3.1.100 \(\int \frac {x^6 (A+B x^2)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=116 \[ -\frac {a^2 x (A b-a B)}{4 b^4 \left (a+b x^2\right )^2}-\frac {5 \sqrt {a} (3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{9/2}}+\frac {a x (9 A b-13 a B)}{8 b^4 \left (a+b x^2\right )}+\frac {x (A b-3 a B)}{b^4}+\frac {B x^3}{3 b^3} \]

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Rubi [A]  time = 0.15, antiderivative size = 116, 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, 1153, 205} \begin {gather*} -\frac {a^2 x (A b-a B)}{4 b^4 \left (a+b x^2\right )^2}+\frac {a x (9 A b-13 a B)}{8 b^4 \left (a+b x^2\right )}+\frac {x (A b-3 a B)}{b^4}-\frac {5 \sqrt {a} (3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{9/2}}+\frac {B x^3}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^6*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

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

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 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -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^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx &=-\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}-\frac {\int \frac {-a^2 (A b-a B)+4 a b (A b-a B) x^2-4 b^2 (A b-a B) x^4-4 b^3 B x^6}{\left (a+b x^2\right )^2} \, dx}{4 b^4}\\ &=-\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}+\frac {\int \frac {-a^2 (7 A b-11 a B)+8 a b (A b-2 a B) x^2+8 a b^2 B x^4}{a+b x^2} \, dx}{8 a b^4}\\ &=-\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}+\frac {\int \left (8 a (A b-3 a B)+8 a b B x^2+\frac {5 \left (-3 a^2 A b+7 a^3 B\right )}{a+b x^2}\right ) \, dx}{8 a b^4}\\ &=\frac {(A b-3 a B) x}{b^4}+\frac {B x^3}{3 b^3}-\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}-\frac {(5 a (3 A b-7 a B)) \int \frac {1}{a+b x^2} \, dx}{8 b^4}\\ &=\frac {(A b-3 a B) x}{b^4}+\frac {B x^3}{3 b^3}-\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}-\frac {5 \sqrt {a} (3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 113, normalized size = 0.97 \begin {gather*} \frac {-105 a^3 B x+5 a^2 b x \left (9 A-35 B x^2\right )+a b^2 x^3 \left (75 A-56 B x^2\right )+8 b^3 x^5 \left (3 A+B x^2\right )}{24 b^4 \left (a+b x^2\right )^2}+\frac {5 \sqrt {a} (7 a B-3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(-105*a^3*B*x + a*b^2*x^3*(75*A - 56*B*x^2) + 5*a^2*b*x*(9*A - 35*B*x^2) + 8*b^3*x^5*(3*A + B*x^2))/(24*b^4*(a
+ b*x^2)^2) + (5*Sqrt[a]*(-3*A*b + 7*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(9/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^6*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

IntegrateAlgebraic[(x^6*(A + B*x^2))/(a + b*x^2)^3, x]

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fricas [A]  time = 0.45, size = 358, normalized size = 3.09 \begin {gather*} \left [\frac {16 \, B b^{3} x^{7} - 16 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{5} - 50 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{3} - 15 \, {\left ({\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{4} + 7 \, B a^{3} - 3 \, A a^{2} b + 2 \, {\left (7 \, B a^{2} b - 3 \, A a 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 ) - 30 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} x}{48 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac {8 \, B b^{3} x^{7} - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{5} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{3} + 15 \, {\left ({\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{4} + 7 \, B a^{3} - 3 \, A a^{2} b + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} x}{24 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/48*(16*B*b^3*x^7 - 16*(7*B*a*b^2 - 3*A*b^3)*x^5 - 50*(7*B*a^2*b - 3*A*a*b^2)*x^3 - 15*((7*B*a*b^2 - 3*A*b^3
)*x^4 + 7*B*a^3 - 3*A*a^2*b + 2*(7*B*a^2*b - 3*A*a*b^2)*x^2)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*
x^2 + a)) - 30*(7*B*a^3 - 3*A*a^2*b)*x)/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4), 1/24*(8*B*b^3*x^7 - 8*(7*B*a*b^2 -
3*A*b^3)*x^5 - 25*(7*B*a^2*b - 3*A*a*b^2)*x^3 + 15*((7*B*a*b^2 - 3*A*b^3)*x^4 + 7*B*a^3 - 3*A*a^2*b + 2*(7*B*a
^2*b - 3*A*a*b^2)*x^2)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) - 15*(7*B*a^3 - 3*A*a^2*b)*x)/(b^6*x^4 + 2*a*b^5*x^2
+ a^2*b^4)]

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giac [A]  time = 0.38, size = 111, normalized size = 0.96 \begin {gather*} \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} - \frac {13 \, B a^{2} b x^{3} - 9 \, A a b^{2} x^{3} + 11 \, B a^{3} x - 7 \, A a^{2} b x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{4}} + \frac {B b^{6} x^{3} - 9 \, B a b^{5} x + 3 \, A b^{6} x}{3 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 147, normalized size = 1.27 \begin {gather*} \frac {9 A a \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}-\frac {13 B \,a^{2} x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{3}}+\frac {7 A \,a^{2} x}{8 \left (b \,x^{2}+a \right )^{2} b^{3}}-\frac {11 B \,a^{3} x}{8 \left (b \,x^{2}+a \right )^{2} b^{4}}+\frac {B \,x^{3}}{3 b^{3}}-\frac {15 A a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{3}}+\frac {35 B \,a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{4}}+\frac {A x}{b^{3}}-\frac {3 B a x}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(B*x^2+A)/(b*x^2+a)^3,x)

[Out]

1/3*B*x^3/b^3+1/b^3*A*x-3/b^4*B*a*x+9/8*a/b^2/(b*x^2+a)^2*A*x^3-13/8*a^2/b^3/(b*x^2+a)^2*B*x^3+7/8*a^2/b^3/(b*
x^2+a)^2*A*x-11/8*a^3/b^4/(b*x^2+a)^2*B*x-15/8*a/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*A+35/8*a^2/b^4/(a*b
)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*B

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maxima [A]  time = 2.34, size = 120, normalized size = 1.03 \begin {gather*} -\frac {{\left (13 \, B a^{2} b - 9 \, A a b^{2}\right )} x^{3} + {\left (11 \, B a^{3} - 7 \, A a^{2} b\right )} x}{8 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} + \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} + \frac {B b x^{3} - 3 \, {\left (3 \, B a - A b\right )} x}{3 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.11, size = 138, normalized size = 1.19 \begin {gather*} \frac {x^3\,\left (\frac {9\,A\,a\,b^2}{8}-\frac {13\,B\,a^2\,b}{8}\right )-x\,\left (\frac {11\,B\,a^3}{8}-\frac {7\,A\,a^2\,b}{8}\right )}{a^2\,b^4+2\,a\,b^5\,x^2+b^6\,x^4}+x\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )+\frac {B\,x^3}{3\,b^3}+\frac {5\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (3\,A\,b-7\,B\,a\right )}{7\,B\,a^2-3\,A\,a\,b}\right )\,\left (3\,A\,b-7\,B\,a\right )}{8\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(A + B*x^2))/(a + b*x^2)^3,x)

[Out]

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

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sympy [A]  time = 1.25, size = 214, normalized size = 1.84 \begin {gather*} \frac {B x^{3}}{3 b^{3}} + x \left (\frac {A}{b^{3}} - \frac {3 B a}{b^{4}}\right ) - \frac {5 \sqrt {- \frac {a}{b^{9}}} \left (- 3 A b + 7 B a\right ) \log {\left (- \frac {5 b^{4} \sqrt {- \frac {a}{b^{9}}} \left (- 3 A b + 7 B a\right )}{- 15 A b + 35 B a} + x \right )}}{16} + \frac {5 \sqrt {- \frac {a}{b^{9}}} \left (- 3 A b + 7 B a\right ) \log {\left (\frac {5 b^{4} \sqrt {- \frac {a}{b^{9}}} \left (- 3 A b + 7 B a\right )}{- 15 A b + 35 B a} + x \right )}}{16} + \frac {x^{3} \left (9 A a b^{2} - 13 B a^{2} b\right ) + x \left (7 A a^{2} b - 11 B a^{3}\right )}{8 a^{2} b^{4} + 16 a b^{5} x^{2} + 8 b^{6} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(B*x**2+A)/(b*x**2+a)**3,x)

[Out]

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

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