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

Optimal. Leaf size=66 \[ \frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d+e}}\right )}{\sqrt {d+e}}-\frac {(A-B) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d-e}}\right )}{\sqrt {d-e}} \]

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Rubi [A]  time = 0.10, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {827, 1166, 206} \begin {gather*} \frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d+e}}\right )}{\sqrt {d+e}}-\frac {(A-B) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d-e}}\right )}{\sqrt {d-e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - B)*ArcTanh[Sqrt[d + e*x]/Sqrt[d - e]])/Sqrt[d - e]) + ((A + B)*ArcTanh[Sqrt[d + e*x]/Sqrt[d + e]])/Sqr
t[d + e]

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 66, normalized size = 1.00 \begin {gather*} \frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d+e}}\right )}{\sqrt {d+e}}-\frac {(A-B) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d-e}}\right )}{\sqrt {d-e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - B)*ArcTanh[Sqrt[d + e*x]/Sqrt[d - e]])/Sqrt[d - e]) + ((A + B)*ArcTanh[Sqrt[d + e*x]/Sqrt[d + e]])/Sqr
t[d + e]

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IntegrateAlgebraic [A]  time = 0.10, size = 85, normalized size = 1.29 \begin {gather*} \frac {(B-A) \tan ^{-1}\left (\frac {\sqrt {e-d} \sqrt {d+e x}}{d-e}\right )}{\sqrt {e-d}}+\frac {(A+B) \tan ^{-1}\left (\frac {\sqrt {-d-e} \sqrt {d+e x}}{d+e}\right )}{\sqrt {-d-e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-A + B)*ArcTan[(Sqrt[-d + e]*Sqrt[d + e*x])/(d - e)])/Sqrt[-d + e] + ((A + B)*ArcTan[(Sqrt[-d - e]*Sqrt[d +
e*x])/(d + e)])/Sqrt[-d - e]

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fricas [A]  time = 0.49, size = 451, normalized size = 6.83 \begin {gather*} \left [-\frac {{\left ({\left (A - B\right )} d + {\left (A - B\right )} e\right )} \sqrt {d - e} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d - e} + 2 \, d - e}{x + 1}\right ) - {\left ({\left (A + B\right )} d - {\left (A + B\right )} e\right )} \sqrt {d + e} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d + e} + 2 \, d + e}{x - 1}\right )}{2 \, {\left (d^{2} - e^{2}\right )}}, -\frac {2 \, {\left ({\left (A - B\right )} d + {\left (A - B\right )} e\right )} \sqrt {-d + e} \arctan \left (-\frac {\sqrt {e x + d} \sqrt {-d + e}}{d - e}\right ) - {\left ({\left (A + B\right )} d - {\left (A + B\right )} e\right )} \sqrt {d + e} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d + e} + 2 \, d + e}{x - 1}\right )}{2 \, {\left (d^{2} - e^{2}\right )}}, -\frac {2 \, {\left ({\left (A + B\right )} d - {\left (A + B\right )} e\right )} \sqrt {-d - e} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d - e}}{d + e}\right ) + {\left ({\left (A - B\right )} d + {\left (A - B\right )} e\right )} \sqrt {d - e} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d - e} + 2 \, d - e}{x + 1}\right )}{2 \, {\left (d^{2} - e^{2}\right )}}, -\frac {{\left ({\left (A - B\right )} d + {\left (A - B\right )} e\right )} \sqrt {-d + e} \arctan \left (-\frac {\sqrt {e x + d} \sqrt {-d + e}}{d - e}\right ) + {\left ({\left (A + B\right )} d - {\left (A + B\right )} e\right )} \sqrt {-d - e} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d - e}}{d + e}\right )}{d^{2} - e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((A - B)*d + (A - B)*e)*sqrt(d - e)*log((e*x + 2*sqrt(e*x + d)*sqrt(d - e) + 2*d - e)/(x + 1)) - ((A +
B)*d - (A + B)*e)*sqrt(d + e)*log((e*x + 2*sqrt(e*x + d)*sqrt(d + e) + 2*d + e)/(x - 1)))/(d^2 - e^2), -1/2*(2
*((A - B)*d + (A - B)*e)*sqrt(-d + e)*arctan(-sqrt(e*x + d)*sqrt(-d + e)/(d - e)) - ((A + B)*d - (A + B)*e)*sq
rt(d + e)*log((e*x + 2*sqrt(e*x + d)*sqrt(d + e) + 2*d + e)/(x - 1)))/(d^2 - e^2), -1/2*(2*((A + B)*d - (A + B
)*e)*sqrt(-d - e)*arctan(sqrt(e*x + d)*sqrt(-d - e)/(d + e)) + ((A - B)*d + (A - B)*e)*sqrt(d - e)*log((e*x +
2*sqrt(e*x + d)*sqrt(d - e) + 2*d - e)/(x + 1)))/(d^2 - e^2), -(((A - B)*d + (A - B)*e)*sqrt(-d + e)*arctan(-s
qrt(e*x + d)*sqrt(-d + e)/(d - e)) + ((A + B)*d - (A + B)*e)*sqrt(-d - e)*arctan(sqrt(e*x + d)*sqrt(-d - e)/(d
+ e)))/(d^2 - e^2)]

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giac [A]  time = 0.19, size = 68, normalized size = 1.03 \begin {gather*} \frac {{\left (A - B\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d + e}}\right )}{\sqrt {-d + e}} - \frac {{\left (A + B\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d - e}}\right )}{\sqrt {-d - e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A - B)*arctan(sqrt(x*e + d)/sqrt(-d + e))/sqrt(-d + e) - (A + B)*arctan(sqrt(x*e + d)/sqrt(-d - e))/sqrt(-d -
e)

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maple [A]  time = 0.09, size = 95, normalized size = 1.44 \begin {gather*} \frac {A \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d +e}}\right )}{\sqrt {d +e}}+\frac {A \arctan \left (\frac {\sqrt {e x +d}}{\sqrt {-d +e}}\right )}{\sqrt {-d +e}}+\frac {B \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d +e}}\right )}{\sqrt {d +e}}-\frac {B \arctan \left (\frac {\sqrt {e x +d}}{\sqrt {-d +e}}\right )}{\sqrt {-d +e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d-4*e>0)', see `assume?` for
more details)Is 4*d-4*e positive or negative?

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mupad [B]  time = 1.99, size = 773, normalized size = 11.71 \begin {gather*} -\frac {\mathrm {atan}\left (\frac {\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A-B\right )\,\left (32\,B\,d\,e^2-32\,A\,e^3+\frac {32\,d\,e^2\,\left (A-B\right )\,\sqrt {d+e\,x}}{\sqrt {d-e}}\right )}{2\,\sqrt {d-e}}\right )\,\left (A-B\right )\,1{}\mathrm {i}}{2\,\sqrt {d-e}}+\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A-B\right )\,\left (32\,A\,e^3-32\,B\,d\,e^2+\frac {32\,d\,e^2\,\left (A-B\right )\,\sqrt {d+e\,x}}{\sqrt {d-e}}\right )}{2\,\sqrt {d-e}}\right )\,\left (A-B\right )\,1{}\mathrm {i}}{2\,\sqrt {d-e}}}{16\,B^3\,e^2-16\,A^2\,B\,e^2+\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A-B\right )\,\left (32\,B\,d\,e^2-32\,A\,e^3+\frac {32\,d\,e^2\,\left (A-B\right )\,\sqrt {d+e\,x}}{\sqrt {d-e}}\right )}{2\,\sqrt {d-e}}\right )\,\left (A-B\right )}{2\,\sqrt {d-e}}-\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A-B\right )\,\left (32\,A\,e^3-32\,B\,d\,e^2+\frac {32\,d\,e^2\,\left (A-B\right )\,\sqrt {d+e\,x}}{\sqrt {d-e}}\right )}{2\,\sqrt {d-e}}\right )\,\left (A-B\right )}{2\,\sqrt {d-e}}}\right )\,\left (A-B\right )\,1{}\mathrm {i}}{\sqrt {d-e}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A+B\right )\,\left (32\,B\,d\,e^2-32\,A\,e^3+\frac {32\,d\,e^2\,\left (A+B\right )\,\sqrt {d+e\,x}}{\sqrt {d+e}}\right )}{2\,\sqrt {d+e}}\right )\,\left (A+B\right )\,1{}\mathrm {i}}{2\,\sqrt {d+e}}+\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A+B\right )\,\left (32\,A\,e^3-32\,B\,d\,e^2+\frac {32\,d\,e^2\,\left (A+B\right )\,\sqrt {d+e\,x}}{\sqrt {d+e}}\right )}{2\,\sqrt {d+e}}\right )\,\left (A+B\right )\,1{}\mathrm {i}}{2\,\sqrt {d+e}}}{16\,B^3\,e^2-16\,A^2\,B\,e^2+\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A+B\right )\,\left (32\,B\,d\,e^2-32\,A\,e^3+\frac {32\,d\,e^2\,\left (A+B\right )\,\sqrt {d+e\,x}}{\sqrt {d+e}}\right )}{2\,\sqrt {d+e}}\right )\,\left (A+B\right )}{2\,\sqrt {d+e}}-\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A+B\right )\,\left (32\,A\,e^3-32\,B\,d\,e^2+\frac {32\,d\,e^2\,\left (A+B\right )\,\sqrt {d+e\,x}}{\sqrt {d+e}}\right )}{2\,\sqrt {d+e}}\right )\,\left (A+B\right )}{2\,\sqrt {d+e}}}\right )\,\left (A+B\right )\,1{}\mathrm {i}}{\sqrt {d+e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (atan(((((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A - B)*(32*B*d*e^2 - 32*A*e^3 + (32*d*e^2*(A - B)*(d
+ e*x)^(1/2))/(d - e)^(1/2)))/(2*(d - e)^(1/2)))*(A - B)*1i)/(2*(d - e)^(1/2)) + (((16*A^2*e^2 + 16*B^2*e^2)*(
d + e*x)^(1/2) - ((A - B)*(32*A*e^3 - 32*B*d*e^2 + (32*d*e^2*(A - B)*(d + e*x)^(1/2))/(d - e)^(1/2)))/(2*(d -
e)^(1/2)))*(A - B)*1i)/(2*(d - e)^(1/2)))/(16*B^3*e^2 - 16*A^2*B*e^2 + (((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(
1/2) - ((A - B)*(32*B*d*e^2 - 32*A*e^3 + (32*d*e^2*(A - B)*(d + e*x)^(1/2))/(d - e)^(1/2)))/(2*(d - e)^(1/2)))
*(A - B))/(2*(d - e)^(1/2)) - (((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A - B)*(32*A*e^3 - 32*B*d*e^2 +
(32*d*e^2*(A - B)*(d + e*x)^(1/2))/(d - e)^(1/2)))/(2*(d - e)^(1/2)))*(A - B))/(2*(d - e)^(1/2))))*(A - B)*1i)
/(d - e)^(1/2) - (atan(((((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A + B)*(32*B*d*e^2 - 32*A*e^3 + (32*d*
e^2*(A + B)*(d + e*x)^(1/2))/(d + e)^(1/2)))/(2*(d + e)^(1/2)))*(A + B)*1i)/(2*(d + e)^(1/2)) + (((16*A^2*e^2
+ 16*B^2*e^2)*(d + e*x)^(1/2) - ((A + B)*(32*A*e^3 - 32*B*d*e^2 + (32*d*e^2*(A + B)*(d + e*x)^(1/2))/(d + e)^(
1/2)))/(2*(d + e)^(1/2)))*(A + B)*1i)/(2*(d + e)^(1/2)))/(16*B^3*e^2 - 16*A^2*B*e^2 + (((16*A^2*e^2 + 16*B^2*e
^2)*(d + e*x)^(1/2) - ((A + B)*(32*B*d*e^2 - 32*A*e^3 + (32*d*e^2*(A + B)*(d + e*x)^(1/2))/(d + e)^(1/2)))/(2*
(d + e)^(1/2)))*(A + B))/(2*(d + e)^(1/2)) - (((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A + B)*(32*A*e^3
- 32*B*d*e^2 + (32*d*e^2*(A + B)*(d + e*x)^(1/2))/(d + e)^(1/2)))/(2*(d + e)^(1/2)))*(A + B))/(2*(d + e)^(1/2)
)))*(A + B)*1i)/(d + e)^(1/2)

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sympy [A]  time = 57.36, size = 78, normalized size = 1.18 \begin {gather*} \frac {\left (- A - B\right ) \operatorname {atan}{\left (\frac {1}{\sqrt {- \frac {1}{d + e}} \sqrt {d + e x}} \right )}}{\sqrt {- \frac {1}{d + e}} \left (d + e\right )} + \frac {\left (A - B\right ) \operatorname {atan}{\left (\frac {1}{\sqrt {- \frac {1}{d - e}} \sqrt {d + e x}} \right )}}{\sqrt {- \frac {1}{d - e}} \left (d - e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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