\(\int \frac {A+C x^2}{(a+b x+c x^2)^{5/2}} \, dx\) [183]

Optimal result

Integrand size = 22, antiderivative size = 114 \[ \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \left (b c \left (A+\frac {a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \left (8 A c+4 a C+\frac {b^2 C}{c}\right ) (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}} \]

[Out]

-2/3*(b*c*(A+a*C/c)+(2*A*c^2+(-2*a*c+b^2)*C)*x)/c/(-4*a*c+b^2)/(c*x^2+b*x+a)^(3/2)+2/3*(8*A*c+4*C*a+b^2*C/c)*(
2*c*x+b)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1674, 12, 627} \[ \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 (b+2 c x) \left (4 a C+8 A c+\frac {b^2 C}{c}\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac {a C}{c}+A\right )\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[In]

Int[(A + C*x^2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (2*(8*A
*c + 4*a*C + (b^2*C)/c)*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + 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 627

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]

Rule 1674

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
]

Rubi steps \begin{align*} \text {integral}& = -\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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {8 A c+4 a C+\frac {b^2 C}{c}}{2 \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\left (8 A c+4 a C+\frac {b^2 C}{c}\right ) \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \left (8 A c+4 a C+\frac {b^2 C}{c}\right ) (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94 \[ \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {-2 A (b+2 c x) \left (b^2-8 b c x-4 c \left (3 a+2 c x^2\right )\right )+2 C \left (8 a^2 b+b^2 x^2 (3 b+2 c x)+4 a x \left (3 b^2+3 b c x+2 c^2 x^2\right )\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

[In]

Integrate[(A + C*x^2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.12

method	result	size
trager	\(\frac {\frac {32}{3} A \,c^{3} x^{3}+\frac {16}{3} C a \,c^{2} x^{3}+\frac {4}{3} C \,b^{2} c \,x^{3}+16 A b \,c^{2} x^{2}+8 C a b c \,x^{2}+2 C \,b^{3} x^{2}+16 a A \,c^{2} x +4 A \,b^{2} c x +8 C a \,b^{2} x +8 A a b c -\frac {2}{3} A \,b^{3}+\frac {16}{3} b \,a^{2} C}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(128\)
gosper	\(\frac {\frac {32}{3} A \,c^{3} x^{3}+\frac {16}{3} C a \,c^{2} x^{3}+\frac {4}{3} C \,b^{2} c \,x^{3}+16 A b \,c^{2} x^{2}+8 C a b c \,x^{2}+2 C \,b^{3} x^{2}+16 a A \,c^{2} x +4 A \,b^{2} c x +8 C a \,b^{2} x +8 A a b c -\frac {2}{3} A \,b^{3}+\frac {16}{3} b \,a^{2} C}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\)	\(137\)
default	\(A \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )+C \left (-\frac {x}{2 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{4 c}+\frac {a \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )\)	\(259\)

[In]

int((C*x^2+A)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3*(16*A*c^3*x^3+8*C*a*c^2*x^3+2*C*b^2*c*x^3+24*A*b*c^2*x^2+12*C*a*b*c*x^2+3*C*b^3*x^2+24*A*a*c^2*x+6*A*b^2*c
*x+12*C*a*b^2*x+12*A*a*b*c-A*b^3+8*C*a^2*b)/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(3/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (106) = 212\).

Time = 0.67 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.12 \[ \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (8 \, C a^{2} b - A b^{3} + 12 \, A a b c + 2 \, {\left (C b^{2} c + 4 \, C a c^{2} + 8 \, A c^{3}\right )} x^{3} + 3 \, {\left (C b^{3} + 4 \, C a b c + 8 \, A b c^{2}\right )} x^{2} + 6 \, {\left (2 \, C a b^{2} + A b^{2} c + 4 \, A a c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \]

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

2/3*(8*C*a^2*b - A*b^3 + 12*A*a*b*c + 2*(C*b^2*c + 4*C*a*c^2 + 8*A*c^3)*x^3 + 3*(C*b^3 + 4*C*a*b*c + 8*A*b*c^2
)*x^2 + 6*(2*C*a*b^2 + A*b^2*c + 4*A*a*c^2)*x)*sqrt(c*x^2 + b*x + a)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^
4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3
*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((C*x**2+A)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.69 \[ \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left ({\left (\frac {2 \, {\left (C b^{2} c + 4 \, C a c^{2} + 8 \, A c^{3}\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (C b^{3} + 4 \, C a b c + 8 \, A b c^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {6 \, {\left (2 \, C a b^{2} + A b^{2} c + 4 \, A a c^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {8 \, C a^{2} b - A b^{3} + 12 \, A a b c}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \]

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

2/3*(((2*(C*b^2*c + 4*C*a*c^2 + 8*A*c^3)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(C*b^3 + 4*C*a*b*c + 8*A*b*c^2)/
(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + 6*(2*C*a*b^2 + A*b^2*c + 4*A*a*c^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + (8
*C*a^2*b - A*b^3 + 12*A*a*b*c)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))/(c*x^2 + b*x + a)^(3/2)

Mupad [B] (verification not implemented)

Time = 13.39 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.11 \[ \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2\,\left (8\,C\,a^2\,b+12\,C\,a\,b^2\,x+12\,C\,a\,b\,c\,x^2+12\,A\,a\,b\,c+8\,C\,a\,c^2\,x^3+24\,A\,a\,c^2\,x+3\,C\,b^3\,x^2-A\,b^3+2\,C\,b^2\,c\,x^3+6\,A\,b^2\,c\,x+24\,A\,b\,c^2\,x^2+16\,A\,c^3\,x^3\right )}{3\,{\left (4\,a\,c-b^2\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \]

[In]

int((A + C*x^2)/(a + b*x + c*x^2)^(5/2),x)

[Out]

(2*(16*A*c^3*x^3 - A*b^3 + 3*C*b^3*x^2 + 8*C*a^2*b + 24*A*a*c^2*x + 6*A*b^2*c*x + 12*C*a*b^2*x + 24*A*b*c^2*x^
2 + 8*C*a*c^2*x^3 + 2*C*b^2*c*x^3 + 12*A*a*b*c + 12*C*a*b*c*x^2))/(3*(4*a*c - b^2)^2*(a + b*x + c*x^2)^(3/2))