Integrand size = 22, antiderivative size = 165 \[ \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx=\frac {\left (\left (5 b^2-4 a c\right ) e^2+16 c d (c d-b e)\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}+\frac {e (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (b^2-4 a c\right ) \left (\left (5 b^2-4 a c\right ) e^2+16 c d (c d-b e)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}} \] Output:
1/64*((-4*a*c+5*b^2)*e^2+16*c*d*(-b*e+c*d))*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/ c^3+1/24*e*(6*c*e*x-5*b*e+16*c*d)*(c*x^2+b*x+a)^(3/2)/c^2-1/128*(-4*a*c+b^ 2)*((-4*a*c+5*b^2)*e^2+16*c*d*(-b*e+c*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c *x^2+b*x+a)^(1/2))/c^(7/2)
Time = 1.62 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.18 \[ \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx=\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^3 e^2-2 b^2 c e (24 d+5 e x)+4 b c \left (-13 a e^2+2 c \left (6 d^2+4 d e x+e^2 x^2\right )\right )+8 c^2 \left (a e (16 d+3 e x)+2 c x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )\right )-3 \left (b^2-4 a c\right ) \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{192 c^{7/2}} \] Input:
Integrate[(d + e*x)^2*Sqrt[a + b*x + c*x^2],x]
Output:
(Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^3*e^2 - 2*b^2*c*e*(24*d + 5*e*x) + 4* b*c*(-13*a*e^2 + 2*c*(6*d^2 + 4*d*e*x + e^2*x^2)) + 8*c^2*(a*e*(16*d + 3*e *x) + 2*c*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2))) - 3*(b^2 - 4*a*c)*(16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/(192*c^(7/2))
Time = 0.34 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1166, 27, 1160, 1087, 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.
| \(\displaystyle \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 1166 |
| \(\displaystyle \frac {\int \frac {1}{2} \left (8 c d^2-e (3 b d+2 a e)+5 e (2 c d-b e) x\right ) \sqrt {c x^2+b x+a}dx}{4 c}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \left (8 c d^2-3 b e d-2 a e^2+5 e (2 c d-b e) x\right ) \sqrt {c x^2+b x+a}dx}{8 c}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {\frac {\left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \int \sqrt {c x^2+b x+a}dx}{2 c}+\frac {5 e \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{3 c}}{8 c}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {\frac {\left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{2 c}+\frac {5 e \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{3 c}}{8 c}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {\frac {\left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{2 c}+\frac {5 e \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{3 c}}{8 c}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right )}{2 c}+\frac {5 e \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{3 c}}{8 c}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
Input:
Int[(d + e*x)^2*Sqrt[a + b*x + c*x^2],x]
Output:
(e*(d + e*x)*(a + b*x + c*x^2)^(3/2))/(4*c) + ((5*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(3/2))/(3*c) + ((16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*( ((b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2* c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(3/2))))/(2*c))/(8*c)
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[(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_))*((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_)*((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[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
Time = 0.97 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.39
| method | result | size |
| risch | \(-\frac {\left (-48 e^{2} c^{3} x^{3}-8 b \,c^{2} e^{2} x^{2}-128 c^{3} d e \,x^{2}-24 c^{2} a \,e^{2} x +10 b^{2} c \,e^{2} x -32 b d x \,c^{2} e -96 c^{3} d^{2} x +52 a b c \,e^{2}-128 a \,c^{2} d e -15 e^{2} b^{3}+48 b^{2} c d e -48 b \,c^{2} d^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{192 c^{3}}-\frac {\left (16 a^{2} c^{2} e^{2}-24 a \,b^{2} c \,e^{2}+64 a b \,c^{2} d e -64 a \,c^{3} d^{2}+5 b^{4} e^{2}-16 b^{3} c d e +16 b^{2} c^{2} d^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}\) | \(229\) |
| default | \(d^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )+e^{2} \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )+2 d e \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(349\) |
Input:
int((e*x+d)^2*(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/192*(-48*c^3*e^2*x^3-8*b*c^2*e^2*x^2-128*c^3*d*e*x^2-24*a*c^2*e^2*x+10* b^2*c*e^2*x-32*b*c^2*d*e*x-96*c^3*d^2*x+52*a*b*c*e^2-128*a*c^2*d*e-15*b^3* e^2+48*b^2*c*d*e-48*b*c^2*d^2)/c^3*(c*x^2+b*x+a)^(1/2)-1/128*(16*a^2*c^2*e ^2-24*a*b^2*c*e^2+64*a*b*c^2*d*e-64*a*c^3*d^2+5*b^4*e^2-16*b^3*c*d*e+16*b^ 2*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 = 0.10 (sec) , antiderivative size = 505, normalized size of antiderivative = 3.06 \[ \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx=\left [\frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} - 16 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} e^{2} x^{3} + 48 \, b c^{3} d^{2} - 16 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} d e + {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} e^{2} + 8 \, {\left (16 \, c^{4} d e + b c^{3} e^{2}\right )} x^{2} + 2 \, {\left (48 \, c^{4} d^{2} + 16 \, b c^{3} d e - {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{4}}, \frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} - 16 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (48 \, c^{4} e^{2} x^{3} + 48 \, b c^{3} d^{2} - 16 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} d e + {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} e^{2} + 8 \, {\left (16 \, c^{4} d e + b c^{3} e^{2}\right )} x^{2} + 2 \, {\left (48 \, c^{4} d^{2} + 16 \, b c^{3} d e - {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{4}}\right ] \] Input:
integrate((e*x+d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
Output:
[1/768*(3*(16*(b^2*c^2 - 4*a*c^3)*d^2 - 16*(b^3*c - 4*a*b*c^2)*d*e + (5*b^ 4 - 24*a*b^2*c + 16*a^2*c^2)*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*(48*c^4*e^2*x^3 + 48*b*c^3*d^2 - 16*(3*b^2*c^2 - 8*a*c^3)*d*e + (15*b^3*c - 52*a*b*c^2)*e^ 2 + 8*(16*c^4*d*e + b*c^3*e^2)*x^2 + 2*(48*c^4*d^2 + 16*b*c^3*d*e - (5*b^2 *c^2 - 12*a*c^3)*e^2)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/384*(3*(16*(b^2*c^2 - 4*a*c^3)*d^2 - 16*(b^3*c - 4*a*b*c^2)*d*e + (5*b^4 - 24*a*b^2*c + 16*a^ 2*c^2)*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*(48*c^4*e^2*x^3 + 48*b*c^3*d^2 - 16*(3*b^2*c ^2 - 8*a*c^3)*d*e + (15*b^3*c - 52*a*b*c^2)*e^2 + 8*(16*c^4*d*e + b*c^3*e^ 2)*x^2 + 2*(48*c^4*d^2 + 16*b*c^3*d*e - (5*b^2*c^2 - 12*a*c^3)*e^2)*x)*sqr t(c*x^2 + b*x + a))/c^4]
Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (155) = 310\).
Time = 0.95 (sec) , antiderivative size = 473, normalized size of antiderivative = 2.87 \[ \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \left (\frac {e^{2} x^{3}}{4} + \frac {x^{2} \left (\frac {b e^{2}}{8} + 2 c d e\right )}{3 c} + \frac {x \left (\frac {a e^{2}}{4} + 2 b d e - \frac {5 b \left (\frac {b e^{2}}{8} + 2 c d e\right )}{6 c} + c d^{2}\right )}{2 c} + \frac {2 a d e - \frac {2 a \left (\frac {b e^{2}}{8} + 2 c d e\right )}{3 c} + b d^{2} - \frac {3 b \left (\frac {a e^{2}}{4} + 2 b d e - \frac {5 b \left (\frac {b e^{2}}{8} + 2 c d e\right )}{6 c} + c d^{2}\right )}{4 c}}{c}\right ) + \left (a d^{2} - \frac {a \left (\frac {a e^{2}}{4} + 2 b d e - \frac {5 b \left (\frac {b e^{2}}{8} + 2 c d e\right )}{6 c} + c d^{2}\right )}{2 c} - \frac {b \left (2 a d e - \frac {2 a \left (\frac {b e^{2}}{8} + 2 c d e\right )}{3 c} + b d^{2} - \frac {3 b \left (\frac {a e^{2}}{4} + 2 b d e - \frac {5 b \left (\frac {b e^{2}}{8} + 2 c d e\right )}{6 c} + c d^{2}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {e^{2} \left (a + b x\right )^{\frac {7}{2}}}{7 b^{2}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- 2 a e^{2} + 2 b d e\right )}{5 b^{2}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (a^{2} e^{2} - 2 a b d e + b^{2} d^{2}\right )}{3 b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\begin {cases} d^{2} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{3}}{3 e} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((e*x+d)**2*(c*x**2+b*x+a)**(1/2),x)
Output:
Piecewise((sqrt(a + b*x + c*x**2)*(e**2*x**3/4 + x**2*(b*e**2/8 + 2*c*d*e) /(3*c) + x*(a*e**2/4 + 2*b*d*e - 5*b*(b*e**2/8 + 2*c*d*e)/(6*c) + c*d**2)/ (2*c) + (2*a*d*e - 2*a*(b*e**2/8 + 2*c*d*e)/(3*c) + b*d**2 - 3*b*(a*e**2/4 + 2*b*d*e - 5*b*(b*e**2/8 + 2*c*d*e)/(6*c) + c*d**2)/(4*c))/c) + (a*d**2 - a*(a*e**2/4 + 2*b*d*e - 5*b*(b*e**2/8 + 2*c*d*e)/(6*c) + c*d**2)/(2*c) - b*(2*a*d*e - 2*a*(b*e**2/8 + 2*c*d*e)/(3*c) + b*d**2 - 3*b*(a*e**2/4 + 2* b*d*e - 5*b*(b*e**2/8 + 2*c*d*e)/(6*c) + c*d**2)/(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*(e**2*(a + b*x)**(7/2)/(7*b**2) + (a + b*x)**(5/2)*(-2*a*e* *2 + 2*b*d*e)/(5*b**2) + (a + b*x)**(3/2)*(a**2*e**2 - 2*a*b*d*e + b**2*d* *2)/(3*b**2))/b, Ne(b, 0)), (sqrt(a)*Piecewise((d**2*x, Eq(e, 0)), ((d + e *x)**3/(3*e), True)), True))
Exception generated. \[ \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x+d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")
Output:
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 = 0.32 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.43 \[ \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, e^{2} x + \frac {16 \, c^{3} d e + b c^{2} e^{2}}{c^{3}}\right )} x + \frac {48 \, c^{3} d^{2} + 16 \, b c^{2} d e - 5 \, b^{2} c e^{2} + 12 \, a c^{2} e^{2}}{c^{3}}\right )} x + \frac {48 \, b c^{2} d^{2} - 48 \, b^{2} c d e + 128 \, a c^{2} d e + 15 \, b^{3} e^{2} - 52 \, a b c e^{2}}{c^{3}}\right )} + \frac {{\left (16 \, b^{2} c^{2} d^{2} - 64 \, a c^{3} d^{2} - 16 \, b^{3} c d e + 64 \, a b c^{2} d e + 5 \, b^{4} e^{2} - 24 \, a b^{2} c e^{2} + 16 \, a^{2} c^{2} e^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {7}{2}}} \] Input:
integrate((e*x+d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
Output:
1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*e^2*x + (16*c^3*d*e + b*c^2*e^2)/c^3) *x + (48*c^3*d^2 + 16*b*c^2*d*e - 5*b^2*c*e^2 + 12*a*c^2*e^2)/c^3)*x + (48 *b*c^2*d^2 - 48*b^2*c*d*e + 128*a*c^2*d*e + 15*b^3*e^2 - 52*a*b*c*e^2)/c^3 ) + 1/128*(16*b^2*c^2*d^2 - 64*a*c^3*d^2 - 16*b^3*c*d*e + 64*a*b*c^2*d*e + 5*b^4*e^2 - 24*a*b^2*c*e^2 + 16*a^2*c^2*e^2)*log(abs(2*(sqrt(c)*x - sqrt( c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)
Time = 6.45 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.01 \[ \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx=d^2\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {e^2\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}-\frac {a\,e^2\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}+\frac {d^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}-\frac {5\,b\,e^2\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}+\frac {d\,e\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{12\,c^2}+\frac {d\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{8\,c^{5/2}} \] Input:
int((d + e*x)^2*(a + b*x + c*x^2)^(1/2),x)
Output:
d^2*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (e^2*x*(a + b*x + c*x^2)^(3/ 2))/(4*c) - (a*e^2*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c) + (d^2*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/ (2*c^(3/2)) - (5*b*e^2*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/ 2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*( a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) + (d*e*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(12*c^2) + (d*e*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(8*c^(5/2))
Time = 1.60 (sec) , antiderivative size = 589, normalized size of antiderivative = 3.57 \[ \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx=\frac {-104 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2} e^{2}+256 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} d e +48 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} e^{2} x +30 \sqrt {c \,x^{2}+b x +a}\, b^{3} c \,e^{2}-96 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} d e -20 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} e^{2} x +96 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} d^{2}+64 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} d e x +16 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} e^{2} x^{2}+192 \sqrt {c \,x^{2}+b x +a}\, c^{4} d^{2} x +256 \sqrt {c \,x^{2}+b x +a}\, c^{4} d e \,x^{2}+96 \sqrt {c \,x^{2}+b x +a}\, c^{4} e^{2} x^{3}-48 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{2} e^{2}+72 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c \,e^{2}-192 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a b \,c^{2} d e +192 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,c^{3} d^{2}-15 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{4} e^{2}+48 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{3} c d e -48 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{2} c^{2} d^{2}}{384 c^{4}} \] Input:
int((e*x+d)^2*(c*x^2+b*x+a)^(1/2),x)
Output:
( - 104*sqrt(a + b*x + c*x**2)*a*b*c**2*e**2 + 256*sqrt(a + b*x + c*x**2)* a*c**3*d*e + 48*sqrt(a + b*x + c*x**2)*a*c**3*e**2*x + 30*sqrt(a + b*x + c *x**2)*b**3*c*e**2 - 96*sqrt(a + b*x + c*x**2)*b**2*c**2*d*e - 20*sqrt(a + b*x + c*x**2)*b**2*c**2*e**2*x + 96*sqrt(a + b*x + c*x**2)*b*c**3*d**2 + 64*sqrt(a + b*x + c*x**2)*b*c**3*d*e*x + 16*sqrt(a + b*x + c*x**2)*b*c**3* e**2*x**2 + 192*sqrt(a + b*x + c*x**2)*c**4*d**2*x + 256*sqrt(a + b*x + c* x**2)*c**4*d*e*x**2 + 96*sqrt(a + b*x + c*x**2)*c**4*e**2*x**3 - 48*sqrt(c )*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a **2*c**2*e**2 + 72*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c *x)/sqrt(4*a*c - b**2))*a*b**2*c*e**2 - 192*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c**2*d*e + 192*sqrt(c )*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a *c**3*d**2 - 15*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x) /sqrt(4*a*c - b**2))*b**4*e**2 + 48*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**3*c*d*e - 48*sqrt(c)*log((2*sq rt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*c**2*d* *2)/(384*c**4)