3.1196 \(\int \frac{A+B x}{(d+e x)^2 \sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=128 \[ \frac{\sqrt{b x+c x^2} (B d-A e)}{d (d+e x) (c d-b e)}-\frac{(A b e-2 A c d+b B d) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 d^{3/2} (c d-b e)^{3/2}} \]

[Out]

((B*d - A*e)*Sqrt[b*x + c*x^2])/(d*(c*d - b*e)*(d + e*x)) - ((b*B*d - 2*A*c*d + A*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 = 0.0848131, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {806, 724, 206} \[ \frac{\sqrt{b x+c x^2} (B d-A e)}{d (d+e x) (c d-b e)}-\frac{(A b e-2 A c d+b B d) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 d^{3/2} (c d-b e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((B*d - A*e)*Sqrt[b*x + c*x^2])/(d*(c*d - b*e)*(d + e*x)) - ((b*B*d - 2*A*c*d + A*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 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(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, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[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{A+B x}{(d+e x)^2 \sqrt{b x+c x^2}} \, dx &=\frac{(B d-A e) \sqrt{b x+c x^2}}{d (c d-b e) (d+e x)}-\frac{(b B d-2 A c d+A b e) \int \frac{1}{(d+e x) \sqrt{b x+c x^2}} \, dx}{2 d (c d-b e)}\\ &=\frac{(B d-A e) \sqrt{b x+c x^2}}{d (c d-b e) (d+e x)}+\frac{(b B d-2 A c d+A 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{(B d-A e) \sqrt{b x+c x^2}}{d (c d-b e) (d+e x)}-\frac{(b B d-2 A c d+A 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 = 0.195715, size = 133, normalized size = 1.04 \[ \frac{\sqrt{x} \left (\frac{\sqrt{d} \sqrt{x} (b+c x) (B d-A e)}{(d+e x) (c d-b e)}+\frac{\sqrt{b+c x} (A b e-2 A c d+b B d) \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{(b e-c d)^{3/2}}\right )}{d^{3/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*((Sqrt[d]*(B*d - A*e)*Sqrt[x]*(b + c*x))/((c*d - b*e)*(d + e*x)) + ((b*B*d - 2*A*c*d + A*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 = 0.012, size = 849, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((B*x+A)/(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 = 1.58358, size = 842, normalized size = 6.58 \begin{align*} \left [-\frac{{\left (A b d e +{\left (B b - 2 \, A c\right )} d^{2} +{\left (A b e^{2} +{\left (B b - 2 \, A c\right )} d e\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 (B c d^{3} + A b d e^{2} -{\left (B b + A c\right )} d^{2} e\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 (A b d e +{\left (B b - 2 \, A c\right )} d^{2} +{\left (A b e^{2} +{\left (B b - 2 \, A c\right )} d e\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 (B c d^{3} + A b d e^{2} -{\left (B b + A c\right )} d^{2} e\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((B*x+A)/(e*x+d)^2/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[-1/2*((A*b*d*e + (B*b - 2*A*c)*d^2 + (A*b*e^2 + (B*b - 2*A*c)*d*e)*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*(B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)*sq
rt(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), -((A*b*d
*e + (B*b - 2*A*c)*d^2 + (A*b*e^2 + (B*b - 2*A*c)*d*e)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sq
rt(c*x^2 + b*x)/((c*d - b*e)*x)) - (B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)*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{A + B x}{\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((B*x+A)/(e*x+d)**2/(c*x**2+b*x)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out