Integrand size = 19, antiderivative size = 97 \[ \int (A+B x) \sqrt {b x+c x^2} \, dx=-\frac {(b B-2 A c) (b+2 c x) \sqrt {b x+c x^2}}{8 c^2}+\frac {B \left (b x+c x^2\right )^{3/2}}{3 c}+\frac {b^2 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{5/2}} \]
1/3*B*(c*x^2+b*x)^(3/2)/c+1/8*b^2*(-2*A*c+B*b)*arctanh(x*c^(1/2)/(c*x^2+b* x)^(1/2))/c^(5/2)-1/8*(-2*A*c+B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^2
Time = 0.39 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26 \[ \int (A+B x) \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \sqrt {x} \left (-3 b^2 B+2 b c (3 A+B x)+4 c^2 x (3 A+2 B x)\right )+\frac {6 b^2 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{\sqrt {b+c x}}\right )}{24 c^{5/2} \sqrt {x}} \]
(Sqrt[x*(b + c*x)]*(Sqrt[c]*Sqrt[x]*(-3*b^2*B + 2*b*c*(3*A + B*x) + 4*c^2* x*(3*A + 2*B*x)) + (6*b^2*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b ] + Sqrt[b + c*x])])/Sqrt[b + c*x]))/(24*c^(5/2)*Sqrt[x])
Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1160, 1087, 1091, 219}
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.
\(\displaystyle \int (A+B x) \sqrt {b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2}}{3 c}-\frac {(b B-2 A c) \int \sqrt {c x^2+b x}dx}{2 c}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2}}{3 c}-\frac {(b B-2 A c) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{2 c}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2}}{3 c}-\frac {(b B-2 A c) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{2 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2}}{3 c}-\frac {(b B-2 A c) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{2 c}\) |
(B*(b*x + c*x^2)^(3/2))/(3*c) - ((b*B - 2*A*c)*(((b + 2*c*x)*Sqrt[b*x + c* x^2])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))))/( 2*c)
3.1.70.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85
method | result | size |
pseudoelliptic | \(\frac {\left (-\frac {1}{2} A \,b^{2} c +\frac {1}{4} B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\left (\frac {\left (\frac {B x}{3}+A \right ) b \,c^{\frac {3}{2}}}{2}+x \left (\frac {2 B x}{3}+A \right ) c^{\frac {5}{2}}-\frac {B \sqrt {c}\, b^{2}}{4}\right ) \sqrt {x \left (c x +b \right )}}{2 c^{\frac {5}{2}}}\) | \(82\) |
risch | \(\frac {\left (8 B \,c^{2} x^{2}+12 A \,c^{2} x +2 B b c x +6 A b c -3 B \,b^{2}\right ) x \left (c x +b \right )}{24 c^{2} \sqrt {x \left (c x +b \right )}}-\frac {b^{2} \left (2 A c -B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {5}{2}}}\) | \(97\) |
default | \(A \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )+B \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(139\) |
1/2/c^(5/2)*((-1/2*A*b^2*c+1/4*B*b^3)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2)) +(1/2*(1/3*B*x+A)*b*c^(3/2)+x*(2/3*B*x+A)*c^(5/2)-1/4*B*c^(1/2)*b^2)*(x*(c *x+b))^(1/2))
Time = 0.28 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.10 \[ \int (A+B x) \sqrt {b x+c x^2} \, dx=\left [-\frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{2} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{3}}, -\frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (8 \, B c^{3} x^{2} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{3}}\right ] \]
[-1/48*(3*(B*b^3 - 2*A*b^2*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)* sqrt(c)) - 2*(8*B*c^3*x^2 - 3*B*b^2*c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A*c^3)* x)*sqrt(c*x^2 + b*x))/c^3, -1/24*(3*(B*b^3 - 2*A*b^2*c)*sqrt(-c)*arctan(sq rt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*B*c^3*x^2 - 3*B*b^2*c + 6*A*b*c^2 + 2 *(B*b*c^2 + 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^3]
Time = 0.42 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.72 \[ \int (A+B x) \sqrt {b x+c x^2} \, dx=\begin {cases} - \frac {b \left (A b - \frac {3 b \left (A c + \frac {B b}{6}\right )}{4 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{2 c} + \sqrt {b x + c x^{2}} \left (\frac {B x^{2}}{3} + \frac {x \left (A c + \frac {B b}{6}\right )}{2 c} + \frac {A b - \frac {3 b \left (A c + \frac {B b}{6}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {3}{2}}}{3} + \frac {B \left (b x\right )^{\frac {5}{2}}}{5 b}\right )}{b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-b*(A*b - 3*b*(A*c + B*b/6)/(4*c))*Piecewise((log(b + 2*sqrt(c) *sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log(b /(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(2*c) + sqrt(b*x + c*x**2)*(B *x**2/3 + x*(A*c + B*b/6)/(2*c) + (A*b - 3*b*(A*c + B*b/6)/(4*c))/c), Ne(c , 0)), (2*(A*(b*x)**(3/2)/3 + B*(b*x)**(5/2)/(5*b))/b, Ne(b, 0)), (0, True ))
Time = 0.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.59 \[ \int (A+B x) \sqrt {b x+c x^2} \, dx=\frac {1}{2} \, \sqrt {c x^{2} + b x} A x - \frac {\sqrt {c x^{2} + b x} B b x}{4 \, c} + \frac {B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} - \frac {\sqrt {c x^{2} + b x} B b^{2}}{8 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{3 \, c} + \frac {\sqrt {c x^{2} + b x} A b}{4 \, c} \]
1/2*sqrt(c*x^2 + b*x)*A*x - 1/4*sqrt(c*x^2 + b*x)*B*b*x/c + 1/16*B*b^3*log (2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 1/8*A*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) - 1/8*sqrt(c*x^2 + b*x)*B*b^2/c^2 + 1/3*(c*x^2 + b*x)^(3/2)*B/c + 1/4*sqrt(c*x^2 + b*x)*A*b/c
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int (A+B x) \sqrt {b x+c x^2} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, B x + \frac {B b c + 6 \, A c^{2}}{c^{2}}\right )} x - \frac {3 \, {\left (B b^{2} - 2 \, A b c\right )}}{c^{2}}\right )} - \frac {{\left (B b^{3} - 2 \, A b^{2} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {5}{2}}} \]
1/24*sqrt(c*x^2 + b*x)*(2*(4*B*x + (B*b*c + 6*A*c^2)/c^2)*x - 3*(B*b^2 - 2 *A*b*c)/c^2) - 1/16*(B*b^3 - 2*A*b^2*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(5/2)
Time = 10.41 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.31 \[ \int (A+B x) \sqrt {b x+c x^2} \, dx=A\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )+\frac {B\,b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {B\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}-\frac {A\,b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}} \]