3.21.35 \(\int \frac {f+g x}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\)

Optimal. Leaf size=131 \[ -\frac {2 (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 \sqrt {2 c d-b e}}-\frac {2 g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 \sqrt {d+e x}} \]

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Rubi [A]  time = 0.20, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {794, 660, 208} \begin {gather*} -\frac {2 (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 \sqrt {2 c d-b e}}-\frac {2 g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)/(Sqrt[d + e*x]*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2]),x]

[Out]

(-2*g*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(c*e^2*Sqrt[d + e*x]) - (2*(e*f - d*g)*ArcTanh[Sqrt[d*(c*d -
b*e) - b*e^2*x - c*e^2*x^2]/(Sqrt[2*c*d - b*e]*Sqrt[d + e*x])])/(e^2*Sqrt[2*c*d - b*e])

Rule 208

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

Rule 660

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int \frac {f+g x}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx &=-\frac {2 g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 \sqrt {d+e x}}-\frac {\left (2 \left (\frac {1}{2} e \left (-2 c e^2 f+b e^2 g\right )+\frac {1}{2} \left (c e^3 f-\left (-c d e^2+b e^3\right ) g\right )\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{c e^3}\\ &=-\frac {2 g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 \sqrt {d+e x}}+(2 (e f-d g)) \operatorname {Subst}\left (\int \frac {1}{-2 c d e^2+b e^3+e^2 x^2} \, dx,x,\frac {\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{\sqrt {d+e x}}\right )\\ &=-\frac {2 g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 \sqrt {d+e x}}-\frac {2 (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {2 c d-b e} \sqrt {d+e x}}\right )}{e^2 \sqrt {2 c d-b e}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 148, normalized size = 1.13 \begin {gather*} \frac {2 \sqrt {d+e x} \left (g (2 c d-b e) (c (d-e x)-b e)-c \sqrt {2 c d-b e} (d g-e f) \sqrt {c (d-e x)-b e} \tanh ^{-1}\left (\frac {\sqrt {-b e+c d-c e x}}{\sqrt {2 c d-b e}}\right )\right )}{c e^2 (b e-2 c d) \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)/(Sqrt[d + e*x]*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2]),x]

[Out]

(2*Sqrt[d + e*x]*((2*c*d - b*e)*g*(-(b*e) + c*(d - e*x)) - c*Sqrt[2*c*d - b*e]*(-(e*f) + d*g)*Sqrt[-(b*e) + c*
(d - e*x)]*ArcTanh[Sqrt[c*d - b*e - c*e*x]/Sqrt[2*c*d - b*e]]))/(c*e^2*(-2*c*d + b*e)*Sqrt[(d + e*x)*(-(b*e) +
c*(d - e*x))])

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IntegrateAlgebraic [A]  time = 0.45, size = 144, normalized size = 1.10 \begin {gather*} -\frac {2 (d g-e f) \tan ^{-1}\left (\frac {\sqrt {b e-2 c d} \sqrt {(d+e x) (2 c d-b e)-c (d+e x)^2}}{\sqrt {d+e x} (b e+c (d+e x)-2 c d)}\right )}{e^2 \sqrt {b e-2 c d}}-\frac {2 g \sqrt {(d+e x) (2 c d-b e)-c (d+e x)^2}}{c e^2 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(f + g*x)/(Sqrt[d + e*x]*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2]),x]

[Out]

(-2*g*Sqrt[(2*c*d - b*e)*(d + e*x) - c*(d + e*x)^2])/(c*e^2*Sqrt[d + e*x]) - (2*(-(e*f) + d*g)*ArcTan[(Sqrt[-2
*c*d + b*e]*Sqrt[(2*c*d - b*e)*(d + e*x) - c*(d + e*x)^2])/(Sqrt[d + e*x]*(-2*c*d + b*e + c*(d + e*x)))])/(e^2
*Sqrt[-2*c*d + b*e])

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fricas [A]  time = 0.43, size = 442, normalized size = 3.37 \begin {gather*} \left [-\frac {2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c d - b e\right )} \sqrt {e x + d} g + {\left (c d e f - c d^{2} g + {\left (c e^{2} f - c d e g\right )} x\right )} \sqrt {2 \, c d - b e} \log \left (-\frac {c e^{2} x^{2} - 3 \, c d^{2} + 2 \, b d e - 2 \, {\left (c d e - b e^{2}\right )} x - 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {2 \, c d - b e} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right )}{2 \, c^{2} d^{2} e^{2} - b c d e^{3} + {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x}, -\frac {2 \, {\left (\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c d - b e\right )} \sqrt {e x + d} g + {\left (c d e f - c d^{2} g + {\left (c e^{2} f - c d e g\right )} x\right )} \sqrt {-2 \, c d + b e} \arctan \left (\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {-2 \, c d + b e} \sqrt {e x + d}}{c e^{2} x^{2} + b e^{2} x - c d^{2} + b d e}\right )\right )}}{2 \, c^{2} d^{2} e^{2} - b c d e^{3} + {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)^(1/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

[-(2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*d - b*e)*sqrt(e*x + d)*g + (c*d*e*f - c*d^2*g + (c*e^2*f
- c*d*e*g)*x)*sqrt(2*c*d - b*e)*log(-(c*e^2*x^2 - 3*c*d^2 + 2*b*d*e - 2*(c*d*e - b*e^2)*x - 2*sqrt(-c*e^2*x^2
- b*e^2*x + c*d^2 - b*d*e)*sqrt(2*c*d - b*e)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)))/(2*c^2*d^2*e^2 - b*c*d
*e^3 + (2*c^2*d*e^3 - b*c*e^4)*x), -2*(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*d - b*e)*sqrt(e*x + d)*
g + (c*d*e*f - c*d^2*g + (c*e^2*f - c*d*e*g)*x)*sqrt(-2*c*d + b*e)*arctan(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 -
b*d*e)*sqrt(-2*c*d + b*e)*sqrt(e*x + d)/(c*e^2*x^2 + b*e^2*x - c*d^2 + b*d*e)))/(2*c^2*d^2*e^2 - b*c*d*e^3 + (
2*c^2*d*e^3 - b*c*e^4)*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {g x + f}{\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)^(1/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="giac")

[Out]

integrate((g*x + f)/(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d)), x)

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maple [A]  time = 0.06, size = 161, normalized size = 1.23 \begin {gather*} -\frac {2 \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (c d g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-c e f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+\sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, g \right )}{\sqrt {e x +d}\, \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)/(e*x+d)^(1/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x)

[Out]

-2/(e*x+d)^(1/2)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*(arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c*d*
g-arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c*e*f+g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2))/(-c*e*x-b
*e+c*d)^(1/2)/c/e^2/(b*e-2*c*d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {g x + f}{\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)^(1/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((g*x + f)/(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {f+g\,x}{\sqrt {d+e\,x}\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)/((d + e*x)^(1/2)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2)),x)

[Out]

int((f + g*x)/((d + e*x)^(1/2)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {f + g x}{\sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )} \sqrt {d + e x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)**(1/2)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(1/2),x)

[Out]

Integral((f + g*x)/(sqrt(-(d + e*x)*(b*e - c*d + c*e*x))*sqrt(d + e*x)), x)

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