3.1.51 \(\int \frac {x^3 (A+B x+C x^2)}{(a+b x^2)^{9/2}} \, dx\)

Optimal. Leaf size=139 \[ \frac {2 B x}{35 a^2 b^2 \sqrt {a+b x^2}}-\frac {2 (4 a C+3 A b)-3 b B x}{105 a b^3 \left (a+b x^2\right )^{3/2}}-\frac {x (x (4 a C+3 A b)+3 a B)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {x^3 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1804, 819, 639, 191} \begin {gather*} \frac {2 B x}{35 a^2 b^2 \sqrt {a+b x^2}}-\frac {x (x (4 a C+3 A b)+3 a B)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {2 (4 a C+3 A b)-3 b B x}{105 a b^3 \left (a+b x^2\right )^{3/2}}-\frac {x^3 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x]

[Out]

-(x^3*(a*B - (A*b - a*C)*x))/(7*a*b*(a + b*x^2)^(7/2)) - (x*(3*a*B + (3*A*b + 4*a*C)*x))/(35*a*b^2*(a + b*x^2)
^(5/2)) - (2*(3*A*b + 4*a*C) - 3*b*B*x)/(105*a*b^3*(a + b*x^2)^(3/2)) + (2*B*x)/(35*a^2*b^2*Sqrt[a + b*x^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 1804

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

Rubi steps

\begin {align*} \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx &=-\frac {x^3 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {\int \frac {x^2 (-3 a B-(3 A b+4 a C) x)}{\left (a+b x^2\right )^{7/2}} \, dx}{7 a b}\\ &=-\frac {x^3 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {x (3 a B+(3 A b+4 a C) x)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {\int \frac {-3 a^2 B-2 a (3 A b+4 a C) x}{\left (a+b x^2\right )^{5/2}} \, dx}{35 a^2 b^2}\\ &=-\frac {x^3 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {x (3 a B+(3 A b+4 a C) x)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {2 (3 A b+4 a C)-3 b B x}{105 a b^3 \left (a+b x^2\right )^{3/2}}+\frac {(2 B) \int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx}{35 a b^2}\\ &=-\frac {x^3 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {x (3 a B+(3 A b+4 a C) x)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {2 (3 A b+4 a C)-3 b B x}{105 a b^3 \left (a+b x^2\right )^{3/2}}+\frac {2 B x}{35 a^2 b^2 \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 84, normalized size = 0.60 \begin {gather*} \frac {-8 a^4 C-2 a^3 b \left (3 A+14 C x^2\right )-7 a^2 b^2 x^2 \left (3 A+5 C x^2\right )+21 a b^3 B x^5+6 b^4 B x^7}{105 a^2 b^3 \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x]

[Out]

(-8*a^4*C + 21*a*b^3*B*x^5 + 6*b^4*B*x^7 - 7*a^2*b^2*x^2*(3*A + 5*C*x^2) - 2*a^3*b*(3*A + 14*C*x^2))/(105*a^2*
b^3*(a + b*x^2)^(7/2))

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IntegrateAlgebraic [A]  time = 1.15, size = 88, normalized size = 0.63 \begin {gather*} \frac {-8 a^4 C-6 a^3 A b-28 a^3 b C x^2-21 a^2 A b^2 x^2-35 a^2 b^2 C x^4+21 a b^3 B x^5+6 b^4 B x^7}{105 a^2 b^3 \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x]

[Out]

(-6*a^3*A*b - 8*a^4*C - 21*a^2*A*b^2*x^2 - 28*a^3*b*C*x^2 - 35*a^2*b^2*C*x^4 + 21*a*b^3*B*x^5 + 6*b^4*B*x^7)/(
105*a^2*b^3*(a + b*x^2)^(7/2))

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fricas [A]  time = 1.01, size = 131, normalized size = 0.94 \begin {gather*} \frac {{\left (6 \, B b^{4} x^{7} + 21 \, B a b^{3} x^{5} - 35 \, C a^{2} b^{2} x^{4} - 8 \, C a^{4} - 6 \, A a^{3} b - 7 \, {\left (4 \, C a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{2} b^{7} x^{8} + 4 \, a^{3} b^{6} x^{6} + 6 \, a^{4} b^{5} x^{4} + 4 \, a^{5} b^{4} x^{2} + a^{6} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="fricas")

[Out]

1/105*(6*B*b^4*x^7 + 21*B*a*b^3*x^5 - 35*C*a^2*b^2*x^4 - 8*C*a^4 - 6*A*a^3*b - 7*(4*C*a^3*b + 3*A*a^2*b^2)*x^2
)*sqrt(b*x^2 + a)/(a^2*b^7*x^8 + 4*a^3*b^6*x^6 + 6*a^4*b^5*x^4 + 4*a^5*b^4*x^2 + a^6*b^3)

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giac [A]  time = 0.59, size = 95, normalized size = 0.68 \begin {gather*} \frac {{\left ({\left (3 \, {\left (\frac {2 \, B b x^{2}}{a^{2}} + \frac {7 \, B}{a}\right )} x - \frac {35 \, C}{b}\right )} x^{2} - \frac {7 \, {\left (4 \, C a^{4} b + 3 \, A a^{3} b^{2}\right )}}{a^{3} b^{3}}\right )} x^{2} - \frac {2 \, {\left (4 \, C a^{5} + 3 \, A a^{4} b\right )}}{a^{3} b^{3}}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="giac")

[Out]

1/105*(((3*(2*B*b*x^2/a^2 + 7*B/a)*x - 35*C/b)*x^2 - 7*(4*C*a^4*b + 3*A*a^3*b^2)/(a^3*b^3))*x^2 - 2*(4*C*a^5 +
3*A*a^4*b)/(a^3*b^3))/(b*x^2 + a)^(7/2)

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maple [A]  time = 0.01, size = 85, normalized size = 0.61 \begin {gather*} -\frac {-6 B \,x^{7} b^{4}-21 B \,x^{5} a \,b^{3}+35 C \,a^{2} b^{2} x^{4}+21 A \,a^{2} b^{2} x^{2}+28 C \,a^{3} b \,x^{2}+6 A \,a^{3} b +8 C \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x)

[Out]

-1/105*(-6*B*b^4*x^7-21*B*a*b^3*x^5+35*C*a^2*b^2*x^4+21*A*a^2*b^2*x^2+28*C*a^3*b*x^2+6*A*a^3*b+8*C*a^4)/(b*x^2
+a)^(7/2)/a^2/b^3

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maxima [A]  time = 1.38, size = 179, normalized size = 1.29 \begin {gather*} -\frac {C x^{4}}{3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {B x^{3}}{4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {4 \, C a x^{2}}{15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {A x^{2}}{5 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {3 \, B x}{140 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {2 \, B x}{35 \, \sqrt {b x^{2} + a} a^{2} b^{2}} + \frac {B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2}} - \frac {3 \, B a x}{28 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {8 \, C a^{2}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} - \frac {2 \, A a}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="maxima")

[Out]

-1/3*C*x^4/((b*x^2 + a)^(7/2)*b) - 1/4*B*x^3/((b*x^2 + a)^(7/2)*b) - 4/15*C*a*x^2/((b*x^2 + a)^(7/2)*b^2) - 1/
5*A*x^2/((b*x^2 + a)^(7/2)*b) + 3/140*B*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*B*x/(sqrt(b*x^2 + a)*a^2*b^2) + 1/35*
B*x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28*B*a*x/((b*x^2 + a)^(7/2)*b^2) - 8/105*C*a^2/((b*x^2 + a)^(7/2)*b^3) - 2/3
5*A*a/((b*x^2 + a)^(7/2)*b^2)

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mupad [B]  time = 1.14, size = 133, normalized size = 0.96 \begin {gather*} \frac {\frac {a\,\left (\frac {A}{7\,b}-\frac {C\,a}{7\,b^2}\right )}{b}+\frac {B\,a\,x}{7\,b^2}}{{\left (b\,x^2+a\right )}^{7/2}}-\frac {\frac {C}{3\,b^3}-\frac {B\,x}{35\,a\,b^2}}{{\left (b\,x^2+a\right )}^{3/2}}+\frac {\frac {a\,\left (\frac {C}{5\,b^2}-\frac {7\,A\,b-7\,C\,a}{35\,a\,b^2}\right )}{b}-\frac {8\,B\,x}{35\,b^2}}{{\left (b\,x^2+a\right )}^{5/2}}+\frac {2\,B\,x}{35\,a^2\,b^2\,\sqrt {b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x)

[Out]

((a*(A/(7*b) - (C*a)/(7*b^2)))/b + (B*a*x)/(7*b^2))/(a + b*x^2)^(7/2) - (C/(3*b^3) - (B*x)/(35*a*b^2))/(a + b*
x^2)^(3/2) + ((a*(C/(5*b^2) - (7*A*b - 7*C*a)/(35*a*b^2)))/b - (8*B*x)/(35*b^2))/(a + b*x^2)^(5/2) + (2*B*x)/(
35*a^2*b^2*(a + b*x^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(C*x**2+B*x+A)/(b*x**2+a)**(9/2),x)

[Out]

Timed out

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