3.2.1 \(\int \frac {(A+B x^2) (a+b x^2+c x^4)^3}{x^3} \, dx\) [101]

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

[Out]

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

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Rubi [A]
time = 0.14, 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 = {1265, 779} \begin {gather*} -\frac {a^3 A}{2 x^2}+a^2 \log (x) (a B+3 A b)+\frac {3}{8} c x^8 \left (a B c+A b c+b^2 B\right )+\frac {3}{2} a x^2 \left (A \left (a c+b^2\right )+a b B\right )+\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 {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 {1}{10} c^2 x^{10} (A c+3 b B)+\frac {1}{12} B c^3 x^{12} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a^3*A)/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 779

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 1265

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^3} \, dx &=\frac {1}{2} \text {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} \text {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*}

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Mathematica [A]
time = 0.05, size = 162, normalized size = 1.00 \begin {gather*} -\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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a^3*A)/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 = 0.02, size = 188, normalized size = 1.16

method	result	size
norman	\(\frac {\left (\frac {1}{10} c^{3} A +\frac {3}{10} B b \,c^{2}\right ) x^{12}+\left (\frac {3}{2} a^{2} c A +\frac {3}{2} A a \,b^{2}+\frac {3}{2} B \,a^{2} b \right ) x^{4}+\left (\frac {3}{8} b \,c^{2} A +\frac {3}{8} c^{2} a B +\frac {3}{8} B \,b^{2} c \right ) x^{10}+\left (\frac {1}{2} c^{2} a A +\frac {1}{2} A \,b^{2} c +a b B c +\frac {1}{6} b^{3} B \right ) x^{8}+\left (\frac {3}{2} A a b c +\frac {1}{4} A \,b^{3}+\frac {3}{4} a^{2} c B +\frac {3}{4} B a \,b^{2}\right ) x^{6}-\frac {a^{3} A}{2}+\frac {B \,c^{3} x^{14}}{12}}{x^{2}}+\left (3 a^{2} b A +a^{3} B \right ) \ln \left (x \right )\)	\(169\)
default	\(\frac {B \,c^{3} x^{12}}{12}+\frac {A \,c^{3} x^{10}}{10}+\frac {3 B b \,c^{2} x^{10}}{10}+\frac {3 A b \,c^{2} x^{8}}{8}+\frac {3 B a \,c^{2} x^{8}}{8}+\frac {3 B \,b^{2} c \,x^{8}}{8}+\frac {x^{6} c^{2} a A}{2}+\frac {A \,b^{2} c \,x^{6}}{2}+B a b c \,x^{6}+\frac {B \,b^{3} x^{6}}{6}+\frac {3 A a b c \,x^{4}}{2}+\frac {A \,b^{3} x^{4}}{4}+\frac {3 a^{2} c B \,x^{4}}{4}+\frac {3 B a \,b^{2} x^{4}}{4}+\frac {3 A \,a^{2} c \,x^{2}}{2}+\frac {3 A a \,b^{2} x^{2}}{2}+\frac {3 B \,a^{2} b \,x^{2}}{2}-\frac {a^{3} A}{2 x^{2}}+a^{2} \left (3 A b +a B \right ) \ln \left (x \right )\)	\(188\)
risch	\(\frac {B \,c^{3} x^{12}}{12}+\frac {A \,c^{3} x^{10}}{10}+\frac {3 B b \,c^{2} x^{10}}{10}+\frac {3 A b \,c^{2} x^{8}}{8}+\frac {3 B a \,c^{2} x^{8}}{8}+\frac {3 B \,b^{2} c \,x^{8}}{8}+\frac {x^{6} c^{2} a A}{2}+\frac {A \,b^{2} c \,x^{6}}{2}+B a b c \,x^{6}+\frac {B \,b^{3} x^{6}}{6}+\frac {3 A a b c \,x^{4}}{2}+\frac {A \,b^{3} x^{4}}{4}+\frac {3 a^{2} c B \,x^{4}}{4}+\frac {3 B a \,b^{2} x^{4}}{4}+\frac {3 A \,a^{2} c \,x^{2}}{2}+\frac {3 A a \,b^{2} x^{2}}{2}+\frac {3 B \,a^{2} b \,x^{2}}{2}-\frac {a^{3} A}{2 x^{2}}+3 A \ln \left (x \right ) a^{2} b +B \ln \left (x \right ) a^{3}\)	\(190\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 167, normalized size = 1.03 \begin {gather*} \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 {gather*}

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 = 0.35, size = 170, normalized size = 1.05 \begin {gather*} \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 {gather*}

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 = 0.18, size = 197, normalized size = 1.22 \begin {gather*} - \frac {A a^{3}}{2 x^{2}} + \frac {B c^{3} x^{12}}{12} + a^{2} \cdot \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} \cdot \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} \cdot \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} \cdot \left (\frac {3 A a^{2} c}{2} + \frac {3 A a b^{2}}{2} + \frac {3 B a^{2} b}{2}\right ) \end {gather*}

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 = 4.21, size = 212, normalized size = 1.31 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.06, size = 166, normalized size = 1.02 \begin {gather*} x^4\,\left (\frac {3\,B\,c\,a^2}{4}+\frac {3\,B\,a\,b^2}{4}+\frac {3\,A\,c\,a\,b}{2}+\frac {A\,b^3}{4}\right )+x^6\,\left (\frac {B\,b^3}{6}+\frac {A\,b^2\,c}{2}+B\,a\,b\,c+\frac {A\,a\,c^2}{2}\right )+x^{10}\,\left (\frac {A\,c^3}{10}+\frac {3\,B\,b\,c^2}{10}\right )+\ln \left (x\right )\,\left (B\,a^3+3\,A\,b\,a^2\right )+x^2\,\left (\frac {3\,B\,a^2\,b}{2}+\frac {3\,A\,c\,a^2}{2}+\frac {3\,A\,a\,b^2}{2}\right )+x^8\,\left (\frac {3\,B\,b^2\,c}{8}+\frac {3\,A\,b\,c^2}{8}+\frac {3\,B\,a\,c^2}{8}\right )-\frac {A\,a^3}{2\,x^2}+\frac {B\,c^3\,x^{12}}{12} \end {gather*}

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^3,x)

[Out]

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

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