3.1.66 \(\int \frac {x^4 (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [66]

Optimal. Leaf size=63 \[ -\frac {(b B-A c) x}{2 b c \left (b+c x^2\right )}+\frac {(b B+A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{3/2} c^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1598, 393, 211} \begin {gather*} \frac {(A c+b B) \text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{3/2} c^{3/2}}-\frac {x (b B-A c)}{2 b c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 393

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 1598

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

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Mathematica [A]
time = 0.03, size = 63, normalized size = 1.00 \begin {gather*} -\frac {(b B-A c) x}{2 b c \left (b+c x^2\right )}+\frac {(b B+A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{3/2} c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.42, size = 57, normalized size = 0.90

method	result	size
default	\(\frac {\left (A c -B b \right ) x}{2 b c \left (c \,x^{2}+b \right )}+\frac {\left (A c +B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 b c \sqrt {b c}}\)	\(57\)
risch	\(\frac {\left (A c -B b \right ) x}{2 b c \left (c \,x^{2}+b \right )}-\frac {\ln \left (c x +\sqrt {-b c}\right ) A}{4 \sqrt {-b c}\, b}-\frac {\ln \left (c x +\sqrt {-b c}\right ) B}{4 \sqrt {-b c}\, c}+\frac {\ln \left (-c x +\sqrt {-b c}\right ) A}{4 \sqrt {-b c}\, b}+\frac {\ln \left (-c x +\sqrt {-b c}\right ) B}{4 \sqrt {-b c}\, c}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 57, normalized size = 0.90 \begin {gather*} -\frac {{\left (B b - A c\right )} x}{2 \, {\left (b c^{2} x^{2} + b^{2} c\right )}} + \frac {{\left (B b + A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.89, size = 182, normalized size = 2.89 \begin {gather*} \left [-\frac {{\left (B b^{2} + A b c + {\left (B b c + A 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^{2} c - A b c^{2}\right )} x}{4 \, {\left (b^{2} c^{3} x^{2} + b^{3} c^{2}\right )}}, \frac {{\left (B b^{2} + A b c + {\left (B b c + A c^{2}\right )} x^{2}\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right ) - {\left (B b^{2} c - A b c^{2}\right )} x}{2 \, {\left (b^{2} c^{3} x^{2} + b^{3} c^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (54) = 108\).
time = 0.21, size = 112, normalized size = 1.78 \begin {gather*} \frac {x \left (A c - B b\right )}{2 b^{2} c + 2 b c^{2} x^{2}} - \frac {\sqrt {- \frac {1}{b^{3} c^{3}}} \left (A c + B b\right ) \log {\left (- b^{2} c \sqrt {- \frac {1}{b^{3} c^{3}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{b^{3} c^{3}}} \left (A c + B b\right ) \log {\left (b^{2} c \sqrt {- \frac {1}{b^{3} c^{3}}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.96, size = 57, normalized size = 0.90 \begin {gather*} \frac {{\left (B b + A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b c} - \frac {B b x - A c x}{2 \, {\left (c x^{2} + b\right )} b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.12, size = 51, normalized size = 0.81 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c+B\,b\right )}{2\,b^{3/2}\,c^{3/2}}+\frac {x\,\left (A\,c-B\,b\right )}{2\,b\,c\,\left (c\,x^2+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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