Integrand size = 20, antiderivative size = 113 \[ \int x (A+B x) \sqrt {b x+c x^2} \, dx=\frac {b (5 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {(5 b B-8 A c-6 B c x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac {b^3 (5 b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}} \]
-1/24*(-6*B*c*x-8*A*c+5*B*b)*(c*x^2+b*x)^(3/2)/c^2-1/64*b^3*(-8*A*c+5*B*b) *arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+1/64*b*(-8*A*c+5*B*b)*(2*c*x +b)*(c*x^2+b*x)^(1/2)/c^3
Time = 0.55 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21 \[ \int x (A+B x) \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^3 B+8 b c^2 x (2 A+B x)+16 c^3 x^2 (4 A+3 B x)-2 b^2 c (12 A+5 B x)\right )+\frac {6 b^3 (5 b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{192 c^{7/2}} \]
(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*B + 8*b*c^2*x*(2*A + B*x) + 16*c^3*x^2 *(4*A + 3*B*x) - 2*b^2*c*(12*A + 5*B*x)) + (6*b^3*(5*b*B - 8*A*c)*ArcTanh[ (Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])])/(Sqrt[x]*Sqrt[b + c*x])))/(1 92*c^(7/2))
Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1225, 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.
\(\displaystyle \int x (A+B x) \sqrt {b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {b (5 b B-8 A c) \int \sqrt {c x^2+b x}dx}{16 c^2}-\frac {\left (b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {b (5 b B-8 A c) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c^2}-\frac {\left (b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {b (5 b B-8 A c) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c^2}-\frac {\left (b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {b (5 b B-8 A c) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c^2}-\frac {\left (b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2}\) |
-1/24*((5*b*B - 8*A*c - 6*B*c*x)*(b*x + c*x^2)^(3/2))/c^2 + (b*(5*b*B - 8* A*c)*(((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqr t[b*x + c*x^2]])/(4*c^(3/2))))/(16*c^2)
3.1.69.3.1 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_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90
method | result | size |
pseudoelliptic | \(\frac {\frac {3 \left (A \,b^{3} c -\frac {5}{8} b^{4} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{2}+\left (-\frac {3 \left (\frac {5 B x}{12}+A \right ) b^{2} c^{\frac {3}{2}}}{2}+b x \left (\frac {B x}{2}+A \right ) c^{\frac {5}{2}}+\left (3 B \,x^{3}+4 A \,x^{2}\right ) c^{\frac {7}{2}}+\frac {15 B \sqrt {c}\, b^{3}}{16}\right ) \sqrt {x \left (c x +b \right )}}{12 c^{\frac {7}{2}}}\) | \(102\) |
risch | \(-\frac {\left (-48 B \,c^{3} x^{3}-64 A \,c^{3} x^{2}-8 B b \,c^{2} x^{2}-16 A b \,c^{2} x +10 B \,b^{2} c x +24 A \,b^{2} c -15 B \,b^{3}\right ) x \left (c x +b \right )}{192 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {b^{3} \left (8 A c -5 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}\) | \(121\) |
default | \(B \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )+A \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(186\) |
1/12*(3/2*(A*b^3*c-5/8*b^4*B)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(-3/2*( 5/12*B*x+A)*b^2*c^(3/2)+b*x*(1/2*B*x+A)*c^(5/2)+(3*B*x^3+4*A*x^2)*c^(7/2)+ 15/16*B*c^(1/2)*b^3)*(x*(c*x+b))^(1/2))/c^(7/2)
Time = 0.28 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.24 \[ \int x (A+B x) \sqrt {b x+c x^2} \, dx=\left [-\frac {3 \, {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{4}}\right ] \]
[-1/384*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b* x)*sqrt(c)) - 2*(48*B*c^4*x^3 + 15*B*b^3*c - 24*A*b^2*c^2 + 8*(B*b*c^3 + 8 *A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/192 *(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x) ) + (48*B*c^4*x^3 + 15*B*b^3*c - 24*A*b^2*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]
Time = 0.54 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.82 \[ \int x (A+B x) \sqrt {b x+c x^2} \, dx=\begin {cases} \frac {3 b^{2} \left (A b - \frac {5 b \left (A c + \frac {B b}{8}\right )}{6 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c^{2}} + \sqrt {b x + c x^{2}} \left (\frac {B x^{3}}{4} - \frac {3 b \left (A b - \frac {5 b \left (A c + \frac {B b}{8}\right )}{6 c}\right )}{4 c^{2}} + \frac {x^{2} \left (A c + \frac {B b}{8}\right )}{3 c} + \frac {x \left (A b - \frac {5 b \left (A c + \frac {B b}{8}\right )}{6 c}\right )}{2 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {5}{2}}}{5} + \frac {B \left (b x\right )^{\frac {7}{2}}}{7 b}\right )}{b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((3*b**2*(A*b - 5*b*(A*c + B*b/8)/(6*c))*Piecewise((log(b + 2*sqr t(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*l og(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(8*c**2) + sqrt(b*x + c*x **2)*(B*x**3/4 - 3*b*(A*b - 5*b*(A*c + B*b/8)/(6*c))/(4*c**2) + x**2*(A*c + B*b/8)/(3*c) + x*(A*b - 5*b*(A*c + B*b/8)/(6*c))/(2*c)), Ne(c, 0)), (2*( A*(b*x)**(5/2)/5 + B*(b*x)**(7/2)/(7*b))/b**2, Ne(b, 0)), (0, True))
Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (97) = 194\).
Time = 0.18 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.75 \[ \int x (A+B x) \sqrt {b x+c x^2} \, dx=\frac {5 \, \sqrt {c x^{2} + b x} B b^{2} x}{32 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B x}{4 \, c} - \frac {\sqrt {c x^{2} + b x} A b x}{4 \, c} - \frac {5 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{24 \, c^{2}} - \frac {\sqrt {c x^{2} + b x} A b^{2}}{8 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{3 \, c} \]
5/32*sqrt(c*x^2 + b*x)*B*b^2*x/c^2 + 1/4*(c*x^2 + b*x)^(3/2)*B*x/c - 1/4*s qrt(c*x^2 + b*x)*A*b*x/c - 5/128*B*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x) *sqrt(c))/c^(7/2) + 1/16*A*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c) )/c^(5/2) + 5/64*sqrt(c*x^2 + b*x)*B*b^3/c^3 - 5/24*(c*x^2 + b*x)^(3/2)*B* b/c^2 - 1/8*sqrt(c*x^2 + b*x)*A*b^2/c^2 + 1/3*(c*x^2 + b*x)^(3/2)*A/c
Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.15 \[ \int x (A+B x) \sqrt {b x+c x^2} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, B x + \frac {B b c^{2} + 8 \, A c^{3}}{c^{3}}\right )} x - \frac {5 \, B b^{2} c - 8 \, A b c^{2}}{c^{3}}\right )} x + \frac {3 \, {\left (5 \, B b^{3} - 8 \, A b^{2} c\right )}}{c^{3}}\right )} + \frac {{\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {7}{2}}} \]
1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x + (B*b*c^2 + 8*A*c^3)/c^3)*x - (5*B*b ^2*c - 8*A*b*c^2)/c^3)*x + 3*(5*B*b^3 - 8*A*b^2*c)/c^3) + 1/128*(5*B*b^4 - 8*A*b^3*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2 )
Time = 10.34 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.46 \[ \int x (A+B x) \sqrt {b x+c x^2} \, dx=\frac {A\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}+\frac {B\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,B\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}+\frac {A\,b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}} \]
(A*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2) + (B*x*(b*x + c*x^2)^(3/2))/(4*c) - (5*B*b*((b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c *x^2)^(1/2)))/(16*c^(5/2)) + ((b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b *c*x))/(24*c^2)))/(8*c) + (A*b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2) ^(1/2)))/(16*c^(5/2))