\(\int \frac {1}{(d+e x) (b x+c x^2)^{3/2}} \, dx\) [168]

Optimal result

Integrand size = 21, antiderivative size = 119 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{b d \sqrt {b x+c x^2}}-\frac {2 c (2 c d-b e) x}{b^2 d (c d-b e) \sqrt {b x+c x^2}}+\frac {2 e^2 \text {arctanh}\left (\frac {\sqrt {c d-b e} x}{\sqrt {d} \sqrt {b x+c x^2}}\right )}{d^{3/2} (c d-b e)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (\sqrt {d} \sqrt {-c d+b e} \left (-b^2 e+2 c^2 d x+b c (d-e x)\right )+b^2 e^2 \sqrt {x} \sqrt {b+c x} \arctan \left (\frac {-e \sqrt {x} \sqrt {b+c x}+\sqrt {c} (d+e x)}{\sqrt {d} \sqrt {-c d+b e}}\right )\right )}{b^2 d^{3/2} (-c d+b e)^{3/2} \sqrt {x (b+c x)}} \] Input:

Integrate[1/((d + e*x)*(b*x + c*x^2)^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1165, 27, 1154, 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.

\(\displaystyle \int \frac {1}{\left (b x+c x^2\right )^{3/2} (d+e x)} \, dx\)

\(\Big \downarrow \) 1165

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {e^2 \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{d (c d-b e)}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (c d-b e)}\)

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

Int[1/((d + e*x)*(b*x + c*x^2)^(3/2)),x]
 

Output:

(-2*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(b^2*d*(c*d - b*e)*Sqrt[b*x + c*x 
^2]) + (e^2*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqr 
t[b*x + c*x^2])])/(d^(3/2)*(c*d - b*e)^(3/2))
 

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 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-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]
 

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) 
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 
2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d 
+ e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p 
+ 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + 
 b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(-\frac {2 \sqrt {x \left (c x +b \right )}}{b^{2} d x}+\frac {2 c^{2} x}{b^{2} \left (b e -c d \right ) \sqrt {x \left (c x +b \right )}}+\frac {2 e^{2} \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )}{\sqrt {d \left (b e -c d \right )}\, d \left (b e -c d \right )}\)	\(107\)
risch	\(-\frac {2 \left (c x +b \right )}{b^{2} d \sqrt {x \left (c x +b \right )}}+\frac {e \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{d \left (b e -c d \right ) \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}+\frac {2 c \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{b^{2} \left (b e -c d \right ) \left (\frac {b}{c}+x \right )}\)	\(214\)
default	\(\frac {-\frac {e^{2}}{d \left (b e -c d \right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}+\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{d \left (b e -c d \right ) \left (-\frac {4 c d \left (b e -c d \right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}+\frac {e^{2} \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{d \left (b e -c d \right ) \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}}{e}\)	\(340\)

Input:

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

Output:

-2/b^2/d*(x*(c*x+b))^(1/2)/x+2/b^2*c^2/(b*e-c*d)/(x*(c*x+b))^(1/2)*x+2*e^2 
/(d*(b*e-c*d))^(1/2)*arctan((x*(c*x+b))^(1/2)/x*d/(d*(b*e-c*d))^(1/2))/d/( 
b*e-c*d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (105) = 210\).

Time = 0.09 (sec) , antiderivative size = 448, normalized size of antiderivative = 3.76 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) + 2 \, {\left (b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + b^{3} d e^{2} + {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e + b^{4} c d^{2} e^{2}\right )} x^{2} + {\left (b^{3} c^{2} d^{4} - 2 \, b^{4} c d^{3} e + b^{5} d^{2} e^{2}\right )} x}, -\frac {2 \, {\left ({\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt {-c d^{2} + b d e} \arctan \left (\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{c d x + b d}\right ) + {\left (b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + b^{3} d e^{2} + {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e + b^{4} c d^{2} e^{2}\right )} x^{2} + {\left (b^{3} c^{2} d^{4} - 2 \, b^{4} c d^{3} e + b^{5} d^{2} e^{2}\right )} x}\right ] \] Input:

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

Output:

[-((b^2*c*e^2*x^2 + b^3*e^2*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e 
)*x - 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) + 2*(b*c^2*d^3 - 
 2*b^2*c*d^2*e + b^3*d*e^2 + (2*c^3*d^3 - 3*b*c^2*d^2*e + b^2*c*d*e^2)*x)* 
sqrt(c*x^2 + b*x))/((b^2*c^3*d^4 - 2*b^3*c^2*d^3*e + b^4*c*d^2*e^2)*x^2 + 
(b^3*c^2*d^4 - 2*b^4*c*d^3*e + b^5*d^2*e^2)*x), -2*((b^2*c*e^2*x^2 + b^3*e 
^2*x)*sqrt(-c*d^2 + b*d*e)*arctan(sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/( 
c*d*x + b*d)) + (b*c^2*d^3 - 2*b^2*c*d^2*e + b^3*d*e^2 + (2*c^3*d^3 - 3*b* 
c^2*d^2*e + b^2*c*d*e^2)*x)*sqrt(c*x^2 + b*x))/((b^2*c^3*d^4 - 2*b^3*c^2*d 
^3*e + b^4*c*d^2*e^2)*x^2 + (b^3*c^2*d^4 - 2*b^4*c*d^3*e + b^5*d^2*e^2)*x) 
]
 

Sympy [F]

\[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(e*x+d)/(c*x^2+b*x)^(3/2),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(b*e-c*d>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.34 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, e^{2} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e}}\right )}{{\left (c d^{2} - b d e\right )} \sqrt {-c d^{2} + b d e}} - \frac {2 \, {\left (\frac {{\left (2 \, c^{2} d^{2} - b c d e\right )} x}{b^{2} c d^{3} - b^{3} d^{2} e} + \frac {b c d^{2} - b^{2} d e}{b^{2} c d^{3} - b^{3} d^{2} e}\right )}}{\sqrt {c x^{2} + b x}} \] Input:

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

Output:

-2*e^2*arctan(((sqrt(c)*x - sqrt(c*x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 
+ b*d*e))/((c*d^2 - b*d*e)*sqrt(-c*d^2 + b*d*e)) - 2*((2*c^2*d^2 - b*c*d*e 
)*x/(b^2*c*d^3 - b^3*d^2*e) + (b*c*d^2 - b^2*d*e)/(b^2*c*d^3 - b^3*d^2*e)) 
/sqrt(c*x^2 + b*x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.44 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {d}\, \sqrt {c x +b}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {b e -c d}-\sqrt {e}\, \sqrt {c x +b}-\sqrt {x}\, \sqrt {e}\, \sqrt {c}}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} e^{2} x +2 \sqrt {d}\, \sqrt {c x +b}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {b e -c d}+\sqrt {e}\, \sqrt {c x +b}+\sqrt {x}\, \sqrt {e}\, \sqrt {c}}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} e^{2} x -2 \sqrt {c}\, \sqrt {c x +b}\, b^{2} d \,e^{2} x +6 \sqrt {c}\, \sqrt {c x +b}\, b c \,d^{2} e x -4 \sqrt {c}\, \sqrt {c x +b}\, c^{2} d^{3} x -2 \sqrt {x}\, b^{3} d \,e^{2}+4 \sqrt {x}\, b^{2} c \,d^{2} e -2 \sqrt {x}\, b^{2} c d \,e^{2} x -2 \sqrt {x}\, b \,c^{2} d^{3}+6 \sqrt {x}\, b \,c^{2} d^{2} e x -4 \sqrt {x}\, c^{3} d^{3} x}{\sqrt {c x +b}\, b^{2} d^{2} x \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )} \] Input:

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

Output:

(2*(sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) - sqrt(e)* 
sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*b**2*e**2*x + 
sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) + sqrt(e)*sqrt 
(b + c*x) + sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*b**2*e**2*x - sqrt 
(c)*sqrt(b + c*x)*b**2*d*e**2*x + 3*sqrt(c)*sqrt(b + c*x)*b*c*d**2*e*x - 2 
*sqrt(c)*sqrt(b + c*x)*c**2*d**3*x - sqrt(x)*b**3*d*e**2 + 2*sqrt(x)*b**2* 
c*d**2*e - sqrt(x)*b**2*c*d*e**2*x - sqrt(x)*b*c**2*d**3 + 3*sqrt(x)*b*c** 
2*d**2*e*x - 2*sqrt(x)*c**3*d**3*x))/(sqrt(b + c*x)*b**2*d**2*x*(b**2*e**2 
 - 2*b*c*d*e + c**2*d**2))