3.185 \(\int \frac{A+C x^2}{(a+b x+c x^2)^{9/2}} \, dx\)

Optimal. Leaf size=220 \[ \frac{256 c (b+2 c x) \left (4 a c C+24 A c^2+5 b^2 C\right )}{105 \left (b^2-4 a c\right )^4 \sqrt{a+b x+c x^2}}-\frac{32 (b+2 c x) \left (4 a c C+24 A c^2+5 b^2 C\right )}{105 \left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )^{3/2}}-\frac{2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )\right )}{7 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}}+\frac{2 (b+2 c x) \left (4 a C+24 A c+\frac{5 b^2 C}{c}\right )}{35 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{5/2}} \]

[Out]

(-2*(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x))/(7*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(7/2)) + (2*(24*
A*c + 4*a*C + (5*b^2*C)/c)*(b + 2*c*x))/(35*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^(5/2)) - (32*(24*A*c^2 + 5*b^2*C
+ 4*a*c*C)*(b + 2*c*x))/(105*(b^2 - 4*a*c)^3*(a + b*x + c*x^2)^(3/2)) + (256*c*(24*A*c^2 + 5*b^2*C + 4*a*c*C)
*(b + 2*c*x))/(105*(b^2 - 4*a*c)^4*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.141816, antiderivative size = 220, 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 = {1660, 12, 614, 613} \[ \frac{256 c (b+2 c x) \left (4 a c C+24 A c^2+5 b^2 C\right )}{105 \left (b^2-4 a c\right )^4 \sqrt{a+b x+c x^2}}-\frac{32 (b+2 c x) \left (4 a c C+24 A c^2+5 b^2 C\right )}{105 \left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )^{3/2}}-\frac{2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )\right )}{7 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}}+\frac{2 (b+2 c x) \left (4 a C+24 A c+\frac{5 b^2 C}{c}\right )}{35 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x))/(7*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(7/2)) + (2*(24*
A*c + 4*a*C + (5*b^2*C)/c)*(b + 2*c*x))/(35*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^(5/2)) - (32*(24*A*c^2 + 5*b^2*C
+ 4*a*c*C)*(b + 2*c*x))/(105*(b^2 - 4*a*c)^3*(a + b*x + c*x^2)^(3/2)) + (256*c*(24*A*c^2 + 5*b^2*C + 4*a*c*C)
*(b + 2*c*x))/(105*(b^2 - 4*a*c)^4*Sqrt[a + b*x + c*x^2])

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && 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 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{A+C x^2}{\left (a+b x+c x^2\right )^{9/2}} \, dx &=-\frac{2 \left (b c \left (A+\frac{a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\right ) x\right )}{7 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}}-\frac{2 \int \frac{24 A c+4 a C+\frac{5 b^2 C}{c}}{2 \left (a+b x+c x^2\right )^{7/2}} \, dx}{7 \left (b^2-4 a c\right )}\\ &=-\frac{2 \left (b c \left (A+\frac{a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\right ) x\right )}{7 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}}-\frac{\left (24 A c+4 a C+\frac{5 b^2 C}{c}\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{7/2}} \, dx}{7 \left (b^2-4 a c\right )}\\ &=-\frac{2 \left (b c \left (A+\frac{a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\right ) x\right )}{7 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}}+\frac{2 \left (24 A c+4 a C+\frac{5 b^2 C}{c}\right ) (b+2 c x)}{35 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{5/2}}+\frac{\left (16 \left (24 A c^2+5 b^2 C+4 a c C\right )\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{5/2}} \, dx}{35 \left (b^2-4 a c\right )^2}\\ &=-\frac{2 \left (b c \left (A+\frac{a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\right ) x\right )}{7 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}}+\frac{2 \left (24 A c+4 a C+\frac{5 b^2 C}{c}\right ) (b+2 c x)}{35 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{5/2}}-\frac{32 \left (24 A c^2+5 b^2 C+4 a c C\right ) (b+2 c x)}{105 \left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )^{3/2}}-\frac{\left (128 c \left (24 A c^2+5 b^2 C+4 a c C\right )\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{105 \left (b^2-4 a c\right )^3}\\ &=-\frac{2 \left (b c \left (A+\frac{a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\right ) x\right )}{7 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}}+\frac{2 \left (24 A c+4 a C+\frac{5 b^2 C}{c}\right ) (b+2 c x)}{35 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{5/2}}-\frac{32 \left (24 A c^2+5 b^2 C+4 a c C\right ) (b+2 c x)}{105 \left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )^{3/2}}+\frac{256 c \left (24 A c^2+5 b^2 C+4 a c C\right ) (b+2 c x)}{105 \left (b^2-4 a c\right )^4 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 1.78204, size = 199, normalized size = 0.9 \[ \frac{2 \left (3 \left (b^2-4 a c\right )^2 (b+2 c x) (a+x (b+c x)) \left (4 a c C+24 A c^2+5 b^2 C\right )-16 c \left (b^2-4 a c\right ) (b+2 c x) (a+x (b+c x))^2 \left (4 a c C+24 A c^2+5 b^2 C\right )+128 c^2 (b+2 c x) (a+x (b+c x))^3 \left (4 a c C+24 A c^2+5 b^2 C\right )-15 \left (b^2-4 a c\right )^3 \left (a C (b-2 c x)+A c (b+2 c x)+b^2 C x\right )\right )}{105 c \left (b^2-4 a c\right )^4 (a+x (b+c x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*(b^2 - 4*a*c)^2*(24*A*c^2 + 5*b^2*C + 4*a*c*C)*(b + 2*c*x)*(a + x*(b + c*x)) - 16*c*(b^2 - 4*a*c)*(24*A*
c^2 + 5*b^2*C + 4*a*c*C)*(b + 2*c*x)*(a + x*(b + c*x))^2 + 128*c^2*(24*A*c^2 + 5*b^2*C + 4*a*c*C)*(b + 2*c*x)*
(a + x*(b + c*x))^3 - 15*(b^2 - 4*a*c)^3*(b^2*C*x + a*C*(b - 2*c*x) + A*c*(b + 2*c*x))))/(105*c*(b^2 - 4*a*c)^
4*(a + x*(b + c*x))^(7/2))

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Maple [B]  time = 0.053, size = 555, normalized size = 2.5 \begin{align*}{\frac{12288\,A{c}^{7}{x}^{7}+2048\,Ca{c}^{6}{x}^{7}+2560\,C{b}^{2}{c}^{5}{x}^{7}+43008\,Ab{c}^{6}{x}^{6}+7168\,Cab{c}^{5}{x}^{6}+8960\,C{b}^{3}{c}^{4}{x}^{6}+43008\,Aa{c}^{6}{x}^{5}+53760\,A{b}^{2}{c}^{5}{x}^{5}+7168\,C{a}^{2}{c}^{5}{x}^{5}+17920\,Ca{b}^{2}{c}^{4}{x}^{5}+11200\,C{b}^{4}{c}^{3}{x}^{5}+107520\,Aab{c}^{5}{x}^{4}+26880\,A{b}^{3}{c}^{4}{x}^{4}+17920\,C{a}^{2}b{c}^{4}{x}^{4}+26880\,Ca{b}^{3}{c}^{3}{x}^{4}+5600\,C{b}^{5}{c}^{2}{x}^{4}+53760\,A{a}^{2}{c}^{5}{x}^{3}+80640\,Aa{b}^{2}{c}^{4}{x}^{3}+3360\,A{b}^{4}{c}^{3}{x}^{3}+8960\,C{a}^{3}{c}^{4}{x}^{3}+24640\,C{a}^{2}{b}^{2}{c}^{3}{x}^{3}+17360\,Ca{b}^{4}{c}^{2}{x}^{3}+700\,C{b}^{6}c{x}^{3}+80640\,A{a}^{2}b{c}^{4}{x}^{2}+13440\,Aa{b}^{3}{c}^{3}{x}^{2}-336\,A{x}^{2}{b}^{5}{c}^{2}+13440\,C{a}^{3}b{c}^{3}{x}^{2}+19040\,C{a}^{2}{b}^{3}{c}^{2}{x}^{2}+2744\,Ca{b}^{5}c{x}^{2}-70\,C{b}^{7}{x}^{2}+26880\,A{a}^{3}{c}^{4}x+20160\,A{a}^{2}{b}^{2}{c}^{3}x-1680\,Aa{b}^{4}{c}^{2}x+84\,A{b}^{6}cx+13440\,C{a}^{3}{b}^{2}{c}^{2}x+2240\,C{a}^{2}{b}^{4}cx-56\,Ca{b}^{6}x+13440\,A{a}^{3}b{c}^{3}-3360\,A{a}^{2}{b}^{3}{c}^{2}+504\,Aa{b}^{5}c-30\,A{b}^{7}+3840\,C{a}^{4}b{c}^{2}+640\,C{a}^{3}{b}^{3}c-16\,C{a}^{2}{b}^{5}}{26880\,{a}^{4}{c}^{4}-26880\,{a}^{3}{b}^{2}{c}^{3}+10080\,{a}^{2}{b}^{4}{c}^{2}-1680\,a{b}^{6}c+105\,{b}^{8}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105/(c*x^2+b*x+a)^(7/2)*(6144*A*c^7*x^7+1024*C*a*c^6*x^7+1280*C*b^2*c^5*x^7+21504*A*b*c^6*x^6+3584*C*a*b*c^5
*x^6+4480*C*b^3*c^4*x^6+21504*A*a*c^6*x^5+26880*A*b^2*c^5*x^5+3584*C*a^2*c^5*x^5+8960*C*a*b^2*c^4*x^5+5600*C*b
^4*c^3*x^5+53760*A*a*b*c^5*x^4+13440*A*b^3*c^4*x^4+8960*C*a^2*b*c^4*x^4+13440*C*a*b^3*c^3*x^4+2800*C*b^5*c^2*x
^4+26880*A*a^2*c^5*x^3+40320*A*a*b^2*c^4*x^3+1680*A*b^4*c^3*x^3+4480*C*a^3*c^4*x^3+12320*C*a^2*b^2*c^3*x^3+868
0*C*a*b^4*c^2*x^3+350*C*b^6*c*x^3+40320*A*a^2*b*c^4*x^2+6720*A*a*b^3*c^3*x^2-168*A*b^5*c^2*x^2+6720*C*a^3*b*c^
3*x^2+9520*C*a^2*b^3*c^2*x^2+1372*C*a*b^5*c*x^2-35*C*b^7*x^2+13440*A*a^3*c^4*x+10080*A*a^2*b^2*c^3*x-840*A*a*b
^4*c^2*x+42*A*b^6*c*x+6720*C*a^3*b^2*c^2*x+1120*C*a^2*b^4*c*x-28*C*a*b^6*x+6720*A*a^3*b*c^3-1680*A*a^2*b^3*c^2
+252*A*a*b^5*c-15*A*b^7+1920*C*a^4*b*c^2+320*C*a^3*b^3*c-8*C*a^2*b^5)/(256*a^4*c^4-256*a^3*b^2*c^3+96*a^2*b^4*
c^2-16*a*b^6*c+b^8)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.35431, size = 1152, normalized size = 5.24 \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+A)/(c*x^2+b*x+a)^(9/2),x, algorithm="giac")

[Out]

1/105*(((2*(8*(2*(4*(2*(5*C*b^2*c^5 + 4*C*a*c^6 + 24*A*c^7)*x/(b^8*c^4 - 16*a*b^6*c^5 + 96*a^2*b^4*c^6 - 256*a
^3*b^2*c^7 + 256*a^4*c^8) + 7*(5*C*b^3*c^4 + 4*C*a*b*c^5 + 24*A*b*c^6)/(b^8*c^4 - 16*a*b^6*c^5 + 96*a^2*b^4*c^
6 - 256*a^3*b^2*c^7 + 256*a^4*c^8))*x + 7*(25*C*b^4*c^3 + 40*C*a*b^2*c^4 + 16*C*a^2*c^5 + 120*A*b^2*c^5 + 96*A
*a*c^6)/(b^8*c^4 - 16*a*b^6*c^5 + 96*a^2*b^4*c^6 - 256*a^3*b^2*c^7 + 256*a^4*c^8))*x + 35*(5*C*b^5*c^2 + 24*C*
a*b^3*c^3 + 16*C*a^2*b*c^4 + 24*A*b^3*c^4 + 96*A*a*b*c^5)/(b^8*c^4 - 16*a*b^6*c^5 + 96*a^2*b^4*c^6 - 256*a^3*b
^2*c^7 + 256*a^4*c^8))*x + 35*(5*C*b^6*c + 124*C*a*b^4*c^2 + 176*C*a^2*b^2*c^3 + 24*A*b^4*c^3 + 64*C*a^3*c^4 +
576*A*a*b^2*c^4 + 384*A*a^2*c^5)/(b^8*c^4 - 16*a*b^6*c^5 + 96*a^2*b^4*c^6 - 256*a^3*b^2*c^7 + 256*a^4*c^8))*x
- 7*(5*C*b^7 - 196*C*a*b^5*c - 1360*C*a^2*b^3*c^2 + 24*A*b^5*c^2 - 960*C*a^3*b*c^3 - 960*A*a*b^3*c^3 - 5760*A
*a^2*b*c^4)/(b^8*c^4 - 16*a*b^6*c^5 + 96*a^2*b^4*c^6 - 256*a^3*b^2*c^7 + 256*a^4*c^8))*x - 14*(2*C*a*b^6 - 80*
C*a^2*b^4*c - 3*A*b^6*c - 480*C*a^3*b^2*c^2 + 60*A*a*b^4*c^2 - 720*A*a^2*b^2*c^3 - 960*A*a^3*c^4)/(b^8*c^4 - 1
6*a*b^6*c^5 + 96*a^2*b^4*c^6 - 256*a^3*b^2*c^7 + 256*a^4*c^8))*x - (8*C*a^2*b^5 + 15*A*b^7 - 320*C*a^3*b^3*c -
252*A*a*b^5*c - 1920*C*a^4*b*c^2 + 1680*A*a^2*b^3*c^2 - 6720*A*a^3*b*c^3)/(b^8*c^4 - 16*a*b^6*c^5 + 96*a^2*b^
4*c^6 - 256*a^3*b^2*c^7 + 256*a^4*c^8))/(c*x^2 + b*x + a)^(7/2)