3.1464 \(\int \frac{A+B x}{\sqrt{\frac{A^2 e-B^2 e}{2 A B}+e x} (1+x^2)} \, dx\)

Optimal. Leaf size=133 \[ \frac{\sqrt{2} \sqrt{A} \sqrt{B} \tan ^{-1}\left (\frac{\sqrt{A} \sqrt{e \left (\frac{A}{B}-\frac{B}{A}+2 x\right )}}{\sqrt{B} \sqrt{e}}+\frac{A}{B}\right )}{\sqrt{e}}-\frac{\sqrt{2} \sqrt{A} \sqrt{B} \tan ^{-1}\left (\frac{A}{B}-\frac{\sqrt{A} \sqrt{e \left (\frac{A}{B}-\frac{B}{A}+2 x\right )}}{\sqrt{B} \sqrt{e}}\right )}{\sqrt{e}} \]

[Out]

-((Sqrt[2]*Sqrt[A]*Sqrt[B]*ArcTan[A/B - (Sqrt[A]*Sqrt[e*(A/B - B/A + 2*x)])/(Sqrt[B]*Sqrt[e])])/Sqrt[e]) + (Sq
rt[2]*Sqrt[A]*Sqrt[B]*ArcTan[A/B + (Sqrt[A]*Sqrt[e*(A/B - B/A + 2*x)])/(Sqrt[B]*Sqrt[e])])/Sqrt[e]

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Rubi [A]  time = 0.430311, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.093, Rules used = {827, 1161, 618, 204} \[ \frac{\sqrt{2} \sqrt{A} \sqrt{B} \tan ^{-1}\left (\frac{\sqrt{A} \sqrt{e \left (\frac{A}{B}-\frac{B}{A}+2 x\right )}}{\sqrt{B} \sqrt{e}}+\frac{A}{B}\right )}{\sqrt{e}}-\frac{\sqrt{2} \sqrt{A} \sqrt{B} \tan ^{-1}\left (\frac{A}{B}-\frac{\sqrt{A} \sqrt{e \left (\frac{A}{B}-\frac{B}{A}+2 x\right )}}{\sqrt{B} \sqrt{e}}\right )}{\sqrt{e}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[(A^2*e - B^2*e)/(2*A*B) + e*x]*(1 + x^2)),x]

[Out]

-((Sqrt[2]*Sqrt[A]*Sqrt[B]*ArcTan[A/B - (Sqrt[A]*Sqrt[e*(A/B - B/A + 2*x)])/(Sqrt[B]*Sqrt[e])])/Sqrt[e]) + (Sq
rt[2]*Sqrt[A]*Sqrt[B]*ArcTan[A/B + (Sqrt[A]*Sqrt[e*(A/B - B/A + 2*x)])/(Sqrt[B]*Sqrt[e])])/Sqrt[e]

Rule 827

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

Rule 1161

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.185069, size = 142, normalized size = 1.07 \[ -\frac{i \sqrt{2} \sqrt{A} \sqrt{B} \sqrt{\frac{A}{B}-\frac{B}{A}+2 x} \left (\tanh ^{-1}\left (\frac{\sqrt{A} \sqrt{B} \sqrt{\frac{A}{B}-\frac{B}{A}+2 x}}{A-i B}\right )-\tanh ^{-1}\left (\frac{\sqrt{A} \sqrt{B} \sqrt{\frac{A}{B}-\frac{B}{A}+2 x}}{A+i B}\right )\right )}{\sqrt{e \left (\frac{A}{B}-\frac{B}{A}+2 x\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(Sqrt[(A^2*e - B^2*e)/(2*A*B) + e*x]*(1 + x^2)),x]

[Out]

((-I)*Sqrt[2]*Sqrt[A]*Sqrt[B]*Sqrt[A/B - B/A + 2*x]*(ArcTanh[(Sqrt[A]*Sqrt[B]*Sqrt[A/B - B/A + 2*x])/(A - I*B)
] - ArcTanh[(Sqrt[A]*Sqrt[B]*Sqrt[A/B - B/A + 2*x])/(A + I*B)]))/Sqrt[e*(A/B - B/A + 2*x)]

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Maple [A]  time = 0.051, size = 128, normalized size = 1. \begin{align*}{\sqrt{2}AB\arctan \left ({\frac{1}{2\,B} \left ( 2\,\sqrt{2\,ex+{\frac{e \left ({A}^{2}-{B}^{2} \right ) }{AB}}}AB+2\,\sqrt{{A}^{3}Be} \right ){\frac{1}{\sqrt{AeB}}}} \right ){\frac{1}{\sqrt{AeB}}}}+{\sqrt{2}AB\arctan \left ({\frac{1}{2\,B} \left ( 2\,\sqrt{2\,ex+{\frac{e \left ({A}^{2}-{B}^{2} \right ) }{AB}}}AB-2\,\sqrt{{A}^{3}Be} \right ){\frac{1}{\sqrt{AeB}}}} \right ){\frac{1}{\sqrt{AeB}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*(B*x+A)/(x^2+1)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2),x)

[Out]

2^(1/2)*A*B/(A*e*B)^(1/2)*arctan(1/2*(2*(2*e*x+e*(A^2-B^2)/A/B)^(1/2)*A*B+2*(A^3*B*e)^(1/2))/B/(A*e*B)^(1/2))+
2^(1/2)*A*B/(A*e*B)^(1/2)*arctan(1/2*(2*(2*e*x+e*(A^2-B^2)/A/B)^(1/2)*A*B-2*(A^3*B*e)^(1/2))/B/(A*e*B)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, \int \frac{B x + A}{\sqrt{4 \, e x + \frac{2 \,{\left (A^{2} e - B^{2} e\right )}}{A B}}{\left (x^{2} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*(B*x+A)/(x^2+1)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2),x, algorithm="maxima")

[Out]

2*integrate((B*x + A)/(sqrt(4*e*x + 2*(A^2*e - B^2*e)/(A*B))*(x^2 + 1)), x)

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Fricas [A]  time = 2.46931, size = 350, normalized size = 2.63 \begin{align*} \left [\frac{1}{2} \, \sqrt{2} \sqrt{-\frac{A B}{e}} \log \left (\frac{A^{2} x^{2} - 4 \, A B x - A^{2} + 2 \, B^{2} + 2 \,{\left (A x - B\right )} \sqrt{-\frac{A B}{e}} \sqrt{\frac{2 \, A B e x +{\left (A^{2} - B^{2}\right )} e}{A B}}}{x^{2} + 1}\right ), \sqrt{2} \sqrt{\frac{A B}{e}} \arctan \left (\frac{{\left (A x - B\right )} \sqrt{\frac{A B}{e}} \sqrt{\frac{2 \, A B e x +{\left (A^{2} - B^{2}\right )} e}{A B}}}{2 \, A B x + A^{2} - B^{2}}\right )\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*(B*x+A)/(x^2+1)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2),x, algorithm="fricas")

[Out]

[1/2*sqrt(2)*sqrt(-A*B/e)*log((A^2*x^2 - 4*A*B*x - A^2 + 2*B^2 + 2*(A*x - B)*sqrt(-A*B/e)*sqrt((2*A*B*e*x + (A
^2 - B^2)*e)/(A*B)))/(x^2 + 1)), sqrt(2)*sqrt(A*B/e)*arctan((A*x - B)*sqrt(A*B/e)*sqrt((2*A*B*e*x + (A^2 - B^2
)*e)/(A*B))/(2*A*B*x + A^2 - B^2))]

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Sympy [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(2*(B*x+A)/(x**2+1)/(2*(A**2*e-B**2*e)/A/B+4*e*x)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{2 \,{\left (B x + A\right )}}{\sqrt{4 \, e x + \frac{2 \,{\left (A^{2} e - B^{2} e\right )}}{A B}}{\left (x^{2} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*(B*x+A)/(x^2+1)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2),x, algorithm="giac")

[Out]

integrate(2*(B*x + A)/(sqrt(4*e*x + 2*(A^2*e - B^2*e)/(A*B))*(x^2 + 1)), x)