Integrand size = 20, antiderivative size = 90 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 (A b-2 a B-(b B-2 A c) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {8 (b B-2 A c) (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}} \] Output:
1/3*(-2*A*b+4*B*a+2*(-2*A*c+B*b)*x)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(3/2)-8/3*( -2*A*c+B*b)*(2*c*x+b)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^(1/2)
Time = 1.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \left (A (b+2 c x) \left (b^2-8 b c x-4 c \left (3 a+2 c x^2\right )\right )+B \left (8 a^2 c+2 a b (b+6 c x)+b x \left (3 b^2+12 b c x+8 c^2 x^2\right )\right )\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \] Input:
Integrate[(A + B*x)/(a + b*x + c*x^2)^(5/2),x]
Output:
(-2*(A*(b + 2*c*x)*(b^2 - 8*b*c*x - 4*c*(3*a + 2*c*x^2)) + B*(8*a^2*c + 2* a*b*(b + 6*c*x) + b*x*(3*b^2 + 12*b*c*x + 8*c^2*x^2))))/(3*(b^2 - 4*a*c)^2 *(a + x*(b + c*x))^(3/2))
Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1159, 1088}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1159 |
\(\displaystyle \frac {4 (b B-2 A c) \int \frac {1}{\left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}-\frac {2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle -\frac {8 (b+2 c x) (b B-2 A c)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\) |
Input:
Int[(A + B*x)/(a + b*x + c*x^2)^(5/2),x]
Output:
(-2*(A*b - 2*a*B - (b*B - 2*A*c)*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3 /2)) - (8*(b*B - 2*A*c)*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x ^2])
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]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* x + c*x^2)^(p + 1), x] - Simp[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 - 4*a* c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] & & LtQ[p, -1] && NeQ[p, -3/2]
Time = 1.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.37
method | result | size |
trager | \(\frac {\frac {32}{3} A \,c^{3} x^{3}-\frac {16}{3} x^{3} B b \,c^{2}+16 A b \,c^{2} x^{2}-8 x^{2} B \,b^{2} c +16 A a \,c^{2} x +4 A \,b^{2} c x -8 B a b c x -2 x B \,b^{3}+8 A a b c -\frac {2}{3} A \,b^{3}-\frac {16}{3} B \,a^{2} c -\frac {4}{3} B a \,b^{2}}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\) | \(123\) |
gosper | \(\frac {\frac {32}{3} A \,c^{3} x^{3}-\frac {16}{3} x^{3} B b \,c^{2}+16 A b \,c^{2} x^{2}-8 x^{2} B \,b^{2} c +16 A a \,c^{2} x +4 A \,b^{2} c x -8 B a b c x -2 x B \,b^{3}+8 A a b c -\frac {2}{3} A \,b^{3}-\frac {16}{3} B \,a^{2} c -\frac {4}{3} B a \,b^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}\) | \(132\) |
orering | \(\frac {\frac {32}{3} A \,c^{3} x^{3}-\frac {16}{3} x^{3} B b \,c^{2}+16 A b \,c^{2} x^{2}-8 x^{2} B \,b^{2} c +16 A a \,c^{2} x +4 A \,b^{2} c x -8 B a b c x -2 x B \,b^{3}+8 A a b c -\frac {2}{3} A \,b^{3}-\frac {16}{3} B \,a^{2} c -\frac {4}{3} B a \,b^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}\) | \(132\) |
default | \(A \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )+B \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )\) | \(162\) |
Input:
int((B*x+A)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/3*(16*A*c^3*x^3-8*B*b*c^2*x^3+24*A*b*c^2*x^2-12*B*b^2*c*x^2+24*A*a*c^2*x +6*A*b^2*c*x-12*B*a*b*c*x-3*B*b^3*x+12*A*a*b*c-A*b^3-8*B*a^2*c-2*B*a*b^2)/ (4*a*c-b^2)^2/(c*x^2+b*x+a)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (82) = 164\).
Time = 0.48 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.72 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (2 \, B a b^{2} + A b^{3} + 8 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x^{3} + 12 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} x^{2} + 4 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} c + 3 \, {\left (B b^{3} - 8 \, A a c^{2} + 2 \, {\left (2 \, B a b - A b^{2}\right )} c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \] Input:
integrate((B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
Output:
-2/3*(2*B*a*b^2 + A*b^3 + 8*(B*b*c^2 - 2*A*c^3)*x^3 + 12*(B*b^2*c - 2*A*b* c^2)*x^2 + 4*(2*B*a^2 - 3*A*a*b)*c + 3*(B*b^3 - 8*A*a*c^2 + 2*(2*B*a*b - A *b^2)*c)*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)
\[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {A + B x}{\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((B*x+A)/(c*x**2+b*x+a)**(5/2),x)
Output:
Integral((A + B*x)/(a + b*x + c*x**2)**(5/2), x)
Exception generated. \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (82) = 164\).
Time = 0.25 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.14 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left ({\left (4 \, {\left (\frac {2 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {3 \, {\left (B b^{3} + 4 \, B a b c - 2 \, A b^{2} c - 8 \, A a c^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {2 \, B a b^{2} + A b^{3} + 8 \, B a^{2} c - 12 \, A a b c}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \] Input:
integrate((B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
Output:
-2/3*((4*(2*(B*b*c^2 - 2*A*c^3)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(B*b^ 2*c - 2*A*b*c^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + 3*(B*b^3 + 4*B*a*b*c - 2*A*b^2*c - 8*A*a*c^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + (2*B*a*b^2 + A*b^3 + 8*B*a^2*c - 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)
Time = 10.70 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.34 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2\,\left (8\,B\,a^2\,c+2\,B\,a\,b^2+12\,B\,a\,b\,c\,x-12\,A\,a\,b\,c-24\,A\,a\,c^2\,x+3\,B\,b^3\,x+A\,b^3+12\,B\,b^2\,c\,x^2-6\,A\,b^2\,c\,x+8\,B\,b\,c^2\,x^3-24\,A\,b\,c^2\,x^2-16\,A\,c^3\,x^3\right )}{3\,{\left (4\,a\,c-b^2\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \] Input:
int((A + B*x)/(a + b*x + c*x^2)^(5/2),x)
Output:
-(2*(A*b^3 - 16*A*c^3*x^3 + 2*B*a*b^2 + 8*B*a^2*c + 3*B*b^3*x - 24*A*a*c^2 *x - 6*A*b^2*c*x - 24*A*b*c^2*x^2 + 12*B*b^2*c*x^2 + 8*B*b*c^2*x^3 - 12*A* a*b*c + 12*B*a*b*c*x))/(3*(4*a*c - b^2)^2*(a + b*x + c*x^2)^(3/2))
Time = 0.36 (sec) , antiderivative size = 435, normalized size of antiderivative = 4.83 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {8 \sqrt {c \,x^{2}+b x +a}\, a^{2} b c +48 \sqrt {c \,x^{2}+b x +a}\, a^{2} c^{2} x -6 \sqrt {c \,x^{2}+b x +a}\, a \,b^{3}-12 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c x +48 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2} x^{2}+32 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} x^{3}-6 \sqrt {c \,x^{2}+b x +a}\, b^{4} x -24 \sqrt {c \,x^{2}+b x +a}\, b^{3} c \,x^{2}-16 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} x^{3}-32 \sqrt {c}\, a^{3} c +16 \sqrt {c}\, a^{2} b^{2}-64 \sqrt {c}\, a^{2} b c x -64 \sqrt {c}\, a^{2} c^{2} x^{2}+32 \sqrt {c}\, a \,b^{3} x -64 \sqrt {c}\, a b \,c^{2} x^{3}-32 \sqrt {c}\, a \,c^{3} x^{4}+16 \sqrt {c}\, b^{4} x^{2}+32 \sqrt {c}\, b^{3} c \,x^{3}+16 \sqrt {c}\, b^{2} c^{2} x^{4}}{48 a^{2} c^{4} x^{4}-24 a \,b^{2} c^{3} x^{4}+3 b^{4} c^{2} x^{4}+96 a^{2} b \,c^{3} x^{3}-48 a \,b^{3} c^{2} x^{3}+6 b^{5} c \,x^{3}+96 a^{3} c^{3} x^{2}-18 a \,b^{4} c \,x^{2}+3 b^{6} x^{2}+96 a^{3} b \,c^{2} x -48 a^{2} b^{3} c x +6 a \,b^{5} x +48 a^{4} c^{2}-24 a^{3} b^{2} c +3 a^{2} b^{4}} \] Input:
int((B*x+A)/(c*x^2+b*x+a)^(5/2),x)
Output:
(2*(4*sqrt(a + b*x + c*x**2)*a**2*b*c + 24*sqrt(a + b*x + c*x**2)*a**2*c** 2*x - 3*sqrt(a + b*x + c*x**2)*a*b**3 - 6*sqrt(a + b*x + c*x**2)*a*b**2*c* x + 24*sqrt(a + b*x + c*x**2)*a*b*c**2*x**2 + 16*sqrt(a + b*x + c*x**2)*a* c**3*x**3 - 3*sqrt(a + b*x + c*x**2)*b**4*x - 12*sqrt(a + b*x + c*x**2)*b* *3*c*x**2 - 8*sqrt(a + b*x + c*x**2)*b**2*c**2*x**3 - 16*sqrt(c)*a**3*c + 8*sqrt(c)*a**2*b**2 - 32*sqrt(c)*a**2*b*c*x - 32*sqrt(c)*a**2*c**2*x**2 + 16*sqrt(c)*a*b**3*x - 32*sqrt(c)*a*b*c**2*x**3 - 16*sqrt(c)*a*c**3*x**4 + 8*sqrt(c)*b**4*x**2 + 16*sqrt(c)*b**3*c*x**3 + 8*sqrt(c)*b**2*c**2*x**4))/ (3*(16*a**4*c**2 - 8*a**3*b**2*c + 32*a**3*b*c**2*x + 32*a**3*c**3*x**2 + a**2*b**4 - 16*a**2*b**3*c*x + 32*a**2*b*c**3*x**3 + 16*a**2*c**4*x**4 + 2 *a*b**5*x - 6*a*b**4*c*x**2 - 16*a*b**3*c**2*x**3 - 8*a*b**2*c**3*x**4 + b **6*x**2 + 2*b**5*c*x**3 + b**4*c**2*x**4))