3.1.50 \(\int \frac {x^4 (A+B x+C x^2)}{(a+b x^2)^{9/2}} \, dx\)

Optimal. Leaf size=149 \[ \frac {x (5 a C+2 A b)}{35 a^2 b^3 \sqrt {a+b x^2}}-\frac {3 x (5 a C+2 A b)+8 a B}{105 a b^3 \left (a+b x^2\right )^{3/2}}-\frac {x^2 (x (5 a C+2 A b)+4 a B)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {x^4 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}} \]

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Rubi [A]  time = 0.18, antiderivative size = 149, 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, 778, 191} \begin {gather*} \frac {x (5 a C+2 A b)}{35 a^2 b^3 \sqrt {a+b x^2}}-\frac {x^2 (x (5 a C+2 A b)+4 a B)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {3 x (5 a C+2 A b)+8 a B}{105 a b^3 \left (a+b x^2\right )^{3/2}}-\frac {x^4 (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^4*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x]

[Out]

-(x^4*(a*B - (A*b - a*C)*x))/(7*a*b*(a + b*x^2)^(7/2)) - (x^2*(4*a*B + (2*A*b + 5*a*C)*x))/(35*a*b^2*(a + b*x^
2)^(5/2)) - (8*a*B + 3*(2*A*b + 5*a*C)*x)/(105*a*b^3*(a + b*x^2)^(3/2)) + ((2*A*b + 5*a*C)*x)/(35*a^2*b^3*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 778

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

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^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx &=-\frac {x^4 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {\int \frac {x^3 (-4 a B-(2 A b+5 a C) x)}{\left (a+b x^2\right )^{7/2}} \, dx}{7 a b}\\ &=-\frac {x^4 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {x^2 (4 a B+(2 A b+5 a C) x)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {\int \frac {x \left (-8 a^2 B-3 a (2 A b+5 a C) x\right )}{\left (a+b x^2\right )^{5/2}} \, dx}{35 a^2 b^2}\\ &=-\frac {x^4 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {x^2 (4 a B+(2 A b+5 a C) x)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {8 a B+3 (2 A b+5 a C) x}{105 a b^3 \left (a+b x^2\right )^{3/2}}+\frac {(2 A b+5 a C) \int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx}{35 a b^3}\\ &=-\frac {x^4 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {x^2 (4 a B+(2 A b+5 a C) x)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {8 a B+3 (2 A b+5 a C) x}{105 a b^3 \left (a+b x^2\right )^{3/2}}+\frac {(2 A b+5 a C) x}{35 a^2 b^3 \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 78, normalized size = 0.52 \begin {gather*} \frac {-8 a^4 B-28 a^3 b B x^2-35 a^2 b^2 B x^4+3 a b^3 x^5 \left (7 A+5 C x^2\right )+6 A b^4 x^7}{105 a^2 b^3 \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.93, size = 79, normalized size = 0.53 \begin {gather*} \frac {-8 a^4 B-28 a^3 b B x^2-35 a^2 b^2 B x^4+21 a A b^3 x^5+15 a b^3 C x^7+6 A b^4 x^7}{105 a^2 b^3 \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x]

[Out]

(-8*a^4*B - 28*a^3*b*B*x^2 - 35*a^2*b^2*B*x^4 + 21*a*A*b^3*x^5 + 6*A*b^4*x^7 + 15*a*b^3*C*x^7)/(105*a^2*b^3*(a
+ b*x^2)^(7/2))

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fricas [A]  time = 0.93, size = 122, normalized size = 0.82 \begin {gather*} \frac {{\left (21 \, A a b^{3} x^{5} - 35 \, B a^{2} b^{2} x^{4} + 3 \, {\left (5 \, C a b^{3} + 2 \, A b^{4}\right )} x^{7} - 28 \, B a^{3} b x^{2} - 8 \, B a^{4}\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^4*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="fricas")

[Out]

1/105*(21*A*a*b^3*x^5 - 35*B*a^2*b^2*x^4 + 3*(5*C*a*b^3 + 2*A*b^4)*x^7 - 28*B*a^3*b*x^2 - 8*B*a^4)*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 = 0.58, size = 81, normalized size = 0.54 \begin {gather*} \frac {{\left ({\left (3 \, x {\left (\frac {7 \, A}{a} + \frac {{\left (5 \, C a^{2} b^{3} + 2 \, A a b^{4}\right )} x^{2}}{a^{3} b^{3}}\right )} - \frac {35 \, B}{b}\right )} x^{2} - \frac {28 \, B a}{b^{2}}\right )} x^{2} - \frac {8 \, B a^{2}}{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^4*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 76, normalized size = 0.51 \begin {gather*} \frac {6 A \,b^{4} x^{7}+15 C a \,b^{3} x^{7}+21 A \,x^{5} a \,b^{3}-35 B \,a^{2} b^{2} x^{4}-28 B \,a^{3} b \,x^{2}-8 B \,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^4*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x)

[Out]

1/105*(6*A*b^4*x^7+15*C*a*b^3*x^7+21*A*a*b^3*x^5-35*B*a^2*b^2*x^4-28*B*a^3*b*x^2-8*B*a^4)/(b*x^2+a)^(7/2)/a^2/
b^3

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maxima [A]  time = 1.42, size = 253, normalized size = 1.70 \begin {gather*} -\frac {C x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {B x^{4}}{3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {5 \, C a x^{3}}{8 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {A x^{3}}{4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {4 \, B a x^{2}}{15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} + \frac {C x}{14 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} + \frac {C x}{7 \, \sqrt {b x^{2} + a} a b^{3}} + \frac {3 \, C a x}{56 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}} - \frac {15 \, C a^{2} x}{56 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} + \frac {3 \, A x}{140 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {2 \, A x}{35 \, \sqrt {b x^{2} + a} a^{2} b^{2}} + \frac {A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2}} - \frac {3 \, A a x}{28 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {8 \, B a^{2}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="maxima")

[Out]

-1/2*C*x^5/((b*x^2 + a)^(7/2)*b) - 1/3*B*x^4/((b*x^2 + a)^(7/2)*b) - 5/8*C*a*x^3/((b*x^2 + a)^(7/2)*b^2) - 1/4
*A*x^3/((b*x^2 + a)^(7/2)*b) - 4/15*B*a*x^2/((b*x^2 + a)^(7/2)*b^2) + 1/14*C*x/((b*x^2 + a)^(3/2)*b^3) + 1/7*C
*x/(sqrt(b*x^2 + a)*a*b^3) + 3/56*C*a*x/((b*x^2 + a)^(5/2)*b^3) - 15/56*C*a^2*x/((b*x^2 + a)^(7/2)*b^3) + 3/14
0*A*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*A*x/(sqrt(b*x^2 + a)*a^2*b^2) + 1/35*A*x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28
*A*a*x/((b*x^2 + a)^(7/2)*b^2) - 8/105*B*a^2/((b*x^2 + a)^(7/2)*b^3)

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mupad [B]  time = 1.19, size = 186, normalized size = 1.25 \begin {gather*} \frac {x\,\left (\frac {C\,a^2-A\,a\,b}{35\,a\,b^3}+\frac {a\,\left (\frac {C}{5\,b^2}-\frac {7\,A\,b^2-7\,C\,a\,b}{35\,a\,b^3}\right )}{b}\right )+\frac {2\,B\,a}{5\,b^3}}{{\left (b\,x^2+a\right )}^{5/2}}-\frac {\frac {B}{3\,b^3}+x\,\left (\frac {C}{3\,b^3}-\frac {3\,A\,b-10\,C\,a}{105\,a\,b^3}\right )}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {\frac {B\,a^2}{7\,b^3}-\frac {a\,x\,\left (\frac {A}{7\,b}-\frac {C\,a}{7\,b^2}\right )}{b}}{{\left (b\,x^2+a\right )}^{7/2}}+\frac {x\,\left (2\,A\,b+5\,C\,a\right )}{35\,a^2\,b^3\,\sqrt {b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x)

[Out]

(x*((C*a^2 - A*a*b)/(35*a*b^3) + (a*(C/(5*b^2) - (7*A*b^2 - 7*C*a*b)/(35*a*b^3)))/b) + (2*B*a)/(5*b^3))/(a + b
*x^2)^(5/2) - (B/(3*b^3) + x*(C/(3*b^3) - (3*A*b - 10*C*a)/(105*a*b^3)))/(a + b*x^2)^(3/2) - ((B*a^2)/(7*b^3)
- (a*x*(A/(7*b) - (C*a)/(7*b^2)))/b)/(a + b*x^2)^(7/2) + (x*(2*A*b + 5*C*a))/(35*a^2*b^3*(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**4*(C*x**2+B*x+A)/(b*x**2+a)**(9/2),x)

[Out]

Timed out

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