Integrand size = 27, antiderivative size = 234 \[ \int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {B (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}+\frac {\left (6 A c e (8 c d-3 b e)+B \left (16 c^2 d^2+15 b^2 e^2-4 c e (9 b d+4 a e)\right )+2 c e (4 B c d-5 b B e+6 A c e) x\right ) \sqrt {a+b x+c x^2}}{24 c^3}-\frac {\left (5 b^3 B e^2-6 b^2 c e (2 B d+A e)-8 c^2 \left (2 A c d^2-2 a B d e-a A e^2\right )+4 b c \left (2 B c d^2+4 A c d e-3 a B e^2\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{7/2}} \]
-1/16*(5*b^3*B*e^2-6*b^2*c*e*(A*e+2*B*d)-8*c^2*(-A*a*e^2+2*A*c*d^2-2*B*a*d *e)+4*b*c*(4*A*c*d*e-3*B*a*e^2+2*B*c*d^2))*arctanh(1/2*(2*c*x+b)/c^(1/2)/( c*x^2+b*x+a)^(1/2))/c^(7/2)+1/3*B*(e*x+d)^2*(c*x^2+b*x+a)^(1/2)/c+1/24*(6* A*c*e*(-3*b*e+8*c*d)+B*(16*c^2*d^2+15*b^2*e^2-4*c*e*(4*a*e+9*b*d))+2*c*e*( 6*A*c*e-5*B*b*e+4*B*c*d)*x)*(c*x^2+b*x+a)^(1/2)/c^3
Time = 0.65 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (6 A c e (8 c d-3 b e+2 c e x)+B \left (15 b^2 e^2-2 c e (18 b d+8 a e+5 b e x)+8 c^2 \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )+3 \left (5 b^3 B e^2-6 b^2 c e (2 B d+A e)+8 c^2 \left (-2 A c d^2+2 a B d e+a A e^2\right )+4 b c \left (2 B c d^2+4 A c d e-3 a B e^2\right )\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{48 c^{7/2}} \]
(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(6*A*c*e*(8*c*d - 3*b*e + 2*c*e*x) + B*(1 5*b^2*e^2 - 2*c*e*(18*b*d + 8*a*e + 5*b*e*x) + 8*c^2*(3*d^2 + 3*d*e*x + e^ 2*x^2))) + 3*(5*b^3*B*e^2 - 6*b^2*c*e*(2*B*d + A*e) + 8*c^2*(-2*A*c*d^2 + 2*a*B*d*e + a*A*e^2) + 4*b*c*(2*B*c*d^2 + 4*A*c*d*e - 3*a*B*e^2))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(48*c^(7/2))
Time = 0.44 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1236, 27, 1225, 1092, 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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {\int -\frac {(d+e x) (b B d-6 A c d+4 a B e-(4 B c d-5 b B e+6 A c e) x)}{2 \sqrt {c x^2+b x+a}}dx}{3 c}+\frac {B (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}-\frac {\int \frac {(d+e x) (b B d-6 A c d+4 a B e-(4 B c d-5 b B e+6 A c e) x)}{\sqrt {c x^2+b x+a}}dx}{6 c}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {B (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}-\frac {\frac {3 \left (4 b c \left (-3 a B e^2+4 A c d e+2 B c d^2\right )-8 c^2 \left (-a A e^2-2 a B d e+2 A c d^2\right )-6 b^2 c e (A e+2 B d)+5 b^3 B e^2\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c^2}-\frac {\sqrt {a+b x+c x^2} \left (2 B \left (-2 c e (4 a e+9 b d)+\frac {15 b^2 e^2}{2}+8 c^2 d^2\right )+2 c e x (6 A c e-5 b B e+4 B c d)+6 A c e (8 c d-3 b e)\right )}{4 c^2}}{6 c}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {B (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}-\frac {\frac {3 \left (4 b c \left (-3 a B e^2+4 A c d e+2 B c d^2\right )-8 c^2 \left (-a A e^2-2 a B d e+2 A c d^2\right )-6 b^2 c e (A e+2 B d)+5 b^3 B e^2\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c^2}-\frac {\sqrt {a+b x+c x^2} \left (2 B \left (-2 c e (4 a e+9 b d)+\frac {15 b^2 e^2}{2}+8 c^2 d^2\right )+2 c e x (6 A c e-5 b B e+4 B c d)+6 A c e (8 c d-3 b e)\right )}{4 c^2}}{6 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}-\frac {\frac {3 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (4 b c \left (-3 a B e^2+4 A c d e+2 B c d^2\right )-8 c^2 \left (-a A e^2-2 a B d e+2 A c d^2\right )-6 b^2 c e (A e+2 B d)+5 b^3 B e^2\right )}{8 c^{5/2}}-\frac {\sqrt {a+b x+c x^2} \left (2 B \left (-2 c e (4 a e+9 b d)+\frac {15 b^2 e^2}{2}+8 c^2 d^2\right )+2 c e x (6 A c e-5 b B e+4 B c d)+6 A c e (8 c d-3 b e)\right )}{4 c^2}}{6 c}\) |
(B*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(3*c) - (-1/4*((6*A*c*e*(8*c*d - 3*b *e) + 2*B*(8*c^2*d^2 + (15*b^2*e^2)/2 - 2*c*e*(9*b*d + 4*a*e)) + 2*c*e*(4* B*c*d - 5*b*B*e + 6*A*c*e)*x)*Sqrt[a + b*x + c*x^2])/c^2 + (3*(5*b^3*B*e^2 - 6*b^2*c*e*(2*B*d + A*e) - 8*c^2*(2*A*c*d^2 - 2*a*B*d*e - a*A*e^2) + 4*b *c*(2*B*c*d^2 + 4*A*c*d*e - 3*a*B*e^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr t[a + b*x + c*x^2])])/(8*c^(5/2)))/(6*c)
3.25.66.3.1 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[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , 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 = 0.57 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(-\frac {\left (-8 B \,e^{2} c^{2} x^{2}-12 A \,c^{2} e^{2} x +10 B b c \,e^{2} x -24 B \,c^{2} d e x +18 A b c \,e^{2}-48 A \,c^{2} d e +16 B \,e^{2} a c -15 B \,b^{2} e^{2}+36 B b c d e -24 B \,c^{2} d^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{24 c^{3}}-\frac {\left (8 A a \,c^{2} e^{2}-6 A \,b^{2} c \,e^{2}+16 A b \,c^{2} d e -16 A \,c^{3} d^{2}-12 B a b c \,e^{2}+16 B a \,c^{2} d e +5 b^{3} B \,e^{2}-12 B \,b^{2} c d e +8 B b \,c^{2} d^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}\) | \(224\) |
| default | \(\frac {A \,d^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+B \,e^{2} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )+\left (A \,e^{2}+2 B d e \right ) \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+\left (2 A d e +B \,d^{2}\right ) \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\) | \(401\) |
-1/24*(-8*B*c^2*e^2*x^2-12*A*c^2*e^2*x+10*B*b*c*e^2*x-24*B*c^2*d*e*x+18*A* b*c*e^2-48*A*c^2*d*e+16*B*a*c*e^2-15*B*b^2*e^2+36*B*b*c*d*e-24*B*c^2*d^2)* (c*x^2+b*x+a)^(1/2)/c^3-1/16*(8*A*a*c^2*e^2-6*A*b^2*c*e^2+16*A*b*c^2*d*e-1 6*A*c^3*d^2-12*B*a*b*c*e^2+16*B*a*c^2*d*e+5*B*b^3*e^2-12*B*b^2*c*d*e+8*B*b *c^2*d^2)/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
Time = 0.34 (sec) , antiderivative size = 487, normalized size of antiderivative = 2.08 \[ \int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\left [\frac {3 \, {\left (8 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 4 \, {\left (3 \, B b^{2} c - 4 \, {\left (B a + A b\right )} c^{2}\right )} d e + {\left (5 \, B b^{3} + 8 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} e^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, B c^{3} e^{2} x^{2} + 24 \, B c^{3} d^{2} - 12 \, {\left (3 \, B b c^{2} - 4 \, A c^{3}\right )} d e + {\left (15 \, B b^{2} c - 2 \, {\left (8 \, B a + 9 \, A b\right )} c^{2}\right )} e^{2} + 2 \, {\left (12 \, B c^{3} d e - {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, c^{4}}, \frac {3 \, {\left (8 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 4 \, {\left (3 \, B b^{2} c - 4 \, {\left (B a + A b\right )} c^{2}\right )} d e + {\left (5 \, B b^{3} + 8 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (8 \, B c^{3} e^{2} x^{2} + 24 \, B c^{3} d^{2} - 12 \, {\left (3 \, B b c^{2} - 4 \, A c^{3}\right )} d e + {\left (15 \, B b^{2} c - 2 \, {\left (8 \, B a + 9 \, A b\right )} c^{2}\right )} e^{2} + 2 \, {\left (12 \, B c^{3} d e - {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{4}}\right ] \]
[1/96*(3*(8*(B*b*c^2 - 2*A*c^3)*d^2 - 4*(3*B*b^2*c - 4*(B*a + A*b)*c^2)*d* e + (5*B*b^3 + 8*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*e^2)*sqrt(c)*log(-8*c^2* x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(8*B*c^3*e^2*x^2 + 24*B*c^3*d^2 - 12*(3*B*b*c^2 - 4*A*c^3)*d*e + (15* B*b^2*c - 2*(8*B*a + 9*A*b)*c^2)*e^2 + 2*(12*B*c^3*d*e - (5*B*b*c^2 - 6*A* c^3)*e^2)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/48*(3*(8*(B*b*c^2 - 2*A*c^3)*d^ 2 - 4*(3*B*b^2*c - 4*(B*a + A*b)*c^2)*d*e + (5*B*b^3 + 8*A*a*c^2 - 6*(2*B* a*b + A*b^2)*c)*e^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b) *sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(8*B*c^3*e^2*x^2 + 24*B*c^3*d^2 - 1 2*(3*B*b*c^2 - 4*A*c^3)*d*e + (15*B*b^2*c - 2*(8*B*a + 9*A*b)*c^2)*e^2 + 2 *(12*B*c^3*d*e - (5*B*b*c^2 - 6*A*c^3)*e^2)*x)*sqrt(c*x^2 + b*x + a))/c^4]
Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (243) = 486\).
Time = 0.96 (sec) , antiderivative size = 527, normalized size of antiderivative = 2.25 \[ \int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \left (\frac {B e^{2} x^{2}}{3 c} + \frac {x \left (A e^{2} - \frac {5 B b e^{2}}{6 c} + 2 B d e\right )}{2 c} + \frac {2 A d e - \frac {2 B a e^{2}}{3 c} + B d^{2} - \frac {3 b \left (A e^{2} - \frac {5 B b e^{2}}{6 c} + 2 B d e\right )}{4 c}}{c}\right ) + \left (A d^{2} - \frac {a \left (A e^{2} - \frac {5 B b e^{2}}{6 c} + 2 B d e\right )}{2 c} - \frac {b \left (2 A d e - \frac {2 B a e^{2}}{3 c} + B d^{2} - \frac {3 b \left (A e^{2} - \frac {5 B b e^{2}}{6 c} + 2 B d e\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {B e^{2} \left (a + b x\right )^{\frac {7}{2}}}{7 b^{3}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (A b e^{2} - 3 B a e^{2} + 2 B b d e\right )}{5 b^{3}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- 2 A a b e^{2} + 2 A b^{2} d e + 3 B a^{2} e^{2} - 4 B a b d e + B b^{2} d^{2}\right )}{3 b^{3}} + \frac {\sqrt {a + b x} \left (A a^{2} b e^{2} - 2 A a b^{2} d e + A b^{3} d^{2} - B a^{3} e^{2} + 2 B a^{2} b d e - B a b^{2} d^{2}\right )}{b^{3}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {A d^{2} x + \frac {B e^{2} x^{4}}{4} + \frac {x^{3} \left (A e^{2} + 2 B d e\right )}{3} + \frac {x^{2} \cdot \left (2 A d e + B d^{2}\right )}{2}}{\sqrt {a}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x + c*x**2)*(B*e**2*x**2/(3*c) + x*(A*e**2 - 5*B*b*e **2/(6*c) + 2*B*d*e)/(2*c) + (2*A*d*e - 2*B*a*e**2/(3*c) + B*d**2 - 3*b*(A *e**2 - 5*B*b*e**2/(6*c) + 2*B*d*e)/(4*c))/c) + (A*d**2 - a*(A*e**2 - 5*B* b*e**2/(6*c) + 2*B*d*e)/(2*c) - b*(2*A*d*e - 2*B*a*e**2/(3*c) + B*d**2 - 3 *b*(A*e**2 - 5*B*b*e**2/(6*c) + 2*B*d*e)/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)), Ne(c, 0 )), (2*(B*e**2*(a + b*x)**(7/2)/(7*b**3) + (a + b*x)**(5/2)*(A*b*e**2 - 3* B*a*e**2 + 2*B*b*d*e)/(5*b**3) + (a + b*x)**(3/2)*(-2*A*a*b*e**2 + 2*A*b** 2*d*e + 3*B*a**2*e**2 - 4*B*a*b*d*e + B*b**2*d**2)/(3*b**3) + sqrt(a + b*x )*(A*a**2*b*e**2 - 2*A*a*b**2*d*e + A*b**3*d**2 - B*a**3*e**2 + 2*B*a**2*b *d*e - B*a*b**2*d**2)/b**3)/b, Ne(b, 0)), ((A*d**2*x + B*e**2*x**4/4 + x** 3*(A*e**2 + 2*B*d*e)/3 + x**2*(2*A*d*e + B*d**2)/2)/sqrt(a), True))
Exception generated. \[ \int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.31 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (\frac {4 \, B e^{2} x}{c} + \frac {12 \, B c^{2} d e - 5 \, B b c e^{2} + 6 \, A c^{2} e^{2}}{c^{3}}\right )} x + \frac {24 \, B c^{2} d^{2} - 36 \, B b c d e + 48 \, A c^{2} d e + 15 \, B b^{2} e^{2} - 16 \, B a c e^{2} - 18 \, A b c e^{2}}{c^{3}}\right )} + \frac {{\left (8 \, B b c^{2} d^{2} - 16 \, A c^{3} d^{2} - 12 \, B b^{2} c d e + 16 \, B a c^{2} d e + 16 \, A b c^{2} d e + 5 \, B b^{3} e^{2} - 12 \, B a b c e^{2} - 6 \, A b^{2} c e^{2} + 8 \, A a c^{2} e^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {7}{2}}} \]
1/24*sqrt(c*x^2 + b*x + a)*(2*(4*B*e^2*x/c + (12*B*c^2*d*e - 5*B*b*c*e^2 + 6*A*c^2*e^2)/c^3)*x + (24*B*c^2*d^2 - 36*B*b*c*d*e + 48*A*c^2*d*e + 15*B* b^2*e^2 - 16*B*a*c*e^2 - 18*A*b*c*e^2)/c^3) + 1/16*(8*B*b*c^2*d^2 - 16*A*c ^3*d^2 - 12*B*b^2*c*d*e + 16*B*a*c^2*d*e + 16*A*b*c^2*d*e + 5*B*b^3*e^2 - 12*B*a*b*c*e^2 - 6*A*b^2*c*e^2 + 8*A*a*c^2*e^2)*log(abs(2*(sqrt(c)*x - sqr t(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)
Timed out. \[ \int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^2}{\sqrt {c\,x^2+b\,x+a}} \,d x \]