### 3.184 $$\int \frac{A+C x^2}{(a+b x+c x^2)^{7/2}} \, dx$$

Optimal. Leaf size=167 $-\frac{16 (b+2 c x) \left (4 a c C+16 A c^2+3 b^2 C\right )}{15 \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2}}-\frac{2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )\right )}{5 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}+\frac{2 (b+2 c x) \left (4 a C+16 A c+\frac{3 b^2 C}{c}\right )}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}$

[Out]

(-2*(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x))/(5*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(5/2)) + (2*(16*
A*c + 4*a*C + (3*b^2*C)/c)*(b + 2*c*x))/(15*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^(3/2)) - (16*(16*A*c^2 + 3*b^2*C
+ 4*a*c*C)*(b + 2*c*x))/(15*(b^2 - 4*a*c)^3*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.106591, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {1660, 12, 614, 613} $-\frac{16 (b+2 c x) \left (4 a c C+16 A c^2+3 b^2 C\right )}{15 \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2}}-\frac{2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )\right )}{5 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}+\frac{2 (b+2 c x) \left (4 a C+16 A c+\frac{3 b^2 C}{c}\right )}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(A + C*x^2)/(a + b*x + c*x^2)^(7/2),x]

[Out]

(-2*(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x))/(5*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(5/2)) + (2*(16*
A*c + 4*a*C + (3*b^2*C)/c)*(b + 2*c*x))/(15*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^(3/2)) - (16*(16*A*c^2 + 3*b^2*C
+ 4*a*c*C)*(b + 2*c*x))/(15*(b^2 - 4*a*c)^3*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 )^{7/2}} \, 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 )}{5 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}-\frac{2 \int \frac{16 A c+4 a C+\frac{3 b^2 C}{c}}{2 \left (a+b x+c x^2\right )^{5/2}} \, dx}{5 \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 )}{5 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}-\frac{\left (16 A c+4 a C+\frac{3 b^2 C}{c}\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{5/2}} \, dx}{5 \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 )}{5 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}+\frac{2 \left (16 A c+4 a C+\frac{3 b^2 C}{c}\right ) (b+2 c x)}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}+\frac{\left (8 \left (16 A c^2+3 b^2 C+4 a c C\right )\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{15 \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 )}{5 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}+\frac{2 \left (16 A c+4 a C+\frac{3 b^2 C}{c}\right ) (b+2 c x)}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}-\frac{16 \left (16 A c^2+3 b^2 C+4 a c C\right ) (b+2 c x)}{15 \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 1.85438, size = 148, normalized size = 0.89 $\frac{2 \left (\left (b^2-4 a c\right ) (b+2 c x) (a+x (b+c x)) \left (4 a c C+16 A c^2+3 b^2 C\right )-8 c (b+2 c x) (a+x (b+c x))^2 \left (4 a c C+16 A c^2+3 b^2 C\right )-3 \left (b^2-4 a c\right )^2 \left (a C (b-2 c x)+A c (b+2 c x)+b^2 C x\right )\right )}{15 c \left (b^2-4 a c\right )^3 (a+x (b+c x))^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(A + C*x^2)/(a + b*x + c*x^2)^(7/2),x]

[Out]

(2*((b^2 - 4*a*c)*(16*A*c^2 + 3*b^2*C + 4*a*c*C)*(b + 2*c*x)*(a + x*(b + c*x)) - 8*c*(16*A*c^2 + 3*b^2*C + 4*a
*c*C)*(b + 2*c*x)*(a + x*(b + c*x))^2 - 3*(b^2 - 4*a*c)^2*(b^2*C*x + a*C*(b - 2*c*x) + A*c*(b + 2*c*x))))/(15*
c*(b^2 - 4*a*c)^3*(a + x*(b + c*x))^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.051, size = 316, normalized size = 1.9 \begin{align*}{\frac{512\,A{c}^{5}{x}^{5}+128\,Ca{c}^{4}{x}^{5}+96\,C{b}^{2}{c}^{3}{x}^{5}+1280\,Ab{c}^{4}{x}^{4}+320\,Cab{c}^{3}{x}^{4}+240\,C{b}^{3}{c}^{2}{x}^{4}+1280\,Aa{c}^{4}{x}^{3}+960\,A{x}^{3}{b}^{2}{c}^{3}+320\,C{a}^{2}{c}^{3}{x}^{3}+480\,Ca{b}^{2}{c}^{2}{x}^{3}+180\,C{b}^{4}c{x}^{3}+1920\,Aab{c}^{3}{x}^{2}+160\,A{x}^{2}{b}^{3}{c}^{2}+480\,C{a}^{2}b{c}^{2}{x}^{2}+400\,Ca{b}^{3}c{x}^{2}+30\,C{b}^{5}{x}^{2}+960\,A{a}^{2}{c}^{3}x+480\,Aa{b}^{2}{c}^{2}x-20\,A{b}^{4}cx+480\,C{a}^{2}{b}^{2}cx+40\,Ca{b}^{4}x+480\,A{a}^{2}b{c}^{2}-80\,Aa{b}^{3}c+6\,A{b}^{5}+192\,C{a}^{3}bc+16\,C{a}^{2}{b}^{3}}{960\,{a}^{3}{c}^{3}-720\,{a}^{2}{b}^{2}{c}^{2}+180\,a{b}^{4}c-15\,{b}^{6}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{5}{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)^(7/2),x)

[Out]

2/15/(c*x^2+b*x+a)^(5/2)*(256*A*c^5*x^5+64*C*a*c^4*x^5+48*C*b^2*c^3*x^5+640*A*b*c^4*x^4+160*C*a*b*c^3*x^4+120*
C*b^3*c^2*x^4+640*A*a*c^4*x^3+480*A*b^2*c^3*x^3+160*C*a^2*c^3*x^3+240*C*a*b^2*c^2*x^3+90*C*b^4*c*x^3+960*A*a*b
*c^3*x^2+80*A*b^3*c^2*x^2+240*C*a^2*b*c^2*x^2+200*C*a*b^3*c*x^2+15*C*b^5*x^2+480*A*a^2*c^3*x+240*A*a*b^2*c^2*x
-10*A*b^4*c*x+240*C*a^2*b^2*c*x+20*C*a*b^4*x+240*A*a^2*b*c^2-40*A*a*b^3*c+3*A*b^5+96*C*a^3*b*c+8*C*a^2*b^3)/(6
4*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)

________________________________________________________________________________________

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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 77.4996, size = 1224, normalized size = 7.33 \begin{align*} -\frac{2 \,{\left (8 \, C a^{2} b^{3} + 3 \, A b^{5} + 240 \, A a^{2} b c^{2} + 16 \,{\left (3 \, C b^{2} c^{3} + 4 \, C a c^{4} + 16 \, A c^{5}\right )} x^{5} + 40 \,{\left (3 \, C b^{3} c^{2} + 4 \, C a b c^{3} + 16 \, A b c^{4}\right )} x^{4} + 10 \,{\left (9 \, C b^{4} c + 24 \, C a b^{2} c^{2} + 64 \, A a c^{4} + 16 \,{\left (C a^{2} + 3 \, A b^{2}\right )} c^{3}\right )} x^{3} + 5 \,{\left (3 \, C b^{5} + 40 \, C a b^{3} c + 192 \, A a b c^{3} + 16 \,{\left (3 \, C a^{2} b + A b^{3}\right )} c^{2}\right )} x^{2} + 8 \,{\left (12 \, C a^{3} b - 5 \, A a b^{3}\right )} c + 10 \,{\left (2 \, C a b^{4} + 24 \, A a b^{2} c^{2} + 48 \, A a^{2} c^{3} +{\left (24 \, C a^{2} b^{2} - A b^{4}\right )} c\right )} x\right )} \sqrt{c x^{2} + b x + a}}{15 \,{\left (a^{3} b^{6} - 12 \, a^{4} b^{4} c + 48 \, a^{5} b^{2} c^{2} - 64 \, a^{6} c^{3} +{\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} x^{6} + 3 \,{\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} x^{5} + 3 \,{\left (b^{8} c - 11 \, a b^{6} c^{2} + 36 \, a^{2} b^{4} c^{3} - 16 \, a^{3} b^{2} c^{4} - 64 \, a^{4} c^{5}\right )} x^{4} +{\left (b^{9} - 6 \, a b^{7} c - 24 \, a^{2} b^{5} c^{2} + 224 \, a^{3} b^{3} c^{3} - 384 \, a^{4} b c^{4}\right )} x^{3} + 3 \,{\left (a b^{8} - 11 \, a^{2} b^{6} c + 36 \, a^{3} b^{4} c^{2} - 16 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4}\right )} x^{2} + 3 \,{\left (a^{2} b^{7} - 12 \, a^{3} b^{5} c + 48 \, a^{4} b^{3} c^{2} - 64 \, a^{5} b c^{3}\right )} x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(8*C*a^2*b^3 + 3*A*b^5 + 240*A*a^2*b*c^2 + 16*(3*C*b^2*c^3 + 4*C*a*c^4 + 16*A*c^5)*x^5 + 40*(3*C*b^3*c^2
+ 4*C*a*b*c^3 + 16*A*b*c^4)*x^4 + 10*(9*C*b^4*c + 24*C*a*b^2*c^2 + 64*A*a*c^4 + 16*(C*a^2 + 3*A*b^2)*c^3)*x^3
+ 5*(3*C*b^5 + 40*C*a*b^3*c + 192*A*a*b*c^3 + 16*(3*C*a^2*b + A*b^3)*c^2)*x^2 + 8*(12*C*a^3*b - 5*A*a*b^3)*c
+ 10*(2*C*a*b^4 + 24*A*a*b^2*c^2 + 48*A*a^2*c^3 + (24*C*a^2*b^2 - A*b^4)*c)*x)*sqrt(c*x^2 + b*x + a)/(a^3*b^6
- 12*a^4*b^4*c + 48*a^5*b^2*c^2 - 64*a^6*c^3 + (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^6 + 3*
(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*x^5 + 3*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a
^3*b^2*c^4 - 64*a^4*c^5)*x^4 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*x^3 + 3*(a
*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*x^2 + 3*(a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b
^3*c^2 - 64*a^5*b*c^3)*x)

________________________________________________________________________________________

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)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.2866, size = 659, normalized size = 3.95 \begin{align*} -\frac{{\left ({\left (2 \,{\left (4 \,{\left (\frac{2 \,{\left (3 \, C b^{2} c^{3} + 4 \, C a c^{4} + 16 \, A c^{5}\right )} x}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}} + \frac{5 \,{\left (3 \, C b^{3} c^{2} + 4 \, C a b c^{3} + 16 \, A b c^{4}\right )}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}\right )} x + \frac{5 \,{\left (9 \, C b^{4} c + 24 \, C a b^{2} c^{2} + 16 \, C a^{2} c^{3} + 48 \, A b^{2} c^{3} + 64 \, A a c^{4}\right )}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}\right )} x + \frac{5 \,{\left (3 \, C b^{5} + 40 \, C a b^{3} c + 48 \, C a^{2} b c^{2} + 16 \, A b^{3} c^{2} + 192 \, A a b c^{3}\right )}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}\right )} x + \frac{10 \,{\left (2 \, C a b^{4} + 24 \, C a^{2} b^{2} c - A b^{4} c + 24 \, A a b^{2} c^{2} + 48 \, A a^{2} c^{3}\right )}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}\right )} x + \frac{8 \, C a^{2} b^{3} + 3 \, A b^{5} + 96 \, C a^{3} b c - 40 \, A a b^{3} c + 240 \, A a^{2} b c^{2}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}}{15 \,{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^(7/2),x, algorithm="giac")

[Out]

-1/15*(((2*(4*(2*(3*C*b^2*c^3 + 4*C*a*c^4 + 16*A*c^5)*x/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)
+ 5*(3*C*b^3*c^2 + 4*C*a*b*c^3 + 16*A*b*c^4)/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6))*x + 5*(9
*C*b^4*c + 24*C*a*b^2*c^2 + 16*C*a^2*c^3 + 48*A*b^2*c^3 + 64*A*a*c^4)/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5
- 64*a^3*c^6))*x + 5*(3*C*b^5 + 40*C*a*b^3*c + 48*C*a^2*b*c^2 + 16*A*b^3*c^2 + 192*A*a*b*c^3)/(b^6*c^3 - 12*a
*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6))*x + 10*(2*C*a*b^4 + 24*C*a^2*b^2*c - A*b^4*c + 24*A*a*b^2*c^2 + 48*A*
a^2*c^3)/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6))*x + (8*C*a^2*b^3 + 3*A*b^5 + 96*C*a^3*b*c - 4
0*A*a*b^3*c + 240*A*a^2*b*c^2)/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6))/(c*x^2 + b*x + a)^(5/2)