\(\int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx\) [2338]

Optimal result

Integrand size = 22, antiderivative size = 160 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx=-\frac {\sqrt {a+b x+c x^2}}{e (d+e x)}+\frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e^2}-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 e^2 \sqrt {c d^2-b d e+a e^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {746, 857, 635, 212, 738} \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^2 \sqrt {a e^2-b d e+c d^2}}+\frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e^2}-\frac {\sqrt {a+b x+c x^2}}{e (d+e x)} \]

[In]

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

[Out]

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

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 746

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 + 1))), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*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] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

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 {\sqrt {a+b x+c x^2}}{e (d+e x)}+\frac {\int \frac {b+2 c x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 e} \\ & = -\frac {\sqrt {a+b x+c x^2}}{e (d+e x)}+\frac {c \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{e^2}-\frac {(2 c d-b e) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 e^2} \\ & = -\frac {\sqrt {a+b x+c x^2}}{e (d+e x)}+\frac {(2 c) \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 )}{e^2}+\frac {(2 c d-b e) \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^2} \\ & = -\frac {\sqrt {a+b x+c x^2}}{e (d+e x)}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e^2}-\frac {(2 c d-b e) \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 )}{2 e^2 \sqrt {c d^2-b d e+a e^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx=\frac {-\frac {e \sqrt {a+x (b+c x)}}{d+e x}+\frac {(2 c d-b e) \arctan \left (\frac {\sqrt {-c d^2+e (b d-a e)} x}{\sqrt {a} (d+e x)-d \sqrt {a+x (b+c x)}}\right )}{\sqrt {-c d^2+e (b d-a e)}}+2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{e^2} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(674\) vs. \(2(140)=280\).

Time = 0.38 (sec) , antiderivative size = 675, normalized size of antiderivative = 4.22

method	result	size
default	\(\frac {-\frac {e^{2} \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}}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}+\frac {\left (b e -2 c d \right ) e \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 )}{2 a \,e^{2}-2 b d e +2 c \,d^{2}}+\frac {2 c \,e^{2} \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 )}{a \,e^{2}-b d e +c \,d^{2}}}{e^{2}}\)	\(675\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (140) = 280\).

Time = 1.31 (sec) , antiderivative size = 1456, normalized size of antiderivative = 9.10 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/4*(2*(c*d^3 - b*d^2*e + a*d*e^2 + (c*d^2*e - b*d*e^2 + a*e^3)*x)*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) - (2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*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*(c*d^2*e - b*d*e^2 + a*e^3)*sqrt(c*x^2 + b*x + a))/
(c*d^3*e^2 - b*d^2*e^3 + a*d*e^4 + (c*d^2*e^3 - b*d*e^4 + a*e^5)*x), -1/2*((2*c*d^2 - b*d*e + (2*c*d*e - b*e^2
)*x)*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)) - (c*d^3 - b*d^2*e + a*d*e^2 + (c*d^2*e - b*d*e^2 + a*e^3)*x)*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) + 2*(c*d^2*e - b*d*e^2 + a*e^3)*sqrt(c*x^2 + b*x + a))
/(c*d^3*e^2 - b*d^2*e^3 + a*d*e^4 + (c*d^2*e^3 - b*d*e^4 + a*e^5)*x), -1/4*(4*(c*d^3 - b*d^2*e + a*d*e^2 + (c*
d^2*e - b*d*e^2 + a*e^3)*x)*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*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*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*(c*d^2*e - b*d*e^2 + a*e^3)*sqrt(c*x^2 + b*x + a))/(c*d^3*e^2 - b*d^2*e^3 + a*d*e^4 + (c*d^2*e^3
- b*d*e^4 + a*e^5)*x), -1/2*((2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*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)) + 2*(c*d^3 - b*d^2*e + a*d*e^2 + (c*
d^2*e - b*d*e^2 + a*e^3)*x)*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*(c*d^2*e - b*d*e^2 + a*e^3)*sqrt(c*x^2 + b*x + a))/(c*d^3*e^2 - b*d^2*e^3 + a*d*e^4 + (c*d^2*e^3 - b
*d*e^4 + a*e^5)*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((c*x^2+b*x+a)^(1/2)/(e*x+d)^2,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. \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c*x^2+b*x+a)^(1/2)/(e*x+d)^2,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. \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (d+e\,x\right )}^2} \,d x \]

[In]

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

[Out]

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