3.1.73 \(\int \frac {x^7 (A+B x^2)}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=104 \[ \frac {a^3 (A b-a B)}{2 b^5 \left (a+b x^2\right )}+\frac {a^2 (3 A b-4 a B) \log \left (a+b x^2\right )}{2 b^5}-\frac {a x^2 (2 A b-3 a B)}{2 b^4}+\frac {x^4 (A b-2 a B)}{4 b^3}+\frac {B x^6}{6 b^2} \]

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Rubi [A]  time = 0.13, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 77} \begin {gather*} \frac {a^3 (A b-a B)}{2 b^5 \left (a+b x^2\right )}+\frac {a^2 (3 A b-4 a B) \log \left (a+b x^2\right )}{2 b^5}+\frac {x^4 (A b-2 a B)}{4 b^3}-\frac {a x^2 (2 A b-3 a B)}{2 b^4}+\frac {B x^6}{6 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^7*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

-(a*(2*A*b - 3*a*B)*x^2)/(2*b^4) + ((A*b - 2*a*B)*x^4)/(4*b^3) + (B*x^6)/(6*b^2) + (a^3*(A*b - a*B))/(2*b^5*(a
+ b*x^2)) + (a^2*(3*A*b - 4*a*B)*Log[a + b*x^2])/(2*b^5)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

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 {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {a (-2 A b+3 a B)}{b^4}+\frac {(A b-2 a B) x}{b^3}+\frac {B x^2}{b^2}+\frac {a^3 (-A b+a B)}{b^4 (a+b x)^2}-\frac {a^2 (-3 A b+4 a B)}{b^4 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {a (2 A b-3 a B) x^2}{2 b^4}+\frac {(A b-2 a B) x^4}{4 b^3}+\frac {B x^6}{6 b^2}+\frac {a^3 (A b-a B)}{2 b^5 \left (a+b x^2\right )}+\frac {a^2 (3 A b-4 a B) \log \left (a+b x^2\right )}{2 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 93, normalized size = 0.89 \begin {gather*} \frac {\frac {6 a^3 (A b-a B)}{a+b x^2}+6 a^2 (3 A b-4 a B) \log \left (a+b x^2\right )+3 b^2 x^4 (A b-2 a B)+6 a b x^2 (3 a B-2 A b)+2 b^3 B x^6}{12 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

(6*a*b*(-2*A*b + 3*a*B)*x^2 + 3*b^2*(A*b - 2*a*B)*x^4 + 2*b^3*B*x^6 + (6*a^3*(A*b - a*B))/(a + b*x^2) + 6*a^2*
(3*A*b - 4*a*B)*Log[a + b*x^2])/(12*b^5)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^7*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

IntegrateAlgebraic[(x^7*(A + B*x^2))/(a + b*x^2)^2, x]

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fricas [A]  time = 0.42, size = 148, normalized size = 1.42 \begin {gather*} \frac {2 \, B b^{4} x^{8} - {\left (4 \, B a b^{3} - 3 \, A b^{4}\right )} x^{6} - 6 \, B a^{4} + 6 \, A a^{3} b + 3 \, {\left (4 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 6 \, {\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2} - 6 \, {\left (4 \, B a^{4} - 3 \, A a^{3} b + {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{6} x^{2} + a b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

1/12*(2*B*b^4*x^8 - (4*B*a*b^3 - 3*A*b^4)*x^6 - 6*B*a^4 + 6*A*a^3*b + 3*(4*B*a^2*b^2 - 3*A*a*b^3)*x^4 + 6*(3*B
*a^3*b - 2*A*a^2*b^2)*x^2 - 6*(4*B*a^4 - 3*A*a^3*b + (4*B*a^3*b - 3*A*a^2*b^2)*x^2)*log(b*x^2 + a))/(b^6*x^2 +
a*b^5)

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giac [A]  time = 0.32, size = 135, normalized size = 1.30 \begin {gather*} -\frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{5}} + \frac {2 \, B b^{4} x^{6} - 6 \, B a b^{3} x^{4} + 3 \, A b^{4} x^{4} + 18 \, B a^{2} b^{2} x^{2} - 12 \, A a b^{3} x^{2}}{12 \, b^{6}} + \frac {4 \, B a^{3} b x^{2} - 3 \, A a^{2} b^{2} x^{2} + 3 \, B a^{4} - 2 \, A a^{3} b}{2 \, {\left (b x^{2} + a\right )} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*(4*B*a^3 - 3*A*a^2*b)*log(abs(b*x^2 + a))/b^5 + 1/12*(2*B*b^4*x^6 - 6*B*a*b^3*x^4 + 3*A*b^4*x^4 + 18*B*a^
2*b^2*x^2 - 12*A*a*b^3*x^2)/b^6 + 1/2*(4*B*a^3*b*x^2 - 3*A*a^2*b^2*x^2 + 3*B*a^4 - 2*A*a^3*b)/((b*x^2 + a)*b^5
)

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maple [A]  time = 0.02, size = 122, normalized size = 1.17 \begin {gather*} \frac {B \,x^{6}}{6 b^{2}}+\frac {A \,x^{4}}{4 b^{2}}-\frac {B a \,x^{4}}{2 b^{3}}-\frac {A a \,x^{2}}{b^{3}}+\frac {3 B \,a^{2} x^{2}}{2 b^{4}}+\frac {A \,a^{3}}{2 \left (b \,x^{2}+a \right ) b^{4}}+\frac {3 A \,a^{2} \ln \left (b \,x^{2}+a \right )}{2 b^{4}}-\frac {B \,a^{4}}{2 \left (b \,x^{2}+a \right ) b^{5}}-\frac {2 B \,a^{3} \ln \left (b \,x^{2}+a \right )}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(B*x^2+A)/(b*x^2+a)^2,x)

[Out]

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

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maxima [A]  time = 0.99, size = 107, normalized size = 1.03 \begin {gather*} -\frac {B a^{4} - A a^{3} b}{2 \, {\left (b^{6} x^{2} + a b^{5}\right )}} + \frac {2 \, B b^{2} x^{6} - 3 \, {\left (2 \, B a b - A b^{2}\right )} x^{4} + 6 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} x^{2}}{12 \, b^{4}} - \frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

-1/2*(B*a^4 - A*a^3*b)/(b^6*x^2 + a*b^5) + 1/12*(2*B*b^2*x^6 - 3*(2*B*a*b - A*b^2)*x^4 + 6*(3*B*a^2 - 2*A*a*b)
*x^2)/b^4 - 1/2*(4*B*a^3 - 3*A*a^2*b)*log(b*x^2 + a)/b^5

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mupad [B]  time = 0.09, size = 121, normalized size = 1.16 \begin {gather*} x^4\,\left (\frac {A}{4\,b^2}-\frac {B\,a}{2\,b^3}\right )-x^2\,\left (\frac {a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{2\,b^4}\right )+\frac {B\,x^6}{6\,b^2}-\frac {\ln \left (b\,x^2+a\right )\,\left (4\,B\,a^3-3\,A\,a^2\,b\right )}{2\,b^5}-\frac {B\,a^4-A\,a^3\,b}{2\,b\,\left (b^5\,x^2+a\,b^4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(A + B*x^2))/(a + b*x^2)^2,x)

[Out]

x^4*(A/(4*b^2) - (B*a)/(2*b^3)) - x^2*((a*(A/b^2 - (2*B*a)/b^3))/b + (B*a^2)/(2*b^4)) + (B*x^6)/(6*b^2) - (log
(a + b*x^2)*(4*B*a^3 - 3*A*a^2*b))/(2*b^5) - (B*a^4 - A*a^3*b)/(2*b*(a*b^4 + b^5*x^2))

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sympy [A]  time = 0.72, size = 104, normalized size = 1.00 \begin {gather*} \frac {B x^{6}}{6 b^{2}} - \frac {a^{2} \left (- 3 A b + 4 B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{5}} + x^{4} \left (\frac {A}{4 b^{2}} - \frac {B a}{2 b^{3}}\right ) + x^{2} \left (- \frac {A a}{b^{3}} + \frac {3 B a^{2}}{2 b^{4}}\right ) + \frac {A a^{3} b - B a^{4}}{2 a b^{5} + 2 b^{6} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(B*x**2+A)/(b*x**2+a)**2,x)

[Out]

B*x**6/(6*b**2) - a**2*(-3*A*b + 4*B*a)*log(a + b*x**2)/(2*b**5) + x**4*(A/(4*b**2) - B*a/(2*b**3)) + x**2*(-A
*a/b**3 + 3*B*a**2/(2*b**4)) + (A*a**3*b - B*a**4)/(2*a*b**5 + 2*b**6*x**2)

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