\(\int \frac {A+B x+C x^2}{(a+c x^2)^{9/2}} \, dx\) [117]

Optimal result

Integrand size = 22, antiderivative size = 127 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx=-\frac {a B-(A c-a C) x}{7 a c \left (a+c x^2\right )^{7/2}}+\frac {(6 A c+a C) x}{35 a^2 c \left (a+c x^2\right )^{5/2}}+\frac {4 (6 A c+a C) x}{105 a^3 c \left (a+c x^2\right )^{3/2}}+\frac {8 (6 A c+a C) x}{105 a^4 c \sqrt {a+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1828, 12, 198, 197} \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx=\frac {8 x (a C+6 A c)}{105 a^4 c \sqrt {a+c x^2}}+\frac {4 x (a C+6 A c)}{105 a^3 c \left (a+c x^2\right )^{3/2}}+\frac {x (a C+6 A c)}{35 a^2 c \left (a+c x^2\right )^{5/2}}-\frac {a B-x (A c-a C)}{7 a c \left (a+c x^2\right )^{7/2}} \]

[In]

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

[Out]

-1/7*(a*B - (A*c - a*C)*x)/(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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 197

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 198

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 1828

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*} \text {integral}& = -\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] (verified)

Time = 0.64 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx=\frac {-15 a^4 B+48 A c^4 x^7+35 a^3 c x \left (3 A+C x^2\right )+8 a c^3 x^5 \left (21 A+C x^2\right )+14 a^2 c^2 x^3 \left (15 A+2 C x^2\right )}{105 a^4 c \left (a+c x^2\right )^{7/2}} \]

[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))

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76

method	result	size
gosper	\(\frac {48 A \,c^{4} x^{7}+8 C a \,c^{3} x^{7}+168 a A \,c^{3} x^{5}+28 C \,a^{2} c^{2} x^{5}+210 a^{2} A \,c^{2} x^{3}+35 C \,a^{3} c \,x^{3}+105 a^{3} A c x -15 B \,a^{4}}{105 \left (c \,x^{2}+a \right )^{\frac {7}{2}} a^{4} c}\)	\(96\)
trager	\(\frac {48 A \,c^{4} x^{7}+8 C a \,c^{3} x^{7}+168 a A \,c^{3} x^{5}+28 C \,a^{2} c^{2} x^{5}+210 a^{2} A \,c^{2} x^{3}+35 C \,a^{3} c \,x^{3}+105 a^{3} A c x -15 B \,a^{4}}{105 \left (c \,x^{2}+a \right )^{\frac {7}{2}} a^{4} c}\)	\(96\)
default	\(A \left (\frac {x}{7 a \left (c \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (c \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {c \,x^{2}+a}}\right )}{7 a}}{a}\right )+C \left (-\frac {x}{6 c \left (c \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (c \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (c \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {c \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 c}\right )-\frac {B}{7 c \left (c \,x^{2}+a \right )^{\frac {7}{2}}}\)	\(189\)

[In]

int((C*x^2+B*x+A)/(c*x^2+a)^(9/2),x,method=_RETURNVERBOSE)

[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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx=\frac {{\left (8 \, {\left (C a c^{3} + 6 \, A c^{4}\right )} x^{7} + 105 \, A a^{3} c x + 28 \, {\left (C a^{2} c^{2} + 6 \, A a c^{3}\right )} x^{5} - 15 \, B a^{4} + 35 \, {\left (C a^{3} c + 6 \, A a^{2} c^{2}\right )} x^{3}\right )} \sqrt {c x^{2} + a}}{105 \, {\left (a^{4} c^{5} x^{8} + 4 \, a^{5} c^{4} x^{6} + 6 \, a^{6} c^{3} x^{4} + 4 \, a^{7} c^{2} x^{2} + a^{8} c\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1266 vs. \(2 (117) = 234\).

Time = 26.21 (sec) , antiderivative size = 1880, normalized size of antiderivative = 14.80 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx=\text {Too large to display} \]

[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)))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx=\frac {16 \, A x}{35 \, \sqrt {c x^{2} + a} a^{4}} + \frac {8 \, A x}{35 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, A x}{35 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {A x}{7 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} a} - \frac {C x}{7 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} c} + \frac {8 \, C x}{105 \, \sqrt {c x^{2} + a} a^{3} c} + \frac {4 \, C x}{105 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a^{2} c} + \frac {C x}{35 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a c} - \frac {B}{7 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} c} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (4 \, x^{2} {\left (\frac {2 \, {\left (C a c^{5} + 6 \, A c^{6}\right )} x^{2}}{a^{4} c^{3}} + \frac {7 \, {\left (C a^{2} c^{4} + 6 \, A a c^{5}\right )}}{a^{4} c^{3}}\right )} + \frac {35 \, {\left (C a^{3} c^{3} + 6 \, A a^{2} c^{4}\right )}}{a^{4} c^{3}}\right )} x^{2} + \frac {105 \, A}{a}\right )} x - \frac {15 \, B}{c}}{105 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}}} \]

[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)

Mupad [B] (verification not implemented)

Time = 12.99 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{9/2}} \, dx=\frac {x\,\left (6\,A\,c+C\,a\right )}{35\,a^2\,c\,{\left (c\,x^2+a\right )}^{5/2}}-\frac {\frac {B}{7\,c}-x\,\left (\frac {A}{7\,a}-\frac {C}{7\,c}\right )}{{\left (c\,x^2+a\right )}^{7/2}}+\frac {x\,\left (24\,A\,c+4\,C\,a\right )}{105\,a^3\,c\,{\left (c\,x^2+a\right )}^{3/2}}+\frac {x\,\left (48\,A\,c+8\,C\,a\right )}{105\,a^4\,c\,\sqrt {c\,x^2+a}} \]

[In]

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

[Out]

(x*(6*A*c + C*a))/(35*a^2*c*(a + c*x^2)^(5/2)) - (B/(7*c) - x*(A/(7*a) - C/(7*c)))/(a + c*x^2)^(7/2) + (x*(24*
A*c + 4*C*a))/(105*a^3*c*(a + c*x^2)^(3/2)) + (x*(48*A*c + 8*C*a))/(105*a^4*c*(a + c*x^2)^(1/2))