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

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

[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))

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Rubi [A]  time = 0.0746605, antiderivative size = 117, normalized size of antiderivative = 1., 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} \[ \frac{(2 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 d^{3/2} (c d-b e)^{3/2}}-\frac{e \sqrt{b x+c x^2}}{d (d+e x) (c d-b e)} \]

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 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]

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 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])

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*}

Mathematica [A]  time = 0.143699, size = 122, normalized size = 1.04 \[ \frac{\sqrt{x} \left (\frac{\sqrt{d} e \sqrt{x} (b+c x)}{(d+e x) (b e-c d)}-\frac{\sqrt{b+c x} (2 c d-b e) \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{(b e-c d)^{3/2}}\right )}{d^{3/2} \sqrt{x (b+c x)}} \]

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

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Maple [B]  time = 0.257, size = 355, normalized size = 3. \begin{align*}{\frac{1}{d \left ( be-cd \right ) }\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \left ({\frac{d}{e}}+x \right ) ^{-1}}-{\frac{b}{2\,d \left ( be-cd \right ) }\ln \left ({ \left ( -2\,{\frac{d \left ( be-cd \right ) }{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}}}}+{\frac{c}{e \left ( be-cd \right ) }\ln \left ({ \left ( -2\,{\frac{d \left ( be-cd \right ) }{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 2.04603, size = 716, normalized size = 6.12 \begin{align*} \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{align*}

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

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]  time = 2.00405, size = 540, normalized size = 4.62 \begin{align*} \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{align*}

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)))