3.21.7 \(\int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx\)
Optimal. Leaf size=160 \[ -\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^2 \sqrt {a e^2-b d e+c d^2}}-\frac {\sqrt {a+b x+c x^2}}{e (d+e x)}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e^2} \]
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Rubi [A] time = 0.13, antiderivative size = 160, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used =
{732, 843, 621, 206, 724} \begin {gather*} -\frac {(2 c d-b e) \tanh ^{-1}\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 {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} \end {gather*}
Antiderivative was successfully verified.
[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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 621
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 724
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 732
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 843
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*} \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 {\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) \operatorname {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) \operatorname {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*}
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Mathematica [A] time = 0.17, size = 152, normalized size = 0.95 \begin {gather*} \frac {\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{2 \sqrt {e (a e-b d)+c d^2}}-\frac {e \sqrt {a+x (b+c x)}}{d+e x}+\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{e^2} \end {gather*}
Antiderivative was successfully verified.
[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)) + Sqrt[c]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + ((2
*c*d - b*e)*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)]
)])/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]))/e^2
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IntegrateAlgebraic [A] time = 0.85, size = 174, normalized size = 1.09 \begin {gather*} -\frac {(2 c d-b e) \sqrt {-a e^2+b d e-c d^2} \tan ^{-1}\left (\frac {-e \sqrt {a+b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2+b d e-c d^2}}\right )}{e^2 \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {a+b x+c x^2}}{e (d+e x)}-\frac {\sqrt {c} \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{e^2} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[Sqrt[a + b*x + c*x^2]/(d + e*x)^2,x]
[Out]
-(Sqrt[a + b*x + c*x^2]/(e*(d + e*x))) - ((2*c*d - b*e)*Sqrt[-(c*d^2) + b*d*e - a*e^2]*ArcTan[(Sqrt[c]*d + Sqr
t[c]*e*x - e*Sqrt[a + b*x + c*x^2])/Sqrt[-(c*d^2) + b*d*e - a*e^2]])/(e^2*(c*d^2 - b*d*e + a*e^2)) - (Sqrt[c]*
Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/e^2
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fricas [B] time = 2.18, size = 1456, normalized size = 9.10
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[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)]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Evaluation time: 0.58Error: Bad Argument Type
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maple [B] time = 0.08, size = 1519, normalized size = 9.49
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c*x^2+b*x+a)^(1/2)/(e*x+d)^2,x)
[Out]
-1/(a*e^2-b*d*e+c*d^2)/(x+d/e)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/(a*e^2-b*d*
e+c*d^2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b-1/e/(a*e^2-b*d*e+c*d^2)*((x+d/e)^
2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*c*d-1/e/(a*e^2-b*d*e+c*d^2)*ln(((x+d/e)*c+1/2*(b*e-2*
c*d)/e)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)*d*b+1/e^2/(a*e^2-b*
d*e+c*d^2)*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2
)^(1/2))*c^(3/2)*d^2-1/2/(a*e^2-b*d*e+c*d^2)/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^
2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^
2)^(1/2))/(x+d/e))*a*b+1/e/(a*e^2-b*d*e+c*d^2)/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*
e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/
e^2)^(1/2))/(x+d/e))*a*c*d+1/2/e/(a*e^2-b*d*e+c*d^2)/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e
+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c
*d^2)/e^2)^(1/2))/(x+d/e))*b^2*d-3/2/e^2/(a*e^2-b*d*e+c*d^2)/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(
x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2
-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b*d^2*c+1/e^3/(a*e^2-b*d*e+c*d^2)/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-
2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/
e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*c^2*d^3+c/(a*e^2-b*d*e+c*d^2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a
*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x+c^(1/2)/(a*e^2-b*d*e+c*d^2)*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2
*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*a
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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 details)Is a*e^2-b*d*e +c*d^2 zero or nonzero?
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x + c x^{2}}}{\left (d + e x\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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