Integrand size = 22, antiderivative size = 90 \[ \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx=-\frac {2 c \sqrt {a x+b x^2}}{5 a x^3}+\frac {2 (4 b c-5 a d) \sqrt {a x+b x^2}}{15 a^2 x^2}-\frac {4 b (4 b c-5 a d) \sqrt {a x+b x^2}}{15 a^3 x} \] Output:
-2/5*c*(b*x^2+a*x)^(1/2)/a/x^3+2/15*(-5*a*d+4*b*c)*(b*x^2+a*x)^(1/2)/a^2/x ^2-4/15*b*(-5*a*d+4*b*c)*(b*x^2+a*x)^(1/2)/a^3/x
Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx=-\frac {2 \sqrt {x (a+b x)} \left (8 b^2 c x^2-2 a b x (2 c+5 d x)+a^2 (3 c+5 d x)\right )}{15 a^3 x^3} \] Input:
Integrate[(c + d*x)/(x^3*Sqrt[a*x + b*x^2]),x]
Output:
(-2*Sqrt[x*(a + b*x)]*(8*b^2*c*x^2 - 2*a*b*x*(2*c + 5*d*x) + a^2*(3*c + 5* d*x)))/(15*a^3*x^3)
Time = 0.38 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1220, 1129, 1123}
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.
\(\displaystyle \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle -\frac {(4 b c-5 a d) \int \frac {1}{x^2 \sqrt {b x^2+a x}}dx}{5 a}-\frac {2 c \sqrt {a x+b x^2}}{5 a x^3}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle -\frac {(4 b c-5 a d) \left (-\frac {2 b \int \frac {1}{x \sqrt {b x^2+a x}}dx}{3 a}-\frac {2 \sqrt {a x+b x^2}}{3 a x^2}\right )}{5 a}-\frac {2 c \sqrt {a x+b x^2}}{5 a x^3}\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle -\frac {\left (\frac {4 b \sqrt {a x+b x^2}}{3 a^2 x}-\frac {2 \sqrt {a x+b x^2}}{3 a x^2}\right ) (4 b c-5 a d)}{5 a}-\frac {2 c \sqrt {a x+b x^2}}{5 a x^3}\) |
Input:
Int[(c + d*x)/(x^3*Sqrt[a*x + b*x^2]),x]
Output:
(-2*c*Sqrt[a*x + b*x^2])/(5*a*x^3) - ((4*b*c - 5*a*d)*((-2*Sqrt[a*x + b*x^ 2])/(3*a*x^2) + (4*b*Sqrt[a*x + b*x^2])/(3*a^2*x)))/(5*a)
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b *e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d - b*e) )) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d , e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[Simplify[m + 2*p + 2], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Time = 0.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.54
method | result | size |
pseudoelliptic | \(-\frac {2 \sqrt {x \left (b x +a \right )}\, \left (\left (\frac {5 d x}{3}+c \right ) a^{2}-\frac {4 x b \left (\frac {5 d x}{2}+c \right ) a}{3}+\frac {8 b^{2} c \,x^{2}}{3}\right )}{5 a^{3} x^{3}}\) | \(49\) |
trager | \(-\frac {2 \left (-10 a b d \,x^{2}+8 b^{2} c \,x^{2}+5 a^{2} d x -4 a b c x +3 a^{2} c \right ) \sqrt {b \,x^{2}+a x}}{15 a^{3} x^{3}}\) | \(57\) |
risch | \(-\frac {2 \left (b x +a \right ) \left (-10 a b d \,x^{2}+8 b^{2} c \,x^{2}+5 a^{2} d x -4 a b c x +3 a^{2} c \right )}{15 a^{3} x^{2} \sqrt {x \left (b x +a \right )}}\) | \(60\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-10 a b d \,x^{2}+8 b^{2} c \,x^{2}+5 a^{2} d x -4 a b c x +3 a^{2} c \right )}{15 x^{2} a^{3} \sqrt {b \,x^{2}+a x}}\) | \(62\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-10 a b d \,x^{2}+8 b^{2} c \,x^{2}+5 a^{2} d x -4 a b c x +3 a^{2} c \right )}{15 x^{2} a^{3} \sqrt {b \,x^{2}+a x}}\) | \(62\) |
default | \(c \left (-\frac {2 \sqrt {b \,x^{2}+a x}}{5 a \,x^{3}}-\frac {4 b \left (-\frac {2 \sqrt {b \,x^{2}+a x}}{3 a \,x^{2}}+\frac {4 b \sqrt {b \,x^{2}+a x}}{3 a^{2} x}\right )}{5 a}\right )+d \left (-\frac {2 \sqrt {b \,x^{2}+a x}}{3 a \,x^{2}}+\frac {4 b \sqrt {b \,x^{2}+a x}}{3 a^{2} x}\right )\) | \(112\) |
Input:
int((d*x+c)/x^3/(b*x^2+a*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/5*(x*(b*x+a))^(1/2)*((5/3*d*x+c)*a^2-4/3*x*b*(5/2*d*x+c)*a+8/3*b^2*c*x^ 2)/a^3/x^3
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.64 \[ \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx=-\frac {2 \, {\left (3 \, a^{2} c + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{2} - {\left (4 \, a b c - 5 \, a^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{15 \, a^{3} x^{3}} \] Input:
integrate((d*x+c)/x^3/(b*x^2+a*x)^(1/2),x, algorithm="fricas")
Output:
-2/15*(3*a^2*c + 2*(4*b^2*c - 5*a*b*d)*x^2 - (4*a*b*c - 5*a^2*d)*x)*sqrt(b *x^2 + a*x)/(a^3*x^3)
\[ \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx=\int \frac {c + d x}{x^{3} \sqrt {x \left (a + b x\right )}}\, dx \] Input:
integrate((d*x+c)/x**3/(b*x**2+a*x)**(1/2),x)
Output:
Integral((c + d*x)/(x**3*sqrt(x*(a + b*x))), x)
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.18 \[ \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx=-\frac {16 \, \sqrt {b x^{2} + a x} b^{2} c}{15 \, a^{3} x} + \frac {4 \, \sqrt {b x^{2} + a x} b d}{3 \, a^{2} x} + \frac {8 \, \sqrt {b x^{2} + a x} b c}{15 \, a^{2} x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} d}{3 \, a x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} c}{5 \, a x^{3}} \] Input:
integrate((d*x+c)/x^3/(b*x^2+a*x)^(1/2),x, algorithm="maxima")
Output:
-16/15*sqrt(b*x^2 + a*x)*b^2*c/(a^3*x) + 4/3*sqrt(b*x^2 + a*x)*b*d/(a^2*x) + 8/15*sqrt(b*x^2 + a*x)*b*c/(a^2*x^2) - 2/3*sqrt(b*x^2 + a*x)*d/(a*x^2) - 2/5*sqrt(b*x^2 + a*x)*c/(a*x^3)
Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.48 \[ \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx=\frac {2 \, {\left (15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} \sqrt {b} d + 20 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} b c + 5 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a d + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a \sqrt {b} c + 3 \, a^{2} c\right )}}{15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{5}} \] Input:
integrate((d*x+c)/x^3/(b*x^2+a*x)^(1/2),x, algorithm="giac")
Output:
2/15*(15*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*sqrt(b)*d + 20*(sqrt(b)*x - sqr t(b*x^2 + a*x))^2*b*c + 5*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a*d + 15*(sqrt (b)*x - sqrt(b*x^2 + a*x))*a*sqrt(b)*c + 3*a^2*c)/(sqrt(b)*x - sqrt(b*x^2 + a*x))^5
Time = 9.55 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx=-\frac {2\,\sqrt {b\,x^2+a\,x}\,\left (5\,d\,a^2\,x+3\,c\,a^2-10\,d\,a\,b\,x^2-4\,c\,a\,b\,x+8\,c\,b^2\,x^2\right )}{15\,a^3\,x^3} \] Input:
int((c + d*x)/(x^3*(a*x + b*x^2)^(1/2)),x)
Output:
-(2*(a*x + b*x^2)^(1/2)*(3*a^2*c + 8*b^2*c*x^2 + 5*a^2*d*x - 10*a*b*d*x^2 - 4*a*b*c*x))/(15*a^3*x^3)
Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.18 \[ \int \frac {c+d x}{x^3 \sqrt {a x+b x^2}} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c}{5}-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} d x}{3}+\frac {8 \sqrt {x}\, \sqrt {b x +a}\, a b c x}{15}+\frac {4 \sqrt {x}\, \sqrt {b x +a}\, a b d \,x^{2}}{3}-\frac {16 \sqrt {x}\, \sqrt {b x +a}\, b^{2} c \,x^{2}}{15}-\frac {4 \sqrt {b}\, a b d \,x^{3}}{3}+\frac {16 \sqrt {b}\, b^{2} c \,x^{3}}{15}}{a^{3} x^{3}} \] Input:
int((d*x+c)/x^3/(b*x^2+a*x)^(1/2),x)
Output:
(2*( - 3*sqrt(x)*sqrt(a + b*x)*a**2*c - 5*sqrt(x)*sqrt(a + b*x)*a**2*d*x + 4*sqrt(x)*sqrt(a + b*x)*a*b*c*x + 10*sqrt(x)*sqrt(a + b*x)*a*b*d*x**2 - 8 *sqrt(x)*sqrt(a + b*x)*b**2*c*x**2 - 10*sqrt(b)*a*b*d*x**3 + 8*sqrt(b)*b** 2*c*x**3))/(15*a**3*x**3)