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

Optimal. Leaf size=162 \[ a^2 \log (x) (a B+3 A b)-\frac{a^3 A}{2 x^2}+\frac{1}{6} x^6 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{3}{8} c x^8 \left (a B c+A b c+b^2 B\right )+\frac{1}{4} x^4 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac{3}{2} a x^2 \left (A \left (a c+b^2\right )+a b B\right )+\frac{1}{10} c^2 x^{10} (A c+3 b B)+\frac{1}{12} B c^3 x^{12} \]

[Out]

-(a^3*A)/(2*x^2) + (3*a*(a*b*B + A*(b^2 + a*c))*x^2)/2 + ((3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^4)/4 + ((b
^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^6)/6 + (3*c*(b^2*B + A*b*c + a*B*c)*x^8)/8 + (c^2*(3*b*B + A*c)*x^
10)/10 + (B*c^3*x^12)/12 + a^2*(3*A*b + a*B)*Log[x]

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Rubi [A]  time = 0.22608, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {1251, 765} \[ a^2 \log (x) (a B+3 A b)-\frac{a^3 A}{2 x^2}+\frac{1}{6} x^6 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{3}{8} c x^8 \left (a B c+A b c+b^2 B\right )+\frac{1}{4} x^4 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac{3}{2} a x^2 \left (A \left (a c+b^2\right )+a b B\right )+\frac{1}{10} c^2 x^{10} (A c+3 b B)+\frac{1}{12} B c^3 x^{12} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*A)/(2*x^2) + (3*a*(a*b*B + A*(b^2 + a*c))*x^2)/2 + ((3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^4)/4 + ((b
^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^6)/6 + (3*c*(b^2*B + A*b*c + a*B*c)*x^8)/8 + (c^2*(3*b*B + A*c)*x^
10)/10 + (B*c^3*x^12)/12 + a^2*(3*A*b + a*B)*Log[x]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (a+b x^2+c x^4\right )^3}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (3 a \left (a b B+A \left (b^2+a c\right )\right )+\frac{a^3 A}{x^2}+\frac{a^2 (3 A b+a B)}{x}+\left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x+\left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^2+3 c \left (b^2 B+A b c+a B c\right ) x^3+c^2 (3 b B+A c) x^4+B c^3 x^5\right ) \, dx,x,x^2\right )\\ &=-\frac{a^3 A}{2 x^2}+\frac{3}{2} a \left (a b B+A \left (b^2+a c\right )\right ) x^2+\frac{1}{4} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^4+\frac{1}{6} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^6+\frac{3}{8} c \left (b^2 B+A b c+a B c\right ) x^8+\frac{1}{10} c^2 (3 b B+A c) x^{10}+\frac{1}{12} B c^3 x^{12}+a^2 (3 A b+a B) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0745245, size = 162, normalized size = 1. \[ a^2 \log (x) (a B+3 A b)-\frac{a^3 A}{2 x^2}+\frac{1}{6} x^6 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{3}{8} c x^8 \left (a B c+A b c+b^2 B\right )+\frac{1}{4} x^4 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac{3}{2} a x^2 \left (A \left (a c+b^2\right )+a b B\right )+\frac{1}{10} c^2 x^{10} (A c+3 b B)+\frac{1}{12} B c^3 x^{12} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*A)/(2*x^2) + (3*a*(a*b*B + A*(b^2 + a*c))*x^2)/2 + ((3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^4)/4 + ((b
^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^6)/6 + (3*c*(b^2*B + A*b*c + a*B*c)*x^8)/8 + (c^2*(3*b*B + A*c)*x^
10)/10 + (B*c^3*x^12)/12 + a^2*(3*A*b + a*B)*Log[x]

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Maple [A]  time = 0.008, size = 190, normalized size = 1.2 \begin{align*}{\frac{B{c}^{3}{x}^{12}}{12}}+{\frac{A{x}^{10}{c}^{3}}{10}}+{\frac{3\,B{x}^{10}b{c}^{2}}{10}}+{\frac{3\,A{x}^{8}b{c}^{2}}{8}}+{\frac{3\,B{x}^{8}a{c}^{2}}{8}}+{\frac{3\,B{x}^{8}{b}^{2}c}{8}}+{\frac{A{x}^{6}a{c}^{2}}{2}}+{\frac{A{x}^{6}{b}^{2}c}{2}}+B{x}^{6}abc+{\frac{B{x}^{6}{b}^{3}}{6}}+{\frac{3\,A{x}^{4}abc}{2}}+{\frac{A{x}^{4}{b}^{3}}{4}}+{\frac{3\,B{x}^{4}{a}^{2}c}{4}}+{\frac{3\,B{x}^{4}a{b}^{2}}{4}}+{\frac{3\,A{x}^{2}{a}^{2}c}{2}}+{\frac{3\,A{x}^{2}a{b}^{2}}{2}}+{\frac{3\,B{x}^{2}{a}^{2}b}{2}}+3\,A\ln \left ( x \right ){a}^{2}b+B\ln \left ( x \right ){a}^{3}-{\frac{A{a}^{3}}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*B*c^3*x^12+1/10*A*x^10*c^3+3/10*B*x^10*b*c^2+3/8*A*x^8*b*c^2+3/8*B*x^8*a*c^2+3/8*B*x^8*b^2*c+1/2*A*x^6*a*
c^2+1/2*A*x^6*b^2*c+B*x^6*a*b*c+1/6*B*x^6*b^3+3/2*A*x^4*a*b*c+1/4*A*x^4*b^3+3/4*B*x^4*a^2*c+3/4*B*x^4*a*b^2+3/
2*A*x^2*a^2*c+3/2*A*x^2*a*b^2+3/2*B*x^2*a^2*b+3*A*ln(x)*a^2*b+B*ln(x)*a^3-1/2*a^3*A/x^2

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Maxima [A]  time = 0.978896, size = 225, normalized size = 1.39 \begin{align*} \frac{1}{12} \, B c^{3} x^{12} + \frac{1}{10} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{10} + \frac{3}{8} \,{\left (B b^{2} c +{\left (B a + A b\right )} c^{2}\right )} x^{8} + \frac{1}{6} \,{\left (B b^{3} + 3 \, A a c^{2} + 3 \,{\left (2 \, B a b + A b^{2}\right )} c\right )} x^{6} + \frac{1}{4} \,{\left (3 \, B a b^{2} + A b^{3} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{4} + \frac{3}{2} \,{\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - \frac{A a^{3}}{2 \, x^{2}} + \frac{1}{2} \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*B*c^3*x^12 + 1/10*(3*B*b*c^2 + A*c^3)*x^10 + 3/8*(B*b^2*c + (B*a + A*b)*c^2)*x^8 + 1/6*(B*b^3 + 3*A*a*c^2
+ 3*(2*B*a*b + A*b^2)*c)*x^6 + 1/4*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^4 + 3/2*(B*a^2*b + A*a*b^2 +
A*a^2*c)*x^2 - 1/2*A*a^3/x^2 + 1/2*(B*a^3 + 3*A*a^2*b)*log(x^2)

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Fricas [A]  time = 1.46054, size = 390, normalized size = 2.41 \begin{align*} \frac{10 \, B c^{3} x^{14} + 12 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{12} + 45 \,{\left (B b^{2} c +{\left (B a + A b\right )} c^{2}\right )} x^{10} + 20 \,{\left (B b^{3} + 3 \, A a c^{2} + 3 \,{\left (2 \, B a b + A b^{2}\right )} c\right )} x^{8} + 30 \,{\left (3 \, B a b^{2} + A b^{3} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{6} + 180 \,{\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{4} - 60 \, A a^{3} + 120 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2} \log \left (x\right )}{120 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(10*B*c^3*x^14 + 12*(3*B*b*c^2 + A*c^3)*x^12 + 45*(B*b^2*c + (B*a + A*b)*c^2)*x^10 + 20*(B*b^3 + 3*A*a*c
^2 + 3*(2*B*a*b + A*b^2)*c)*x^8 + 30*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^6 + 180*(B*a^2*b + A*a*b^2
+ A*a^2*c)*x^4 - 60*A*a^3 + 120*(B*a^3 + 3*A*a^2*b)*x^2*log(x))/x^2

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Sympy [A]  time = 0.573669, size = 197, normalized size = 1.22 \begin{align*} - \frac{A a^{3}}{2 x^{2}} + \frac{B c^{3} x^{12}}{12} + a^{2} \left (3 A b + B a\right ) \log{\left (x \right )} + x^{10} \left (\frac{A c^{3}}{10} + \frac{3 B b c^{2}}{10}\right ) + x^{8} \left (\frac{3 A b c^{2}}{8} + \frac{3 B a c^{2}}{8} + \frac{3 B b^{2} c}{8}\right ) + x^{6} \left (\frac{A a c^{2}}{2} + \frac{A b^{2} c}{2} + B a b c + \frac{B b^{3}}{6}\right ) + x^{4} \left (\frac{3 A a b c}{2} + \frac{A b^{3}}{4} + \frac{3 B a^{2} c}{4} + \frac{3 B a b^{2}}{4}\right ) + x^{2} \left (\frac{3 A a^{2} c}{2} + \frac{3 A a b^{2}}{2} + \frac{3 B a^{2} b}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a**3/(2*x**2) + B*c**3*x**12/12 + a**2*(3*A*b + B*a)*log(x) + x**10*(A*c**3/10 + 3*B*b*c**2/10) + x**8*(3*A
*b*c**2/8 + 3*B*a*c**2/8 + 3*B*b**2*c/8) + x**6*(A*a*c**2/2 + A*b**2*c/2 + B*a*b*c + B*b**3/6) + x**4*(3*A*a*b
*c/2 + A*b**3/4 + 3*B*a**2*c/4 + 3*B*a*b**2/4) + x**2*(3*A*a**2*c/2 + 3*A*a*b**2/2 + 3*B*a**2*b/2)

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Giac [A]  time = 1.11929, size = 286, normalized size = 1.77 \begin{align*} \frac{1}{12} \, B c^{3} x^{12} + \frac{3}{10} \, B b c^{2} x^{10} + \frac{1}{10} \, A c^{3} x^{10} + \frac{3}{8} \, B b^{2} c x^{8} + \frac{3}{8} \, B a c^{2} x^{8} + \frac{3}{8} \, A b c^{2} x^{8} + \frac{1}{6} \, B b^{3} x^{6} + B a b c x^{6} + \frac{1}{2} \, A b^{2} c x^{6} + \frac{1}{2} \, A a c^{2} x^{6} + \frac{3}{4} \, B a b^{2} x^{4} + \frac{1}{4} \, A b^{3} x^{4} + \frac{3}{4} \, B a^{2} c x^{4} + \frac{3}{2} \, A a b c x^{4} + \frac{3}{2} \, B a^{2} b x^{2} + \frac{3}{2} \, A a b^{2} x^{2} + \frac{3}{2} \, A a^{2} c x^{2} + \frac{1}{2} \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left (x^{2}\right ) - \frac{B a^{3} x^{2} + 3 \, A a^{2} b x^{2} + A a^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*B*c^3*x^12 + 3/10*B*b*c^2*x^10 + 1/10*A*c^3*x^10 + 3/8*B*b^2*c*x^8 + 3/8*B*a*c^2*x^8 + 3/8*A*b*c^2*x^8 +
1/6*B*b^3*x^6 + B*a*b*c*x^6 + 1/2*A*b^2*c*x^6 + 1/2*A*a*c^2*x^6 + 3/4*B*a*b^2*x^4 + 1/4*A*b^3*x^4 + 3/4*B*a^2*
c*x^4 + 3/2*A*a*b*c*x^4 + 3/2*B*a^2*b*x^2 + 3/2*A*a*b^2*x^2 + 3/2*A*a^2*c*x^2 + 1/2*(B*a^3 + 3*A*a^2*b)*log(x^
2) - 1/2*(B*a^3*x^2 + 3*A*a^2*b*x^2 + A*a^3)/x^2