\(\int \frac {(d+e x)^3}{(b x+c x^2)^{3/2}} \, dx\) [164]

Optimal result

Integrand size = 21, antiderivative size = 134 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 d^3}{b \sqrt {b x+c x^2}}-\frac {2 (2 c d-b e) \left (c^2 d^2-b c d e+b^2 e^2\right ) x}{b^2 c^2 \sqrt {b x+c x^2}}+\frac {e^3 \sqrt {b x+c x^2}}{c^2}+\frac {3 e^2 (2 c d-b e) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {x \left (\frac {\sqrt {c} (b+c x) \left (-4 c^3 d^3 x+3 b^3 e^3 x-2 b c^2 d^2 (d-3 e x)+b^2 c e^2 x (-6 d+e x)\right )}{b^2}+6 e^2 (2 c d-b e) \sqrt {x} (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{c^{5/2} (x (b+c x))^{3/2}} \] Input:

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

Output:

(x*((Sqrt[c]*(b + c*x)*(-4*c^3*d^3*x + 3*b^3*e^3*x - 2*b*c^2*d^2*(d - 3*e* 
x) + b^2*c*e^2*x*(-6*d + e*x)))/b^2 + 6*e^2*(2*c*d - b*e)*Sqrt[x]*(b + c*x 
)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(c^(5/2)*( 
x*(b + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1164, 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.

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

\(\Big \downarrow \) 1164

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 e \int \frac {(d+e x) (b d+(2 c d-b e) x)}{\sqrt {c x^2+b x}}dx}{b^2}-\frac {2 (d+e x)^2 (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1225

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-2*(d + e*x)^2*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (4*e*(( 
(8*c^2*d^2 - 6*b*c*d*e + 3*b^2*e^2 + 2*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x 
^2])/(4*c^2) + (3*b^2*e*(2*c*d - b*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2 
]])/(4*c^(5/2))))/b^2
 

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 1164

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x 
+ c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* 
c))   Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* 
c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p 
+ 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int 
QuadraticQ[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 = 0.75 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(-\frac {3 \left (b^{2} \sqrt {x \left (c x +b \right )}\, e^{2} \left (b e -2 c d \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+2 e^{2} \left (-\frac {e x}{6}+d \right ) x \,b^{2} c^{\frac {3}{2}}+\frac {2 d^{2} b \left (-3 e x +d \right ) c^{\frac {5}{2}}}{3}-\left (b^{3} e^{3} \sqrt {c}-\frac {4 c^{\frac {7}{2}} d^{3}}{3}\right ) x \right )}{\sqrt {x \left (c x +b \right )}\, c^{\frac {5}{2}} b^{2}}\)	\(115\)
risch	\(\frac {\left (c x +b \right ) \left (e^{3} x \,b^{2}-2 c^{2} d^{3}\right )}{b^{2} \sqrt {x \left (c x +b \right )}\, c^{2}}-\frac {\frac {2 \left (-2 b^{3} e^{3}+6 d \,e^{2} b^{2} c -6 d^{2} e b \,c^{2}+2 d^{3} c^{3}\right ) \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{c b \left (\frac {b}{c}+x \right )}+\frac {3 e^{3} b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}-6 b d \,e^{2} \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{2} b}\)	\(196\)
default	\(-\frac {2 d^{3} \left (2 c x +b \right )}{b^{2} \sqrt {c \,x^{2}+b x}}+e^{3} \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )+3 d \,e^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )+3 d^{2} e \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )\)	\(295\)

Input:

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

Output:

-3/(x*(c*x+b))^(1/2)/c^(5/2)*(b^2*(x*(c*x+b))^(1/2)*e^2*(b*e-2*c*d)*arctan 
h((x*(c*x+b))^(1/2)/x/c^(1/2))+2*e^2*(-1/6*e*x+d)*x*b^2*c^(3/2)+2/3*d^2*b* 
(-3*e*x+d)*c^(5/2)-(b^3*e^3*c^(1/2)-4/3*c^(7/2)*d^3)*x)/b^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.71 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, b^{2} c^{2} d e^{2} - b^{3} c e^{3}\right )} x^{2} + {\left (2 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (b^{2} c^{2} e^{3} x^{2} - 2 \, b c^{3} d^{3} - {\left (4 \, c^{4} d^{3} - 6 \, b c^{3} d^{2} e + 6 \, b^{2} c^{2} d e^{2} - 3 \, b^{3} c e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (b^{2} c^{4} x^{2} + b^{3} c^{3} x\right )}}, -\frac {3 \, {\left ({\left (2 \, b^{2} c^{2} d e^{2} - b^{3} c e^{3}\right )} x^{2} + {\left (2 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) - {\left (b^{2} c^{2} e^{3} x^{2} - 2 \, b c^{3} d^{3} - {\left (4 \, c^{4} d^{3} - 6 \, b c^{3} d^{2} e + 6 \, b^{2} c^{2} d e^{2} - 3 \, b^{3} c e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{b^{2} c^{4} x^{2} + b^{3} c^{3} x}\right ] \] Input:

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

Output:

[-1/2*(3*((2*b^2*c^2*d*e^2 - b^3*c*e^3)*x^2 + (2*b^3*c*d*e^2 - b^4*e^3)*x) 
*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(b^2*c^2*e^3*x^2 
 - 2*b*c^3*d^3 - (4*c^4*d^3 - 6*b*c^3*d^2*e + 6*b^2*c^2*d*e^2 - 3*b^3*c*e^ 
3)*x)*sqrt(c*x^2 + b*x))/(b^2*c^4*x^2 + b^3*c^3*x), -(3*((2*b^2*c^2*d*e^2 
- b^3*c*e^3)*x^2 + (2*b^3*c*d*e^2 - b^4*e^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 
 + b*x)*sqrt(-c)/(c*x + b)) - (b^2*c^2*e^3*x^2 - 2*b*c^3*d^3 - (4*c^4*d^3 
- 6*b*c^3*d^2*e + 6*b^2*c^2*d*e^2 - 3*b^3*c*e^3)*x)*sqrt(c*x^2 + b*x))/(b^ 
2*c^4*x^2 + b^3*c^3*x)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.41 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {e^{3} x^{2}}{\sqrt {c x^{2} + b x} c} - \frac {4 \, c d^{3} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {6 \, d^{2} e x}{\sqrt {c x^{2} + b x} b} - \frac {6 \, d e^{2} x}{\sqrt {c x^{2} + b x} c} + \frac {3 \, b e^{3} x}{\sqrt {c x^{2} + b x} c^{2}} + \frac {3 \, d e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} - \frac {3 \, b e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} - \frac {2 \, d^{3}}{\sqrt {c x^{2} + b x} b} \] Input:

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

Output:

e^3*x^2/(sqrt(c*x^2 + b*x)*c) - 4*c*d^3*x/(sqrt(c*x^2 + b*x)*b^2) + 6*d^2* 
e*x/(sqrt(c*x^2 + b*x)*b) - 6*d*e^2*x/(sqrt(c*x^2 + b*x)*c) + 3*b*e^3*x/(s 
qrt(c*x^2 + b*x)*c^2) + 3*d*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c 
))/c^(3/2) - 3/2*b*e^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2 
) - 2*d^3/(sqrt(c*x^2 + b*x)*b)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {\frac {2 \, d^{3}}{b} - {\left (\frac {e^{3} x}{c} - \frac {4 \, c^{3} d^{3} - 6 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} - 3 \, b^{3} e^{3}}{b^{2} c^{2}}\right )} x}{\sqrt {c x^{2} + b x}} - \frac {3 \, {\left (2 \, c d e^{2} - b e^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {5}{2}}} \] Input:

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

Output:

-(2*d^3/b - (e^3*x/c - (4*c^3*d^3 - 6*b*c^2*d^2*e + 6*b^2*c*d*e^2 - 3*b^3* 
e^3)/(b^2*c^2))*x)/sqrt(c*x^2 + b*x) - 3/2*(2*c*d*e^2 - b*e^3)*log(abs(2*( 
sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.79 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {-3 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{3} e^{3} x +6 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{2} c d \,e^{2} x +2 \sqrt {c}\, \sqrt {c x +b}\, b^{3} e^{3} x -6 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c d \,e^{2} x +6 \sqrt {c}\, \sqrt {c x +b}\, b \,c^{2} d^{2} e x -4 \sqrt {c}\, \sqrt {c x +b}\, c^{3} d^{3} x +3 \sqrt {x}\, b^{3} c \,e^{3} x -6 \sqrt {x}\, b^{2} c^{2} d \,e^{2} x +\sqrt {x}\, b^{2} c^{2} e^{3} x^{2}-2 \sqrt {x}\, b \,c^{3} d^{3}+6 \sqrt {x}\, b \,c^{3} d^{2} e x -4 \sqrt {x}\, c^{4} d^{3} x}{\sqrt {c x +b}\, b^{2} c^{3} x} \] Input:

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

Output:

( - 3*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b)) 
*b**3*e**3*x + 6*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c 
))/sqrt(b))*b**2*c*d*e**2*x + 2*sqrt(c)*sqrt(b + c*x)*b**3*e**3*x - 6*sqrt 
(c)*sqrt(b + c*x)*b**2*c*d*e**2*x + 6*sqrt(c)*sqrt(b + c*x)*b*c**2*d**2*e* 
x - 4*sqrt(c)*sqrt(b + c*x)*c**3*d**3*x + 3*sqrt(x)*b**3*c*e**3*x - 6*sqrt 
(x)*b**2*c**2*d*e**2*x + sqrt(x)*b**2*c**2*e**3*x**2 - 2*sqrt(x)*b*c**3*d* 
*3 + 6*sqrt(x)*b*c**3*d**2*e*x - 4*sqrt(x)*c**4*d**3*x)/(sqrt(b + c*x)*b** 
2*c**3*x)