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

Optimal. Leaf size=162 \[ a^3 A \log (x)+\frac {1}{2} a^2 x^2 (a B+3 A b)+\frac {3}{10} c x^{10} \left (a B c+A b c+b^2 B\right )+\frac {3}{4} a x^4 \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{8} x^8 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {1}{6} x^6 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{12} c^2 x^{12} (A c+3 b B)+\frac {1}{14} B c^3 x^{14} \]

[Out]

1/2*a^2*(3*A*b+B*a)*x^2+3/4*a*(a*b*B+A*(a*c+b^2))*x^4+1/6*(3*a*B*(a*c+b^2)+A*(6*a*b*c+b^3))*x^6+1/8*(3*A*a*c^2
+3*A*b^2*c+6*B*a*b*c+B*b^3)*x^8+3/10*c*(A*b*c+B*a*c+B*b^2)*x^10+1/12*c^2*(A*c+3*B*b)*x^12+1/14*B*c^3*x^14+a^3*
A*ln(x)

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Rubi [A]  time = 0.23, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1251, 765} \[ \frac {1}{2} a^2 x^2 (a B+3 A b)+a^3 A \log (x)+\frac {1}{8} x^8 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {3}{10} c x^{10} \left (a B c+A b c+b^2 B\right )+\frac {1}{6} x^6 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {3}{4} a x^4 \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{12} c^2 x^{12} (A c+3 b B)+\frac {1}{14} B c^3 x^{14} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3*A*b + a*B)*x^2)/2 + (3*a*(a*b*B + A*(b^2 + a*c))*x^4)/4 + ((3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^6
)/6 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^8)/8 + (3*c*(b^2*B + A*b*c + a*B*c)*x^10)/10 + (c^2*(3*b*
B + A*c)*x^12)/12 + (B*c^3*x^14)/14 + a^3*A*Log[x]

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]))

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]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (a+b x^2+c x^4\right )^3}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (a^2 (3 A b+a B)+\frac {a^3 A}{x}+3 a \left (a b B+A \left (b^2+a c\right )\right ) x+\left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^2+\left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^3+3 c \left (b^2 B+A b c+a B c\right ) x^4+c^2 (3 b B+A c) x^5+B c^3 x^6\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2} a^2 (3 A b+a B) x^2+\frac {3}{4} a \left (a b B+A \left (b^2+a c\right )\right ) x^4+\frac {1}{6} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^6+\frac {1}{8} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^8+\frac {3}{10} c \left (b^2 B+A b c+a B c\right ) x^{10}+\frac {1}{12} c^2 (3 b B+A c) x^{12}+\frac {1}{14} B c^3 x^{14}+a^3 A \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 162, normalized size = 1.00 \[ a^3 A \log (x)+\frac {1}{2} a^2 x^2 (a B+3 A b)+\frac {3}{10} c x^{10} \left (a B c+A b c+b^2 B\right )+\frac {3}{4} a x^4 \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{8} x^8 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {1}{6} x^6 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{12} c^2 x^{12} (A c+3 b B)+\frac {1}{14} B c^3 x^{14} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3*A*b + a*B)*x^2)/2 + (3*a*(a*b*B + A*(b^2 + a*c))*x^4)/4 + ((3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^6
)/6 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^8)/8 + (3*c*(b^2*B + A*b*c + a*B*c)*x^10)/10 + (c^2*(3*b*
B + A*c)*x^12)/12 + (B*c^3*x^14)/14 + a^3*A*Log[x]

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fricas [A]  time = 0.62, size = 164, normalized size = 1.01 \[ \frac {1}{14} \, B c^{3} x^{14} + \frac {1}{12} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{12} + \frac {3}{10} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{10} + \frac {1}{8} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{8} + \frac {1}{6} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{6} + \frac {3}{4} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{4} + A a^{3} \log \relax (x) + \frac {1}{2} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2} \]

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,x, algorithm="fricas")

[Out]

1/14*B*c^3*x^14 + 1/12*(3*B*b*c^2 + A*c^3)*x^12 + 3/10*(B*b^2*c + (B*a + A*b)*c^2)*x^10 + 1/8*(B*b^3 + 3*A*a*c
^2 + 3*(2*B*a*b + A*b^2)*c)*x^8 + 1/6*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^6 + 3/4*(B*a^2*b + A*a*b^2
+ A*a^2*c)*x^4 + A*a^3*log(x) + 1/2*(B*a^3 + 3*A*a^2*b)*x^2

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giac [A]  time = 0.34, size = 193, normalized size = 1.19 \[ \frac {1}{14} \, B c^{3} x^{14} + \frac {1}{4} \, B b c^{2} x^{12} + \frac {1}{12} \, A c^{3} x^{12} + \frac {3}{10} \, B b^{2} c x^{10} + \frac {3}{10} \, B a c^{2} x^{10} + \frac {3}{10} \, A b c^{2} x^{10} + \frac {1}{8} \, B b^{3} x^{8} + \frac {3}{4} \, B a b c x^{8} + \frac {3}{8} \, A b^{2} c x^{8} + \frac {3}{8} \, A a c^{2} x^{8} + \frac {1}{2} \, B a b^{2} x^{6} + \frac {1}{6} \, A b^{3} x^{6} + \frac {1}{2} \, B a^{2} c x^{6} + A a b c x^{6} + \frac {3}{4} \, B a^{2} b x^{4} + \frac {3}{4} \, A a b^{2} x^{4} + \frac {3}{4} \, A a^{2} c x^{4} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{2} \, A a^{2} b x^{2} + \frac {1}{2} \, A a^{3} \log \left (x^{2}\right ) \]

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,x, algorithm="giac")

[Out]

1/14*B*c^3*x^14 + 1/4*B*b*c^2*x^12 + 1/12*A*c^3*x^12 + 3/10*B*b^2*c*x^10 + 3/10*B*a*c^2*x^10 + 3/10*A*b*c^2*x^
10 + 1/8*B*b^3*x^8 + 3/4*B*a*b*c*x^8 + 3/8*A*b^2*c*x^8 + 3/8*A*a*c^2*x^8 + 1/2*B*a*b^2*x^6 + 1/6*A*b^3*x^6 + 1
/2*B*a^2*c*x^6 + A*a*b*c*x^6 + 3/4*B*a^2*b*x^4 + 3/4*A*a*b^2*x^4 + 3/4*A*a^2*c*x^4 + 1/2*B*a^3*x^2 + 3/2*A*a^2
*b*x^2 + 1/2*A*a^3*log(x^2)

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maple [A]  time = 0.00, size = 191, normalized size = 1.18 \[ \frac {B \,c^{3} x^{14}}{14}+\frac {A \,c^{3} x^{12}}{12}+\frac {B b \,c^{2} x^{12}}{4}+\frac {3 A b \,c^{2} x^{10}}{10}+\frac {3 B a \,c^{2} x^{10}}{10}+\frac {3 B \,b^{2} c \,x^{10}}{10}+\frac {3 A a \,c^{2} x^{8}}{8}+\frac {3 A \,b^{2} c \,x^{8}}{8}+\frac {3 B a b c \,x^{8}}{4}+\frac {B \,b^{3} x^{8}}{8}+A a b c \,x^{6}+\frac {A \,b^{3} x^{6}}{6}+\frac {B \,a^{2} c \,x^{6}}{2}+\frac {B a \,b^{2} x^{6}}{2}+\frac {3 A \,a^{2} c \,x^{4}}{4}+\frac {3 A a \,b^{2} x^{4}}{4}+\frac {3 B \,a^{2} b \,x^{4}}{4}+\frac {3 A \,a^{2} b \,x^{2}}{2}+\frac {B \,a^{3} x^{2}}{2}+A \,a^{3} \ln \relax (x ) \]

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,x)

[Out]

1/14*B*c^3*x^14+1/12*A*x^12*c^3+1/4*B*x^12*b*c^2+3/10*A*x^10*b*c^2+3/10*B*x^10*a*c^2+3/10*B*x^10*b^2*c+3/8*A*x
^8*a*c^2+3/8*A*x^8*b^2*c+3/4*B*x^8*a*b*c+1/8*B*x^8*b^3+A*x^6*a*b*c+1/6*A*x^6*b^3+1/2*B*x^6*a^2*c+1/2*B*x^6*a*b
^2+3/4*A*x^4*a^2*c+3/4*A*x^4*a*b^2+3/4*B*x^4*a^2*b+3/2*A*x^2*a^2*b+1/2*B*x^2*a^3+a^3*A*ln(x)

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maxima [A]  time = 0.79, size = 167, normalized size = 1.03 \[ \frac {1}{14} \, B c^{3} x^{14} + \frac {1}{12} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{12} + \frac {3}{10} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{10} + \frac {1}{8} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{8} + \frac {1}{6} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{6} + \frac {3}{4} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{4} + \frac {1}{2} \, A a^{3} \log \left (x^{2}\right ) + \frac {1}{2} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2} \]

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,x, algorithm="maxima")

[Out]

1/14*B*c^3*x^14 + 1/12*(3*B*b*c^2 + A*c^3)*x^12 + 3/10*(B*b^2*c + (B*a + A*b)*c^2)*x^10 + 1/8*(B*b^3 + 3*A*a*c
^2 + 3*(2*B*a*b + A*b^2)*c)*x^8 + 1/6*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^6 + 3/4*(B*a^2*b + A*a*b^2
+ A*a^2*c)*x^4 + 1/2*A*a^3*log(x^2) + 1/2*(B*a^3 + 3*A*a^2*b)*x^2

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mupad [B]  time = 0.10, size = 166, normalized size = 1.02 \[ x^6\,\left (\frac {B\,c\,a^2}{2}+\frac {B\,a\,b^2}{2}+A\,c\,a\,b+\frac {A\,b^3}{6}\right )+x^8\,\left (\frac {B\,b^3}{8}+\frac {3\,A\,b^2\,c}{8}+\frac {3\,B\,a\,b\,c}{4}+\frac {3\,A\,a\,c^2}{8}\right )+x^2\,\left (\frac {B\,a^3}{2}+\frac {3\,A\,b\,a^2}{2}\right )+x^{12}\,\left (\frac {A\,c^3}{12}+\frac {B\,b\,c^2}{4}\right )+x^4\,\left (\frac {3\,B\,a^2\,b}{4}+\frac {3\,A\,c\,a^2}{4}+\frac {3\,A\,a\,b^2}{4}\right )+x^{10}\,\left (\frac {3\,B\,b^2\,c}{10}+\frac {3\,A\,b\,c^2}{10}+\frac {3\,B\,a\,c^2}{10}\right )+\frac {B\,c^3\,x^{14}}{14}+A\,a^3\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^6*((A*b^3)/6 + (B*a*b^2)/2 + (B*a^2*c)/2 + A*a*b*c) + x^8*((B*b^3)/8 + (3*A*a*c^2)/8 + (3*A*b^2*c)/8 + (3*B*
a*b*c)/4) + x^2*((B*a^3)/2 + (3*A*a^2*b)/2) + x^12*((A*c^3)/12 + (B*b*c^2)/4) + x^4*((3*A*a*b^2)/4 + (3*A*a^2*
c)/4 + (3*B*a^2*b)/4) + x^10*((3*A*b*c^2)/10 + (3*B*a*c^2)/10 + (3*B*b^2*c)/10) + (B*c^3*x^14)/14 + A*a^3*log(
x)

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sympy [A]  time = 0.31, size = 199, normalized size = 1.23 \[ A a^{3} \log {\relax (x )} + \frac {B c^{3} x^{14}}{14} + x^{12} \left (\frac {A c^{3}}{12} + \frac {B b c^{2}}{4}\right ) + x^{10} \left (\frac {3 A b c^{2}}{10} + \frac {3 B a c^{2}}{10} + \frac {3 B b^{2} c}{10}\right ) + x^{8} \left (\frac {3 A a c^{2}}{8} + \frac {3 A b^{2} c}{8} + \frac {3 B a b c}{4} + \frac {B b^{3}}{8}\right ) + x^{6} \left (A a b c + \frac {A b^{3}}{6} + \frac {B a^{2} c}{2} + \frac {B a b^{2}}{2}\right ) + x^{4} \left (\frac {3 A a^{2} c}{4} + \frac {3 A a b^{2}}{4} + \frac {3 B a^{2} b}{4}\right ) + x^{2} \left (\frac {3 A a^{2} b}{2} + \frac {B a^{3}}{2}\right ) \]

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,x)

[Out]

A*a**3*log(x) + B*c**3*x**14/14 + x**12*(A*c**3/12 + B*b*c**2/4) + x**10*(3*A*b*c**2/10 + 3*B*a*c**2/10 + 3*B*
b**2*c/10) + x**8*(3*A*a*c**2/8 + 3*A*b**2*c/8 + 3*B*a*b*c/4 + B*b**3/8) + x**6*(A*a*b*c + A*b**3/6 + B*a**2*c
/2 + B*a*b**2/2) + x**4*(3*A*a**2*c/4 + 3*A*a*b**2/4 + 3*B*a**2*b/4) + x**2*(3*A*a**2*b/2 + B*a**3/2)

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