∫ A+Bx______∘ A2e−B2e 2AB +ex (1+x2) dx

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

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

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Rubi [A] time = 0.43, antiderivative size = 133, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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])

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 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]))

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*}

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Mathematica [C] time = 0.18, size = 142, normalized size = 1.07 \begin {gather*} -\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 )}} \end {gather*}

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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IntegrateAlgebraic [C] time = 0.35, size = 145, normalized size = 1.09 \begin {gather*} \frac {i \sqrt {2} \sqrt {A} \sqrt {B} \tanh ^{-1}\left (\frac {\sqrt {A} \sqrt {B} \sqrt {-\frac {B e}{A}+\frac {A e}{B}+2 e x}}{\sqrt {e} (A+i B)}\right )}{\sqrt {e}}-\frac {i \sqrt {2} \sqrt {A} \sqrt {B} \tanh ^{-1}\left (\frac {\sqrt {A} \sqrt {B} \sqrt {-\frac {B e}{A}+\frac {A e}{B}+2 e x}}{\sqrt {e} (A-i B)}\right )}{\sqrt {e}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

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

[e])])/Sqrt[e]

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fricas [A] time = 0.44, size = 164, normalized size = 1.23 \begin {gather*} \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 {gather*}

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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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Undef/Unsigned Inf encountered in limitU

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