\(\int \frac {(a+b x+c x^2)^{3/2}}{d+e x} \, dx\) [2347]

Optimal result

Integrand size = 22, antiderivative size = 252 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 c e^3}+\frac {\left (a+b x+c x^2\right )^{3/2}}{3 e}-\frac {(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{3/2} e^4}+\frac {\left (c d^2-b d e+a e^2\right )^{3/2} \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {748, 828, 857, 635, 212, 738} \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right )}{16 c^{3/2} e^4}+\frac {\left (a e^2-b d e+c d^2\right )^{3/2} \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^4}+\frac {\sqrt {a+b x+c x^2} \left (-2 c e (5 b d-4 a e)+b^2 e^2-2 c e x (2 c d-b e)+8 c^2 d^2\right )}{8 c e^3}+\frac {\left (a+b x+c x^2\right )^{3/2}}{3 e} \]

[In]

Int[(a + b*x + c*x^2)^(3/2)/(d + e*x),x]

[Out]

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 738

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

Rule 748

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 828

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

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x+c x^2\right )^{3/2}}{3 e}-\frac {\int \frac {(b d-2 a e+(2 c d-b e) x) \sqrt {a+b x+c x^2}}{d+e x} \, dx}{2 e} \\ & = \frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 c e^3}+\frac {\left (a+b x+c x^2\right )^{3/2}}{3 e}+\frac {\int \frac {\frac {1}{2} \left (4 c e (b d-2 a e)^2-d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )-\frac {1}{2} (2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{8 c e^3} \\ & = \frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 c e^3}+\frac {\left (a+b x+c x^2\right )^{3/2}}{3 e}+\frac {\left (c d^2-b d e+a e^2\right )^2 \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e^4}-\frac {\left ((2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c e^4} \\ & = \frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 c e^3}+\frac {\left (a+b x+c x^2\right )^{3/2}}{3 e}-\frac {\left (2 \left (c d^2-b d e+a e^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{e^4}-\frac {\left ((2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right )\right ) \text {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 e^4} \\ & = \frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 c e^3}+\frac {\left (a+b x+c x^2\right )^{3/2}}{3 e}-\frac {(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{3/2} e^4}+\frac {\left (c d^2-b d e+a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\frac {\frac {e \sqrt {a+x (b+c x)} \left (3 b^2 e^2+2 c e (-15 b d+16 a e+7 b e x)+4 c^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )}{c}+48 \sqrt {-c d^2+e (b d-a e)} \left (c d^2+e (-b d+a e)\right ) \arctan \left (\frac {\sqrt {-c d^2+e (b d-a e)} x}{\sqrt {a} (d+e x)-d \sqrt {a+x (b+c x)}}\right )-\frac {3 (2 c d-b e) \left (8 c^2 d^2-b^2 e^2+4 c e (-2 b d+3 a e)\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{c^{3/2}}}{24 e^4} \]

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(d + e*x),x]

[Out]

((e*Sqrt[a + x*(b + c*x)]*(3*b^2*e^2 + 2*c*e*(-15*b*d + 16*a*e + 7*b*e*x) + 4*c^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2
)))/c + 48*Sqrt[-(c*d^2) + e*(b*d - a*e)]*(c*d^2 + e*(-(b*d) + a*e))*ArcTan[(Sqrt[-(c*d^2) + e*(b*d - a*e)]*x)
/(Sqrt[a]*(d + e*x) - d*Sqrt[a + x*(b + c*x)])] - (3*(2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 + 4*c*e*(-2*b*d + 3*a*
e))*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/c^(3/2))/(24*e^4)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.52

method	result	size
risch	\(\frac {\left (8 c^{2} e^{2} x^{2}+14 b c \,e^{2} x -12 c^{2} d e x +32 a c \,e^{2}+3 b^{2} e^{2}-30 b c d e +24 c^{2} d^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{24 c \,e^{3}}+\frac {\frac {\left (12 a b c \,e^{3}-24 a \,c^{2} d \,e^{2}-b^{3} e^{3}-6 b^{2} c d \,e^{2}+24 b \,c^{2} d^{2} e -16 c^{3} d^{3}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{e \sqrt {c}}-\frac {16 \left (a^{2} e^{4}-2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-2 b c \,d^{3} e +c^{2} d^{4}\right ) c \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{16 e^{3} c}\)	\(383\)
default	\(\frac {\frac {\left (\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{3}+\frac {\left (b e -2 c d \right ) \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{2 e}+\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}+\frac {\left (b e -2 c d \right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\right )}{2 e \sqrt {c}}-\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{2}}}{e}\)	\(625\)

[In]

int((c*x^2+b*x+a)^(3/2)/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

1/24/c*(8*c^2*e^2*x^2+14*b*c*e^2*x-12*c^2*d*e*x+32*a*c*e^2+3*b^2*e^2-30*b*c*d*e+24*c^2*d^2)*(c*x^2+b*x+a)^(1/2
)/e^3+1/16/e^3/c*((12*a*b*c*e^3-24*a*c^2*d*e^2-b^3*e^3-6*b^2*c*d*e^2+24*b*c^2*d^2*e-16*c^3*d^3)/e*ln((1/2*b+c*
x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-16*(a^2*e^4-2*a*b*d*e^3+2*a*c*d^2*e^2+b^2*d^2*e^2-2*b*c*d^3*e+c^2*d^4)
*c/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d
^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))

Fricas [A] (verification not implemented)

none

Time = 132.30 (sec) , antiderivative size = 1523, normalized size of antiderivative = 6.04 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\text {Too large to display} \]

[In]

integrate((c*x^2+b*x+a)^(3/2)/(e*x+d),x, algorithm="fricas")

[Out]

[-1/96*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*(b^2*c + 4*a*c^2)*d*e^2 + (b^3 - 12*a*b*c)*e^3)*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) - 48*(c^3*d^2 - b*c^2*d*e + a*c^2*e
^2)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2
+ 4*a*c)*e^2)*x^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(
4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2)) - 4*(8*c^3*e^3*x^2 + 24*c^3*d^2*e -
30*b*c^2*d*e^2 + (3*b^2*c + 32*a*c^2)*e^3 - 2*(6*c^3*d*e^2 - 7*b*c^2*e^3)*x)*sqrt(c*x^2 + b*x + a))/(c^2*e^4)
, 1/48*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*(b^2*c + 4*a*c^2)*d*e^2 + (b^3 - 12*a*b*c)*e^3)*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)) + 24*(c^3*d^2 - b*c^2*d*e + a*c^2*e^2)*sq
rt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a
*c)*e^2)*x^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*
d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(8*c^3*e^3*x^2 + 24*c^3*d^2*e - 30*b*
c^2*d*e^2 + (3*b^2*c + 32*a*c^2)*e^3 - 2*(6*c^3*d*e^2 - 7*b*c^2*e^3)*x)*sqrt(c*x^2 + b*x + a))/(c^2*e^4), 1/96
*(96*(c^3*d^2 - b*c^2*d*e + a*c^2*e^2)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*s
qrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e
^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) - 3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*(b^2*c + 4*a*c^2)*d*e^2 + (b^
3 - 12*a*b*c)*e^3)*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*c^3*e^3*x^2 + 24*c^3*d^2*e - 30*b*c^2*d*e^2 + (3*b^2*c + 32*a*c^2)*e^3 - 2*(6*c^3*d*e^2 - 7*b*c^2*e^
3)*x)*sqrt(c*x^2 + b*x + a))/(c^2*e^4), 1/48*(48*(c^3*d^2 - b*c^2*d*e + a*c^2*e^2)*sqrt(-c*d^2 + b*d*e - a*e^2
)*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*
b*d*e + a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) + 3*(16*c^3*d^3 - 24*b
*c^2*d^2*e + 6*(b^2*c + 4*a*c^2)*d*e^2 + (b^3 - 12*a*b*c)*e^3)*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*c^3*e^3*x^2 + 24*c^3*d^2*e - 30*b*c^2*d*e^2 + (3*b^2*c + 32*a*
c^2)*e^3 - 2*(6*c^3*d*e^2 - 7*b*c^2*e^3)*x)*sqrt(c*x^2 + b*x + a))/(c^2*e^4)]

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{d + e x}\, dx \]

[In]

integrate((c*x**2+b*x+a)**(3/2)/(e*x+d),x)

[Out]

Integral((a + b*x + c*x**2)**(3/2)/(d + e*x), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^2+b*x+a)^(3/2)/(e*x+d),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(a*e^2-b*d*e>0)', see `assume?`
for more de

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c*x^2+b*x+a)^(3/2)/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{d+e\,x} \,d x \]

[In]

int((a + b*x + c*x^2)^(3/2)/(d + e*x),x)

[Out]

int((a + b*x + c*x^2)^(3/2)/(d + e*x), x)