3.1.18 \(\int \frac {(A+B x^2) (b x^2+c x^4)^2}{x^6} \, dx\) [18]

Optimal. Leaf size=48 \[ -\frac {A b^2}{x}+b (b B+2 A c) x+\frac {1}{3} c (2 b B+A c) x^3+\frac {1}{5} B c^2 x^5 \]

[Out]

-A*b^2/x+b*(2*A*c+B*b)*x+1/3*c*(A*c+2*B*b)*x^3+1/5*B*c^2*x^5

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Rubi [A]
time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1598, 459} \begin {gather*} -\frac {A b^2}{x}+\frac {1}{3} c x^3 (A c+2 b B)+b x (2 A c+b B)+\frac {1}{5} B c^2 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^6,x]

[Out]

-((A*b^2)/x) + b*(b*B + 2*A*c)*x + (c*(2*b*B + A*c)*x^3)/3 + (B*c^2*x^5)/5

Rule 459

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

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^6} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^2}{x^2} \, dx\\ &=\int \left (b (b B+2 A c)+\frac {A b^2}{x^2}+c (2 b B+A c) x^2+B c^2 x^4\right ) \, dx\\ &=-\frac {A b^2}{x}+b (b B+2 A c) x+\frac {1}{3} c (2 b B+A c) x^3+\frac {1}{5} B c^2 x^5\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 48, normalized size = 1.00 \begin {gather*} -\frac {A b^2}{x}+b (b B+2 A c) x+\frac {1}{3} c (2 b B+A c) x^3+\frac {1}{5} B c^2 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^6,x]

[Out]

-((A*b^2)/x) + b*(b*B + 2*A*c)*x + (c*(2*b*B + A*c)*x^3)/3 + (B*c^2*x^5)/5

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Maple [A]
time = 0.35, size = 49, normalized size = 1.02

method	result	size
default	\(\frac {B \,c^{2} x^{5}}{5}+\frac {A \,c^{2} x^{3}}{3}+\frac {2 B b c \,x^{3}}{3}+2 A b c x +b^{2} B x -\frac {A \,b^{2}}{x}\)	\(49\)
risch	\(\frac {B \,c^{2} x^{5}}{5}+\frac {A \,c^{2} x^{3}}{3}+\frac {2 B b c \,x^{3}}{3}+2 A b c x +b^{2} B x -\frac {A \,b^{2}}{x}\)	\(49\)
norman	\(\frac {\left (\frac {1}{3} A \,c^{2}+\frac {2}{3} b B c \right ) x^{8}+\left (2 A b c +b^{2} B \right ) x^{6}-A \,b^{2} x^{4}+\frac {B \,c^{2} x^{10}}{5}}{x^{5}}\)	\(55\)
gosper	\(-\frac {-3 B \,c^{2} x^{6}-5 A \,c^{2} x^{4}-10 x^{4} b B c -30 A b c \,x^{2}-15 b^{2} B \,x^{2}+15 b^{2} A}{15 x}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^2/x^6,x,method=_RETURNVERBOSE)

[Out]

1/5*B*c^2*x^5+1/3*A*c^2*x^3+2/3*B*b*c*x^3+2*A*b*c*x+b^2*B*x-A*b^2/x

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Maxima [A]
time = 0.28, size = 48, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, B c^{2} x^{5} + \frac {1}{3} \, {\left (2 \, B b c + A c^{2}\right )} x^{3} - \frac {A b^{2}}{x} + {\left (B b^{2} + 2 \, A b c\right )} x \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^6,x, algorithm="maxima")

[Out]

1/5*B*c^2*x^5 + 1/3*(2*B*b*c + A*c^2)*x^3 - A*b^2/x + (B*b^2 + 2*A*b*c)*x

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Fricas [A]
time = 2.10, size = 53, normalized size = 1.10 \begin {gather*} \frac {3 \, B c^{2} x^{6} + 5 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} - 15 \, A b^{2} + 15 \, {\left (B b^{2} + 2 \, A b c\right )} x^{2}}{15 \, x} \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^6,x, algorithm="fricas")

[Out]

1/15*(3*B*c^2*x^6 + 5*(2*B*b*c + A*c^2)*x^4 - 15*A*b^2 + 15*(B*b^2 + 2*A*b*c)*x^2)/x

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Sympy [A]
time = 0.04, size = 48, normalized size = 1.00 \begin {gather*} - \frac {A b^{2}}{x} + \frac {B c^{2} x^{5}}{5} + x^{3} \left (\frac {A c^{2}}{3} + \frac {2 B b c}{3}\right ) + x \left (2 A b c + B 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)**2/x**6,x)

[Out]

-A*b**2/x + B*c**2*x**5/5 + x**3*(A*c**2/3 + 2*B*b*c/3) + x*(2*A*b*c + B*b**2)

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Giac [A]
time = 0.80, size = 48, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, B c^{2} x^{5} + \frac {2}{3} \, B b c x^{3} + \frac {1}{3} \, A c^{2} x^{3} + B b^{2} x + 2 \, A b c x - \frac {A b^{2}}{x} \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^6,x, algorithm="giac")

[Out]

1/5*B*c^2*x^5 + 2/3*B*b*c*x^3 + 1/3*A*c^2*x^3 + B*b^2*x + 2*A*b*c*x - A*b^2/x

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Mupad [B]
time = 0.05, size = 48, normalized size = 1.00 \begin {gather*} x^3\,\left (\frac {A\,c^2}{3}+\frac {2\,B\,b\,c}{3}\right )+x\,\left (B\,b^2+2\,A\,c\,b\right )-\frac {A\,b^2}{x}+\frac {B\,c^2\,x^5}{5} \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^6,x)

[Out]

x^3*((A*c^2)/3 + (2*B*b*c)/3) + x*(B*b^2 + 2*A*b*c) - (A*b^2)/x + (B*c^2*x^5)/5

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