3.1.64 \(\int \frac {x^6 (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [64]

Optimal. Leaf size=68 \[ \frac {B x}{c^2}+\frac {(b B-A c) x}{2 c^2 \left (b+c x^2\right )}-\frac {(3 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 \sqrt {b} c^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 68, 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 = {1598, 466, 396, 211} \begin {gather*} -\frac {(3 b B-A c) \text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 \sqrt {b} c^{5/2}}+\frac {x (b B-A c)}{2 c^2 \left (b+c x^2\right )}+\frac {B x}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x)/c^2 + ((b*B - A*c)*x)/(2*c^2*(b + c*x^2)) - ((3*b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*Sqrt[b]*c^(5/
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 396

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

Rule 466

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[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 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^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^2 \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=\frac {(b B-A c) x}{2 c^2 \left (b+c x^2\right )}-\frac {\int \frac {b B-A c-2 B c x^2}{b+c x^2} \, dx}{2 c^2}\\ &=\frac {B x}{c^2}+\frac {(b B-A c) x}{2 c^2 \left (b+c x^2\right )}-\frac {(3 b B-A c) \int \frac {1}{b+c x^2} \, dx}{2 c^2}\\ &=\frac {B x}{c^2}+\frac {(b B-A c) x}{2 c^2 \left (b+c x^2\right )}-\frac {(3 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 \sqrt {b} c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 68, normalized size = 1.00 \begin {gather*} \frac {B x}{c^2}-\frac {(-b B+A c) x}{2 c^2 \left (b+c x^2\right )}-\frac {(3 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 \sqrt {b} c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.39, size = 57, normalized size = 0.84

method	result	size
default	\(\frac {B x}{c^{2}}+\frac {\frac {\left (-\frac {A c}{2}+\frac {B b}{2}\right ) x}{c \,x^{2}+b}+\frac {\left (A c -3 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}}{c^{2}}\)	\(57\)
risch	\(\frac {B x}{c^{2}}+\frac {\left (-\frac {A c}{2}+\frac {B b}{2}\right ) x}{c^{2} \left (c \,x^{2}+b \right )}-\frac {\ln \left (c x +\sqrt {-b c}\right ) A}{4 c \sqrt {-b c}}+\frac {3 \ln \left (c x +\sqrt {-b c}\right ) B b}{4 c^{2} \sqrt {-b c}}+\frac {\ln \left (-c x +\sqrt {-b c}\right ) A}{4 c \sqrt {-b c}}-\frac {3 \ln \left (-c x +\sqrt {-b c}\right ) B b}{4 c^{2} \sqrt {-b c}}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 61, normalized size = 0.90 \begin {gather*} \frac {{\left (B b - A c\right )} x}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} + \frac {B x}{c^{2}} - \frac {{\left (3 \, B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.73, size = 208, normalized size = 3.06 \begin {gather*} \left [\frac {4 \, B b c^{2} x^{3} + {\left (3 \, B b^{2} - A b c + {\left (3 \, 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 (3 \, B b^{2} c - A b c^{2}\right )} x}{4 \, {\left (b c^{4} x^{2} + b^{2} c^{3}\right )}}, \frac {2 \, B b c^{2} x^{3} - {\left (3 \, B b^{2} - A b c + {\left (3 \, B b c - A c^{2}\right )} x^{2}\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right ) + {\left (3 \, B b^{2} c - A b c^{2}\right )} x}{2 \, {\left (b c^{4} x^{2} + b^{2} c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.28, size = 114, normalized size = 1.68 \begin {gather*} \frac {B x}{c^{2}} + \frac {x \left (- A c + B b\right )}{2 b c^{2} + 2 c^{3} x^{2}} + \frac {\sqrt {- \frac {1}{b c^{5}}} \left (- A c + 3 B b\right ) \log {\left (- b c^{2} \sqrt {- \frac {1}{b c^{5}}} + x \right )}}{4} - \frac {\sqrt {- \frac {1}{b c^{5}}} \left (- A c + 3 B b\right ) \log {\left (b c^{2} \sqrt {- \frac {1}{b c^{5}}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.47, size = 59, normalized size = 0.87 \begin {gather*} \frac {B x}{c^{2}} - \frac {{\left (3 \, B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{2}} + \frac {B b x - A c x}{2 \, {\left (c x^{2} + b\right )} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.13, size = 59, normalized size = 0.87 \begin {gather*} \frac {B\,x}{c^2}-\frac {x\,\left (\frac {A\,c}{2}-\frac {B\,b}{2}\right )}{c^3\,x^2+b\,c^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c-3\,B\,b\right )}{2\,\sqrt {b}\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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