3.1.22 \(\int \frac {(a+b x^2)^2 (A+B x^2)}{x^9} \, dx\) [22]

Optimal. Leaf size=48 \[ -\frac {A \left (a+b x^2\right )^3}{8 a x^8}+\frac {(A b-4 a B) \left (a+b x^2\right )^3}{24 a^2 x^6} \]

[Out]

-1/8*A*(b*x^2+a)^3/a/x^8+1/24*(A*b-4*B*a)*(b*x^2+a)^3/a^2/x^6

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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 = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {457, 79, 37} \begin {gather*} \frac {\left (a+b x^2\right )^3 (A b-4 a B)}{24 a^2 x^6}-\frac {A \left (a+b x^2\right )^3}{8 a x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(A + B*x^2))/x^9,x]

[Out]

-1/8*(A*(a + b*x^2)^3)/(a*x^8) + ((A*b - 4*a*B)*(a + b*x^2)^3)/(24*a^2*x^6)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 457

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]]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^9} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2 (A+B x)}{x^5} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^3}{8 a x^8}+\frac {(-A b+4 a B) \text {Subst}\left (\int \frac {(a+b x)^2}{x^4} \, dx,x,x^2\right )}{8 a}\\ &=-\frac {A \left (a+b x^2\right )^3}{8 a x^8}+\frac {(A b-4 a B) \left (a+b x^2\right )^3}{24 a^2 x^6}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 55, normalized size = 1.15 \begin {gather*} -\frac {6 b^2 x^4 \left (A+2 B x^2\right )+4 a b x^2 \left (2 A+3 B x^2\right )+a^2 \left (3 A+4 B x^2\right )}{24 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(A + B*x^2))/x^9,x]

[Out]

-1/24*(6*b^2*x^4*(A + 2*B*x^2) + 4*a*b*x^2*(2*A + 3*B*x^2) + a^2*(3*A + 4*B*x^2))/x^8

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Maple [A]
time = 0.06, size = 48, normalized size = 1.00

method	result	size
default	\(-\frac {b \left (A b +2 B a \right )}{4 x^{4}}-\frac {a \left (2 A b +B a \right )}{6 x^{6}}-\frac {b^{2} B}{2 x^{2}}-\frac {a^{2} A}{8 x^{8}}\)	\(48\)
norman	\(\frac {-\frac {b^{2} B \,x^{6}}{2}+\left (-\frac {1}{4} b^{2} A -\frac {1}{2} a b B \right ) x^{4}+\left (-\frac {1}{3} a b A -\frac {1}{6} a^{2} B \right ) x^{2}-\frac {a^{2} A}{8}}{x^{8}}\)	\(53\)
risch	\(\frac {-\frac {b^{2} B \,x^{6}}{2}+\left (-\frac {1}{4} b^{2} A -\frac {1}{2} a b B \right ) x^{4}+\left (-\frac {1}{3} a b A -\frac {1}{6} a^{2} B \right ) x^{2}-\frac {a^{2} A}{8}}{x^{8}}\)	\(53\)
gosper	\(-\frac {12 b^{2} B \,x^{6}+6 A \,b^{2} x^{4}+12 B a b \,x^{4}+8 a A b \,x^{2}+4 B \,a^{2} x^{2}+3 a^{2} A}{24 x^{8}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(B*x^2+A)/x^9,x,method=_RETURNVERBOSE)

[Out]

-1/4*b*(A*b+2*B*a)/x^4-1/6*a*(2*A*b+B*a)/x^6-1/2*b^2*B/x^2-1/8*a^2*A/x^8

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Maxima [A]
time = 0.30, size = 53, normalized size = 1.10 \begin {gather*} -\frac {12 \, B b^{2} x^{6} + 6 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + 3 \, A a^{2} + 4 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^9,x, algorithm="maxima")

[Out]

-1/24*(12*B*b^2*x^6 + 6*(2*B*a*b + A*b^2)*x^4 + 3*A*a^2 + 4*(B*a^2 + 2*A*a*b)*x^2)/x^8

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Fricas [A]
time = 0.92, size = 53, normalized size = 1.10 \begin {gather*} -\frac {12 \, B b^{2} x^{6} + 6 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + 3 \, A a^{2} + 4 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^9,x, algorithm="fricas")

[Out]

-1/24*(12*B*b^2*x^6 + 6*(2*B*a*b + A*b^2)*x^4 + 3*A*a^2 + 4*(B*a^2 + 2*A*a*b)*x^2)/x^8

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Sympy [A]
time = 0.86, size = 58, normalized size = 1.21 \begin {gather*} \frac {- 3 A a^{2} - 12 B b^{2} x^{6} + x^{4} \left (- 6 A b^{2} - 12 B a b\right ) + x^{2} \left (- 8 A a b - 4 B a^{2}\right )}{24 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(B*x**2+A)/x**9,x)

[Out]

(-3*A*a**2 - 12*B*b**2*x**6 + x**4*(-6*A*b**2 - 12*B*a*b) + x**2*(-8*A*a*b - 4*B*a**2))/(24*x**8)

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Giac [A]
time = 1.13, size = 55, normalized size = 1.15 \begin {gather*} -\frac {12 \, B b^{2} x^{6} + 12 \, B a b x^{4} + 6 \, A b^{2} x^{4} + 4 \, B a^{2} x^{2} + 8 \, A a b x^{2} + 3 \, A a^{2}}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^9,x, algorithm="giac")

[Out]

-1/24*(12*B*b^2*x^6 + 12*B*a*b*x^4 + 6*A*b^2*x^4 + 4*B*a^2*x^2 + 8*A*a*b*x^2 + 3*A*a^2)/x^8

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Mupad [B]
time = 0.03, size = 53, normalized size = 1.10 \begin {gather*} -\frac {x^2\,\left (\frac {B\,a^2}{6}+\frac {A\,b\,a}{3}\right )+x^4\,\left (\frac {A\,b^2}{4}+\frac {B\,a\,b}{2}\right )+\frac {A\,a^2}{8}+\frac {B\,b^2\,x^6}{2}}{x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^2)/x^9,x)

[Out]

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

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