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

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

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Rubi [A]  time = 0.21, antiderivative size = 155, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 1.23, size = 302, normalized size = 1.95 \begin {gather*} -\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 )} \end {gather*}

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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IntegrateAlgebraic [C]  time = 0.94, size = 273, normalized size = 1.76 \begin {gather*} \frac {\left (-\sqrt {A e-2 B d}+i \sqrt {A} \sqrt {e}\right ) \tan ^{-1}\left (\frac {\sqrt {B} \sqrt {d+e x}}{\sqrt {-B d-i \sqrt {A} \sqrt {e} \sqrt {A e-2 B d}}}\right )}{\sqrt {B} e \sqrt {A e-2 B d} \sqrt {-B d-i \sqrt {A} \sqrt {e} \sqrt {A e-2 B d}}}+\frac {\left (-\sqrt {A e-2 B d}-i \sqrt {A} \sqrt {e}\right ) \tan ^{-1}\left (\frac {\sqrt {B} \sqrt {d+e x}}{\sqrt {-B d+i \sqrt {A} \sqrt {e} \sqrt {A e-2 B d}}}\right )}{\sqrt {B} e \sqrt {A e-2 B d} \sqrt {-B d+i \sqrt {A} \sqrt {e} \sqrt {A e-2 B d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I*Sqrt[A]*Sqrt[e] - Sqrt[-2*B*d + A*e])*ArcTan[(Sqrt[B]*Sqrt[d + e*x])/Sqrt[-(B*d) - I*Sqrt[A]*Sqrt[e]*Sqrt[
-2*B*d + A*e]]])/(Sqrt[B]*e*Sqrt[-2*B*d + A*e]*Sqrt[-(B*d) - I*Sqrt[A]*Sqrt[e]*Sqrt[-2*B*d + A*e]]) + (((-I)*S
qrt[A]*Sqrt[e] - Sqrt[-2*B*d + A*e])*ArcTan[(Sqrt[B]*Sqrt[d + e*x])/Sqrt[-(B*d) + I*Sqrt[A]*Sqrt[e]*Sqrt[-2*B*
d + A*e]]])/(Sqrt[B]*e*Sqrt[-2*B*d + A*e]*Sqrt[-(B*d) + I*Sqrt[A]*Sqrt[e]*Sqrt[-2*B*d + A*e]])

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fricas [A]  time = 0.44, size = 236, normalized size = 1.52 \begin {gather*} \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 {gather*}

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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giac [B]  time = 1.85, size = 2892, normalized size = 18.66

result too large to display

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]

-(4*A*B^6*d^2*e - 2*A^2*B^5*d*e^2 - 8*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B
^4*d^2*e - 4*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*B^5*d^2 + 4*sqrt(2*A*B*d*e -
A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B^3*d*e^2 - 2*(2*A*B*d*e - A^2*e^2)*B^5*d - sqrt(2*A*
B*d*e - A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*B^5*d - (8*A*B^4*d^2*e - 8*A^2*B^3*d*e^2 - 16*sqrt
(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^2*d^2*e - 8*sqrt(2*A*B*d*e - A^2*e^2)*sqr
t(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*B^3*d^2 + 2*A^3*B^2*e^3 + 16*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d - s
qrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B*d*e^2 + 4*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*
B)*A*B^2*d*e - 4*(2*A*B*d*e - A^2*e^2)*B^3*d - 2*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*
e^2)*B)*B^3*d - 4*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A^3*e^3 + 2*(2*A*B*d*e
- A^2*e^2)*A*B^2*e + sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^2*e)*B^2 - (16*A
*B^5*d^3*e + 16*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^4*d^3*e - 32*A^2*B^4*d^2*e^2 + 8*sqrt(-B^2*d -
sqrt(2*A*B*d*e - A^2*e^2)*B)*B^5*d^3 - 32*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B^3*d^2*e^2 - 12*sqrt
(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^4*d^2*e + 20*A^3*B^3*d*e^3 + 2*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e
^2)*B)*B^5*d^2 + 20*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A^3*B^2*d*e^3 + 4*sqrt(-B^2*d - sqrt(2*A*B*d*e
- A^2*e^2)*B)*A^2*B^3*d*e^2 - 3*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^4*d*e - 8*(2*A*B*d*e - A^2*e^2)
*B^4*d^2 - 4*A^4*B^2*e^4 + 12*(2*A*B*d*e - A^2*e^2)*A*B^3*d*e - 4*sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A
^4*B*e^4 + sqrt(-B^2*d - sqrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B^3*e^2 - 4*(2*A*B*d*e - A^2*e^2)*A^2*B^2*e^2)*abs(B
))*arctan(sqrt(x*e + d)/sqrt(-(B^2*d*e + sqrt(B^4*d^2*e^2 - (B^2*d^2*e - 2*A*B*d*e^2 + A^2*e^3)*B^2*e))*e^(-1)
/B^2))/(16*A*B^7*d^4*e^2 + 8*B^8*d^4*e - 48*A^2*B^6*d^3*e^3 - 20*A*B^7*d^3*e^2 + 2*B^8*d^3*e + 52*A^3*B^5*d^2*
e^4 + 16*A^2*B^6*d^2*e^3 - 5*A*B^7*d^2*e^2 - 24*A^4*B^4*d*e^5 - 4*A^3*B^5*d*e^4 + 4*A^2*B^6*d*e^3 + 4*A^5*B^3*
e^6 - A^3*B^5*e^4) + (4*A*B^6*d^2*e - 2*A^2*B^5*d*e^2 - 8*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2*A*B*d
*e - A^2*e^2)*B)*A*B^4*d^2*e - 4*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*B^5*d^2
+ 4*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B^3*d*e^2 - 2*(2*A*B*d*e - A^2*e^
2)*B^5*d - sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*B^5*d - (8*A*B^4*d^2*e - 8*A^2
*B^3*d*e^2 - 16*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^2*d^2*e - 8*sqrt(2*A*
B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*B^3*d^2 + 2*A^3*B^2*e^3 + 16*sqrt(2*A*B*d*e - A^2*
e^2)*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B*d*e^2 + 4*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2
*A*B*d*e - A^2*e^2)*B)*A*B^2*d*e - 4*(2*A*B*d*e - A^2*e^2)*B^3*d - 2*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + s
qrt(2*A*B*d*e - A^2*e^2)*B)*B^3*d - 4*sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A^3
*e^3 + 2*(2*A*B*d*e - A^2*e^2)*A*B^2*e + sqrt(2*A*B*d*e - A^2*e^2)*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*
A*B^2*e)*B^2 - (16*A*B^5*d^3*e - 16*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^4*d^3*e - 32*A^2*B^4*d^2*e^
2 - 8*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*B^5*d^3 + 32*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B
^3*d^2*e^2 + 12*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^4*d^2*e + 20*A^3*B^3*d*e^3 - 2*sqrt(-B^2*d + sq
rt(2*A*B*d*e - A^2*e^2)*B)*B^5*d^2 - 20*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A^3*B^2*d*e^3 - 4*sqrt(-B^2
*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B^3*d*e^2 + 3*sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A*B^4*d*e - 8*(
2*A*B*d*e - A^2*e^2)*B^4*d^2 - 4*A^4*B^2*e^4 + 12*(2*A*B*d*e - A^2*e^2)*A*B^3*d*e + 4*sqrt(-B^2*d + sqrt(2*A*B
*d*e - A^2*e^2)*B)*A^4*B*e^4 - sqrt(-B^2*d + sqrt(2*A*B*d*e - A^2*e^2)*B)*A^2*B^3*e^2 - 4*(2*A*B*d*e - A^2*e^2
)*A^2*B^2*e^2)*abs(B))*arctan(sqrt(x*e + d)/sqrt(-(B^2*d*e - sqrt(B^4*d^2*e^2 - (B^2*d^2*e - 2*A*B*d*e^2 + A^2
*e^3)*B^2*e))*e^(-1)/B^2))/(16*A*B^7*d^4*e^2 + 8*B^8*d^4*e - 48*A^2*B^6*d^3*e^3 - 20*A*B^7*d^3*e^2 + 2*B^8*d^3
*e + 52*A^3*B^5*d^2*e^4 + 16*A^2*B^6*d^2*e^3 - 5*A*B^7*d^2*e^2 - 24*A^4*B^4*d*e^5 - 4*A^3*B^5*d*e^4 + 4*A^2*B^
6*d*e^3 + 4*A^5*B^3*e^6 - A^3*B^5*e^4)

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maple [A]  time = 0.19, size = 223, normalized size = 1.44 \begin {gather*} -\frac {2 \left (-\frac {\left (-A B e +\sqrt {-\left (A e -2 B d \right ) A \,B^{2} e}\right ) \arctanh \left (\frac {\sqrt {e x +d}\, B}{\sqrt {B^{2} d -\sqrt {-\left (A e -2 B d \right ) A \,B^{2} e}}}\right )}{2 \sqrt {-\left (A e -2 B d \right ) A \,B^{2} e}\, \sqrt {B^{2} d -\sqrt {-\left (A e -2 B d \right ) A \,B^{2} e}}\, B^{2}}-\frac {\left (A B e +\sqrt {-\left (A e -2 B d \right ) A \,B^{2} e}\right ) \arctanh \left (\frac {\sqrt {e x +d}\, B}{\sqrt {B^{2} d +\sqrt {-\left (A e -2 B d \right ) A \,B^{2} e}}}\right )}{2 \sqrt {-\left (A e -2 B d \right ) A \,B^{2} e}\, \sqrt {B^{2} d +\sqrt {-\left (A e -2 B d \right ) A \,B^{2} e}}\, B^{2}}\right ) B^{2}}{e} \end {gather*}

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 0.31, size = 310, normalized size = 2.00 \begin {gather*} \frac {\sqrt {2}\,\left (\mathrm {atan}\left (\frac {\left (A\,e^2\,\sqrt {A\,B\,e-2\,B^2\,d}-2\,B\,d\,e\,\sqrt {A\,B\,e-2\,B^2\,d}\right )\,\left (\left (\frac {\sqrt {2}\,\left (\frac {2\,B^4\,d-2\,A\,B^3\,e}{e^2}-\frac {4\,A^2\,B^4\,e^4-8\,A\,B^5\,d\,e^3+4\,B^6\,d^2\,e^2}{e^4\,\left (2\,B^2\,d-A\,B\,e\right )}\right )}{\left (A\,e-B\,d\right )\,\left (A\,e-2\,B\,d\right )}+\frac {4\,\sqrt {2}\,A\,B^4}{e\,\left (2\,B^2\,d-A\,B\,e\right )\,\left (A\,e-B\,d\right )}\right )\,\sqrt {d+e\,x}-\frac {\sqrt {2}\,\left (\frac {2\,B^4}{e^2}-\frac {4\,B^6\,d}{e^2\,\left (2\,B^2\,d-A\,B\,e\right )}\right )\,{\left (d+e\,x\right )}^{3/2}}{\left (A\,e-B\,d\right )\,\left (A\,e-2\,B\,d\right )}\right )}{4\,A\,B^3}\right )-\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {A\,B\,e-2\,B^2\,d}\,\sqrt {d+e\,x}}{2\,\left (A\,e-2\,B\,d\right )}\right )\right )}{e\,\sqrt {A\,B\,e-2\,B^2\,d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*(atan(((A*e^2*(A*B*e - 2*B^2*d)^(1/2) - 2*B*d*e*(A*B*e - 2*B^2*d)^(1/2))*(((2^(1/2)*((2*B^4*d - 2*A*B
^3*e)/e^2 - (4*A^2*B^4*e^4 + 4*B^6*d^2*e^2 - 8*A*B^5*d*e^3)/(e^4*(2*B^2*d - A*B*e))))/((A*e - B*d)*(A*e - 2*B*
d)) + (4*2^(1/2)*A*B^4)/(e*(2*B^2*d - A*B*e)*(A*e - B*d)))*(d + e*x)^(1/2) - (2^(1/2)*((2*B^4)/e^2 - (4*B^6*d)
/(e^2*(2*B^2*d - A*B*e)))*(d + e*x)^(3/2))/((A*e - B*d)*(A*e - 2*B*d))))/(4*A*B^3)) - atan((2^(1/2)*(A*B*e - 2
*B^2*d)^(1/2)*(d + e*x)^(1/2))/(2*(A*e - 2*B*d)))))/(e*(A*B*e - 2*B^2*d)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {gather*}

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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