\(\int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx\) [609]

Optimal result

Integrand size = 22, antiderivative size = 174 \[ \int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx=\frac {e (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}+\frac {e \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)+10 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{24 c^3}+\frac {(2 c d-b e) \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{7/2}} \] Output:

1/3*e*(e*x+d)^2*(c*x^2+b*x+a)^(1/2)/c+1/24*e*(64*c^2*d^2+15*b^2*e^2-2*c*e* 
(8*a*e+27*b*d)+10*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x+a)^(1/2)/c^3+1/16*(-b*e+2 
*c*d)*(8*c^2*d^2+5*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^(7/2)
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {c} e \sqrt {a+x (b+c x)} \left (15 b^2 e^2-2 c e (27 b d+8 a e+5 b e x)+4 c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )-3 (2 c d-b e) \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{48 c^{7/2}} \] Input:

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

Output:

(2*Sqrt[c]*e*Sqrt[a + x*(b + c*x)]*(15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e + 5 
*b*e*x) + 4*c^2*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) - 3*(2*c*d - b*e)*(8*c^2*d 
^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + 
 x*(b + c*x)]])/(48*c^(7/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1166, 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.

\(\displaystyle \int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx\)

\(\Big \downarrow \) 1166

\(\displaystyle \frac {\int \frac {(d+e x) \left (6 c d^2-e (b d+4 a e)+5 e (2 c d-b e) x\right )}{2 \sqrt {c x^2+b x+a}}dx}{3 c}+\frac {e (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d+e x) \left (6 c d^2-e (b d+4 a e)+5 e (2 c d-b e) x\right )}{\sqrt {c x^2+b x+a}}dx}{6 c}+\frac {e (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}\)

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\frac {3 (2 c d-b e) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c^2}+\frac {e \sqrt {a+b x+c x^2} \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+10 c e x (2 c d-b e)+64 c^2 d^2\right )}{4 c^2}}{6 c}+\frac {e (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\frac {3 (2 c d-b e) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c^2}+\frac {e \sqrt {a+b x+c x^2} \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+10 c e x (2 c d-b e)+64 c^2 d^2\right )}{4 c^2}}{6 c}+\frac {e (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right )}{8 c^{5/2}}+\frac {e \sqrt {a+b x+c x^2} \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+10 c e x (2 c d-b e)+64 c^2 d^2\right )}{4 c^2}}{6 c}+\frac {e (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c}\)

Input:

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

Output:

(e*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(3*c) + ((e*(64*c^2*d^2 + 15*b^2*e^2 
 - 2*c*e*(27*b*d + 8*a*e) + 10*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2]) 
/(4*c^2) + (3*(2*c*d - b*e)*(8*c^2*d^2 + 5*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])])/(8*c^(5/2)))/(6* 
c)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 1092

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]
 

rule 1166

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]
 

rule 1225

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]
 

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {e \left (-8 c^{2} e^{2} x^{2}+10 e^{2} x b c -36 c^{2} d e x +16 a c \,e^{2}-15 b^{2} e^{2}+54 b c d e -72 c^{2} d^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{24 c^{3}}+\frac {\left (12 a b c \,e^{3}-24 d \,e^{2} a \,c^{2}-5 b^{3} e^{3}+18 d \,e^{2} b^{2} c -24 d^{2} e b \,c^{2}+16 d^{3} c^{3}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}\)	\(162\)
default	\(\frac {d^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+e^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )+3 d \,e^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+3 d^{2} e \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\)	\(387\)

Input:

int((e*x+d)^3/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*e*(-8*c^2*e^2*x^2+10*b*c*e^2*x-36*c^2*d*e*x+16*a*c*e^2-15*b^2*e^2+54 
*b*c*d*e-72*c^2*d^2)/c^3*(c*x^2+b*x+a)^(1/2)+1/16*(12*a*b*c*e^3-24*a*c^2*d 
*e^2-5*b^3*e^3+18*b^2*c*d*e^2-24*b*c^2*d^2*e+16*c^3*d^3)/c^(7/2)*ln((1/2*b 
+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.25 \[ \int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx=\left [\frac {3 \, {\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, {\left (3 \, b^{2} c - 4 \, a c^{2}\right )} d e^{2} - {\left (5 \, b^{3} - 12 \, a b c\right )} e^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, c^{3} e^{3} x^{2} + 72 \, c^{3} d^{2} e - 54 \, b c^{2} d e^{2} + {\left (15 \, b^{2} c - 16 \, a c^{2}\right )} e^{3} + 2 \, {\left (18 \, c^{3} d e^{2} - 5 \, b c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, c^{4}}, -\frac {3 \, {\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, {\left (3 \, b^{2} c - 4 \, a c^{2}\right )} d e^{2} - {\left (5 \, b^{3} - 12 \, a b c\right )} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (8 \, c^{3} e^{3} x^{2} + 72 \, c^{3} d^{2} e - 54 \, b c^{2} d e^{2} + {\left (15 \, b^{2} c - 16 \, a c^{2}\right )} e^{3} + 2 \, {\left (18 \, c^{3} d e^{2} - 5 \, b c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{4}}\right ] \] Input:

integrate((e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/96*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*(3*b^2*c - 4*a*c^2)*d*e^2 - (5*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 + 72*c^3*d^2*e 
- 54*b*c^2*d*e^2 + (15*b^2*c - 16*a*c^2)*e^3 + 2*(18*c^3*d*e^2 - 5*b*c^2*e 
^3)*x)*sqrt(c*x^2 + b*x + a))/c^4, -1/48*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 
 6*(3*b^2*c - 4*a*c^2)*d*e^2 - (5*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 + 72*c^3*d^2*e - 54*b*c^2*d*e^2 + (15*b^2*c - 16*a*c^2)*e^3 
+ 2*(18*c^3*d*e^2 - 5*b*c^2*e^3)*x)*sqrt(c*x^2 + b*x + a))/c^4]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (167) = 334\).

Time = 0.96 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.26 \[ \int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \left (\frac {e^{3} x^{2}}{3 c} + \frac {x \left (- \frac {5 b e^{3}}{6 c} + 3 d e^{2}\right )}{2 c} + \frac {- \frac {2 a e^{3}}{3 c} - \frac {3 b \left (- \frac {5 b e^{3}}{6 c} + 3 d e^{2}\right )}{4 c} + 3 d^{2} e}{c}\right ) + \left (- \frac {a \left (- \frac {5 b e^{3}}{6 c} + 3 d e^{2}\right )}{2 c} - \frac {b \left (- \frac {2 a e^{3}}{3 c} - \frac {3 b \left (- \frac {5 b e^{3}}{6 c} + 3 d e^{2}\right )}{4 c} + 3 d^{2} e\right )}{2 c} + d^{3}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {e^{3} \left (a + b x\right )^{\frac {7}{2}}}{7 b^{3}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- 3 a e^{3} + 3 b d e^{2}\right )}{5 b^{3}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \cdot \left (3 a^{2} e^{3} - 6 a b d e^{2} + 3 b^{2} d^{2} e\right )}{3 b^{3}} + \frac {\sqrt {a + b x} \left (- a^{3} e^{3} + 3 a^{2} b d e^{2} - 3 a b^{2} d^{2} e + b^{3} d^{3}\right )}{b^{3}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {\begin {cases} d^{3} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{4}}{4 e} & \text {otherwise} \end {cases}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate((e*x+d)**3/(c*x**2+b*x+a)**(1/2),x)
 

Output:

Piecewise((sqrt(a + b*x + c*x**2)*(e**3*x**2/(3*c) + x*(-5*b*e**3/(6*c) + 
3*d*e**2)/(2*c) + (-2*a*e**3/(3*c) - 3*b*(-5*b*e**3/(6*c) + 3*d*e**2)/(4*c 
) + 3*d**2*e)/c) + (-a*(-5*b*e**3/(6*c) + 3*d*e**2)/(2*c) - b*(-2*a*e**3/( 
3*c) - 3*b*(-5*b*e**3/(6*c) + 3*d*e**2)/(4*c) + 3*d**2*e)/(2*c) + d**3)*Pi 
ecewise((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**3*(a + b*x)**(7/2)/(7*b**3) + (a + b*x)**(5/2) 
*(-3*a*e**3 + 3*b*d*e**2)/(5*b**3) + (a + b*x)**(3/2)*(3*a**2*e**3 - 6*a*b 
*d*e**2 + 3*b**2*d**2*e)/(3*b**3) + sqrt(a + b*x)*(-a**3*e**3 + 3*a**2*b*d 
*e**2 - 3*a*b**2*d**2*e + b**3*d**3)/b**3)/b, Ne(b, 0)), (Piecewise((d**3* 
x, Eq(e, 0)), ((d + e*x)**4/(4*e), True))/sqrt(a), True))
 

Maxima [F(-2)]

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

integrate((e*x+d)^3/(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
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (\frac {4 \, e^{3} x}{c} + \frac {18 \, c^{2} d e^{2} - 5 \, b c e^{3}}{c^{3}}\right )} x + \frac {72 \, c^{2} d^{2} e - 54 \, b c d e^{2} + 15 \, b^{2} e^{3} - 16 \, a c e^{3}}{c^{3}}\right )} - \frac {{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - 24 \, a c^{2} d e^{2} - 5 \, b^{3} e^{3} + 12 \, a b c e^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {7}{2}}} \] Input:

integrate((e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*e^3*x/c + (18*c^2*d*e^2 - 5*b*c*e^3)/c^3) 
*x + (72*c^2*d^2*e - 54*b*c*d*e^2 + 15*b^2*e^3 - 16*a*c*e^3)/c^3) - 1/16*( 
16*c^3*d^3 - 24*b*c^2*d^2*e + 18*b^2*c*d*e^2 - 24*a*c^2*d*e^2 - 5*b^3*e^3 
+ 12*a*b*c*e^3)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b) 
)/c^(7/2)
 

Mupad [F(-1)]

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

int((d + e*x)^3/(a + b*x + c*x^2)^(1/2),x)
 

Output:

int((d + e*x)^3/(a + b*x + c*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 2.72 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.49 \[ \int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx=\frac {-32 \sqrt {c \,x^{2}+b x +a}\, a \,c^{2} e^{3}+30 \sqrt {c \,x^{2}+b x +a}\, b^{2} c \,e^{3}-108 \sqrt {c \,x^{2}+b x +a}\, b \,c^{2} d \,e^{2}-20 \sqrt {c \,x^{2}+b x +a}\, b \,c^{2} e^{3} x +144 \sqrt {c \,x^{2}+b x +a}\, c^{3} d^{2} e +72 \sqrt {c \,x^{2}+b x +a}\, c^{3} d \,e^{2} x +16 \sqrt {c \,x^{2}+b x +a}\, c^{3} e^{3} x^{2}+36 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a b c \,e^{3}-72 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,c^{2} d \,e^{2}-15 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{3} e^{3}+54 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{2} c d \,e^{2}-72 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b \,c^{2} d^{2} e +48 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) c^{3} d^{3}}{48 c^{4}} \] Input:

int((e*x+d)^3/(c*x^2+b*x+a)^(1/2),x)
 

Output:

( - 32*sqrt(a + b*x + c*x**2)*a*c**2*e**3 + 30*sqrt(a + b*x + c*x**2)*b**2 
*c*e**3 - 108*sqrt(a + b*x + c*x**2)*b*c**2*d*e**2 - 20*sqrt(a + b*x + c*x 
**2)*b*c**2*e**3*x + 144*sqrt(a + b*x + c*x**2)*c**3*d**2*e + 72*sqrt(a + 
b*x + c*x**2)*c**3*d*e**2*x + 16*sqrt(a + b*x + c*x**2)*c**3*e**3*x**2 + 3 
6*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*e**3 - 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*c**2*d*e**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**3*e**3 + 54*sqrt(c) 
*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b* 
*2*c*d*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))*b*c**2*d**2*e + 48*sqrt(c)*log((2*sqrt(c)*sqrt(a + b 
*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*c**3*d**3)/(48*c**4)