3.1.23 \(\int \frac {(A+B x^2) (b x^2+c x^4)^2}{x^{11}} \, dx\) [23]

Optimal. Leaf size=51 \[ -\frac {A b^2}{6 x^6}-\frac {b (b B+2 A c)}{4 x^4}-\frac {c (2 b B+A c)}{2 x^2}+B c^2 \log (x) \]

[Out]

-1/6*A*b^2/x^6-1/4*b*(2*A*c+B*b)/x^4-1/2*c*(A*c+2*B*b)/x^2+B*c^2*ln(x)

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Rubi [A]
time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1598, 457, 77} \begin {gather*} -\frac {A b^2}{6 x^6}-\frac {b (2 A c+b B)}{4 x^4}-\frac {c (A c+2 b B)}{2 x^2}+B c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^11,x]

[Out]

-1/6*(A*b^2)/x^6 - (b*(b*B + 2*A*c))/(4*x^4) - (c*(2*b*B + A*c))/(2*x^2) + B*c^2*Log[x]

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 457

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

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 {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^2}{x^{11}} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^2}{x^7} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {(A+B x) (b+c x)^2}{x^4} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {A b^2}{x^4}+\frac {b (b B+2 A c)}{x^3}+\frac {c (2 b B+A c)}{x^2}+\frac {B c^2}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac {A b^2}{6 x^6}-\frac {b (b B+2 A c)}{4 x^4}-\frac {c (2 b B+A c)}{2 x^2}+B c^2 \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 53, normalized size = 1.04 \begin {gather*} -\frac {3 b B x^2 \left (b+4 c x^2\right )+2 A \left (b^2+3 b c x^2+3 c^2 x^4\right )}{12 x^6}+B c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^11,x]

[Out]

-1/12*(3*b*B*x^2*(b + 4*c*x^2) + 2*A*(b^2 + 3*b*c*x^2 + 3*c^2*x^4))/x^6 + B*c^2*Log[x]

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Maple [A]
time = 0.36, size = 46, normalized size = 0.90

method	result	size
default	\(-\frac {A \,b^{2}}{6 x^{6}}-\frac {b \left (2 A c +B b \right )}{4 x^{4}}-\frac {c \left (A c +2 B b \right )}{2 x^{2}}+B \,c^{2} \ln \left (x \right )\)	\(46\)
risch	\(\frac {\left (-\frac {1}{2} A \,c^{2}-b B c \right ) x^{4}+\left (-\frac {1}{2} A b c -\frac {1}{4} b^{2} B \right ) x^{2}-\frac {b^{2} A}{6}}{x^{6}}+B \,c^{2} \ln \left (x \right )\)	\(52\)
norman	\(\frac {\left (-\frac {1}{2} A \,c^{2}-b B c \right ) x^{8}+\left (-\frac {1}{2} A b c -\frac {1}{4} b^{2} B \right ) x^{6}-\frac {A \,b^{2} x^{4}}{6}}{x^{10}}+B \,c^{2} \ln \left (x \right )\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^2/x^11,x,method=_RETURNVERBOSE)

[Out]

-1/6*A*b^2/x^6-1/4*b*(2*A*c+B*b)/x^4-1/2*c*(A*c+2*B*b)/x^2+B*c^2*ln(x)

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Maxima [A]
time = 0.28, size = 55, normalized size = 1.08 \begin {gather*} \frac {1}{2} \, B c^{2} \log \left (x^{2}\right ) - \frac {6 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 2 \, A b^{2} + 3 \, {\left (B b^{2} + 2 \, A b c\right )} x^{2}}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2/x^11,x, algorithm="maxima")

[Out]

1/2*B*c^2*log(x^2) - 1/12*(6*(2*B*b*c + A*c^2)*x^4 + 2*A*b^2 + 3*(B*b^2 + 2*A*b*c)*x^2)/x^6

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Fricas [A]
time = 1.64, size = 55, normalized size = 1.08 \begin {gather*} \frac {12 \, B c^{2} x^{6} \log \left (x\right ) - 6 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} - 2 \, A b^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} x^{2}}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2/x^11,x, algorithm="fricas")

[Out]

1/12*(12*B*c^2*x^6*log(x) - 6*(2*B*b*c + A*c^2)*x^4 - 2*A*b^2 - 3*(B*b^2 + 2*A*b*c)*x^2)/x^6

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Sympy [A]
time = 0.53, size = 56, normalized size = 1.10 \begin {gather*} B c^{2} \log {\left (x \right )} + \frac {- 2 A b^{2} + x^{4} \left (- 6 A c^{2} - 12 B b c\right ) + x^{2} \left (- 6 A b c - 3 B b^{2}\right )}{12 x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**2/x**11,x)

[Out]

B*c**2*log(x) + (-2*A*b**2 + x**4*(-6*A*c**2 - 12*B*b*c) + x**2*(-6*A*b*c - 3*B*b**2))/(12*x**6)

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Giac [A]
time = 1.79, size = 66, normalized size = 1.29 \begin {gather*} \frac {1}{2} \, B c^{2} \log \left (x^{2}\right ) - \frac {11 \, B c^{2} x^{6} + 12 \, B b c x^{4} + 6 \, A c^{2} x^{4} + 3 \, B b^{2} x^{2} + 6 \, A b c x^{2} + 2 \, A b^{2}}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2/x^11,x, algorithm="giac")

[Out]

1/2*B*c^2*log(x^2) - 1/12*(11*B*c^2*x^6 + 12*B*b*c*x^4 + 6*A*c^2*x^4 + 3*B*b^2*x^2 + 6*A*b*c*x^2 + 2*A*b^2)/x^
6

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Mupad [B]
time = 0.09, size = 51, normalized size = 1.00 \begin {gather*} B\,c^2\,\ln \left (x\right )-\frac {x^2\,\left (\frac {B\,b^2}{4}+\frac {A\,c\,b}{2}\right )+x^4\,\left (\frac {A\,c^2}{2}+B\,b\,c\right )+\frac {A\,b^2}{6}}{x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^11,x)

[Out]

B*c^2*log(x) - (x^2*((B*b^2)/4 + (A*b*c)/2) + x^4*((A*c^2)/2 + B*b*c) + (A*b^2)/6)/x^6

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