Integrand size = 24, antiderivative size = 105 \[ \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx=-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}-\frac {(4 b B-3 A c) \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}+\frac {c (4 b B-3 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{5/2}} \] Output:
-1/2*A*(c*x^2+b*x)^(1/2)/b/x^(5/2)-1/4*(-3*A*c+4*B*b)*(c*x^2+b*x)^(1/2)/b^ 2/x^(3/2)+1/4*c*(-3*A*c+4*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/ b^(5/2)
Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx=\frac {-\sqrt {b} (b+c x) (2 A b+4 b B x-3 A c x)+c (4 b B-3 A c) x^2 \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{4 b^{5/2} x^{3/2} \sqrt {x (b+c x)}} \] Input:
Integrate[(A + B*x)/(x^(5/2)*Sqrt[b*x + c*x^2]),x]
Output:
(-(Sqrt[b]*(b + c*x)*(2*A*b + 4*b*B*x - 3*A*c*x)) + c*(4*b*B - 3*A*c)*x^2* Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/(4*b^(5/2)*x^(3/2)*Sqrt[x*(b + c*x)])
Time = 0.41 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1220, 1135, 1136, 220}
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.
\(\displaystyle \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(4 b B-3 A c) \int \frac {1}{x^{3/2} \sqrt {c x^2+b x}}dx}{4 b}-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle \frac {(4 b B-3 A c) \left (-\frac {c \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{2 b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle \frac {(4 b B-3 A c) \left (-\frac {c \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {(4 b B-3 A c) \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}\) |
Input:
Int[(A + B*x)/(x^(5/2)*Sqrt[b*x + c*x^2]),x]
Output:
-1/2*(A*Sqrt[b*x + c*x^2])/(b*x^(5/2)) + ((4*b*B - 3*A*c)*(-(Sqrt[b*x + c* x^2]/(b*x^(3/2))) + (c*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/ 2)))/(4*b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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*((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, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I ntegerQ[2*p]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x _Symbol] :> Simp[2*e Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^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 = 1.01 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {\left (c x +b \right ) \left (-3 A c x +4 B b x +2 A b \right )}{4 b^{2} x^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}-\frac {c \left (3 A c -4 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{4 b^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}\) | \(85\) |
default | \(-\frac {\sqrt {x \left (c x +b \right )}\, \left (3 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{2} x^{2}-4 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b c \,x^{2}-3 A c x \sqrt {c x +b}\, \sqrt {b}+4 B \,b^{\frac {3}{2}} x \sqrt {c x +b}+2 A \,b^{\frac {3}{2}} \sqrt {c x +b}\right )}{4 b^{\frac {5}{2}} x^{\frac {5}{2}} \sqrt {c x +b}}\) | \(109\) |
Input:
int((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/4*(c*x+b)*(-3*A*c*x+4*B*b*x+2*A*b)/b^2/x^(3/2)/(x*(c*x+b))^(1/2)-1/4*c* (3*A*c-4*B*b)/b^(5/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^(1/2) /(x*(c*x+b))^(1/2)
Time = 0.12 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.82 \[ \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx=\left [-\frac {{\left (4 \, B b c - 3 \, A c^{2}\right )} \sqrt {b} x^{3} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (2 \, A b^{2} + {\left (4 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, b^{3} x^{3}}, -\frac {{\left (4 \, B b c - 3 \, A c^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (2 \, A b^{2} + {\left (4 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, b^{3} x^{3}}\right ] \] Input:
integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")
Output:
[-1/8*((4*B*b*c - 3*A*c^2)*sqrt(b)*x^3*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(2*A*b^2 + (4*B*b^2 - 3*A*b*c)*x)*sqrt(c* x^2 + b*x)*sqrt(x))/(b^3*x^3), -1/4*((4*B*b*c - 3*A*c^2)*sqrt(-b)*x^3*arct an(sqrt(c*x^2 + b*x)*sqrt(-b)/(b*sqrt(x))) + (2*A*b^2 + (4*B*b^2 - 3*A*b*c )*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^3)]
\[ \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx=\int \frac {A + B x}{x^{\frac {5}{2}} \sqrt {x \left (b + c x\right )}}\, dx \] Input:
integrate((B*x+A)/x**(5/2)/(c*x**2+b*x)**(1/2),x)
Output:
Integral((A + B*x)/(x**(5/2)*sqrt(x*(b + c*x))), x)
\[ \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx=\int { \frac {B x + A}{\sqrt {c x^{2} + b x} x^{\frac {5}{2}}} \,d x } \] Input:
integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")
Output:
integrate((B*x + A)/(sqrt(c*x^2 + b*x)*x^(5/2)), x)
Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx=-\frac {\frac {{\left (4 \, B b c^{2} - 3 \, A c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c^{2} - 4 \, \sqrt {c x + b} B b^{2} c^{2} - 3 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{3} + 5 \, \sqrt {c x + b} A b c^{3}}{b^{2} c^{2} x^{2}}}{4 \, c} \] Input:
integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")
Output:
-1/4*((4*B*b*c^2 - 3*A*c^3)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (4*(c*x + b)^(3/2)*B*b*c^2 - 4*sqrt(c*x + b)*B*b^2*c^2 - 3*(c*x + b)^(3/ 2)*A*c^3 + 5*sqrt(c*x + b)*A*b*c^3)/(b^2*c^2*x^2))/c
Timed out. \[ \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx=\int \frac {A+B\,x}{x^{5/2}\,\sqrt {c\,x^2+b\,x}} \,d x \] Input:
int((A + B*x)/(x^(5/2)*(b*x + c*x^2)^(1/2)),x)
Output:
int((A + B*x)/(x^(5/2)*(b*x + c*x^2)^(1/2)), x)
Time = 0.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx=\frac {-4 \sqrt {c x +b}\, a \,b^{2}+6 \sqrt {c x +b}\, a b c x -8 \sqrt {c x +b}\, b^{3} x +3 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{2} x^{2}-4 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} c \,x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{2} x^{2}+4 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} c \,x^{2}}{8 b^{3} x^{2}} \] Input:
int((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x)
Output:
( - 4*sqrt(b + c*x)*a*b**2 + 6*sqrt(b + c*x)*a*b*c*x - 8*sqrt(b + c*x)*b** 3*x + 3*sqrt(b)*log(sqrt(b + c*x) - sqrt(b))*a*c**2*x**2 - 4*sqrt(b)*log(s qrt(b + c*x) - sqrt(b))*b**2*c*x**2 - 3*sqrt(b)*log(sqrt(b + c*x) + sqrt(b ))*a*c**2*x**2 + 4*sqrt(b)*log(sqrt(b + c*x) + sqrt(b))*b**2*c*x**2)/(8*b* *3*x**2)