3.4.20 \(\int \frac {1}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx\) [320]

Optimal. Leaf size=117 \[ -\frac {e \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}+\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{2 d^{3/2} (c d-b e)^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {744, 738, 212} \begin {gather*} \frac {(2 c d-b e) \tanh ^{-1}\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 )}{2 d^{3/2} (c d-b e)^{3/2}}-\frac {e \sqrt {b x+c x^2}}{d (d+e x) (c d-b e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*Sqrt[b*x + c*x^2])/(d*(c*d - b*e)*(d + e*x))) + ((2*c*d - b*e)*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]
*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(2*d^(3/2)*(c*d - b*e)^(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 {b x+c x^2}} \, dx &=-\frac {e \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}+\frac {(2 c d-b e) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{2 d (c d-b e)}\\ &=-\frac {e \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}-\frac {(2 c d-b e) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac {-b d-(2 c d-b e) x}{\sqrt {b x+c x^2}}\right )}{d (c d-b e)}\\ &=-\frac {e \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}+\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{2 d^{3/2} (c d-b e)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 137, normalized size = 1.17 \begin {gather*} \frac {\sqrt {x} \left (-\frac {\sqrt {d} e \sqrt {x} (b+c x)}{(c d-b e) (d+e x)}+\frac {(2 c d-b e) \sqrt {b+c x} \tan ^{-1}\left (\frac {-e \sqrt {x} \sqrt {b+c x}+\sqrt {c} (d+e x)}{\sqrt {d} \sqrt {-c d+b e}}\right )}{(-c d+b e)^{3/2}}\right )}{d^{3/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(228\) vs. \(2(101)=202\).
time = 0.44, size = 229, normalized size = 1.96

method	result	size
default	\(\frac {\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{d \left (b e -c d \right ) \left (x +\frac {d}{e}\right )}-\frac {e \left (b e -2 c d \right ) \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{2 d \left (b e -c d \right ) \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}}{e^{2}}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e^2*(1/d/(b*e-c*d)*e^2/(x+d/e)*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)-1/2*e*(b*e-2*c*d)
/d/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+1/e*(b*e-2*c*d)*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2
)*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)-d*(b*e-c*d)/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)^(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-%e*b>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 1.61, size = 355, normalized size = 3.03 \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} \log \left (\frac {2 \, c d x - b x e + b d + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{x e + d}\right ) - 2 \, {\left (c d^{2} e - b d e^{2}\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c^{2} d^{5} + b^{2} d^{2} x e^{3} - {\left (2 \, b c d^{3} x - b^{2} 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} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{c d x - b x e}\right ) - {\left (c d^{2} e - b d e^{2}\right )} \sqrt {c x^{2} + b x}}{c^{2} d^{5} + b^{2} d^{2} x e^{3} - {\left (2 \, b c d^{3} x - b^{2} d^{3}\right )} e^{2} + {\left (c^{2} d^{4} x - 2 \, b c d^{4}\right )} e}\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)^(1/2),x, algorithm="fricas")

[Out]

[1/2*((2*c*d^2 - b*x*e^2 + (2*c*d*x - b*d)*e)*sqrt(c*d^2 - b*d*e)*log((2*c*d*x - b*x*e + b*d + 2*sqrt(c*d^2 -
b*d*e)*sqrt(c*x^2 + b*x))/(x*e + d)) - 2*(c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c^2*d^5 + b^2*d^2*x*e^3 - (2*
b*c*d^3*x - b^2*d^3)*e^2 + (c^2*d^4*x - 2*b*c*d^4)*e), ((2*c*d^2 - b*x*e^2 + (2*c*d*x - b*d)*e)*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*x*e)) - (c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x)
)/(c^2*d^5 + b^2*d^2*x*e^3 - (2*b*c*d^3*x - b^2*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}{\sqrt {x \left (b + c x\right )} \left (d + e x\right )^{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)**(1/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (108) = 216\).
time = 1.50, size = 400, normalized size = 3.42 \begin {gather*} \frac {{\left (2 \, c d \log \left ({\left | 2 \, c d - b e - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c} \right |}\right ) - b e \log \left ({\left | 2 \, c d - b e - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c} \right |}\right ) + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c}\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{2 \, {\left (\sqrt {c d^{2} - b d e} c d^{2} - \sqrt {c d^{2} - b d e} b d e\right )}} - \frac {\sqrt {c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {b e}{x e + d} - \frac {b d e}{{\left (x e + d\right )}^{2}}}}{c d^{2} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - b d e \mathrm {sgn}\left (\frac {1}{x e + d}\right )} - \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left ({\left | 2 \, c d - b e - 2 \, \sqrt {c d^{2} - b d e} {\left (\sqrt {c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {b e}{x e + d} - \frac {b d e}{{\left (x e + d\right )}^{2}}} + \frac {\sqrt {c d^{2} e^{2} - b d e^{3}} e^{\left (-1\right )}}{x e + d}\right )} \right |}\right )}{2 \, {\left (c d^{2} e - b d e^{2}\right )} \sqrt {c d^{2} - b d e} \mathrm {sgn}\left (\frac {1}{x e + d}\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)^(1/2),x, algorithm="giac")

[Out]

1/2*(2*c*d*log(abs(2*c*d - b*e - 2*sqrt(c*d^2 - b*d*e)*sqrt(c))) - b*e*log(abs(2*c*d - b*e - 2*sqrt(c*d^2 - b*
d*e)*sqrt(c))) + 2*sqrt(c*d^2 - b*d*e)*sqrt(c))*sgn(1/(x*e + d))/(sqrt(c*d^2 - b*d*e)*c*d^2 - sqrt(c*d^2 - b*d
*e)*b*d*e) - sqrt(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + b*e/(x*e + d) - b*d*e/(x*e + d)^2)/(c*d^2*sgn(1/(x
*e + d)) - b*d*e*sgn(1/(x*e + d))) - 1/2*(2*c*d*e - b*e^2)*log(abs(2*c*d - b*e - 2*sqrt(c*d^2 - b*d*e)*(sqrt(c
- 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + b*e/(x*e + d) - b*d*e/(x*e + d)^2) + sqrt(c*d^2*e^2 - b*d*e^3)*e^(-1)
/(x*e + d))))/((c*d^2*e - b*d*e^2)*sqrt(c*d^2 - b*d*e)*sgn(1/(x*e + d)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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