3.1233 \(\int \frac{A+B x}{\sqrt{d+e x} (b x+c x^2)} \, dx\)

Optimal. Leaf size=86 \[ -\frac{2 (b B-A c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b \sqrt{c} \sqrt{c d-b e}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b \sqrt{d}} \]

[Out]

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

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Rubi [A]  time = 0.180074, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {826, 1166, 208} \[ -\frac{2 (b B-A c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b \sqrt{c} \sqrt{c d-b e}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b \sqrt{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{-B d+A e+B x^2}{c d^2-b d e+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )\\ &=\frac{(2 A c) \operatorname{Subst}\left (\int \frac{1}{-\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b}+\left (2 \left (\frac{B}{2}-\frac{2 c (-B d+A e)-B (-2 c d+b e)}{2 b e}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )\\ &=-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b \sqrt{d}}-\frac{2 (b B-A c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b \sqrt{c} \sqrt{c d-b e}}\\ \end{align*}

Mathematica [A]  time = 0.156371, size = 84, normalized size = 0.98 \[ \frac{2 \left (\frac{(A c-b B) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{\sqrt{c} \sqrt{c d-b e}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{\sqrt{d}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 101, normalized size = 1.2 \begin{align*} -2\,{\frac{A}{b\sqrt{d}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-2\,{\frac{Ac}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+2\,{\frac{B}{\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07095, size = 1085, normalized size = 12.62 \begin{align*} \left [-\frac{\sqrt{c^{2} d - b c e}{\left (B b - A c\right )} d \log \left (\frac{c e x + 2 \, c d - b e + 2 \, \sqrt{c^{2} d - b c e} \sqrt{e x + d}}{c x + b}\right ) -{\left (A c^{2} d - A b c e\right )} \sqrt{d} \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right )}{b c^{2} d^{2} - b^{2} c d e}, \frac{2 \, \sqrt{-c^{2} d + b c e}{\left (B b - A c\right )} d \arctan \left (\frac{\sqrt{-c^{2} d + b c e} \sqrt{e x + d}}{c e x + c d}\right ) +{\left (A c^{2} d - A b c e\right )} \sqrt{d} \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right )}{b c^{2} d^{2} - b^{2} c d e}, -\frac{\sqrt{c^{2} d - b c e}{\left (B b - A c\right )} d \log \left (\frac{c e x + 2 \, c d - b e + 2 \, \sqrt{c^{2} d - b c e} \sqrt{e x + d}}{c x + b}\right ) - 2 \,{\left (A c^{2} d - A b c e\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right )}{b c^{2} d^{2} - b^{2} c d e}, \frac{2 \,{\left (\sqrt{-c^{2} d + b c e}{\left (B b - A c\right )} d \arctan \left (\frac{\sqrt{-c^{2} d + b c e} \sqrt{e x + d}}{c e x + c d}\right ) +{\left (A c^{2} d - A b c e\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right )\right )}}{b c^{2} d^{2} - b^{2} c d e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(sqrt(c^2*d - b*c*e)*(B*b - A*c)*d*log((c*e*x + 2*c*d - b*e + 2*sqrt(c^2*d - b*c*e)*sqrt(e*x + d))/(c*x + b)
) - (A*c^2*d - A*b*c*e)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x))/(b*c^2*d^2 - b^2*c*d*e), (2*sqrt
(-c^2*d + b*c*e)*(B*b - A*c)*d*arctan(sqrt(-c^2*d + b*c*e)*sqrt(e*x + d)/(c*e*x + c*d)) + (A*c^2*d - A*b*c*e)*
sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x))/(b*c^2*d^2 - b^2*c*d*e), -(sqrt(c^2*d - b*c*e)*(B*b - A*
c)*d*log((c*e*x + 2*c*d - b*e + 2*sqrt(c^2*d - b*c*e)*sqrt(e*x + d))/(c*x + b)) - 2*(A*c^2*d - A*b*c*e)*sqrt(-
d)*arctan(sqrt(e*x + d)*sqrt(-d)/d))/(b*c^2*d^2 - b^2*c*d*e), 2*(sqrt(-c^2*d + b*c*e)*(B*b - A*c)*d*arctan(sqr
t(-c^2*d + b*c*e)*sqrt(e*x + d)/(c*e*x + c*d)) + (A*c^2*d - A*b*c*e)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d)
)/(b*c^2*d^2 - b^2*c*d*e)]

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Sympy [A]  time = 33.2428, size = 87, normalized size = 1.01 \begin{align*} \frac{2 A \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{d}} \sqrt{d + e x}} \right )}}{b d \sqrt{- \frac{1}{d}}} - \frac{2 \left (- A c + B b\right ) \operatorname{atan}{\left (\frac{1}{\sqrt{\frac{c}{b e - c d}} \sqrt{d + e x}} \right )}}{b \sqrt{\frac{c}{b e - c d}} \left (b e - c d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*atan(1/(sqrt(-1/d)*sqrt(d + e*x)))/(b*d*sqrt(-1/d)) - 2*(-A*c + B*b)*atan(1/(sqrt(c/(b*e - c*d))*sqrt(d +
e*x)))/(b*sqrt(c/(b*e - c*d))*(b*e - c*d))

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Giac [A]  time = 1.30507, size = 107, normalized size = 1.24 \begin{align*} \frac{2 \,{\left (B b - A c\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b} + \frac{2 \, A \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(B*b - A*c)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b) + 2*A*arctan(sqrt(x*e + d)
/sqrt(-d))/(b*sqrt(-d))