Integrand size = 29, antiderivative size = 129 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x (c+d x)} \, dx=\frac {B \sqrt {a x+b x^2}}{d}-\frac {(2 b B c-2 A b d-a B d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{\sqrt {b} d^2}+\frac {2 \sqrt {b c-a d} (B c-A d) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a x+b x^2}}\right )}{\sqrt {c} d^2} \] Output:
B*(b*x^2+a*x)^(1/2)/d-(-2*A*b*d-B*a*d+2*B*b*c)*arctanh(b^(1/2)*x/(b*x^2+a* x)^(1/2))/b^(1/2)/d^2+2*(-a*d+b*c)^(1/2)*(-A*d+B*c)*arctanh((-a*d+b*c)^(1/ 2)*x/c^(1/2)/(b*x^2+a*x)^(1/2))/c^(1/2)/d^2
Time = 0.42 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x (c+d x)} \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (2 \sqrt {b} \sqrt {-b c+a d} (B c-A d) \arctan \left (\frac {-d \sqrt {x} \sqrt {a+b x}+\sqrt {b} (c+d x)}{\sqrt {c} \sqrt {-b c+a d}}\right )+\sqrt {c} \left (\sqrt {b} B d \sqrt {x} \sqrt {a+b x}+(2 b B c-2 A b d-a B d) \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )\right )}{\sqrt {b} \sqrt {c} d^2 \sqrt {x (a+b x)}} \] Input:
Integrate[((A + B*x)*Sqrt[a*x + b*x^2])/(x*(c + d*x)),x]
Output:
(Sqrt[x]*Sqrt[a + b*x]*(2*Sqrt[b]*Sqrt[-(b*c) + a*d]*(B*c - A*d)*ArcTan[(- (d*Sqrt[x]*Sqrt[a + b*x]) + Sqrt[b]*(c + d*x))/(Sqrt[c]*Sqrt[-(b*c) + a*d] )] + Sqrt[c]*(Sqrt[b]*B*d*Sqrt[x]*Sqrt[a + b*x] + (2*b*B*c - 2*A*b*d - a*B *d)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])))/(Sqrt[b]*Sqrt[c]*d^2*Sqrt[x *(a + b*x)])
Time = 0.92 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.62, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2153, 2009}
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.
| \(\displaystyle \int \frac {\sqrt {a x+b x^2} (A+B x)}{x (c+d x)} \, dx\) |
| \(\Big \downarrow \) 2153 |
| \(\displaystyle \int \left (\frac {\sqrt {a x+b x^2} (B c-A d)}{c (c+d x)}+\frac {A \sqrt {a x+b x^2}}{c x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (2 b c-a d) (B c-A d)}{\sqrt {b} c d^2}+\frac {\sqrt {b c-a d} (B c-A d) \text {arctanh}\left (\frac {x (2 b c-a d)+a c}{2 \sqrt {c} \sqrt {a x+b x^2} \sqrt {b c-a d}}\right )}{\sqrt {c} d^2}+\frac {a A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{\sqrt {b} c}+\frac {\sqrt {a x+b x^2} (B c-A d)}{c d}+\frac {A \sqrt {a x+b x^2}}{c}\) |
Input:
Int[((A + B*x)*Sqrt[a*x + b*x^2])/(x*(c + d*x)),x]
Output:
(A*Sqrt[a*x + b*x^2])/c + ((B*c - A*d)*Sqrt[a*x + b*x^2])/(c*d) + (a*A*Arc Tanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(Sqrt[b]*c) - ((2*b*c - a*d)*(B*c - A *d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(Sqrt[b]*c*d^2) + (Sqrt[b*c - a*d]*(B*c - A*d)*ArcTanh[(a*c + (2*b*c - a*d)*x)/(2*Sqrt[c]*Sqrt[b*c - a*d ]*Sqrt[a*x + b*x^2])])/(Sqrt[c]*d^2)
Int[(Px_)*((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b _.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e* x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && PolyQ[Px, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ [m] && ILtQ[n, 0])) && !(IGtQ[m, 0] && IGtQ[n, 0])
Time = 0.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.88
| method | result | size |
| pseudoelliptic | \(-\frac {-B \sqrt {x \left (b x +a \right )}\, d -\frac {\left (2 A b d +B a d -2 B b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )}{\sqrt {b}}+\frac {2 \left (a d -b c \right ) \left (A d -B c \right ) \arctan \left (\frac {\sqrt {x \left (b x +a \right )}\, c}{x \sqrt {c \left (a d -b c \right )}}\right )}{\sqrt {c \left (a d -b c \right )}}}{d^{2}}\) | \(114\) |
| risch | \(\frac {B x \left (b x +a \right )}{d \sqrt {x \left (b x +a \right )}}+\frac {\frac {\left (2 A b d +B a d -2 B b c \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{d \sqrt {b}}-\frac {2 \left (A a \,d^{2}-A b c d -B a c d +B b \,c^{2}\right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}}{2 d}\) | \(230\) |
| default | \(\frac {A \left (\sqrt {b \,x^{2}+a x}+\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b}}\right )}{c}-\frac {\left (A d -B c \right ) \left (\sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}+\frac {\left (a d -2 b c \right ) \ln \left (\frac {\frac {a d -2 b c}{2 d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}\right )}{2 d \sqrt {b}}+\frac {c \left (a d -b c \right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}\right )}{c d}\) | \(344\) |
Input:
int((B*x+A)*(b*x^2+a*x)^(1/2)/x/(d*x+c),x,method=_RETURNVERBOSE)
Output:
-1/d^2*(-B*(x*(b*x+a))^(1/2)*d-(2*A*b*d+B*a*d-2*B*b*c)/b^(1/2)*arctanh((x* (b*x+a))^(1/2)/x/b^(1/2))+2*(a*d-b*c)*(A*d-B*c)/(c*(a*d-b*c))^(1/2)*arctan ((x*(b*x+a))^(1/2)/x*c/(c*(a*d-b*c))^(1/2)))
Time = 0.17 (sec) , antiderivative size = 579, normalized size of antiderivative = 4.49 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x (c+d x)} \, dx=\left [\frac {2 \, \sqrt {b x^{2} + a x} B b d - {\left (2 \, B b c - {\left (B a + 2 \, A b\right )} d\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (B b c - A b d\right )} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x - 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right )}{2 \, b d^{2}}, \frac {2 \, \sqrt {b x^{2} + a x} B b d + 4 \, {\left (B b c - A b d\right )} \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right ) - {\left (2 \, B b c - {\left (B a + 2 \, A b\right )} d\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b d^{2}}, \frac {\sqrt {b x^{2} + a x} B b d + {\left (2 \, B b c - {\left (B a + 2 \, A b\right )} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - {\left (B b c - A b d\right )} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x - 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right )}{b d^{2}}, \frac {\sqrt {b x^{2} + a x} B b d + {\left (2 \, B b c - {\left (B a + 2 \, A b\right )} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + 2 \, {\left (B b c - A b d\right )} \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right )}{b d^{2}}\right ] \] Input:
integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x/(d*x+c),x, algorithm="fricas")
Output:
[1/2*(2*sqrt(b*x^2 + a*x)*B*b*d - (2*B*b*c - (B*a + 2*A*b)*d)*sqrt(b)*log( 2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2*(B*b*c - A*b*d)*sqrt((b*c - a *d)/c)*log((a*c + (2*b*c - a*d)*x - 2*sqrt(b*x^2 + a*x)*c*sqrt((b*c - a*d) /c))/(d*x + c)))/(b*d^2), 1/2*(2*sqrt(b*x^2 + a*x)*B*b*d + 4*(B*b*c - A*b* d)*sqrt(-(b*c - a*d)/c)*arctan(-sqrt(b*x^2 + a*x)*c*sqrt(-(b*c - a*d)/c)/( (b*c - a*d)*x)) - (2*B*b*c - (B*a + 2*A*b)*d)*sqrt(b)*log(2*b*x + a + 2*sq rt(b*x^2 + a*x)*sqrt(b)))/(b*d^2), (sqrt(b*x^2 + a*x)*B*b*d + (2*B*b*c - ( B*a + 2*A*b)*d)*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x + a)) - (B *b*c - A*b*d)*sqrt((b*c - a*d)/c)*log((a*c + (2*b*c - a*d)*x - 2*sqrt(b*x^ 2 + a*x)*c*sqrt((b*c - a*d)/c))/(d*x + c)))/(b*d^2), (sqrt(b*x^2 + a*x)*B* b*d + (2*B*b*c - (B*a + 2*A*b)*d)*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(- b)/(b*x + a)) + 2*(B*b*c - A*b*d)*sqrt(-(b*c - a*d)/c)*arctan(-sqrt(b*x^2 + a*x)*c*sqrt(-(b*c - a*d)/c)/((b*c - a*d)*x)))/(b*d^2)]
\[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x (c+d x)} \, dx=\int \frac {\sqrt {x \left (a + b x\right )} \left (A + B x\right )}{x \left (c + d x\right )}\, dx \] Input:
integrate((B*x+A)*(b*x**2+a*x)**(1/2)/x/(d*x+c),x)
Output:
Integral(sqrt(x*(a + b*x))*(A + B*x)/(x*(c + d*x)), x)
Exception generated. \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x/(d*x+c),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(a*d-b*c>0)', see `assume?` for m ore detail
Exception generated. \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x/(d*x+c),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x (c+d x)} \, dx=\int \frac {\sqrt {b\,x^2+a\,x}\,\left (A+B\,x\right )}{x\,\left (c+d\,x\right )} \,d x \] Input:
int(((a*x + b*x^2)^(1/2)*(A + B*x))/(x*(c + d*x)),x)
Output:
int(((a*x + b*x^2)^(1/2)*(A + B*x))/(x*(c + d*x)), x)
Time = 0.22 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.16 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x (c+d x)} \, dx=\frac {-2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a d +2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) b c -2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a d +2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) b c +\sqrt {x}\, \sqrt {b x +a}\, b c d +3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a c d -2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) b \,c^{2}}{c \,d^{2}} \] Input:
int((B*x+A)*(b*x^2+a*x)^(1/2)/x/(d*x+c),x)
Output:
( - 2*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x ) - sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a*d + 2*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*sqrt (b))/(sqrt(c)*sqrt(b)))*b*c - 2*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b *c) + sqrt(d)*sqrt(a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))* a*d + 2*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b *x) + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*b*c + sqrt(x)*sqrt(a + b *x)*b*c*d + 3*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a*c*d - 2*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*b*c**2)/(c*d** 2)