\(\int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx\) [157]

Optimal result

Integrand size = 21, antiderivative size = 166 \[ \int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx=\frac {e \left (24 c^2 d^2-18 b c d e+5 b^2 e^2\right ) \sqrt {b x+c x^2}}{8 c^3}+\frac {e^2 (18 c d-5 b e) x \sqrt {b x+c x^2}}{12 c^2}+\frac {e^3 x^2 \sqrt {b x+c x^2}}{3 c}+\frac {(2 c d-b e) \left (8 c^2 d^2-8 b c d e+5 b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}} \] Output:

1/8*e*(5*b^2*e^2-18*b*c*d*e+24*c^2*d^2)*(c*x^2+b*x)^(1/2)/c^3+1/12*e^2*(-5 
*b*e+18*c*d)*x*(c*x^2+b*x)^(1/2)/c^2+1/3*e^3*x^2*(c*x^2+b*x)^(1/2)/c+1/8*( 
-b*e+2*c*d)*(5*b^2*e^2-8*b*c*d*e+8*c^2*d^2)*arctanh(c^(1/2)*x/(c*x^2+b*x)^ 
(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} e x (b+c x) \left (15 b^2 e^2-2 b c e (27 d+5 e x)+4 c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+3 \left (-16 c^3 d^3+24 b c^2 d^2 e-18 b^2 c d e^2+5 b^3 e^3\right ) \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{24 c^{7/2} \sqrt {x (b+c x)}} \] Input:

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

Output:

(Sqrt[c]*e*x*(b + c*x)*(15*b^2*e^2 - 2*b*c*e*(27*d + 5*e*x) + 4*c^2*(18*d^ 
2 + 9*d*e*x + 2*e^2*x^2)) + 3*(-16*c^3*d^3 + 24*b*c^2*d^2*e - 18*b^2*c*d*e 
^2 + 5*b^3*e^3)*Sqrt[x]*Sqrt[b + c*x]*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c* 
x]])/(24*c^(7/2)*Sqrt[x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1166, 27, 1225, 1091, 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.

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

\(\Big \downarrow \) 1166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\frac {3 (2 c d-b e) \left (5 b^2 e^2-8 b c d e+8 c^2 d^2\right ) \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c^2}+\frac {e \sqrt {b x+c x^2} \left (15 b^2 e^2+10 c e x (2 c d-b e)-54 b c d e+64 c^2 d^2\right )}{4 c^2}}{6 c}+\frac {e \sqrt {b x+c x^2} (d+e x)^2}{3 c}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {\frac {3 (2 c d-b e) \left (5 b^2 e^2-8 b c d e+8 c^2 d^2\right ) \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c^2}+\frac {e \sqrt {b x+c x^2} \left (15 b^2 e^2+10 c e x (2 c d-b e)-54 b c d e+64 c^2 d^2\right )}{4 c^2}}{6 c}+\frac {e \sqrt {b x+c x^2} (d+e x)^2}{3 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) (2 c d-b e) \left (5 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{4 c^{5/2}}+\frac {e \sqrt {b x+c x^2} \left (15 b^2 e^2+10 c e x (2 c d-b e)-54 b c d e+64 c^2 d^2\right )}{4 c^2}}{6 c}+\frac {e \sqrt {b x+c x^2} (d+e x)^2}{3 c}\)

Input:

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

Output:

(e*(d + e*x)^2*Sqrt[b*x + c*x^2])/(3*c) + ((e*(64*c^2*d^2 - 54*b*c*d*e + 1 
5*b^2*e^2 + 10*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(4*c^2) + (3*(2*c*d 
 - b*e)*(8*c^2*d^2 - 8*b*c*d*e + 5*b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + 
 c*x^2]])/(4*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 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{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 = 0.71 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(-\frac {5 \left (\left (b e -2 c d \right ) \left (b^{2} e^{2}-\frac {8}{5} b c d e +\frac {8}{5} c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )-e \sqrt {x \left (c x +b \right )}\, \left (\left (\frac {8}{15} e^{2} x^{2}+\frac {12}{5} d e x +\frac {24}{5} d^{2}\right ) c^{\frac {5}{2}}+e \left (\left (-\frac {2 e x}{3}-\frac {18 d}{5}\right ) c^{\frac {3}{2}}+b e \sqrt {c}\right ) b \right )\right )}{8 c^{\frac {7}{2}}}\)	\(113\)
risch	\(\frac {\left (8 c^{2} e^{2} x^{2}-10 e^{2} x b c +36 c^{2} d e x +15 b^{2} e^{2}-54 b c d e +72 c^{2} d^{2}\right ) e x \left (c x +b \right )}{24 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {\left (5 b^{3} e^{3}-18 d \,e^{2} b^{2} c +24 d^{2} e b \,c^{2}-16 d^{3} c^{3}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {7}{2}}}\)	\(139\)
default	\(\frac {d^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}+e^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )+3 d \,e^{2} \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+3 d^{2} e \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )\)	\(261\)

Input:

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

Output:

-5/8/c^(7/2)*((b*e-2*c*d)*(b^2*e^2-8/5*b*c*d*e+8/5*c^2*d^2)*arctanh((x*(c* 
x+b))^(1/2)/x/c^(1/2))-e*(x*(c*x+b))^(1/2)*((8/15*e^2*x^2+12/5*d*e*x+24/5* 
d^2)*c^(5/2)+e*((-2/3*e*x-18/5*d)*c^(3/2)+b*e*c^(1/2))*b))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.79 \[ \int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx=\left [-\frac {3 \, {\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - 5 \, b^{3} e^{3}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, c^{3} e^{3} x^{2} + 72 \, c^{3} d^{2} e - 54 \, b c^{2} d e^{2} + 15 \, b^{2} c e^{3} + 2 \, {\left (18 \, c^{3} d e^{2} - 5 \, b c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{4}}, -\frac {3 \, {\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - 5 \, b^{3} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) - {\left (8 \, c^{3} e^{3} x^{2} + 72 \, c^{3} d^{2} e - 54 \, b c^{2} d e^{2} + 15 \, b^{2} c e^{3} + 2 \, {\left (18 \, c^{3} d e^{2} - 5 \, b c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{4}}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.55 \[ \int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx=\begin {cases} \sqrt {b x + c x^{2}} \left (\frac {e^{3} x^{2}}{3 c} + \frac {x \left (- \frac {5 b e^{3}}{6 c} + 3 d e^{2}\right )}{2 c} + \frac {- \frac {3 b \left (- \frac {5 b e^{3}}{6 c} + 3 d e^{2}\right )}{4 c} + 3 d^{2} e}{c}\right ) + \left (- \frac {b \left (- \frac {3 b \left (- \frac {5 b e^{3}}{6 c} + 3 d e^{2}\right )}{4 c} + 3 d^{2} e\right )}{2 c} + d^{3}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (d^{3} \sqrt {b x} + \frac {d^{2} e \left (b x\right )^{\frac {3}{2}}}{b} + \frac {3 d e^{2} \left (b x\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {e^{3} \left (b x\right )^{\frac {7}{2}}}{7 b^{3}}\right )}{b} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (\begin {cases} d^{3} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{4}}{4 e} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(b*x + c*x**2)*(e**3*x**2/(3*c) + x*(-5*b*e**3/(6*c) + 3*d* 
e**2)/(2*c) + (-3*b*(-5*b*e**3/(6*c) + 3*d*e**2)/(4*c) + 3*d**2*e)/c) + (- 
b*(-3*b*(-5*b*e**3/(6*c) + 3*d*e**2)/(4*c) + 3*d**2*e)/(2*c) + d**3)*Piece 
wise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0) 
), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)), Ne(c, 
 0)), (2*(d**3*sqrt(b*x) + d**2*e*(b*x)**(3/2)/b + 3*d*e**2*(b*x)**(5/2)/( 
5*b**2) + e**3*(b*x)**(7/2)/(7*b**3))/b, Ne(b, 0)), (zoo*Piecewise((d**3*x 
, Eq(e, 0)), ((d + e*x)**4/(4*e), True)), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.57 \[ \int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} e^{3} x^{2}}{3 \, c} + \frac {3 \, \sqrt {c x^{2} + b x} d e^{2} x}{2 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} b e^{3} x}{12 \, c^{2}} + \frac {d^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{\sqrt {c}} - \frac {3 \, b d^{2} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} + \frac {9 \, b^{2} d e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} - \frac {5 \, b^{3} e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} + \frac {3 \, \sqrt {c x^{2} + b x} d^{2} e}{c} - \frac {9 \, \sqrt {c x^{2} + b x} b d e^{2}}{4 \, c^{2}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{2} e^{3}}{8 \, c^{3}} \] Input:

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

Output:

1/3*sqrt(c*x^2 + b*x)*e^3*x^2/c + 3/2*sqrt(c*x^2 + b*x)*d*e^2*x/c - 5/12*s 
qrt(c*x^2 + b*x)*b*e^3*x/c^2 + d^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqr 
t(c))/sqrt(c) - 3/2*b*d^2*e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c 
^(3/2) + 9/8*b^2*d*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2 
) - 5/16*b^3*e^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 3* 
sqrt(c*x^2 + b*x)*d^2*e/c - 9/4*sqrt(c*x^2 + b*x)*b*d*e^2/c^2 + 5/8*sqrt(c 
*x^2 + b*x)*b^2*e^3/c^3
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (\frac {4 \, e^{3} x}{c} + \frac {18 \, c^{2} d e^{2} - 5 \, b c e^{3}}{c^{3}}\right )} x + \frac {3 \, {\left (24 \, c^{2} d^{2} e - 18 \, b c d e^{2} + 5 \, b^{2} e^{3}\right )}}{c^{3}}\right )} - \frac {{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - 5 \, b^{3} e^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/24*sqrt(c*x^2 + b*x)*(2*(4*e^3*x/c + (18*c^2*d*e^2 - 5*b*c*e^3)/c^3)*x + 
 3*(24*c^2*d^2*e - 18*b*c*d*e^2 + 5*b^2*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 - 5*b^3*e^3)*log(abs(2*(sqrt(c)*x - sqrt(c*x^ 
2 + b*x))*sqrt(c) + b))/c^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.38 \[ \int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx=\frac {15 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c \,e^{3}-54 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{2} d \,e^{2}-10 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{2} e^{3} x +72 \sqrt {x}\, \sqrt {c x +b}\, c^{3} d^{2} e +36 \sqrt {x}\, \sqrt {c x +b}\, c^{3} d \,e^{2} x +8 \sqrt {x}\, \sqrt {c x +b}\, c^{3} e^{3} x^{2}-15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{3} e^{3}+54 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{2} c d \,e^{2}-72 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b \,c^{2} d^{2} e +48 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) c^{3} d^{3}}{24 c^{4}} \] Input:

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

Output:

(15*sqrt(x)*sqrt(b + c*x)*b**2*c*e**3 - 54*sqrt(x)*sqrt(b + c*x)*b*c**2*d* 
e**2 - 10*sqrt(x)*sqrt(b + c*x)*b*c**2*e**3*x + 72*sqrt(x)*sqrt(b + c*x)*c 
**3*d**2*e + 36*sqrt(x)*sqrt(b + c*x)*c**3*d*e**2*x + 8*sqrt(x)*sqrt(b + c 
*x)*c**3*e**3*x**2 - 15*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt 
(b))*b**3*e**3 + 54*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b)) 
*b**2*c*d*e**2 - 72*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b)) 
*b*c**2*d**2*e + 48*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b)) 
*c**3*d**3)/(24*c**4)