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

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

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Rubi [A]  time = 0.07, 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 = {730, 724, 206} \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 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 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 730

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) \operatorname {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 = 0.09, size = 122, normalized size = 1.04 \begin {gather*} \frac {\sqrt {x} \left (\frac {\sqrt {d} e \sqrt {x} (b+c x)}{d+e x}-\frac {\sqrt {b+c x} (2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {c d-b e}}\right )}{d^{3/2} \sqrt {x (b+c x)} (b e-c d)} \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))/(d + e*x) - ((2*c*d - b*e)*Sqrt[b + c*x]*ArcTanh[(Sqrt[c*d - b*e]*Sqrt
[x])/(Sqrt[d]*Sqrt[b + c*x])])/Sqrt[c*d - b*e]))/(d^(3/2)*(-(c*d) + b*e)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.58, size = 115, normalized size = 0.98 \begin {gather*} \frac {(2 c d-b e) \tanh ^{-1}\left (\frac {-e \sqrt {b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {d} \sqrt {c d-b e}}\right )}{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]

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

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fricas [A]  time = 0.44, size = 348, normalized size = 2.97 \begin {gather*} \left [\frac {{\left (2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} 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 (c d^{2} e - b d e^{2}\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e + b^{2} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}\right )} x\right )}}, \frac {{\left (2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} 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}}{{\left (c d - b e\right )} x}\right ) - {\left (c d^{2} e - b d e^{2}\right )} \sqrt {c x^{2} + b x}}{c^{2} d^{5} - 2 \, b c d^{4} e + b^{2} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}\right )} x}\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*d*e + (2*c*d*e - b*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*(c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c^2*d^5 - 2*b*c*d^4*e + b^2*d
^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*x), ((2*c*d^2 - b*d*e + (2*c*d*e - b*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 - b*e)*x)) - (c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x)
)/(c^2*d^5 - 2*b*c*d^4*e + b^2*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*x)]

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giac [B]  time = 1.35, 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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maple [B]  time = 0.05, size = 355, normalized size = 3.03 \begin {gather*} -\frac {b \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{2 \left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, d}+\frac {c \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e}+\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{\left (b e -c d \right ) \left (x +\frac {d}{e}\right ) d} \end {gather*}

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)

[Out]

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

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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(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d positive, negative or zero?

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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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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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