Integrand size = 16, antiderivative size = 83 \[ \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx=\frac {4 \left (2 a+b \sqrt {x}\right )}{3 \left (b^2-4 a c\right ) \left (a+b \sqrt {x}+c x\right )^{3/2}}-\frac {16 b \left (b+2 c \sqrt {x}\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b \sqrt {x}+c x}} \] Output:
4/3*(2*a+b*x^(1/2))/(-4*a*c+b^2)/(a+b*x^(1/2)+c*x)^(3/2)-16/3*b*(b+2*c*x^( 1/2))/(-4*a*c+b^2)^2/(a+b*x^(1/2)+c*x)^(1/2)
Time = 0.80 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx=-\frac {4 \left (2 a b^2+8 a^2 c+3 b^3 \sqrt {x}+12 a b c \sqrt {x}+12 b^2 c x+8 b c^2 x^{3/2}\right )}{3 \left (b^2-4 a c\right )^2 \left (a+b \sqrt {x}+c x\right )^{3/2}} \] Input:
Integrate[(a + b*Sqrt[x] + c*x)^(-5/2),x]
Output:
(-4*(2*a*b^2 + 8*a^2*c + 3*b^3*Sqrt[x] + 12*a*b*c*Sqrt[x] + 12*b^2*c*x + 8 *b*c^2*x^(3/2)))/(3*(b^2 - 4*a*c)^2*(a + b*Sqrt[x] + c*x)^(3/2))
Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1680, 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.
\(\displaystyle \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1680 |
\(\displaystyle 2 \int \frac {\sqrt {x}}{\left (a+c x+b \sqrt {x}\right )^{5/2}}d\sqrt {x}\) |
\(\Big \downarrow \) 1159 |
\(\displaystyle 2 \left (\frac {4 b \int \frac {1}{\left (a+c x+b \sqrt {x}\right )^{3/2}}d\sqrt {x}}{3 \left (b^2-4 a c\right )}+\frac {2 \left (2 a+b \sqrt {x}\right )}{3 \left (b^2-4 a c\right ) \left (a+b \sqrt {x}+c x\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle 2 \left (\frac {2 \left (2 a+b \sqrt {x}\right )}{3 \left (b^2-4 a c\right ) \left (a+b \sqrt {x}+c x\right )^{3/2}}-\frac {8 b \left (b+2 c \sqrt {x}\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b \sqrt {x}+c x}}\right )\) |
Input:
Int[(a + b*Sqrt[x] + c*x)^(-5/2),x]
Output:
2*((2*(2*a + b*Sqrt[x]))/(3*(b^2 - 4*a*c)*(a + b*Sqrt[x] + c*x)^(3/2)) - ( 8*b*(b + 2*c*Sqrt[x]))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*Sqrt[x] + c*x]))
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]
Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k* n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && Fr actionQ[n]
Time = 0.02 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(-\frac {2}{3 c \left (a +b \sqrt {x}+c x \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {2 b}{3}+\frac {4 c \sqrt {x}}{3}}{\left (4 a c -b^{2}\right ) \left (a +b \sqrt {x}+c x \right )^{\frac {3}{2}}}+\frac {16 c \left (b +2 c \sqrt {x}\right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {a +b \sqrt {x}+c x}}\right )}{c}\) | \(95\) |
default | \(-\frac {2}{3 c \left (a +b \sqrt {x}+c x \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {2 b}{3}+\frac {4 c \sqrt {x}}{3}}{\left (4 a c -b^{2}\right ) \left (a +b \sqrt {x}+c x \right )^{\frac {3}{2}}}+\frac {16 c \left (b +2 c \sqrt {x}\right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {a +b \sqrt {x}+c x}}\right )}{c}\) | \(95\) |
Input:
int(1/(a+b*x^(1/2)+c*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3/c/(a+b*x^(1/2)+c*x)^(3/2)-b/c*(2/3*(b+2*c*x^(1/2))/(4*a*c-b^2)/(a+b*x ^(1/2)+c*x)^(3/2)+16/3*c/(4*a*c-b^2)^2*(b+2*c*x^(1/2))/(a+b*x^(1/2)+c*x)^( 1/2))
Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (67) = 134\).
Time = 0.34 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.12 \[ \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx=\frac {4 \, {\left (4 \, b^{2} c^{3} x^{3} - 2 \, a^{3} b^{2} - 8 \, a^{4} c - 2 \, {\left (3 \, b^{4} c - 7 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} x^{2} + 4 \, {\left (a b^{4} - 4 \, a^{3} c^{2}\right )} x - {\left (8 \, b c^{4} x^{3} - a^{2} b^{3} - 4 \, a^{3} b c - {\left (13 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2} + {\left (3 \, b^{5} - 10 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \sqrt {x}\right )} \sqrt {c x + b \sqrt {x} + a}}{3 \, {\left (a^{4} b^{4} - 8 \, a^{5} b^{2} c + 16 \, a^{6} c^{2} + {\left (b^{4} c^{4} - 8 \, a b^{2} c^{5} + 16 \, a^{2} c^{6}\right )} x^{4} - 2 \, {\left (b^{6} c^{2} - 10 \, a b^{4} c^{3} + 32 \, a^{2} b^{2} c^{4} - 32 \, a^{3} c^{5}\right )} x^{3} + {\left (b^{8} - 12 \, a b^{6} c + 54 \, a^{2} b^{4} c^{2} - 112 \, a^{3} b^{2} c^{3} + 96 \, a^{4} c^{4}\right )} x^{2} - 2 \, {\left (a^{2} b^{6} - 10 \, a^{3} b^{4} c + 32 \, a^{4} b^{2} c^{2} - 32 \, a^{5} c^{3}\right )} x\right )}} \] Input:
integrate(1/(a+b*x^(1/2)+c*x)^(5/2),x, algorithm="fricas")
Output:
4/3*(4*b^2*c^3*x^3 - 2*a^3*b^2 - 8*a^4*c - 2*(3*b^4*c - 7*a*b^2*c^2 + 4*a^ 2*c^3)*x^2 + 4*(a*b^4 - 4*a^3*c^2)*x - (8*b*c^4*x^3 - a^2*b^3 - 4*a^3*b*c - (13*b^3*c^2 - 28*a*b*c^3)*x^2 + (3*b^5 - 10*a*b^3*c + 16*a^2*b*c^2)*x)*s qrt(x))*sqrt(c*x + b*sqrt(x) + a)/(a^4*b^4 - 8*a^5*b^2*c + 16*a^6*c^2 + (b ^4*c^4 - 8*a*b^2*c^5 + 16*a^2*c^6)*x^4 - 2*(b^6*c^2 - 10*a*b^4*c^3 + 32*a^ 2*b^2*c^4 - 32*a^3*c^5)*x^3 + (b^8 - 12*a*b^6*c + 54*a^2*b^4*c^2 - 112*a^3 *b^2*c^3 + 96*a^4*c^4)*x^2 - 2*(a^2*b^6 - 10*a^3*b^4*c + 32*a^4*b^2*c^2 - 32*a^5*c^3)*x)
\[ \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \sqrt {x} + c x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a+b*x**(1/2)+c*x)**(5/2),x)
Output:
Integral((a + b*sqrt(x) + c*x)**(-5/2), x)
\[ \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx=\int { \frac {1}{{\left (c x + b \sqrt {x} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a+b*x^(1/2)+c*x)^(5/2),x, algorithm="maxima")
Output:
integrate((c*x + b*sqrt(x) + a)^(-5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (67) = 134\).
Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.81 \[ \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx=-\frac {4 \, {\left ({\left (4 \, {\left (\frac {2 \, b c^{2} \sqrt {x}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, b^{2} c}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} \sqrt {x} + \frac {3 \, {\left (b^{3} + 4 \, a b c\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} \sqrt {x} + \frac {2 \, {\left (a b^{2} + 4 \, a^{2} c\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x + b \sqrt {x} + a\right )}^{\frac {3}{2}}} \] Input:
integrate(1/(a+b*x^(1/2)+c*x)^(5/2),x, algorithm="giac")
Output:
-4/3*((4*(2*b*c^2*sqrt(x)/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*b^2*c/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*sqrt(x) + 3*(b^3 + 4*a*b*c)/(b^4 - 8*a*b^2*c + 16 *a^2*c^2))*sqrt(x) + 2*(a*b^2 + 4*a^2*c)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))/( c*x + b*sqrt(x) + a)^(3/2)
Timed out. \[ \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+c\,x+b\,\sqrt {x}\right )}^{5/2}} \,d x \] Input:
int(1/(a + c*x + b*x^(1/2))^(5/2),x)
Output:
int(1/(a + c*x + b*x^(1/2))^(5/2), x)
Time = 0.25 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.72 \[ \int \frac {1}{\left (a+b \sqrt {x}+c x\right )^{5/2}} \, dx=\frac {-48 \sqrt {x}\, \sqrt {\sqrt {x}\, b +a +c x}\, a b c -12 \sqrt {x}\, \sqrt {\sqrt {x}\, b +a +c x}\, b^{3}-32 \sqrt {x}\, \sqrt {\sqrt {x}\, b +a +c x}\, b \,c^{2} x -32 \sqrt {\sqrt {x}\, b +a +c x}\, a^{2} c -8 \sqrt {\sqrt {x}\, b +a +c x}\, a \,b^{2}-48 \sqrt {\sqrt {x}\, b +a +c x}\, b^{2} c x +64 \sqrt {x}\, \sqrt {c}\, a \,b^{2}+64 \sqrt {x}\, \sqrt {c}\, b^{2} c x +32 \sqrt {c}\, a^{2} b +64 \sqrt {c}\, a b c x +32 \sqrt {c}\, b^{3} x +32 \sqrt {c}\, b \,c^{2} x^{2}}{96 \sqrt {x}\, a^{3} b \,c^{2}-48 \sqrt {x}\, a^{2} b^{3} c +96 \sqrt {x}\, a^{2} b \,c^{3} x +6 \sqrt {x}\, a \,b^{5}-48 \sqrt {x}\, a \,b^{3} c^{2} x +6 \sqrt {x}\, b^{5} c x +48 a^{4} c^{2}-24 a^{3} b^{2} c +96 a^{3} c^{3} x +3 a^{2} b^{4}+48 a^{2} c^{4} x^{2}-18 a \,b^{4} c x -24 a \,b^{2} c^{3} x^{2}+3 b^{6} x +3 b^{4} c^{2} x^{2}} \] Input:
int(1/(a+b*x^(1/2)+c*x)^(5/2),x)
Output:
(4*( - 12*sqrt(x)*sqrt(sqrt(x)*b + a + c*x)*a*b*c - 3*sqrt(x)*sqrt(sqrt(x) *b + a + c*x)*b**3 - 8*sqrt(x)*sqrt(sqrt(x)*b + a + c*x)*b*c**2*x - 8*sqrt (sqrt(x)*b + a + c*x)*a**2*c - 2*sqrt(sqrt(x)*b + a + c*x)*a*b**2 - 12*sqr t(sqrt(x)*b + a + c*x)*b**2*c*x + 16*sqrt(x)*sqrt(c)*a*b**2 + 16*sqrt(x)*s qrt(c)*b**2*c*x + 8*sqrt(c)*a**2*b + 16*sqrt(c)*a*b*c*x + 8*sqrt(c)*b**3*x + 8*sqrt(c)*b*c**2*x**2))/(3*(32*sqrt(x)*a**3*b*c**2 - 16*sqrt(x)*a**2*b* *3*c + 32*sqrt(x)*a**2*b*c**3*x + 2*sqrt(x)*a*b**5 - 16*sqrt(x)*a*b**3*c** 2*x + 2*sqrt(x)*b**5*c*x + 16*a**4*c**2 - 8*a**3*b**2*c + 32*a**3*c**3*x + a**2*b**4 + 16*a**2*c**4*x**2 - 6*a*b**4*c*x - 8*a*b**2*c**3*x**2 + b**6* x + b**4*c**2*x**2))