}+\frac {3 (b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2} \sqrt {c}}-\frac {A}{b^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(b^3*x)) + ((b*B - A*c)*x)/(4*b^2*(b + c*x^2)^2) + ((3*b*B - 7*A*c)*x)/(8*b^3*(b + c*x^2)) + (3*(b*B - 5*A
*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(7/2)*Sqrt[c])

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 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 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 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^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {A+B x^2}{x^2 \left (b+c x^2\right )^3} \, dx\\ &=\frac {(b B-A c) x}{4 b^2 \left (b+c x^2\right )^2}-\frac {1}{4} \int \frac {-\frac {4 A}{b}-\frac {3 (b B-A c) x^2}{b^2}}{x^2 \left (b+c x^2\right )^2} \, dx\\ &=\frac {(b B-A c) x}{4 b^2 \left (b+c x^2\right )^2}+\frac {(3 b B-7 A c) x}{8 b^3 \left (b+c x^2\right )}+\frac {1}{8} \int \frac {\frac {8 A}{b^2}+\frac {(3 b B-7 A c) x^2}{b^3}}{x^2 \left (b+c x^2\right )} \, dx\\ &=-\frac {A}{b^3 x}+\frac {(b B-A c) x}{4 b^2 \left (b+c x^2\right )^2}+\frac {(3 b B-7 A c) x}{8 b^3 \left (b+c x^2\right )}+\frac {(3 (b B-5 A c)) \int \frac {1}{b+c x^2} \, dx}{8 b^3}\\ &=-\frac {A}{b^3 x}+\frac {(b B-A c) x}{4 b^2 \left (b+c x^2\right )^2}+\frac {(3 b B-7 A c) x}{8 b^3 \left (b+c x^2\right )}+\frac {3 (b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2} \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 96, normalized size = 1.00 \[ \frac {3 (b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2} \sqrt {c}}+\frac {x (3 b B-7 A c)}{8 b^3 \left (b+c x^2\right )}-\frac {A}{b^3 x}+\frac {x (b B-A c)}{4 b^2 \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(b^3*x)) + ((b*B - A*c)*x)/(4*b^2*(b + c*x^2)^2) + ((3*b*B - 7*A*c)*x)/(8*b^3*(b + c*x^2)) + (3*(b*B - 5*A
*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(7/2)*Sqrt[c])

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fricas [A]  time = 1.08, size = 324, normalized size = 3.38 \[ \left [-\frac {16 \, A b^{3} c - 6 \, {\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} - 10 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2} - 3 \, {\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{5} + 2 \, {\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{3} + {\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right )}{16 \, {\left (b^{4} c^{3} x^{5} + 2 \, b^{5} c^{2} x^{3} + b^{6} c x\right )}}, -\frac {8 \, A b^{3} c - 3 \, {\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} - 5 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2} - 3 \, {\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{5} + 2 \, {\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{3} + {\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right )}{8 \, {\left (b^{4} c^{3} x^{5} + 2 \, b^{5} c^{2} x^{3} + b^{6} c x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(16*A*b^3*c - 6*(B*b^2*c^2 - 5*A*b*c^3)*x^4 - 10*(B*b^3*c - 5*A*b^2*c^2)*x^2 - 3*((B*b*c^2 - 5*A*c^3)*x
^5 + 2*(B*b^2*c - 5*A*b*c^2)*x^3 + (B*b^3 - 5*A*b^2*c)*x)*sqrt(-b*c)*log((c*x^2 + 2*sqrt(-b*c)*x - b)/(c*x^2 +
b)))/(b^4*c^3*x^5 + 2*b^5*c^2*x^3 + b^6*c*x), -1/8*(8*A*b^3*c - 3*(B*b^2*c^2 - 5*A*b*c^3)*x^4 - 5*(B*b^3*c -
5*A*b^2*c^2)*x^2 - 3*((B*b*c^2 - 5*A*c^3)*x^5 + 2*(B*b^2*c - 5*A*b*c^2)*x^3 + (B*b^3 - 5*A*b^2*c)*x)*sqrt(b*c)
*arctan(sqrt(b*c)*x/b))/(b^4*c^3*x^5 + 2*b^5*c^2*x^3 + b^6*c*x)]

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giac [A]  time = 0.17, size = 82, normalized size = 0.85 \[ \frac {3 \, {\left (B b - 5 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{3}} - \frac {A}{b^{3} x} + \frac {3 \, B b c x^{3} - 7 \, A c^{2} x^{3} + 5 \, B b^{2} x - 9 \, A b c x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(B*b - 5*A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3) - A/(b^3*x) + 1/8*(3*B*b*c*x^3 - 7*A*c^2*x^3 + 5*B*b^2
*x - 9*A*b*c*x)/((c*x^2 + b)^2*b^3)

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maple [A]  time = 0.06, size = 125, normalized size = 1.30 \[ -\frac {7 A \,c^{2} x^{3}}{8 \left (c \,x^{2}+b \right )^{2} b^{3}}+\frac {3 B c \,x^{3}}{8 \left (c \,x^{2}+b \right )^{2} b^{2}}-\frac {9 A c x}{8 \left (c \,x^{2}+b \right )^{2} b^{2}}+\frac {5 B x}{8 \left (c \,x^{2}+b \right )^{2} b}-\frac {15 A c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b^{3}}+\frac {3 B \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b^{2}}-\frac {A}{b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8/b^3/(c*x^2+b)^2*A*x^3*c^2+3/8/b^2/(c*x^2+b)^2*B*x^3*c-9/8/b^2/(c*x^2+b)^2*A*c*x+5/8/b/(c*x^2+b)^2*B*x-15/
8/b^3/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*A*c+3/8/b^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*B-A/b^3/x

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maxima [A]  time = 2.97, size = 96, normalized size = 1.00 \[ \frac {3 \, {\left (B b c - 5 \, A c^{2}\right )} x^{4} - 8 \, A b^{2} + 5 \, {\left (B b^{2} - 5 \, A b c\right )} x^{2}}{8 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}} + \frac {3 \, {\left (B b - 5 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.18, size = 113, normalized size = 1.18 \[ -\frac {\frac {A}{b}+\frac {5\,x^2\,\left (5\,A\,c-B\,b\right )}{8\,b^2}+\frac {3\,c\,x^4\,\left (5\,A\,c-B\,b\right )}{8\,b^3}}{b^2\,x+2\,b\,c\,x^3+c^2\,x^5}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {c}\,x\,\left (5\,A\,c-B\,b\right )}{\sqrt {b}\,\left (15\,A\,c-3\,B\,b\right )}\right )\,\left (5\,A\,c-B\,b\right )}{8\,b^{7/2}\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (A/b + (5*x^2*(5*A*c - B*b))/(8*b^2) + (3*c*x^4*(5*A*c - B*b))/(8*b^3))/(b^2*x + c^2*x^5 + 2*b*c*x^3) - (3*a
tan((3*c^(1/2)*x*(5*A*c - B*b))/(b^(1/2)*(15*A*c - 3*B*b)))*(5*A*c - B*b))/(8*b^(7/2)*c^(1/2))

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sympy [B]  time = 0.73, size = 194, normalized size = 2.02 \[ - \frac {3 \sqrt {- \frac {1}{b^{7} c}} \left (- 5 A c + B b\right ) \log {\left (- \frac {3 b^{4} \sqrt {- \frac {1}{b^{7} c}} \left (- 5 A c + B b\right )}{- 15 A c + 3 B b} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{b^{7} c}} \left (- 5 A c + B b\right ) \log {\left (\frac {3 b^{4} \sqrt {- \frac {1}{b^{7} c}} \left (- 5 A c + B b\right )}{- 15 A c + 3 B b} + x \right )}}{16} + \frac {- 8 A b^{2} + x^{4} \left (- 15 A c^{2} + 3 B b c\right ) + x^{2} \left (- 25 A b c + 5 B b^{2}\right )}{8 b^{5} x + 16 b^{4} c x^{3} + 8 b^{3} c^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(-1/(b**7*c))*(-5*A*c + B*b)*log(-3*b**4*sqrt(-1/(b**7*c))*(-5*A*c + B*b)/(-15*A*c + 3*B*b) + x)/16 + 3
*sqrt(-1/(b**7*c))*(-5*A*c + B*b)*log(3*b**4*sqrt(-1/(b**7*c))*(-5*A*c + B*b)/(-15*A*c + 3*B*b) + x)/16 + (-8*
A*b**2 + x**4*(-15*A*c**2 + 3*B*b*c) + x**2*(-25*A*b*c + 5*B*b**2))/(8*b**5*x + 16*b**4*c*x**3 + 8*b**3*c**2*x
**5)

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