Integrand size = 22, antiderivative size = 127 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {b (5 b B-6 A c) \sqrt {b x+c x^2}}{8 c^3}-\frac {(5 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^2}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {b^2 (5 b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}} \]
-1/8*b^2*(-6*A*c+5*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+1/8*b *(-6*A*c+5*B*b)*(c*x^2+b*x)^(1/2)/c^3-1/12*(-6*A*c+5*B*b)*x*(c*x^2+b*x)^(1 /2)/c^2+1/3*B*x^2*(c*x^2+b*x)^(1/2)/c
Time = 0.41 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} x (b+c x) \left (15 b^2 B+4 c^2 x (3 A+2 B x)-2 b c (9 A+5 B x)\right )+6 b^2 (5 b B-6 A c) \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{24 c^{7/2} \sqrt {x (b+c x)}} \]
(Sqrt[c]*x*(b + c*x)*(15*b^2*B + 4*c^2*x*(3*A + 2*B*x) - 2*b*c*(9*A + 5*B* x)) + 6*b^2*(5*b*B - 6*A*c)*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x] )/(Sqrt[b] - Sqrt[b + c*x])])/(24*c^(7/2)*Sqrt[x*(b + c*x)])
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1221, 1134, 1160, 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.
\(\displaystyle \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \int \frac {x^2}{\sqrt {c x^2+b x}}dx}{6 c}\) |
\(\Big \downarrow \) 1134 |
\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \int \frac {x}{\sqrt {c x^2+b x}}dx}{4 c}\right )}{6 c}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \int \frac {1}{\sqrt {c x^2+b x}}dx}{2 c}\right )}{4 c}\right )}{6 c}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{c}\right )}{4 c}\right )}{6 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}\right )}{4 c}\right )}{6 c}\) |
(B*x^2*Sqrt[b*x + c*x^2])/(3*c) - ((5*b*B - 6*A*c)*((x*Sqrt[b*x + c*x^2])/ (2*c) - (3*b*(Sqrt[b*x + c*x^2]/c - (b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^ 2]])/c^(3/2)))/(4*c)))/(6*c)
3.2.11.3.1 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[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_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))) 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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 *p]
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]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 )/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0]
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(\frac {\left (\frac {3}{2} A \,b^{2} c -\frac {5}{4} B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\sqrt {x \left (c x +b \right )}\, \left (-\frac {3 \left (\frac {5 B x}{9}+A \right ) b \,c^{\frac {3}{2}}}{2}+x \left (\frac {2 B x}{3}+A \right ) c^{\frac {5}{2}}+\frac {5 B \sqrt {c}\, b^{2}}{4}\right )}{2 c^{\frac {7}{2}}}\) | \(82\) |
risch | \(-\frac {\left (-8 B \,c^{2} x^{2}-12 A \,c^{2} x +10 B b c x +18 A b c -15 B \,b^{2}\right ) x \left (c x +b \right )}{24 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {b^{2} \left (6 A c -5 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {7}{2}}}\) | \(97\) |
default | \(B \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )+A \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )\) | \(172\) |
1/2*((3/2*A*b^2*c-5/4*B*b^3)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(x*(c*x+ b))^(1/2)*(-3/2*(5/9*B*x+A)*b*c^(3/2)+x*(2/3*B*x+A)*c^(5/2)+5/4*B*c^(1/2)* b^2))/c^(7/2)
Time = 0.28 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.63 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\left [-\frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{4}}\right ] \]
[-1/48*(3*(5*B*b^3 - 6*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 + 15*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A *c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/24*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(-c)*ar ctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^3*x^2 + 15*B*b^2*c - 18*A* b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]
Time = 0.49 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.42 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\begin {cases} \frac {3 b^{2} \left (A - \frac {5 B b}{6 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c^{2}} + \sqrt {b x + c x^{2}} \left (\frac {B x^{2}}{3 c} - \frac {3 b \left (A - \frac {5 B b}{6 c}\right )}{4 c^{2}} + \frac {x \left (A - \frac {5 B b}{6 c}\right )}{2 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {5}{2}}}{5} + \frac {B \left (b x\right )^{\frac {7}{2}}}{7 b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]
Piecewise((3*b**2*(A - 5*B*b/(6*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))/(8*c**2) + sqrt(b*x + c*x**2)*(B*x**2/( 3*c) - 3*b*(A - 5*B*b/(6*c))/(4*c**2) + x*(A - 5*B*b/(6*c))/(2*c)), Ne(c, 0)), (2*(A*(b*x)**(5/2)/5 + B*(b*x)**(7/2)/(7*b))/b**3, Ne(b, 0)), (zoo*(A *x**3/3 + B*x**4/4), True))
Time = 0.19 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.26 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} B x^{2}}{3 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} B b x}{12 \, c^{2}} + \frac {\sqrt {c x^{2} + b x} A x}{2 \, c} - \frac {5 \, B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} + \frac {3 \, A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{2}}{8 \, c^{3}} - \frac {3 \, \sqrt {c x^{2} + b x} A b}{4 \, c^{2}} \]
1/3*sqrt(c*x^2 + b*x)*B*x^2/c - 5/12*sqrt(c*x^2 + b*x)*B*b*x/c^2 + 1/2*sqr t(c*x^2 + b*x)*A*x/c - 5/16*B*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt (c))/c^(7/2) + 3/8*A*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5 /2) + 5/8*sqrt(c*x^2 + b*x)*B*b^2/c^3 - 3/4*sqrt(c*x^2 + b*x)*A*b/c^2
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (\frac {4 \, B x}{c} - \frac {5 \, B b c - 6 \, A c^{2}}{c^{3}}\right )} x + \frac {3 \, {\left (5 \, B b^{2} - 6 \, A b c\right )}}{c^{3}}\right )} + \frac {{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {7}{2}}} \]
1/24*sqrt(c*x^2 + b*x)*(2*(4*B*x/c - (5*B*b*c - 6*A*c^2)/c^3)*x + 3*(5*B*b ^2 - 6*A*b*c)/c^3) + 1/16*(5*B*b^3 - 6*A*b^2*c)*log(abs(2*(sqrt(c)*x - sqr t(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2)
Timed out. \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^2\,\left (A+B\,x\right )}{\sqrt {c\,x^2+b\,x}} \,d x \]