3.24.79 \(\int \frac {1}{(d+e x)^2 \sqrt {a+b x+c x^2}} \, dx\) [2379]

Optimal. Leaf size=134 \[ -\frac {e \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 \left (c d^2-b d e+a e^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {744, 738, 212} \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 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {e \sqrt {a+b x+c x^2}}{(d+e x) \left (a e^2-b d e+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*
((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, 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] && EqQ[m + 2*p + 3, 0]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^2 \sqrt {a+b x+c x^2}} \, dx &=-\frac {e \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac {(2 c d-b e) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {e \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\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 )}{c d^2-b d e+a e^2}\\ &=-\frac {e \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\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 \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.56, size = 125, normalized size = 0.93 \begin {gather*} -\frac {\frac {e \sqrt {a+x (b+c x)}}{d+e x}+\frac {(2 c d-b e) \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{\sqrt {-c d^2+e (b d-a e)}}}{c d^2+e (-b d+a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(270\) vs. \(2(122)=244\).
time = 0.80, size = 271, normalized size = 2.02

method	result	size
default	\(\frac {-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{\left (e^{2} a -b d e +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}+\frac {e \left (b e -2 c d \right ) \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{2 \left (e^{2} a -b d e +c \,d^{2}\right ) \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}}{e^{2}}\)	\(271\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/(e*x+d)^2/(c*x^2+b*x+a)^(1/2),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(c*d^2-%e*b*d+%e^2*a>0)', see `
assume?` for

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (127) = 254\).
time = 3.30, size = 653, normalized size = 4.87 \begin {gather*} \left [-\frac {{\left (2 \, c d^{2} - b x e^{2} + {\left (2 \, c d x - b d\right )} e\right )} \sqrt {c d^{2} - b d e + a e^{2}} \log \left (-\frac {8 \, c^{2} d^{2} x^{2} + 8 \, b c d^{2} x + {\left (b^{2} + 4 \, a c\right )} d^{2} - 4 \, \sqrt {c d^{2} - b d e + a e^{2}} {\left (2 \, c d x + b d - {\left (b x + 2 \, a\right )} e\right )} \sqrt {c x^{2} + b x + a} + {\left (8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 8 \, a^{2}\right )} e^{2} - 2 \, {\left (4 \, b c d x^{2} + 4 \, a b d + {\left (3 \, b^{2} + 4 \, a c\right )} d x\right )} e}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 4 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left (c^{2} d^{5} + a^{2} x e^{5} - {\left (2 \, a b d x - a^{2} d\right )} e^{4} - {\left (2 \, a b d^{2} - {\left (b^{2} + 2 \, a c\right )} d^{2} x\right )} e^{3} - {\left (2 \, b c d^{3} x - {\left (b^{2} + 2 \, a c\right )} d^{3}\right )} e^{2} + {\left (c^{2} d^{4} x - 2 \, b c d^{4}\right )} e\right )}}, \frac {{\left (2 \, c d^{2} - b x e^{2} + {\left (2 \, c d x - b d\right )} e\right )} \sqrt {-c d^{2} + b d e - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e - a e^{2}} {\left (2 \, c d x + b d - {\left (b x + 2 \, a\right )} e\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (c^{2} d^{2} x^{2} + b c d^{2} x + a c d^{2} + {\left (a c x^{2} + a b x + a^{2}\right )} e^{2} - {\left (b c d x^{2} + b^{2} d x + a b d\right )} e\right )}}\right ) - 2 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (c^{2} d^{5} + a^{2} x e^{5} - {\left (2 \, a b d x - a^{2} d\right )} e^{4} - {\left (2 \, a b d^{2} - {\left (b^{2} + 2 \, a c\right )} d^{2} x\right )} e^{3} - {\left (2 \, b c d^{3} x - {\left (b^{2} + 2 \, a c\right )} d^{3}\right )} e^{2} + {\left (c^{2} d^{4} x - 2 \, b c d^{4}\right )} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((2*c*d^2 - b*x*e^2 + (2*c*d*x - b*d)*e)*sqrt(c*d^2 - b*d*e + a*e^2)*log(-(8*c^2*d^2*x^2 + 8*b*c*d^2*x +
(b^2 + 4*a*c)*d^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*(2*c*d*x + b*d - (b*x + 2*a)*e)*sqrt(c*x^2 + b*x + a) + (8*
a*b*x + (b^2 + 4*a*c)*x^2 + 8*a^2)*e^2 - 2*(4*b*c*d*x^2 + 4*a*b*d + (3*b^2 + 4*a*c)*d*x)*e)/(x^2*e^2 + 2*d*x*e
+ d^2)) + 4*(c*d^2*e - b*d*e^2 + a*e^3)*sqrt(c*x^2 + b*x + a))/(c^2*d^5 + a^2*x*e^5 - (2*a*b*d*x - a^2*d)*e^4
- (2*a*b*d^2 - (b^2 + 2*a*c)*d^2*x)*e^3 - (2*b*c*d^3*x - (b^2 + 2*a*c)*d^3)*e^2 + (c^2*d^4*x - 2*b*c*d^4)*e),
1/2*((2*c*d^2 - b*x*e^2 + (2*c*d*x - b*d)*e)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a
*e^2)*(2*c*d*x + b*d - (b*x + 2*a)*e)*sqrt(c*x^2 + b*x + a)/(c^2*d^2*x^2 + b*c*d^2*x + a*c*d^2 + (a*c*x^2 + a*
b*x + a^2)*e^2 - (b*c*d*x^2 + b^2*d*x + a*b*d)*e)) - 2*(c*d^2*e - b*d*e^2 + a*e^3)*sqrt(c*x^2 + b*x + a))/(c^2
*d^5 + a^2*x*e^5 - (2*a*b*d*x - a^2*d)*e^4 - (2*a*b*d^2 - (b^2 + 2*a*c)*d^2*x)*e^3 - (2*b*c*d^3*x - (b^2 + 2*a
*c)*d^3)*e^2 + (c^2*d^4*x - 2*b*c*d^4)*e)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x\right )^{2} \sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (d+e\,x\right )}^2\,\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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