3.23.25 \(\int \frac {(A+B x) (d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=234 \[ \frac {\sqrt {a+b x+c x^2} \left (B \left (-4 c e (4 a e+9 b d)+15 b^2 e^2+16 c^2 d^2\right )+2 c e x (6 A c e-5 b B e+4 B c d)+6 A c e (8 c d-3 b e)\right )}{24 c^3}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (4 b c \left (-3 a B e^2+4 A c d e+2 B c d^2\right )-8 c^2 \left (-a A e^2-2 a B d e+2 A c d^2\right )-6 b^2 c e (A e+2 B d)+5 b^3 B e^2\right )}{16 c^{7/2}}+\frac {B (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {832, 779, 621, 206} \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (B \left (-4 c e (4 a e+9 b d)+15 b^2 e^2+16 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 )}{24 c^3}-\frac {\tanh ^{-1}\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 )}{16 c^{7/2}}+\frac {B (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(3*c) + ((6*A*c*e*(8*c*d - 3*b*e) + B*(16*c^2*d^2 + 15*b^2*e^2 - 4*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])/(24*c^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))*Ar
cTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*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 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 \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 {\int \frac {(d+e x) \left (\frac {1}{2} (-b B d+6 A c d-4 a B e)+\frac {1}{2} (4 B c d-5 b B e+6 A c e) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{3 c}\\ &=\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 ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^3}\\ &=\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 ) \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 )}{8 c^3}\\ &=\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 ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 225, normalized size = 0.96 \begin {gather*} \frac {\frac {\sqrt {a+x (b+c x)} \left (B \left (-2 c e (8 a e+18 b d+5 b e x)+15 b^2 e^2+8 c^2 d (2 d+e x)\right )+6 A c e (-3 b e+8 c d+2 c e x)\right )}{8 c^2}-\frac {3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\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 )}{16 c^{5/2}}+B (d+e x)^2 \sqrt {a+x (b+c x)}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(d + e*x)^2*Sqrt[a + x*(b + c*x)] + (Sqrt[a + x*(b + c*x)]*(6*A*c*e*(8*c*d - 3*b*e + 2*c*e*x) + B*(15*b^2*e
^2 + 8*c^2*d*(2*d + e*x) - 2*c*e*(18*b*d + 8*a*e + 5*b*e*x))))/(8*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]*Sqrt[a + x*(b + c*x)])])/(16*c^(5/2)))/(3*c)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.87, size = 234, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-16 a B c e^2-18 A b c e^2+48 A c^2 d e+12 A c^2 e^2 x+15 b^2 B e^2-36 b B c d e-10 b B c e^2 x+24 B c^2 d^2+24 B c^2 d e x+8 B c^2 e^2 x^2\right )}{24 c^3}+\frac {\log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right ) \left (8 a A c^2 e^2-12 a b B c e^2+16 a B c^2 d e-6 A b^2 c e^2+16 A b c^2 d e-16 A c^3 d^2+5 b^3 B e^2-12 b^2 B c d e+8 b B c^2 d^2\right )}{16 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^2)/Sqrt[a + b*x + c*x^2],x]

[Out]

(Sqrt[a + b*x + c*x^2]*(24*B*c^2*d^2 - 36*b*B*c*d*e + 48*A*c^2*d*e + 15*b^2*B*e^2 - 18*A*b*c*e^2 - 16*a*B*c*e^
2 + 24*B*c^2*d*e*x - 10*b*B*c*e^2*x + 12*A*c^2*e^2*x + 8*B*c^2*e^2*x^2))/(24*c^3) + ((8*b*B*c^2*d^2 - 16*A*c^3
*d^2 - 12*b^2*B*c*d*e + 16*A*b*c^2*d*e + 16*a*B*c^2*d*e + 5*b^3*B*e^2 - 6*A*b^2*c*e^2 - 12*a*b*B*c*e^2 + 8*a*A
*c^2*e^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(16*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 487, normalized size = 2.08 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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)*ar
ctan(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 - 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]

________________________________________________________________________________________

giac [A]  time = 0.26, size = 231, normalized size = 0.99 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (\frac {4 \, B x e^{2}}{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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*B*x*e^2/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 - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 537, normalized size = 2.29 \begin {gather*} \frac {\sqrt {c \,x^{2}+b x +a}\, B \,e^{2} x^{2}}{3 c}-\frac {A a \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}+\frac {3 A \,b^{2} 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 d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {A \,d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+\frac {3 B a b \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {5}{2}}}-\frac {B a d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}-\frac {5 B \,b^{3} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}+\frac {3 B \,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 {5}{2}}}-\frac {B b \,d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{2}+b x +a}\, A \,e^{2} x}{2 c}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, B b \,e^{2} x}{12 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B d e x}{c}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A b \,e^{2}}{4 c^{2}}+\frac {2 \sqrt {c \,x^{2}+b x +a}\, A d e}{c}-\frac {2 \sqrt {c \,x^{2}+b x +a}\, B a \,e^{2}}{3 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, B \,b^{2} e^{2}}{8 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B b d e}{2 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B \,d^{2}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*e^2*x^2/c*(c*x^2+b*x+a)^(1/2)-5/12*B*e^2*b/c^2*x*(c*x^2+b*x+a)^(1/2)+5/8*B*e^2*b^2/c^3*(c*x^2+b*x+a)^(1/
2)-5/16*B*e^2*b^3/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/4*B*e^2*b/c^(5/2)*a*ln((c*x+1/2*b)/c^(
1/2)+(c*x^2+b*x+a)^(1/2))-2/3*B*e^2*a/c^2*(c*x^2+b*x+a)^(1/2)+1/2*x/c*(c*x^2+b*x+a)^(1/2)*A*e^2+x/c*(c*x^2+b*x
+a)^(1/2)*B*d*e-3/4*b/c^2*(c*x^2+b*x+a)^(1/2)*A*e^2-3/2*b/c^2*(c*x^2+b*x+a)^(1/2)*B*d*e+3/8*b^2/c^(5/2)*ln((c*
x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*A*e^2+3/4*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d*e-
1/2*a/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*A*e^2-a/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)
^(1/2))*B*d*e+2/c*(c*x^2+b*x+a)^(1/2)*A*d*e+1/c*(c*x^2+b*x+a)^(1/2)*B*d^2-b/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*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d^2+A*d^2*ln((c*x+1/2*b)/c
^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(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 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^2}{\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________