3.1463 \(\int \frac{A+B x}{\sqrt{d+e x} (2 A B d-A^2 e-B^2 e x^2)} \, dx\)

Optimal. Leaf size=155 \[ \frac{\log \left (\sqrt{2} \sqrt{B} \sqrt{d+e x} \sqrt{2 B d-A e}-A e+B (d+e x)+B d\right )}{\sqrt{2} \sqrt{B} e \sqrt{2 B d-A e}}-\frac{\log \left (-\sqrt{2} \sqrt{B} \sqrt{d+e x} \sqrt{2 B d-A e}-A e+B (d+e x)+B d\right )}{\sqrt{2} \sqrt{B} e \sqrt{2 B d-A e}} \]

[Out]

-(Log[B*d - A*e - Sqrt[2]*Sqrt[B]*Sqrt[2*B*d - A*e]*Sqrt[d + e*x] + B*(d + e*x)]/(Sqrt[2]*Sqrt[B]*e*Sqrt[2*B*d
- A*e])) + Log[B*d - A*e + Sqrt[2]*Sqrt[B]*Sqrt[2*B*d - A*e]*Sqrt[d + e*x] + B*(d + e*x)]/(Sqrt[2]*Sqrt[B]*e*
Sqrt[2*B*d - A*e])

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Rubi [A]  time = 0.211479, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {827, 1164, 628} \[ \frac{\log \left (\sqrt{2} \sqrt{B} \sqrt{d+e x} \sqrt{2 B d-A e}-A e+B (d+e x)+B d\right )}{\sqrt{2} \sqrt{B} e \sqrt{2 B d-A e}}-\frac{\log \left (-\sqrt{2} \sqrt{B} \sqrt{d+e x} \sqrt{2 B d-A e}-A e+B (d+e x)+B d\right )}{\sqrt{2} \sqrt{B} e \sqrt{2 B d-A e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[B*d - A*e - Sqrt[2]*Sqrt[B]*Sqrt[2*B*d - A*e]*Sqrt[d + e*x] + B*(d + e*x)]/(Sqrt[2]*Sqrt[B]*e*Sqrt[2*B*d
- A*e])) + Log[B*d - A*e + Sqrt[2]*Sqrt[B]*Sqrt[2*B*d - A*e]*Sqrt[d + e*x] + B*(d + e*x)]/(Sqrt[2]*Sqrt[B]*e*
Sqrt[2*B*d - A*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 1164

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*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/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[b^2
- 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 1.26635, size = 302, normalized size = 1.95 \[ -\frac{\left (\sqrt{A} \sqrt{e} \sqrt{2 B d-A e}+A e-2 B d\right ) \left (\sqrt{B d-\sqrt{A} \sqrt{e} \sqrt{2 B d-A e}} \sqrt{\sqrt{A} \sqrt{e} \sqrt{2 B d-A e}+B d} \tanh ^{-1}\left (\frac{\sqrt{B} \sqrt{d+e x}}{\sqrt{B d-\sqrt{A} \sqrt{e} \sqrt{2 B d-A e}}}\right )+(B d-A e) \tanh ^{-1}\left (\frac{\sqrt{B} \sqrt{d+e x}}{\sqrt{\sqrt{A} \sqrt{e} \sqrt{2 B d-A e}+B d}}\right )\right )}{\sqrt{B} \sqrt{2 B d-A e} \sqrt{\sqrt{A} \sqrt{e} \sqrt{2 B d-A e}+B d} \left (A^{3/2} e^{5/2}-2 \sqrt{A} B d e^{3/2}+B d e \sqrt{2 B d-A e}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((-2*B*d + A*e + Sqrt[A]*Sqrt[e]*Sqrt[2*B*d - A*e])*(Sqrt[B*d - Sqrt[A]*Sqrt[e]*Sqrt[2*B*d - A*e]]*Sqrt[B*d
+ Sqrt[A]*Sqrt[e]*Sqrt[2*B*d - A*e]]*ArcTanh[(Sqrt[B]*Sqrt[d + e*x])/Sqrt[B*d - Sqrt[A]*Sqrt[e]*Sqrt[2*B*d - A
*e]]] + (B*d - A*e)*ArcTanh[(Sqrt[B]*Sqrt[d + e*x])/Sqrt[B*d + Sqrt[A]*Sqrt[e]*Sqrt[2*B*d - A*e]]]))/(Sqrt[B]*
Sqrt[2*B*d - A*e]*Sqrt[B*d + Sqrt[A]*Sqrt[e]*Sqrt[2*B*d - A*e]]*(-2*Sqrt[A]*B*d*e^(3/2) + A^(3/2)*e^(5/2) + B*
d*e*Sqrt[2*B*d - A*e])))

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Maple [A]  time = 0.065, size = 223, normalized size = 1.4 \begin{align*} -2\,{\frac{{B}^{2}}{e} \left ( -1/2\,{\frac{-AeB+\sqrt{-Ae{B}^{2} \left ( Ae-2\,Bd \right ) }}{\sqrt{-Ae{B}^{2} \left ( Ae-2\,Bd \right ) }{B}^{2}\sqrt{{B}^{2}d-\sqrt{-Ae{B}^{2} \left ( Ae-2\,Bd \right ) }}}{\it Artanh} \left ({\frac{B\sqrt{ex+d}}{\sqrt{{B}^{2}d-\sqrt{-Ae{B}^{2} \left ( Ae-2\,Bd \right ) }}}} \right ) }-1/2\,{\frac{AeB+\sqrt{-Ae{B}^{2} \left ( Ae-2\,Bd \right ) }}{\sqrt{-Ae{B}^{2} \left ( Ae-2\,Bd \right ) }{B}^{2}\sqrt{{B}^{2}d+\sqrt{-Ae{B}^{2} \left ( Ae-2\,Bd \right ) }}}{\it Artanh} \left ({\frac{B\sqrt{ex+d}}{\sqrt{{B}^{2}d+\sqrt{-Ae{B}^{2} \left ( Ae-2\,Bd \right ) }}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.54598, size = 527, normalized size = 3.4 \begin{align*} \left [\frac{\sqrt{2} \log \left (\frac{B^{2} e^{2} x^{2} + 8 \, B^{2} d^{2} - 6 \, A B d e + A^{2} e^{2} + 4 \,{\left (2 \, B^{2} d e - A B e^{2}\right )} x + \frac{2 \, \sqrt{2}{\left (4 \, B^{3} d^{2} - 4 \, A B^{2} d e + A^{2} B e^{2} +{\left (2 \, B^{3} d e - A B^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{\sqrt{2 \, B^{2} d - A B e}}}{B^{2} e x^{2} - 2 \, A B d + A^{2} e}\right )}{2 \, \sqrt{2 \, B^{2} d - A B e} e}, -\frac{\sqrt{2} \sqrt{-\frac{1}{2 \, B^{2} d - A B e}} \arctan \left (\frac{\sqrt{2}{\left (B e x + 2 \, B d - A e\right )} \sqrt{-\frac{1}{2 \, B^{2} d - A B e}}}{2 \, \sqrt{e x + d}}\right )}{e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(2)*log((B^2*e^2*x^2 + 8*B^2*d^2 - 6*A*B*d*e + A^2*e^2 + 4*(2*B^2*d*e - A*B*e^2)*x + 2*sqrt(2)*(4*B^3
*d^2 - 4*A*B^2*d*e + A^2*B*e^2 + (2*B^3*d*e - A*B^2*e^2)*x)*sqrt(e*x + d)/sqrt(2*B^2*d - A*B*e))/(B^2*e*x^2 -
2*A*B*d + A^2*e))/(sqrt(2*B^2*d - A*B*e)*e), -sqrt(2)*sqrt(-1/(2*B^2*d - A*B*e))*arctan(1/2*sqrt(2)*(B*e*x + 2
*B*d - A*e)*sqrt(-1/(2*B^2*d - A*B*e))/sqrt(e*x + d))/e]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{A}{A^{2} e \sqrt{d + e x} - 2 A B d \sqrt{d + e x} + B^{2} e x^{2} \sqrt{d + e x}}\, dx - \int \frac{B x}{A^{2} e \sqrt{d + e x} - 2 A B d \sqrt{d + e x} + B^{2} e x^{2} \sqrt{d + e x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError