Integrand size = 26, antiderivative size = 267 \[ \int (A+B x) (d+e x)^2 \sqrt {b x+c x^2} \, dx=\frac {\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac {b^2 \left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}} \]
1/5*B*(e*x+d)^2*(c*x^2+b*x)^(3/2)/c+1/240*(10*A*c*e*(-5*b*e+16*c*d)+B*(35* b^2*e^2-100*b*c*d*e+32*c^2*d^2)+6*c*e*(10*A*c*e-7*B*b*e+4*B*c*d)*x)*(c*x^2 +b*x)^(3/2)/c^3-1/128*b^2*(32*A*c^3*d^2-7*b^3*B*e^2+10*b^2*c*e*(A*e+2*B*d) -16*b*c^2*d*(2*A*e+B*d))*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)+1/12 8*(32*A*c^3*d^2-7*b^3*B*e^2+10*b^2*c*e*(A*e+2*B*d)-16*b*c^2*d*(2*A*e+B*d)) *(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^4
Time = 1.68 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.13 \[ \int (A+B x) (d+e x)^2 \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-105 b^4 B e^2+10 b^3 c e (30 B d+15 A e+7 B e x)+16 b c^3 \left (5 A \left (6 d^2+4 d e x+e^2 x^2\right )+B x \left (10 d^2+10 d e x+3 e^2 x^2\right )\right )+32 c^4 x \left (5 A \left (6 d^2+8 d e x+3 e^2 x^2\right )+2 B x \left (10 d^2+15 d e x+6 e^2 x^2\right )\right )-4 b^2 c^2 \left (5 A e (24 d+5 e x)+2 B \left (30 d^2+25 d e x+7 e^2 x^2\right )\right )\right )+\frac {30 b^2 \left (-32 A c^3 d^2+7 b^3 B e^2-10 b^2 c e (2 B d+A e)+16 b c^2 d (B d+2 A e)\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{1920 c^{9/2}} \]
(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4*B*e^2 + 10*b^3*c*e*(30*B*d + 15*A*e + 7*B*e*x) + 16*b*c^3*(5*A*(6*d^2 + 4*d*e*x + e^2*x^2) + B*x*(10*d^2 + 10* d*e*x + 3*e^2*x^2)) + 32*c^4*x*(5*A*(6*d^2 + 8*d*e*x + 3*e^2*x^2) + 2*B*x* (10*d^2 + 15*d*e*x + 6*e^2*x^2)) - 4*b^2*c^2*(5*A*e*(24*d + 5*e*x) + 2*B*( 30*d^2 + 25*d*e*x + 7*e^2*x^2))) + (30*b^2*(-32*A*c^3*d^2 + 7*b^3*B*e^2 - 10*b^2*c*e*(2*B*d + A*e) + 16*b*c^2*d*(B*d + 2*A*e))*ArcTanh[(Sqrt[c]*Sqrt [x])/(-Sqrt[b] + Sqrt[b + c*x])])/(Sqrt[x]*Sqrt[b + c*x])))/(1920*c^(9/2))
Time = 0.42 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1236, 27, 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.
| \(\displaystyle \int (A+B x) \sqrt {b x+c x^2} (d+e x)^2 \, dx\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {\int -\frac {1}{2} (d+e x) ((3 b B-10 A c) d-(4 B c d-7 b B e+10 A c e) x) \sqrt {c x^2+b x}dx}{5 c}+\frac {B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}-\frac {\int (d+e x) ((3 b B-10 A c) d-(4 B c d-7 b B e+10 A c e) x) \sqrt {c x^2+b x}dx}{10 c}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}-\frac {-\frac {5 \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right ) \int \sqrt {c x^2+b x}dx}{16 c^2}-\frac {\left (b x+c x^2\right )^{3/2} \left (6 c e x (10 A c e-7 b B e+4 B c d)+10 A c e (16 c d-5 b e)+2 B \left (\frac {35 b^2 e^2}{2}-50 b c d e+16 c^2 d^2\right )\right )}{24 c^2}}{10 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}-\frac {-\frac {5 \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right ) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c^2}-\frac {\left (b x+c x^2\right )^{3/2} \left (6 c e x (10 A c e-7 b B e+4 B c d)+10 A c e (16 c d-5 b e)+2 B \left (\frac {35 b^2 e^2}{2}-50 b c d e+16 c^2 d^2\right )\right )}{24 c^2}}{10 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}-\frac {-\frac {5 \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right ) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c^2}-\frac {\left (b x+c x^2\right )^{3/2} \left (6 c e x (10 A c e-7 b B e+4 B c d)+10 A c e (16 c d-5 b e)+2 B \left (\frac {35 b^2 e^2}{2}-50 b c d e+16 c^2 d^2\right )\right )}{24 c^2}}{10 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}-\frac {-\frac {5 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right ) \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{16 c^2}-\frac {\left (b x+c x^2\right )^{3/2} \left (6 c e x (10 A c e-7 b B e+4 B c d)+10 A c e (16 c d-5 b e)+2 B \left (\frac {35 b^2 e^2}{2}-50 b c d e+16 c^2 d^2\right )\right )}{24 c^2}}{10 c}\) |
(B*(d + e*x)^2*(b*x + c*x^2)^(3/2))/(5*c) - (-1/24*((10*A*c*e*(16*c*d - 5* b*e) + 2*B*(16*c^2*d^2 - 50*b*c*d*e + (35*b^2*e^2)/2) + 6*c*e*(4*B*c*d - 7 *b*B*e + 10*A*c*e)*x)*(b*x + c*x^2)^(3/2))/c^2 - (5*(32*A*c^3*d^2 - 7*b^3* B*e^2 + 10*b^2*c*e*(2*B*d + A*e) - 16*b*c^2*d*(B*d + 2*A*e))*(((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))))/(16*c^2))/(10*c)
3.12.63.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1 )*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m *(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ {a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
Time = 0.44 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.90
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {5 \left (-\frac {7 b^{3} B \,e^{2}}{10}+b^{2} c e \left (A e +2 B d \right )-\frac {16 c^{2} \left (A e +\frac {B d}{2}\right ) d b}{5}+\frac {16 A \,c^{3} d^{2}}{5}\right ) b^{2} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{64}+\frac {5 \left (\frac {16 \left (\frac {x^{2} \left (\frac {3 B x}{5}+A \right ) e^{2}}{6}+\frac {2 \left (\frac {B x}{2}+A \right ) x d e}{3}+d^{2} \left (\frac {B x}{3}+A \right )\right ) b \,c^{\frac {7}{2}}}{5}+\frac {32 x \left (\frac {\left (\frac {4 B x}{5}+A \right ) x^{2} e^{2}}{2}+\frac {4 \left (\frac {3 B x}{4}+A \right ) x d e}{3}+d^{2} \left (\frac {2 B x}{3}+A \right )\right ) c^{\frac {9}{2}}}{5}+\left (\left (-\frac {2 \left (\frac {14 B x}{25}+A \right ) x \,e^{2}}{3}-\frac {16 d \left (\frac {5 B x}{12}+A \right ) e}{5}-\frac {8 B \,d^{2}}{5}\right ) c^{\frac {5}{2}}+\left (\left (\left (\frac {7 B x}{15}+A \right ) e +2 B d \right ) c^{\frac {3}{2}}-\frac {7 B b e \sqrt {c}}{10}\right ) e b \right ) b^{2}\right ) \sqrt {x \left (c x +b \right )}}{64}}{c^{\frac {9}{2}}}\) | \(239\) |
| risch | \(\frac {\left (384 B \,c^{4} e^{2} x^{4}+480 A \,c^{4} e^{2} x^{3}+48 B b \,c^{3} e^{2} x^{3}+960 B \,c^{4} d e \,x^{3}+80 A b \,c^{3} e^{2} x^{2}+1280 A \,c^{4} d e \,x^{2}-56 B \,b^{2} c^{2} e^{2} x^{2}+160 B b \,c^{3} d e \,x^{2}+640 B \,c^{4} d^{2} x^{2}-100 A \,b^{2} c^{2} e^{2} x +320 A b \,c^{3} d e x +960 A \,c^{4} d^{2} x +70 B \,b^{3} c \,e^{2} x -200 B \,b^{2} c^{2} d e x +160 B b \,c^{3} d^{2} x +150 A \,b^{3} c \,e^{2}-480 A \,b^{2} c^{2} d e +480 A b \,c^{3} d^{2}-105 B \,b^{4} e^{2}+300 B \,b^{3} c d e -240 B \,b^{2} c^{2} d^{2}\right ) x \left (c x +b \right )}{1920 c^{4} \sqrt {x \left (c x +b \right )}}-\frac {b^{2} \left (10 A \,b^{2} c \,e^{2}-32 A b \,c^{2} d e +32 A \,c^{3} d^{2}-7 b^{3} B \,e^{2}+20 B \,b^{2} c d e -16 B b \,c^{2} d^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}\) | \(351\) |
| default | \(A \,d^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )+B \,e^{2} \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )+\left (A \,e^{2}+2 B d e \right ) \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )+\left (2 A d e +B \,d^{2}\right ) \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(399\) |
5/64*(-(-7/10*b^3*B*e^2+b^2*c*e*(A*e+2*B*d)-16/5*c^2*(A*e+1/2*B*d)*d*b+16/ 5*A*c^3*d^2)*b^2*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(16/5*(1/6*x^2*(3/5* B*x+A)*e^2+2/3*(1/2*B*x+A)*x*d*e+d^2*(1/3*B*x+A))*b*c^(7/2)+32/5*x*(1/2*(4 /5*B*x+A)*x^2*e^2+4/3*(3/4*B*x+A)*x*d*e+d^2*(2/3*B*x+A))*c^(9/2)+((-2/3*(1 4/25*B*x+A)*x*e^2-16/5*d*(5/12*B*x+A)*e-8/5*B*d^2)*c^(5/2)+(((7/15*B*x+A)* e+2*B*d)*c^(3/2)-7/10*B*b*e*c^(1/2))*e*b)*b^2)*(x*(c*x+b))^(1/2))/c^(9/2)
Time = 0.36 (sec) , antiderivative size = 684, normalized size of antiderivative = 2.56 \[ \int (A+B x) (d+e x)^2 \sqrt {b x+c x^2} \, dx=\left [-\frac {15 \, {\left (16 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{2} - 4 \, {\left (5 \, B b^{4} c - 8 \, A b^{3} c^{2}\right )} d e + {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} e^{2}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} e^{2} x^{4} + 48 \, {\left (20 \, B c^{5} d e + {\left (B b c^{4} + 10 \, A c^{5}\right )} e^{2}\right )} x^{3} - 240 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{2} + 60 \, {\left (5 \, B b^{3} c^{2} - 8 \, A b^{2} c^{3}\right )} d e - 15 \, {\left (7 \, B b^{4} c - 10 \, A b^{3} c^{2}\right )} e^{2} + 8 \, {\left (80 \, B c^{5} d^{2} + 20 \, {\left (B b c^{4} + 8 \, A c^{5}\right )} d e - {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} e^{2}\right )} x^{2} + 10 \, {\left (16 \, {\left (B b c^{4} + 6 \, A c^{5}\right )} d^{2} - 4 \, {\left (5 \, B b^{2} c^{3} - 8 \, A b c^{4}\right )} d e + {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{5}}, -\frac {15 \, {\left (16 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{2} - 4 \, {\left (5 \, B b^{4} c - 8 \, A b^{3} c^{2}\right )} d e + {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (384 \, B c^{5} e^{2} x^{4} + 48 \, {\left (20 \, B c^{5} d e + {\left (B b c^{4} + 10 \, A c^{5}\right )} e^{2}\right )} x^{3} - 240 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{2} + 60 \, {\left (5 \, B b^{3} c^{2} - 8 \, A b^{2} c^{3}\right )} d e - 15 \, {\left (7 \, B b^{4} c - 10 \, A b^{3} c^{2}\right )} e^{2} + 8 \, {\left (80 \, B c^{5} d^{2} + 20 \, {\left (B b c^{4} + 8 \, A c^{5}\right )} d e - {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} e^{2}\right )} x^{2} + 10 \, {\left (16 \, {\left (B b c^{4} + 6 \, A c^{5}\right )} d^{2} - 4 \, {\left (5 \, B b^{2} c^{3} - 8 \, A b c^{4}\right )} d e + {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{5}}\right ] \]
[-1/3840*(15*(16*(B*b^3*c^2 - 2*A*b^2*c^3)*d^2 - 4*(5*B*b^4*c - 8*A*b^3*c^ 2)*d*e + (7*B*b^5 - 10*A*b^4*c)*e^2)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*B*c^5*e^2*x^4 + 48*(20*B*c^5*d*e + (B*b*c^4 + 10* A*c^5)*e^2)*x^3 - 240*(B*b^2*c^3 - 2*A*b*c^4)*d^2 + 60*(5*B*b^3*c^2 - 8*A* b^2*c^3)*d*e - 15*(7*B*b^4*c - 10*A*b^3*c^2)*e^2 + 8*(80*B*c^5*d^2 + 20*(B *b*c^4 + 8*A*c^5)*d*e - (7*B*b^2*c^3 - 10*A*b*c^4)*e^2)*x^2 + 10*(16*(B*b* c^4 + 6*A*c^5)*d^2 - 4*(5*B*b^2*c^3 - 8*A*b*c^4)*d*e + (7*B*b^3*c^2 - 10*A *b^2*c^3)*e^2)*x)*sqrt(c*x^2 + b*x))/c^5, -1/1920*(15*(16*(B*b^3*c^2 - 2*A *b^2*c^3)*d^2 - 4*(5*B*b^4*c - 8*A*b^3*c^2)*d*e + (7*B*b^5 - 10*A*b^4*c)*e ^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (384*B*c^5*e^2*x^4 + 48*(20*B*c^5*d*e + (B*b*c^4 + 10*A*c^5)*e^2)*x^3 - 240*(B*b^2*c^3 - 2*A *b*c^4)*d^2 + 60*(5*B*b^3*c^2 - 8*A*b^2*c^3)*d*e - 15*(7*B*b^4*c - 10*A*b^ 3*c^2)*e^2 + 8*(80*B*c^5*d^2 + 20*(B*b*c^4 + 8*A*c^5)*d*e - (7*B*b^2*c^3 - 10*A*b*c^4)*e^2)*x^2 + 10*(16*(B*b*c^4 + 6*A*c^5)*d^2 - 4*(5*B*b^2*c^3 - 8*A*b*c^4)*d*e + (7*B*b^3*c^2 - 10*A*b^2*c^3)*e^2)*x)*sqrt(c*x^2 + b*x))/c ^5]
Leaf count of result is larger than twice the leaf count of optimal. 578 vs. \(2 (275) = 550\).
Time = 0.72 (sec) , antiderivative size = 578, normalized size of antiderivative = 2.16 \[ \int (A+B x) (d+e x)^2 \sqrt {b x+c x^2} \, dx=\begin {cases} - \frac {b \left (A b d^{2} - \frac {3 b \left (2 A b d e + A c d^{2} + B b d^{2} - \frac {5 b \left (A b e^{2} + 2 A c d e + 2 B b d e + B c d^{2} - \frac {7 b \left (A c e^{2} + \frac {B b e^{2}}{10} + 2 B c d e\right )}{8 c}\right )}{6 c}\right )}{4 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{2 c} + \sqrt {b x + c x^{2}} \left (\frac {B e^{2} x^{4}}{5} + \frac {x^{3} \left (A c e^{2} + \frac {B b e^{2}}{10} + 2 B c d e\right )}{4 c} + \frac {x^{2} \left (A b e^{2} + 2 A c d e + 2 B b d e + B c d^{2} - \frac {7 b \left (A c e^{2} + \frac {B b e^{2}}{10} + 2 B c d e\right )}{8 c}\right )}{3 c} + \frac {x \left (2 A b d e + A c d^{2} + B b d^{2} - \frac {5 b \left (A b e^{2} + 2 A c d e + 2 B b d e + B c d^{2} - \frac {7 b \left (A c e^{2} + \frac {B b e^{2}}{10} + 2 B c d e\right )}{8 c}\right )}{6 c}\right )}{2 c} + \frac {A b d^{2} - \frac {3 b \left (2 A b d e + A c d^{2} + B b d^{2} - \frac {5 b \left (A b e^{2} + 2 A c d e + 2 B b d e + B c d^{2} - \frac {7 b \left (A c e^{2} + \frac {B b e^{2}}{10} + 2 B c d e\right )}{8 c}\right )}{6 c}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A d^{2} \left (b x\right )^{\frac {3}{2}}}{3} + \frac {B e^{2} \left (b x\right )^{\frac {9}{2}}}{9 b^{3}} + \frac {\left (b x\right )^{\frac {5}{2}} \cdot \left (2 A d e + B d^{2}\right )}{5 b} + \frac {\left (b x\right )^{\frac {7}{2}} \left (A e^{2} + 2 B d e\right )}{7 b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-b*(A*b*d**2 - 3*b*(2*A*b*d*e + A*c*d**2 + B*b*d**2 - 5*b*(A*b* e**2 + 2*A*c*d*e + 2*B*b*d*e + B*c*d**2 - 7*b*(A*c*e**2 + B*b*e**2/10 + 2* B*c*d*e)/(8*c))/(6*c))/(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*e**2*x**4/5 + x **3*(A*c*e**2 + B*b*e**2/10 + 2*B*c*d*e)/(4*c) + x**2*(A*b*e**2 + 2*A*c*d* e + 2*B*b*d*e + B*c*d**2 - 7*b*(A*c*e**2 + B*b*e**2/10 + 2*B*c*d*e)/(8*c)) /(3*c) + x*(2*A*b*d*e + A*c*d**2 + B*b*d**2 - 5*b*(A*b*e**2 + 2*A*c*d*e + 2*B*b*d*e + B*c*d**2 - 7*b*(A*c*e**2 + B*b*e**2/10 + 2*B*c*d*e)/(8*c))/(6* c))/(2*c) + (A*b*d**2 - 3*b*(2*A*b*d*e + A*c*d**2 + B*b*d**2 - 5*b*(A*b*e* *2 + 2*A*c*d*e + 2*B*b*d*e + B*c*d**2 - 7*b*(A*c*e**2 + B*b*e**2/10 + 2*B* c*d*e)/(8*c))/(6*c))/(4*c))/c), Ne(c, 0)), (2*(A*d**2*(b*x)**(3/2)/3 + B*e **2*(b*x)**(9/2)/(9*b**3) + (b*x)**(5/2)*(2*A*d*e + B*d**2)/(5*b) + (b*x)* *(7/2)*(A*e**2 + 2*B*d*e)/(7*b**2))/b, Ne(b, 0)), (0, True))
Leaf count of result is larger than twice the leaf count of optimal. 512 vs. \(2 (247) = 494\).
Time = 0.20 (sec) , antiderivative size = 512, normalized size of antiderivative = 1.92 \[ \int (A+B x) (d+e x)^2 \sqrt {b x+c x^2} \, dx=\frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B e^{2} x^{2}}{5 \, c} + \frac {1}{2} \, \sqrt {c x^{2} + b x} A d^{2} x - \frac {7 \, \sqrt {c x^{2} + b x} B b^{3} e^{2} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b e^{2} x}{40 \, c^{2}} - \frac {A b^{2} d^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {7 \, B b^{5} e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} + \frac {\sqrt {c x^{2} + b x} A b d^{2}}{4 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} B b^{4} e^{2}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2} e^{2}}{48 \, c^{3}} + \frac {5 \, {\left (2 \, B d e + A e^{2}\right )} \sqrt {c x^{2} + b x} b^{2} x}{32 \, c^{2}} + \frac {{\left (2 \, B d e + A e^{2}\right )} {\left (c x^{2} + b x\right )}^{\frac {3}{2}} x}{4 \, c} - \frac {{\left (B d^{2} + 2 \, A d e\right )} \sqrt {c x^{2} + b x} b x}{4 \, c} - \frac {5 \, {\left (2 \, B d e + A e^{2}\right )} b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {{\left (B d^{2} + 2 \, A d e\right )} b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} + \frac {5 \, {\left (2 \, B d e + A e^{2}\right )} \sqrt {c x^{2} + b x} b^{3}}{64 \, c^{3}} - \frac {5 \, {\left (2 \, B d e + A e^{2}\right )} {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b}{24 \, c^{2}} - \frac {{\left (B d^{2} + 2 \, A d e\right )} \sqrt {c x^{2} + b x} b^{2}}{8 \, c^{2}} + \frac {{\left (B d^{2} + 2 \, A d e\right )} {\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{3 \, c} \]
1/5*(c*x^2 + b*x)^(3/2)*B*e^2*x^2/c + 1/2*sqrt(c*x^2 + b*x)*A*d^2*x - 7/64 *sqrt(c*x^2 + b*x)*B*b^3*e^2*x/c^3 - 7/40*(c*x^2 + b*x)^(3/2)*B*b*e^2*x/c^ 2 - 1/8*A*b^2*d^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + 7 /256*B*b^5*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) + 1/4* sqrt(c*x^2 + b*x)*A*b*d^2/c - 7/128*sqrt(c*x^2 + b*x)*B*b^4*e^2/c^4 + 7/48 *(c*x^2 + b*x)^(3/2)*B*b^2*e^2/c^3 + 5/32*(2*B*d*e + A*e^2)*sqrt(c*x^2 + b *x)*b^2*x/c^2 + 1/4*(2*B*d*e + A*e^2)*(c*x^2 + b*x)^(3/2)*x/c - 1/4*(B*d^2 + 2*A*d*e)*sqrt(c*x^2 + b*x)*b*x/c - 5/128*(2*B*d*e + A*e^2)*b^4*log(2*c* x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 1/16*(B*d^2 + 2*A*d*e)*b^3* log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + 5/64*(2*B*d*e + A*e ^2)*sqrt(c*x^2 + b*x)*b^3/c^3 - 5/24*(2*B*d*e + A*e^2)*(c*x^2 + b*x)^(3/2) *b/c^2 - 1/8*(B*d^2 + 2*A*d*e)*sqrt(c*x^2 + b*x)*b^2/c^2 + 1/3*(B*d^2 + 2* A*d*e)*(c*x^2 + b*x)^(3/2)/c
Time = 0.28 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.31 \[ \int (A+B x) (d+e x)^2 \sqrt {b x+c x^2} \, dx=\frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B e^{2} x + \frac {20 \, B c^{4} d e + B b c^{3} e^{2} + 10 \, A c^{4} e^{2}}{c^{4}}\right )} x + \frac {80 \, B c^{4} d^{2} + 20 \, B b c^{3} d e + 160 \, A c^{4} d e - 7 \, B b^{2} c^{2} e^{2} + 10 \, A b c^{3} e^{2}}{c^{4}}\right )} x + \frac {5 \, {\left (16 \, B b c^{3} d^{2} + 96 \, A c^{4} d^{2} - 20 \, B b^{2} c^{2} d e + 32 \, A b c^{3} d e + 7 \, B b^{3} c e^{2} - 10 \, A b^{2} c^{2} e^{2}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (16 \, B b^{2} c^{2} d^{2} - 32 \, A b c^{3} d^{2} - 20 \, B b^{3} c d e + 32 \, A b^{2} c^{2} d e + 7 \, B b^{4} e^{2} - 10 \, A b^{3} c e^{2}\right )}}{c^{4}}\right )} - \frac {{\left (16 \, B b^{3} c^{2} d^{2} - 32 \, A b^{2} c^{3} d^{2} - 20 \, B b^{4} c d e + 32 \, A b^{3} c^{2} d e + 7 \, B b^{5} e^{2} - 10 \, A b^{4} c e^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {9}{2}}} \]
1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*e^2*x + (20*B*c^4*d*e + B*b*c^3*e^2 + 10*A*c^4*e^2)/c^4)*x + (80*B*c^4*d^2 + 20*B*b*c^3*d*e + 160*A*c^4*d*e - 7*B*b^2*c^2*e^2 + 10*A*b*c^3*e^2)/c^4)*x + 5*(16*B*b*c^3*d^2 + 96*A*c^4*d ^2 - 20*B*b^2*c^2*d*e + 32*A*b*c^3*d*e + 7*B*b^3*c*e^2 - 10*A*b^2*c^2*e^2) /c^4)*x - 15*(16*B*b^2*c^2*d^2 - 32*A*b*c^3*d^2 - 20*B*b^3*c*d*e + 32*A*b^ 2*c^2*d*e + 7*B*b^4*e^2 - 10*A*b^3*c*e^2)/c^4) - 1/256*(16*B*b^3*c^2*d^2 - 32*A*b^2*c^3*d^2 - 20*B*b^4*c*d*e + 32*A*b^3*c^2*d*e + 7*B*b^5*e^2 - 10*A *b^4*c*e^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(9/2 )
Time = 11.67 (sec) , antiderivative size = 537, normalized size of antiderivative = 2.01 \[ \int (A+B x) (d+e x)^2 \sqrt {b x+c x^2} \, dx=A\,d^2\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )+\frac {A\,e^2\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,A\,b\,e^2\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}-\frac {7\,B\,b\,e^2\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}\right )}{10\,c}+\frac {B\,e^2\,x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}-\frac {A\,b^2\,d^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}+\frac {B\,b^3\,d^2\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {B\,d^2\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}+\frac {B\,d\,e\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{2\,c}-\frac {5\,B\,b\,d\,e\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{4\,c}+\frac {A\,b^3\,d\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{8\,c^{5/2}}+\frac {A\,d\,e\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{12\,c^2} \]
A*d^2*(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) + (A*e^2*x*(b*x + c*x^2)^(3/2))/ (4*c) - (5*A*b*e^2*((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) - (7*B*b*e^2*((x*(b*x + c*x^2)^(3/2))/(4*c) - (5*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)))/(10*c) + (B*e^2*x^ 2*(b*x + c*x^2)^(3/2))/(5*c) - (A*b^2*d^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2)) + (B*b^3*d^2*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + (B*d^2*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3* b^2 + 2*b*c*x))/(24*c^2) + (B*d*e*x*(b*x + c*x^2)^(3/2))/(2*c) - (5*B*b*d* e*((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)))/(4*c) + (A*b ^3*d*e*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(8*c^(5/2)) + (A* d*e*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(12*c^2)