3.99 \(\int \frac {x^8 (A+B x^2)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=138 \[ \frac {7 a^{3/2} (5 A b-9 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}}+\frac {a^3 x (A b-a B)}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 x (13 A b-17 a B)}{8 b^5 \left (a+b x^2\right )}-\frac {3 a x (A b-2 a B)}{b^5}+\frac {x^3 (A b-3 a B)}{3 b^4}+\frac {B x^5}{5 b^3} \]

[Out]

-3*a*(A*b-2*B*a)*x/b^5+1/3*(A*b-3*B*a)*x^3/b^4+1/5*B*x^5/b^3+1/4*a^3*(A*b-B*a)*x/b^5/(b*x^2+a)^2-1/8*a^2*(13*A
*b-17*B*a)*x/b^5/(b*x^2+a)+7/8*a^(3/2)*(5*A*b-9*B*a)*arctan(x*b^(1/2)/a^(1/2))/b^(11/2)

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Rubi [A]  time = 0.22, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {455, 1814, 1810, 205} \[ -\frac {a^2 x (13 A b-17 a B)}{8 b^5 \left (a+b x^2\right )}+\frac {a^3 x (A b-a B)}{4 b^5 \left (a+b x^2\right )^2}+\frac {7 a^{3/2} (5 A b-9 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}}+\frac {x^3 (A b-3 a B)}{3 b^4}-\frac {3 a x (A b-2 a B)}{b^5}+\frac {B x^5}{5 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(-3*a*(A*b - 2*a*B)*x)/b^5 + ((A*b - 3*a*B)*x^3)/(3*b^4) + (B*x^5)/(5*b^3) + (a^3*(A*b - a*B)*x)/(4*b^5*(a + b
*x^2)^2) - (a^2*(13*A*b - 17*a*B)*x)/(8*b^5*(a + b*x^2)) + (7*a^(3/2)*(5*A*b - 9*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[
a]])/(8*b^(11/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx &=\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {\int \frac {a^3 (A b-a B)-4 a^2 b (A b-a B) x^2+4 a b^2 (A b-a B) x^4-4 b^3 (A b-a B) x^6-4 b^4 B x^8}{\left (a+b x^2\right )^2} \, dx}{4 b^5}\\ &=\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {\int \frac {a^3 (11 A b-15 a B)-8 a^2 b (2 A b-3 a B) x^2+8 a b^2 (A b-2 a B) x^4+8 a b^3 B x^6}{a+b x^2} \, dx}{8 a b^5}\\ &=\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {\int \left (-24 a^2 (A b-2 a B)+8 a b (A b-3 a B) x^2+8 a b^2 B x^4-\frac {7 \left (-5 a^3 A b+9 a^4 B\right )}{a+b x^2}\right ) \, dx}{8 a b^5}\\ &=-\frac {3 a (A b-2 a B) x}{b^5}+\frac {(A b-3 a B) x^3}{3 b^4}+\frac {B x^5}{5 b^3}+\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {\left (7 a^2 (5 A b-9 a B)\right ) \int \frac {1}{a+b x^2} \, dx}{8 b^5}\\ &=-\frac {3 a (A b-2 a B) x}{b^5}+\frac {(A b-3 a B) x^3}{3 b^4}+\frac {B x^5}{5 b^3}+\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {7 a^{3/2} (5 A b-9 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 133, normalized size = 0.96 \[ \frac {x \left (945 a^4 B-525 a^3 b \left (A-3 B x^2\right )+7 a^2 b^2 x^2 \left (72 B x^2-125 A\right )-8 a b^3 x^4 \left (35 A+9 B x^2\right )+8 b^4 x^6 \left (5 A+3 B x^2\right )\right )}{120 b^5 \left (a+b x^2\right )^2}-\frac {7 a^{3/2} (9 a B-5 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(x*(945*a^4*B - 525*a^3*b*(A - 3*B*x^2) + 8*b^4*x^6*(5*A + 3*B*x^2) - 8*a*b^3*x^4*(35*A + 9*B*x^2) + 7*a^2*b^2
*x^2*(-125*A + 72*B*x^2)))/(120*b^5*(a + b*x^2)^2) - (7*a^(3/2)*(-5*A*b + 9*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/
(8*b^(11/2))

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fricas [A]  time = 0.49, size = 416, normalized size = 3.01 \[ \left [\frac {48 \, B b^{4} x^{9} - 16 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{7} + 112 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{5} + 350 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3} - 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 210 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b\right )} x}{240 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}, \frac {24 \, B b^{4} x^{9} - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{7} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{5} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3} - 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b\right )} x}{120 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/240*(48*B*b^4*x^9 - 16*(9*B*a*b^3 - 5*A*b^4)*x^7 + 112*(9*B*a^2*b^2 - 5*A*a*b^3)*x^5 + 350*(9*B*a^3*b - 5*A
*a^2*b^2)*x^3 - 105*(9*B*a^4 - 5*A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^4 + 2*(9*B*a^3*b - 5*A*a^2*b^2)*x^2)*sq
rt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 210*(9*B*a^4 - 5*A*a^3*b)*x)/(b^7*x^4 + 2*a*b^6*x^2
+ a^2*b^5), 1/120*(24*B*b^4*x^9 - 8*(9*B*a*b^3 - 5*A*b^4)*x^7 + 56*(9*B*a^2*b^2 - 5*A*a*b^3)*x^5 + 175*(9*B*a
^3*b - 5*A*a^2*b^2)*x^3 - 105*(9*B*a^4 - 5*A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^4 + 2*(9*B*a^3*b - 5*A*a^2*b^
2)*x^2)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) + 105*(9*B*a^4 - 5*A*a^3*b)*x)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5)]

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giac [A]  time = 0.36, size = 138, normalized size = 1.00 \[ -\frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {17 \, B a^{3} b x^{3} - 13 \, A a^{2} b^{2} x^{3} + 15 \, B a^{4} x - 11 \, A a^{3} b x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{5}} + \frac {3 \, B b^{12} x^{5} - 15 \, B a b^{11} x^{3} + 5 \, A b^{12} x^{3} + 90 \, B a^{2} b^{10} x - 45 \, A a b^{11} x}{15 \, b^{15}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-7/8*(9*B*a^3 - 5*A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) + 1/8*(17*B*a^3*b*x^3 - 13*A*a^2*b^2*x^3 + 15
*B*a^4*x - 11*A*a^3*b*x)/((b*x^2 + a)^2*b^5) + 1/15*(3*B*b^12*x^5 - 15*B*a*b^11*x^3 + 5*A*b^12*x^3 + 90*B*a^2*
b^10*x - 45*A*a*b^11*x)/b^15

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maple [A]  time = 0.04, size = 174, normalized size = 1.26 \[ -\frac {13 A \,a^{2} x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{3}}+\frac {17 B \,a^{3} x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{4}}+\frac {B \,x^{5}}{5 b^{3}}-\frac {11 A \,a^{3} x}{8 \left (b \,x^{2}+a \right )^{2} b^{4}}+\frac {A \,x^{3}}{3 b^{3}}+\frac {15 B \,a^{4} x}{8 \left (b \,x^{2}+a \right )^{2} b^{5}}-\frac {B a \,x^{3}}{b^{4}}+\frac {35 A \,a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{4}}-\frac {63 B \,a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{5}}-\frac {3 A a x}{b^{4}}+\frac {6 B \,a^{2} x}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(B*x^2+A)/(b*x^2+a)^3,x)

[Out]

1/5*B*x^5/b^3+1/3/b^3*A*x^3-1/b^4*B*x^3*a-3/b^4*a*A*x+6/b^5*a^2*B*x-13/8*a^2/b^3/(b*x^2+a)^2*A*x^3+17/8*a^3/b^
4/(b*x^2+a)^2*B*x^3-11/8*a^3/b^4/(b*x^2+a)^2*A*x+15/8*a^4/b^5/(b*x^2+a)^2*B*x+35/8*a^2/b^4/(a*b)^(1/2)*arctan(
1/(a*b)^(1/2)*b*x)*A-63/8*a^3/b^5/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*B

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maxima [A]  time = 2.39, size = 147, normalized size = 1.07 \[ \frac {{\left (17 \, B a^{3} b - 13 \, A a^{2} b^{2}\right )} x^{3} + {\left (15 \, B a^{4} - 11 \, A a^{3} b\right )} x}{8 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} - \frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {3 \, B b^{2} x^{5} - 5 \, {\left (3 \, B a b - A b^{2}\right )} x^{3} + 45 \, {\left (2 \, B a^{2} - A a b\right )} x}{15 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/8*((17*B*a^3*b - 13*A*a^2*b^2)*x^3 + (15*B*a^4 - 11*A*a^3*b)*x)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5) - 7/8*(9*B
*a^3 - 5*A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) + 1/15*(3*B*b^2*x^5 - 5*(3*B*a*b - A*b^2)*x^3 + 45*(2*
B*a^2 - A*a*b)*x)/b^5

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mupad [B]  time = 0.10, size = 177, normalized size = 1.28 \[ \frac {x\,\left (\frac {15\,B\,a^4}{8}-\frac {11\,A\,a^3\,b}{8}\right )-x^3\,\left (\frac {13\,A\,a^2\,b^2}{8}-\frac {17\,B\,a^3\,b}{8}\right )}{a^2\,b^5+2\,a\,b^6\,x^2+b^7\,x^4}-x\,\left (\frac {3\,a\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )}{b}+\frac {3\,B\,a^2}{b^5}\right )+x^3\,\left (\frac {A}{3\,b^3}-\frac {B\,a}{b^4}\right )+\frac {B\,x^5}{5\,b^3}-\frac {7\,a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,x\,\left (5\,A\,b-9\,B\,a\right )}{9\,B\,a^3-5\,A\,a^2\,b}\right )\,\left (5\,A\,b-9\,B\,a\right )}{8\,b^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8*(A + B*x^2))/(a + b*x^2)^3,x)

[Out]

(x*((15*B*a^4)/8 - (11*A*a^3*b)/8) - x^3*((13*A*a^2*b^2)/8 - (17*B*a^3*b)/8))/(a^2*b^5 + b^7*x^4 + 2*a*b^6*x^2
) - x*((3*a*(A/b^3 - (3*B*a)/b^4))/b + (3*B*a^2)/b^5) + x^3*(A/(3*b^3) - (B*a)/b^4) + (B*x^5)/(5*b^3) - (7*a^(
3/2)*atan((a^(3/2)*b^(1/2)*x*(5*A*b - 9*B*a))/(9*B*a^3 - 5*A*a^2*b))*(5*A*b - 9*B*a))/(8*b^(11/2))

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sympy [A]  time = 1.36, size = 252, normalized size = 1.83 \[ \frac {B x^{5}}{5 b^{3}} + x^{3} \left (\frac {A}{3 b^{3}} - \frac {B a}{b^{4}}\right ) + x \left (- \frac {3 A a}{b^{4}} + \frac {6 B a^{2}}{b^{5}}\right ) + \frac {7 \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right ) \log {\left (- \frac {7 b^{5} \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right )}{- 35 A a b + 63 B a^{2}} + x \right )}}{16} - \frac {7 \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right ) \log {\left (\frac {7 b^{5} \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right )}{- 35 A a b + 63 B a^{2}} + x \right )}}{16} + \frac {x^{3} \left (- 13 A a^{2} b^{2} + 17 B a^{3} b\right ) + x \left (- 11 A a^{3} b + 15 B a^{4}\right )}{8 a^{2} b^{5} + 16 a b^{6} x^{2} + 8 b^{7} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(B*x**2+A)/(b*x**2+a)**3,x)

[Out]

B*x**5/(5*b**3) + x**3*(A/(3*b**3) - B*a/b**4) + x*(-3*A*a/b**4 + 6*B*a**2/b**5) + 7*sqrt(-a**3/b**11)*(-5*A*b
+ 9*B*a)*log(-7*b**5*sqrt(-a**3/b**11)*(-5*A*b + 9*B*a)/(-35*A*a*b + 63*B*a**2) + x)/16 - 7*sqrt(-a**3/b**11)
*(-5*A*b + 9*B*a)*log(7*b**5*sqrt(-a**3/b**11)*(-5*A*b + 9*B*a)/(-35*A*a*b + 63*B*a**2) + x)/16 + (x**3*(-13*A
*a**2*b**2 + 17*B*a**3*b) + x*(-11*A*a**3*b + 15*B*a**4))/(8*a**2*b**5 + 16*a*b**6*x**2 + 8*b**7*x**4)

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