3.1548 \(\int (b+2 c x) (d+e x)^2 \sqrt {a+b x+c x^2} \, dx\)

Optimal. Leaf size=195 \[ -\frac {e \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{7/2}}+\frac {e \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} (2 c d-b e)}{32 c^3}+\frac {\left (a+b x+c x^2\right )^{3/2} \left (-2 c e (8 a e+5 b d)+5 b^2 e^2+6 c e x (2 c d-b e)+16 c^2 d^2\right )}{60 c^2}+\frac {2}{5} (d+e x)^2 \left (a+b x+c x^2\right )^{3/2} \]

[Out]

2/5*(e*x+d)^2*(c*x^2+b*x+a)^(3/2)+1/60*(16*c^2*d^2+5*b^2*e^2-2*c*e*(8*a*e+5*b*d)+6*c*e*(-b*e+2*c*d)*x)*(c*x^2+
b*x+a)^(3/2)/c^2-1/64*(-4*a*c+b^2)^2*e*(-b*e+2*c*d)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)
+1/32*(-4*a*c+b^2)*e*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {832, 779, 612, 621, 206} \[ \frac {\left (a+b x+c x^2\right )^{3/2} \left (-2 c e (8 a e+5 b d)+5 b^2 e^2+6 c e x (2 c d-b e)+16 c^2 d^2\right )}{60 c^2}+\frac {e \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} (2 c d-b e)}{32 c^3}-\frac {e \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{7/2}}+\frac {2}{5} (d+e x)^2 \left (a+b x+c x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(b + 2*c*x)*(d + e*x)^2*Sqrt[a + b*x + c*x^2],x]

[Out]

((b^2 - 4*a*c)*e*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(32*c^3) + (2*(d + e*x)^2*(a + b*x + c*x^2)^
(3/2))/5 + ((16*c^2*d^2 + 5*b^2*e^2 - 2*c*e*(5*b*d + 8*a*e) + 6*c*e*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(3/2))/
(60*c^2) - ((b^2 - 4*a*c)^2*e*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(7/2
))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 612

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] - Dist[(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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 779

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] + Dist[(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] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rubi steps

\begin {align*} \int (b+2 c x) (d+e x)^2 \sqrt {a+b x+c x^2} \, dx &=\frac {2}{5} (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}+\frac {\int (d+e x) (2 c (b d-2 a e)+2 c (2 c d-b e) x) \sqrt {a+b x+c x^2} \, dx}{5 c}\\ &=\frac {2}{5} (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}+\frac {\left (16 c^2 d^2+5 b^2 e^2-2 c e (5 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{60 c^2}+\frac {\left (\left (b^2-4 a c\right ) e (2 c d-b e)\right ) \int \sqrt {a+b x+c x^2} \, dx}{8 c^2}\\ &=\frac {\left (b^2-4 a c\right ) e (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^3}+\frac {2}{5} (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}+\frac {\left (16 c^2 d^2+5 b^2 e^2-2 c e (5 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{60 c^2}-\frac {\left (\left (b^2-4 a c\right )^2 e (2 c d-b e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{64 c^3}\\ &=\frac {\left (b^2-4 a c\right ) e (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^3}+\frac {2}{5} (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}+\frac {\left (16 c^2 d^2+5 b^2 e^2-2 c e (5 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{60 c^2}-\frac {\left (\left (b^2-4 a c\right )^2 e (2 c d-b e)\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{32 c^3}\\ &=\frac {\left (b^2-4 a c\right ) e (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^3}+\frac {2}{5} (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}+\frac {\left (16 c^2 d^2+5 b^2 e^2-2 c e (5 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{60 c^2}-\frac {\left (b^2-4 a c\right )^2 e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 177, normalized size = 0.91 \[ \frac {e \left (b^2-4 a c\right ) (b e-2 c d) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}\right )}{64 c^{7/2}}+\frac {(a+x (b+c x))^{3/2} \left (-2 c e (8 a e+5 b d+3 b e x)+5 b^2 e^2+4 c^2 d (4 d+3 e x)\right )}{60 c^2}+\frac {2}{5} (d+e x)^2 (a+x (b+c x))^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b + 2*c*x)*(d + e*x)^2*Sqrt[a + b*x + c*x^2],x]

[Out]

(2*(d + e*x)^2*(a + x*(b + c*x))^(3/2))/5 + ((a + x*(b + c*x))^(3/2)*(5*b^2*e^2 + 4*c^2*d*(4*d + 3*e*x) - 2*c*
e*(5*b*d + 8*a*e + 3*b*e*x)))/(60*c^2) + ((b^2 - 4*a*c)*e*(-2*c*d + b*e)*(-2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b
+ c*x)] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(64*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 617, normalized size = 3.16 \[ \left [-\frac {15 \, {\left (2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d e - {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (192 \, c^{5} e^{2} x^{4} + 320 \, a c^{4} d^{2} + 48 \, {\left (10 \, c^{5} d e + 3 \, b c^{4} e^{2}\right )} x^{3} + 10 \, {\left (3 \, b^{3} c^{2} - 20 \, a b c^{3}\right )} d e - {\left (15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}\right )} e^{2} + 8 \, {\left (40 \, c^{5} d^{2} + 50 \, b c^{4} d e - {\left (b^{2} c^{3} - 8 \, a c^{4}\right )} e^{2}\right )} x^{2} + 2 \, {\left (160 \, b c^{4} d^{2} - 10 \, {\left (b^{2} c^{3} - 12 \, a c^{4}\right )} d e + {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1920 \, c^{4}}, \frac {15 \, {\left (2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d e - {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (192 \, c^{5} e^{2} x^{4} + 320 \, a c^{4} d^{2} + 48 \, {\left (10 \, c^{5} d e + 3 \, b c^{4} e^{2}\right )} x^{3} + 10 \, {\left (3 \, b^{3} c^{2} - 20 \, a b c^{3}\right )} d e - {\left (15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}\right )} e^{2} + 8 \, {\left (40 \, c^{5} d^{2} + 50 \, b c^{4} d e - {\left (b^{2} c^{3} - 8 \, a c^{4}\right )} e^{2}\right )} x^{2} + 2 \, {\left (160 \, b c^{4} d^{2} - 10 \, {\left (b^{2} c^{3} - 12 \, a c^{4}\right )} d e + {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{960 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/1920*(15*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*e - (b^5 - 8*a*b^3*c + 16*a^2*b*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*(192*c^5*e^2*x^4 + 320*a*c^
4*d^2 + 48*(10*c^5*d*e + 3*b*c^4*e^2)*x^3 + 10*(3*b^3*c^2 - 20*a*b*c^3)*d*e - (15*b^4*c - 100*a*b^2*c^2 + 128*
a^2*c^3)*e^2 + 8*(40*c^5*d^2 + 50*b*c^4*d*e - (b^2*c^3 - 8*a*c^4)*e^2)*x^2 + 2*(160*b*c^4*d^2 - 10*(b^2*c^3 -
12*a*c^4)*d*e + (5*b^3*c^2 - 28*a*b*c^3)*e^2)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/960*(15*(2*(b^4*c - 8*a*b^2*c^2
+ 16*a^2*c^3)*d*e - (b^5 - 8*a*b^3*c + 16*a^2*b*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*(192*c^5*e^2*x^4 + 320*a*c^4*d^2 + 48*(10*c^5*d*e + 3*b*c^4*e^2)*x^3
+ 10*(3*b^3*c^2 - 20*a*b*c^3)*d*e - (15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3)*e^2 + 8*(40*c^5*d^2 + 50*b*c^4*d*
e - (b^2*c^3 - 8*a*c^4)*e^2)*x^2 + 2*(160*b*c^4*d^2 - 10*(b^2*c^3 - 12*a*c^4)*d*e + (5*b^3*c^2 - 28*a*b*c^3)*e
^2)*x)*sqrt(c*x^2 + b*x + a))/c^4]

________________________________________________________________________________________

giac [A]  time = 0.24, size = 308, normalized size = 1.58 \[ \frac {1}{480} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (4 \, c x e^{2} + \frac {10 \, c^{5} d e + 3 \, b c^{4} e^{2}}{c^{4}}\right )} x + \frac {40 \, c^{5} d^{2} + 50 \, b c^{4} d e - b^{2} c^{3} e^{2} + 8 \, a c^{4} e^{2}}{c^{4}}\right )} x + \frac {160 \, b c^{4} d^{2} - 10 \, b^{2} c^{3} d e + 120 \, a c^{4} d e + 5 \, b^{3} c^{2} e^{2} - 28 \, a b c^{3} e^{2}}{c^{4}}\right )} x + \frac {320 \, a c^{4} d^{2} + 30 \, b^{3} c^{2} d e - 200 \, a b c^{3} d e - 15 \, b^{4} c e^{2} + 100 \, a b^{2} c^{2} e^{2} - 128 \, a^{2} c^{3} e^{2}}{c^{4}}\right )} + \frac {{\left (2 \, b^{4} c d e - 16 \, a b^{2} c^{2} d e + 32 \, a^{2} c^{3} d e - b^{5} e^{2} + 8 \, a b^{3} c e^{2} - 16 \, a^{2} b c^{2} e^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{64 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/480*sqrt(c*x^2 + b*x + a)*(2*(4*(6*(4*c*x*e^2 + (10*c^5*d*e + 3*b*c^4*e^2)/c^4)*x + (40*c^5*d^2 + 50*b*c^4*d
*e - b^2*c^3*e^2 + 8*a*c^4*e^2)/c^4)*x + (160*b*c^4*d^2 - 10*b^2*c^3*d*e + 120*a*c^4*d*e + 5*b^3*c^2*e^2 - 28*
a*b*c^3*e^2)/c^4)*x + (320*a*c^4*d^2 + 30*b^3*c^2*d*e - 200*a*b*c^3*d*e - 15*b^4*c*e^2 + 100*a*b^2*c^2*e^2 - 1
28*a^2*c^3*e^2)/c^4) + 1/64*(2*b^4*c*d*e - 16*a*b^2*c^2*d*e + 32*a^2*c^3*d*e - b^5*e^2 + 8*a*b^3*c*e^2 - 16*a^
2*b*c^2*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

________________________________________________________________________________________

maple [B]  time = 0.11, size = 535, normalized size = 2.74 \[ \frac {a^{2} b \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}-\frac {a^{2} d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\frac {a \,b^{3} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}+\frac {a \,b^{2} d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}+\frac {b^{5} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{64 c^{\frac {7}{2}}}-\frac {b^{4} d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 c^{\frac {5}{2}}}+\frac {\sqrt {c \,x^{2}+b x +a}\, a b \,e^{2} x}{4 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, a d e x}{2}-\frac {\sqrt {c \,x^{2}+b x +a}\, b^{3} e^{2} x}{16 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, b^{2} d e x}{8 c}+\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} e^{2} x^{2}}{5}+\frac {\sqrt {c \,x^{2}+b x +a}\, a \,b^{2} e^{2}}{8 c^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, a b d e}{4 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, b^{4} e^{2}}{32 c^{3}}+\frac {\sqrt {c \,x^{2}+b x +a}\, b^{3} d e}{16 c^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b \,e^{2} x}{10 c}+\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} d e x -\frac {4 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,e^{2}}{15 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} e^{2}}{12 c^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b d e}{6 c}+\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} d^{2}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(e*x+d)^2*(c*x^2+b*x+a)^(1/2),x)

[Out]

1/4*b^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a*d*e-1/2*a^2/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^
2+b*x+a)^(1/2))*d*e-1/4*a/c*(c*x^2+b*x+a)^(1/2)*b*d*e+1/8*b^2/c*x*(c*x^2+b*x+a)^(1/2)*d*e-1/32*b^4/c^(5/2)*ln(
(c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e+1/4/c*e^2*b*a*x*(c*x^2+b*x+a)^(1/2)+1/4/c^(3/2)*e^2*b*a^2*ln((c*x
+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/8/c^(5/2)*e^2*b^3*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+2/5*e^2
*x^2*(c*x^2+b*x+a)^(3/2)+x*(c*x^2+b*x+a)^(3/2)*d*e+1/12/c^2*e^2*b^2*(c*x^2+b*x+a)^(3/2)-1/32/c^3*e^2*b^4*(c*x^
2+b*x+a)^(1/2)-4/15/c*e^2*a*(c*x^2+b*x+a)^(3/2)-1/6*b/c*(c*x^2+b*x+a)^(3/2)*d*e+1/16*b^3/c^2*(c*x^2+b*x+a)^(1/
2)*d*e-1/2*a*x*(c*x^2+b*x+a)^(1/2)*d*e+2/3*(c*x^2+b*x+a)^(3/2)*d^2-1/10/c*e^2*b*x*(c*x^2+b*x+a)^(3/2)-1/16/c^2
*e^2*b^3*x*(c*x^2+b*x+a)^(1/2)+1/64/c^(7/2)*e^2*b^5*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/8/c^2*e^2*b^
2*a*(c*x^2+b*x+a)^(1/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

________________________________________________________________________________________

mupad [B]  time = 3.63, size = 876, normalized size = 4.49 \[ \frac {7\,b\,e^2\,\left (\frac {5\,b\,\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 {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {a\,\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}\right )}{5}-\frac {4\,a\,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 )}{5}+\frac {2\,e^2\,x^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{5}+\frac {d^2\,\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}+b\,d^2\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}-\frac {5\,b\,d\,e\,\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 )}{2}-\frac {5\,b^2\,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}+d\,e\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}+\frac {d^2\,\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^{3/2}}-a\,d\,e\,\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 )-\frac {a\,b\,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 {b\,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 {b\,e^2\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {b\,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}}+\frac {b\,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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + 2*c*x)*(d + e*x)^2*(a + b*x + c*x^2)^(1/2),x)

[Out]

(7*b*e^2*((5*b*((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) - (x*(a + b*x + c*x^2)^(3/2))/(4*c) + (
a*((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)))/5 - (4*a*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)))/5 + (2*e^2*x^2*(a + b*x
+ c*x^2)^(3/2))/5 + (d^2*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(12*c) + b*d^2*(x/2 + b
/(4*c))*(a + b*x + c*x^2)^(1/2) - (5*b*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))/(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)))/2 - (5*b^2*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*x*(a + b*x + c*x^2)^(3/2) + (d^2*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^(3/2)) - a*d*e*((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))) - (a*b*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) + (b*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)) + (b*e^2*x*(a +
b*x + c*x^2)^(3/2))/(4*c) + (b*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)) + (b*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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b + 2 c x\right ) \left (d + e x\right )^{2} \sqrt {a + b x + c x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)**2*(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((b + 2*c*x)*(d + e*x)**2*sqrt(a + b*x + c*x**2), x)

________________________________________________________________________________________