3.117 \(\int \frac{A+B x+C x^2}{(a+c x^2)^{9/2}} \, dx\)
Optimal. Leaf size=127 \[ \frac{8 x (a C+6 A c)}{105 a^4 c \sqrt{a+c x^2}}+\frac{4 x (a C+6 A c)}{105 a^3 c \left (a+c x^2\right )^{3/2}}+\frac{x (a C+6 A c)}{35 a^2 c \left (a+c x^2\right )^{5/2}}-\frac{a B-x (A c-a C)}{7 a c \left (a+c x^2\right )^{7/2}} \]
[Out]
-(a*B - (A*c - a*C)*x)/(7*a*c*(a + c*x^2)^(7/2)) + ((6*A*c + a*C)*x)/(35*a^2*c*(a + c*x^2)^(5/2)) + (4*(6*A*c
+ a*C)*x)/(105*a^3*c*(a + c*x^2)^(3/2)) + (8*(6*A*c + a*C)*x)/(105*a^4*c*Sqrt[a + c*x^2])
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Rubi [A] time = 0.0871571, antiderivative size = 127, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used =
{1814, 12, 192, 191} \[ \frac{8 x (a C+6 A c)}{105 a^4 c \sqrt{a+c x^2}}+\frac{4 x (a C+6 A c)}{105 a^3 c \left (a+c x^2\right )^{3/2}}+\frac{x (a C+6 A c)}{35 a^2 c \left (a+c x^2\right )^{5/2}}-\frac{a B-x (A c-a C)}{7 a c \left (a+c x^2\right )^{7/2}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*x + C*x^2)/(a + c*x^2)^(9/2),x]
[Out]
-(a*B - (A*c - a*C)*x)/(7*a*c*(a + c*x^2)^(7/2)) + ((6*A*c + a*C)*x)/(35*a^2*c*(a + c*x^2)^(5/2)) + (4*(6*A*c
+ a*C)*x)/(105*a^3*c*(a + c*x^2)^(3/2)) + (8*(6*A*c + a*C)*x)/(105*a^4*c*Sqrt[a + c*x^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]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 192
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]
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]
Rubi steps
\begin{align*} \int \frac{A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx &=-\frac{a B-(A c-a C) x}{7 a c \left (a+c x^2\right )^{7/2}}-\frac{\int \frac{-6 A-\frac{a C}{c}}{\left (a+c x^2\right )^{7/2}} \, dx}{7 a}\\ &=-\frac{a B-(A c-a C) x}{7 a c \left (a+c x^2\right )^{7/2}}+\frac{(6 A c+a C) \int \frac{1}{\left (a+c x^2\right )^{7/2}} \, dx}{7 a c}\\ &=-\frac{a B-(A c-a C) x}{7 a c \left (a+c x^2\right )^{7/2}}+\frac{(6 A c+a C) x}{35 a^2 c \left (a+c x^2\right )^{5/2}}+\frac{(4 (6 A c+a C)) \int \frac{1}{\left (a+c x^2\right )^{5/2}} \, dx}{35 a^2 c}\\ &=-\frac{a B-(A c-a C) x}{7 a c \left (a+c x^2\right )^{7/2}}+\frac{(6 A c+a C) x}{35 a^2 c \left (a+c x^2\right )^{5/2}}+\frac{4 (6 A c+a C) x}{105 a^3 c \left (a+c x^2\right )^{3/2}}+\frac{(8 (6 A c+a C)) \int \frac{1}{\left (a+c x^2\right )^{3/2}} \, dx}{105 a^3 c}\\ &=-\frac{a B-(A c-a C) x}{7 a c \left (a+c x^2\right )^{7/2}}+\frac{(6 A c+a C) x}{35 a^2 c \left (a+c x^2\right )^{5/2}}+\frac{4 (6 A c+a C) x}{105 a^3 c \left (a+c x^2\right )^{3/2}}+\frac{8 (6 A c+a C) x}{105 a^4 c \sqrt{a+c x^2}}\\ \end{align*}
Mathematica [A] time = 0.0670119, size = 92, normalized size = 0.72 \[ \frac{14 a^2 c^2 x^3 \left (15 A+2 C x^2\right )+35 a^3 c x \left (3 A+C x^2\right )-15 a^4 B+8 a c^3 x^5 \left (21 A+C x^2\right )+48 A c^4 x^7}{105 a^4 c \left (a+c x^2\right )^{7/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*x + C*x^2)/(a + c*x^2)^(9/2),x]
[Out]
(-15*a^4*B + 48*A*c^4*x^7 + 35*a^3*c*x*(3*A + C*x^2) + 8*a*c^3*x^5*(21*A + C*x^2) + 14*a^2*c^2*x^3*(15*A + 2*C
*x^2))/(105*a^4*c*(a + c*x^2)^(7/2))
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Maple [A] time = 0.052, size = 96, normalized size = 0.8 \begin{align*}{\frac{48\,A{c}^{4}{x}^{7}+8\,Ca{c}^{3}{x}^{7}+168\,Aa{c}^{3}{x}^{5}+28\,C{a}^{2}{c}^{2}{x}^{5}+210\,A{a}^{2}{c}^{2}{x}^{3}+35\,C{a}^{3}c{x}^{3}+105\,Ax{a}^{3}c-15\,B{a}^{4}}{105\,{a}^{4}c} \left ( c{x}^{2}+a \right ) ^{-{\frac{7}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((C*x^2+B*x+A)/(c*x^2+a)^(9/2),x)
[Out]
1/105*(48*A*c^4*x^7+8*C*a*c^3*x^7+168*A*a*c^3*x^5+28*C*a^2*c^2*x^5+210*A*a^2*c^2*x^3+35*C*a^3*c*x^3+105*A*a^3*
c*x-15*B*a^4)/(c*x^2+a)^(7/2)/a^4/c
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Maxima [A] time = 0.99926, size = 207, normalized size = 1.63 \begin{align*} \frac{16 \, A x}{35 \, \sqrt{c x^{2} + a} a^{4}} + \frac{8 \, A x}{35 \,{\left (c x^{2} + a\right )}^{\frac{3}{2}} a^{3}} + \frac{6 \, A x}{35 \,{\left (c x^{2} + a\right )}^{\frac{5}{2}} a^{2}} + \frac{A x}{7 \,{\left (c x^{2} + a\right )}^{\frac{7}{2}} a} - \frac{C x}{7 \,{\left (c x^{2} + a\right )}^{\frac{7}{2}} c} + \frac{8 \, C x}{105 \, \sqrt{c x^{2} + a} a^{3} c} + \frac{4 \, C x}{105 \,{\left (c x^{2} + a\right )}^{\frac{3}{2}} a^{2} c} + \frac{C x}{35 \,{\left (c x^{2} + a\right )}^{\frac{5}{2}} a c} - \frac{B}{7 \,{\left (c x^{2} + a\right )}^{\frac{7}{2}} c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((C*x^2+B*x+A)/(c*x^2+a)^(9/2),x, algorithm="maxima")
[Out]
16/35*A*x/(sqrt(c*x^2 + a)*a^4) + 8/35*A*x/((c*x^2 + a)^(3/2)*a^3) + 6/35*A*x/((c*x^2 + a)^(5/2)*a^2) + 1/7*A*
x/((c*x^2 + a)^(7/2)*a) - 1/7*C*x/((c*x^2 + a)^(7/2)*c) + 8/105*C*x/(sqrt(c*x^2 + a)*a^3*c) + 4/105*C*x/((c*x^
2 + a)^(3/2)*a^2*c) + 1/35*C*x/((c*x^2 + a)^(5/2)*a*c) - 1/7*B/((c*x^2 + a)^(7/2)*c)
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Fricas [A] time = 1.71249, size = 289, normalized size = 2.28 \begin{align*} \frac{{\left (8 \,{\left (C a c^{3} + 6 \, A c^{4}\right )} x^{7} + 105 \, A a^{3} c x + 28 \,{\left (C a^{2} c^{2} + 6 \, A a c^{3}\right )} x^{5} - 15 \, B a^{4} + 35 \,{\left (C a^{3} c + 6 \, A a^{2} c^{2}\right )} x^{3}\right )} \sqrt{c x^{2} + a}}{105 \,{\left (a^{4} c^{5} x^{8} + 4 \, a^{5} c^{4} x^{6} + 6 \, a^{6} c^{3} x^{4} + 4 \, a^{7} c^{2} x^{2} + a^{8} c\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((C*x^2+B*x+A)/(c*x^2+a)^(9/2),x, algorithm="fricas")
[Out]
1/105*(8*(C*a*c^3 + 6*A*c^4)*x^7 + 105*A*a^3*c*x + 28*(C*a^2*c^2 + 6*A*a*c^3)*x^5 - 15*B*a^4 + 35*(C*a^3*c + 6
*A*a^2*c^2)*x^3)*sqrt(c*x^2 + a)/(a^4*c^5*x^8 + 4*a^5*c^4*x^6 + 6*a^6*c^3*x^4 + 4*a^7*c^2*x^2 + a^8*c)
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Sympy [B] time = 122.907, size = 1880, normalized size = 14.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((C*x**2+B*x+A)/(c*x**2+a)**(9/2),x)
[Out]
A*(35*a**14*x/(35*a**(37/2)*sqrt(1 + c*x**2/a) + 210*a**(35/2)*c*x**2*sqrt(1 + c*x**2/a) + 525*a**(33/2)*c**2*
x**4*sqrt(1 + c*x**2/a) + 700*a**(31/2)*c**3*x**6*sqrt(1 + c*x**2/a) + 525*a**(29/2)*c**4*x**8*sqrt(1 + c*x**2
/a) + 210*a**(27/2)*c**5*x**10*sqrt(1 + c*x**2/a) + 35*a**(25/2)*c**6*x**12*sqrt(1 + c*x**2/a)) + 175*a**13*c*
x**3/(35*a**(37/2)*sqrt(1 + c*x**2/a) + 210*a**(35/2)*c*x**2*sqrt(1 + c*x**2/a) + 525*a**(33/2)*c**2*x**4*sqrt
(1 + c*x**2/a) + 700*a**(31/2)*c**3*x**6*sqrt(1 + c*x**2/a) + 525*a**(29/2)*c**4*x**8*sqrt(1 + c*x**2/a) + 210
*a**(27/2)*c**5*x**10*sqrt(1 + c*x**2/a) + 35*a**(25/2)*c**6*x**12*sqrt(1 + c*x**2/a)) + 371*a**12*c**2*x**5/(
35*a**(37/2)*sqrt(1 + c*x**2/a) + 210*a**(35/2)*c*x**2*sqrt(1 + c*x**2/a) + 525*a**(33/2)*c**2*x**4*sqrt(1 + c
*x**2/a) + 700*a**(31/2)*c**3*x**6*sqrt(1 + c*x**2/a) + 525*a**(29/2)*c**4*x**8*sqrt(1 + c*x**2/a) + 210*a**(2
7/2)*c**5*x**10*sqrt(1 + c*x**2/a) + 35*a**(25/2)*c**6*x**12*sqrt(1 + c*x**2/a)) + 429*a**11*c**3*x**7/(35*a**
(37/2)*sqrt(1 + c*x**2/a) + 210*a**(35/2)*c*x**2*sqrt(1 + c*x**2/a) + 525*a**(33/2)*c**2*x**4*sqrt(1 + c*x**2/
a) + 700*a**(31/2)*c**3*x**6*sqrt(1 + c*x**2/a) + 525*a**(29/2)*c**4*x**8*sqrt(1 + c*x**2/a) + 210*a**(27/2)*c
**5*x**10*sqrt(1 + c*x**2/a) + 35*a**(25/2)*c**6*x**12*sqrt(1 + c*x**2/a)) + 286*a**10*c**4*x**9/(35*a**(37/2)
*sqrt(1 + c*x**2/a) + 210*a**(35/2)*c*x**2*sqrt(1 + c*x**2/a) + 525*a**(33/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 7
00*a**(31/2)*c**3*x**6*sqrt(1 + c*x**2/a) + 525*a**(29/2)*c**4*x**8*sqrt(1 + c*x**2/a) + 210*a**(27/2)*c**5*x*
*10*sqrt(1 + c*x**2/a) + 35*a**(25/2)*c**6*x**12*sqrt(1 + c*x**2/a)) + 104*a**9*c**5*x**11/(35*a**(37/2)*sqrt(
1 + c*x**2/a) + 210*a**(35/2)*c*x**2*sqrt(1 + c*x**2/a) + 525*a**(33/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 700*a**
(31/2)*c**3*x**6*sqrt(1 + c*x**2/a) + 525*a**(29/2)*c**4*x**8*sqrt(1 + c*x**2/a) + 210*a**(27/2)*c**5*x**10*sq
rt(1 + c*x**2/a) + 35*a**(25/2)*c**6*x**12*sqrt(1 + c*x**2/a)) + 16*a**8*c**6*x**13/(35*a**(37/2)*sqrt(1 + c*x
**2/a) + 210*a**(35/2)*c*x**2*sqrt(1 + c*x**2/a) + 525*a**(33/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 700*a**(31/2)*
c**3*x**6*sqrt(1 + c*x**2/a) + 525*a**(29/2)*c**4*x**8*sqrt(1 + c*x**2/a) + 210*a**(27/2)*c**5*x**10*sqrt(1 +
c*x**2/a) + 35*a**(25/2)*c**6*x**12*sqrt(1 + c*x**2/a))) + B*Piecewise((-1/(7*a**3*c*sqrt(a + c*x**2) + 21*a**
2*c**2*x**2*sqrt(a + c*x**2) + 21*a*c**3*x**4*sqrt(a + c*x**2) + 7*c**4*x**6*sqrt(a + c*x**2)), Ne(c, 0)), (x*
*2/(2*a**(9/2)), True)) + C*(35*a**5*x**3/(105*a**(19/2)*sqrt(1 + c*x**2/a) + 420*a**(17/2)*c*x**2*sqrt(1 + c*
x**2/a) + 630*a**(15/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 420*a**(13/2)*c**3*x**6*sqrt(1 + c*x**2/a) + 105*a**(11
/2)*c**4*x**8*sqrt(1 + c*x**2/a)) + 63*a**4*c*x**5/(105*a**(19/2)*sqrt(1 + c*x**2/a) + 420*a**(17/2)*c*x**2*sq
rt(1 + c*x**2/a) + 630*a**(15/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 420*a**(13/2)*c**3*x**6*sqrt(1 + c*x**2/a) + 1
05*a**(11/2)*c**4*x**8*sqrt(1 + c*x**2/a)) + 36*a**3*c**2*x**7/(105*a**(19/2)*sqrt(1 + c*x**2/a) + 420*a**(17/
2)*c*x**2*sqrt(1 + c*x**2/a) + 630*a**(15/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 420*a**(13/2)*c**3*x**6*sqrt(1 + c
*x**2/a) + 105*a**(11/2)*c**4*x**8*sqrt(1 + c*x**2/a)) + 8*a**2*c**3*x**9/(105*a**(19/2)*sqrt(1 + c*x**2/a) +
420*a**(17/2)*c*x**2*sqrt(1 + c*x**2/a) + 630*a**(15/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 420*a**(13/2)*c**3*x**6
*sqrt(1 + c*x**2/a) + 105*a**(11/2)*c**4*x**8*sqrt(1 + c*x**2/a)))
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Giac [A] time = 1.21873, size = 151, normalized size = 1.19 \begin{align*} \frac{{\left ({\left (4 \, x^{2}{\left (\frac{2 \,{\left (C a c^{5} + 6 \, A c^{6}\right )} x^{2}}{a^{4} c^{3}} + \frac{7 \,{\left (C a^{2} c^{4} + 6 \, A a c^{5}\right )}}{a^{4} c^{3}}\right )} + \frac{35 \,{\left (C a^{3} c^{3} + 6 \, A a^{2} c^{4}\right )}}{a^{4} c^{3}}\right )} x^{2} + \frac{105 \, A}{a}\right )} x - \frac{15 \, B}{c}}{105 \,{\left (c x^{2} + a\right )}^{\frac{7}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((C*x^2+B*x+A)/(c*x^2+a)^(9/2),x, algorithm="giac")
[Out]
1/105*(((4*x^2*(2*(C*a*c^5 + 6*A*c^6)*x^2/(a^4*c^3) + 7*(C*a^2*c^4 + 6*A*a*c^5)/(a^4*c^3)) + 35*(C*a^3*c^3 + 6
*A*a^2*c^4)/(a^4*c^3))*x^2 + 105*A/a)*x - 15*B/c)/(c*x^2 + a)^(7/2)