3.36 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^{13}} \, dx\)

Optimal. Leaf size=71 \[ -\frac{b^2 (3 A c+b B)}{4 x^4}-\frac{A b^3}{6 x^6}+c^2 \log (x) (A c+3 b B)-\frac{3 b c (A c+b B)}{2 x^2}+\frac{1}{2} B c^3 x^2 \]

[Out]

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

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Rubi [A]  time = 0.0640838, antiderivative size = 71, normalized size of antiderivative = 1., 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 = {1584, 446, 76} \[ -\frac{b^2 (3 A c+b B)}{4 x^4}-\frac{A b^3}{6 x^6}+c^2 \log (x) (A c+3 b B)-\frac{3 b c (A c+b B)}{2 x^2}+\frac{1}{2} B c^3 x^2 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^13,x]

[Out]

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

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]

Rule 446

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 76

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])

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^7} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) (b+c x)^3}{x^4} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (B c^3+\frac{A b^3}{x^4}+\frac{b^2 (b B+3 A c)}{x^3}+\frac{3 b c (b B+A c)}{x^2}+\frac{c^2 (3 b B+A c)}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac{A b^3}{6 x^6}-\frac{b^2 (b B+3 A c)}{4 x^4}-\frac{3 b c (b B+A c)}{2 x^2}+\frac{1}{2} B c^3 x^2+c^2 (3 b B+A c) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.037125, size = 71, normalized size = 1. \[ -\frac{b^2 (3 A c+b B)}{4 x^4}-\frac{A b^3}{6 x^6}+c^2 \log (x) (A c+3 b B)-\frac{3 b c (A c+b B)}{2 x^2}+\frac{1}{2} B c^3 x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^13,x]

[Out]

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

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Maple [A]  time = 0.006, size = 75, normalized size = 1.1 \begin{align*}{\frac{B{c}^{3}{x}^{2}}{2}}+A\ln \left ( x \right ){c}^{3}+3\,B\ln \left ( x \right ) b{c}^{2}-{\frac{3\,A{b}^{2}c}{4\,{x}^{4}}}-{\frac{B{b}^{3}}{4\,{x}^{4}}}-{\frac{3\,Ab{c}^{2}}{2\,{x}^{2}}}-{\frac{3\,B{b}^{2}c}{2\,{x}^{2}}}-{\frac{A{b}^{3}}{6\,{x}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x)

[Out]

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

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Maxima [A]  time = 1.05915, size = 104, normalized size = 1.46 \begin{align*} \frac{1}{2} \, B c^{3} x^{2} + \frac{1}{2} \,{\left (3 \, B b c^{2} + A c^{3}\right )} \log \left (x^{2}\right ) - \frac{18 \,{\left (B b^{2} c + A b c^{2}\right )} x^{4} + 2 \, A b^{3} + 3 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{12 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x, algorithm="maxima")

[Out]

1/2*B*c^3*x^2 + 1/2*(3*B*b*c^2 + A*c^3)*log(x^2) - 1/12*(18*(B*b^2*c + A*b*c^2)*x^4 + 2*A*b^3 + 3*(B*b^3 + 3*A
*b^2*c)*x^2)/x^6

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Fricas [A]  time = 0.469042, size = 171, normalized size = 2.41 \begin{align*} \frac{6 \, B c^{3} x^{8} + 12 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} \log \left (x\right ) - 18 \,{\left (B b^{2} c + A b c^{2}\right )} x^{4} - 2 \, A b^{3} - 3 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{12 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x, algorithm="fricas")

[Out]

1/12*(6*B*c^3*x^8 + 12*(3*B*b*c^2 + A*c^3)*x^6*log(x) - 18*(B*b^2*c + A*b*c^2)*x^4 - 2*A*b^3 - 3*(B*b^3 + 3*A*
b^2*c)*x^2)/x^6

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Sympy [A]  time = 1.41572, size = 75, normalized size = 1.06 \begin{align*} \frac{B c^{3} x^{2}}{2} + c^{2} \left (A c + 3 B b\right ) \log{\left (x \right )} - \frac{2 A b^{3} + x^{4} \left (18 A b c^{2} + 18 B b^{2} c\right ) + x^{2} \left (9 A b^{2} c + 3 B b^{3}\right )}{12 x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**13,x)

[Out]

B*c**3*x**2/2 + c**2*(A*c + 3*B*b)*log(x) - (2*A*b**3 + x**4*(18*A*b*c**2 + 18*B*b**2*c) + x**2*(9*A*b**2*c +
3*B*b**3))/(12*x**6)

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Giac [A]  time = 1.18589, size = 134, normalized size = 1.89 \begin{align*} \frac{1}{2} \, B c^{3} x^{2} + \frac{1}{2} \,{\left (3 \, B b c^{2} + A c^{3}\right )} \log \left (x^{2}\right ) - \frac{33 \, B b c^{2} x^{6} + 11 \, A c^{3} x^{6} + 18 \, B b^{2} c x^{4} + 18 \, A b c^{2} x^{4} + 3 \, B b^{3} x^{2} + 9 \, A b^{2} c x^{2} + 2 \, A b^{3}}{12 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x, algorithm="giac")

[Out]

1/2*B*c^3*x^2 + 1/2*(3*B*b*c^2 + A*c^3)*log(x^2) - 1/12*(33*B*b*c^2*x^6 + 11*A*c^3*x^6 + 18*B*b^2*c*x^4 + 18*A
*b*c^2*x^4 + 3*B*b^3*x^2 + 9*A*b^2*c*x^2 + 2*A*b^3)/x^6