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

Optimal. Leaf size=192 \[ \frac {(2 c d-b e) (-b e g-2 c d g+4 c e f) \tan ^{-1}\left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{8 c^{3/2} e^2}+\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-b e g-2 c d g+4 c e f)}{4 c e^2}-\frac {g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 c e^2 (d+e x)} \]

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Rubi [A]  time = 0.22, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {794, 664, 621, 204} \begin {gather*} \frac {(2 c d-b e) (-b e g-2 c d g+4 c e f) \tan ^{-1}\left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{8 c^{3/2} e^2}+\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-b e g-2 c d g+4 c e f)}{4 c e^2}-\frac {g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 c e^2 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*c*e*f - 2*c*d*g - b*e*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(4*c*e^2) - (g*(d*(c*d - b*e) - b*e^2*
x - c*e^2*x^2)^(3/2))/(2*c*e^2*(d + e*x)) + ((2*c*d - b*e)*(4*c*e*f - 2*c*d*g - b*e*g)*ArcTan[(e*(b + 2*c*x))/
(2*Sqrt[c]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])])/(8*c^(3/2)*e^2)

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 621

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

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), 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] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

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 {c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx &=-\frac {g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 c e^2 (d+e x)}-\frac {\left (c e^3 f-\left (-c d e^2+b e^3\right ) g+\frac {3}{2} e \left (-2 c e^2 f+b e^2 g\right )\right ) \int \frac {\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx}{2 c e^3}\\ &=\frac {(4 c e f-2 c d g-b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c e^2}-\frac {g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 c e^2 (d+e x)}+\frac {((2 c d-b e) (4 c e f-2 c d g-b e g)) \int \frac {1}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{8 c e}\\ &=\frac {(4 c e f-2 c d g-b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c e^2}-\frac {g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 c e^2 (d+e x)}+\frac {((2 c d-b e) (4 c e f-2 c d g-b e g)) \operatorname {Subst}\left (\int \frac {1}{-4 c e^2-x^2} \, dx,x,\frac {-b e^2-2 c e^2 x}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}\right )}{4 c e}\\ &=\frac {(4 c e f-2 c d g-b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c e^2}-\frac {g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 c e^2 (d+e x)}+\frac {(2 c d-b e) (4 c e f-2 c d g-b e g) \tan ^{-1}\left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{8 c^{3/2} e^2}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 174, normalized size = 0.91 \begin {gather*} \frac {\sqrt {(d+e x) (c (d-e x)-b e)} \left (\sqrt {c} \sqrt {e} (b e g+2 c (-2 d g+2 e f+e g x))+\frac {\sqrt {e (2 c d-b e)} (-b e g-2 c d g+4 c e f) \sin ^{-1}\left (\frac {\sqrt {c} \sqrt {e} \sqrt {d+e x}}{\sqrt {e (2 c d-b e)}}\right )}{\sqrt {d+e x} \sqrt {\frac {b e-c d+c e x}{b e-2 c d}}}\right )}{4 c^{3/2} e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))]*(Sqrt[c]*Sqrt[e]*(b*e*g + 2*c*(2*e*f - 2*d*g + e*g*x)) + (Sqrt[e*(2*c*
d - b*e)]*(4*c*e*f - 2*c*d*g - b*e*g)*ArcSin[(Sqrt[c]*Sqrt[e]*Sqrt[d + e*x])/Sqrt[e*(2*c*d - b*e)]])/(Sqrt[d +
e*x]*Sqrt[(-(c*d) + b*e + c*e*x)/(-2*c*d + b*e)])))/(4*c^(3/2)*e^(5/2))

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IntegrateAlgebraic [A]  time = 2.69, size = 312, normalized size = 1.62 \begin {gather*} -\frac {\sqrt {-c e^2} \left (-b^2 e^2 g+4 b c e^2 f+4 c^2 d^2 g-8 c^2 d e f\right ) \log \left (b^2 e^2-8 c x \sqrt {-c e^2} \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2}-4 b c d e-4 b c e^2 x+4 c^2 d^2-8 c^2 e^2 x^2\right )}{16 c^2 e^3}+\frac {\left (-b^2 e^2 g+4 b c e^2 f+4 c^2 d^2 g-8 c^2 d e f\right ) \tan ^{-1}\left (\frac {\sqrt {c} \left (2 \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2}-2 x \sqrt {-c e^2}\right )}{b e}\right )}{8 c^{3/2} e^2}+\frac {\sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2} (b e g-4 c d g+4 c e f+2 c e g x)}{4 c e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*c*e*f - 4*c*d*g + b*e*g + 2*c*e*g*x)*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2])/(4*c*e^2) + ((-8*c^2*d*e*f
+ 4*b*c*e^2*f + 4*c^2*d^2*g - b^2*e^2*g)*ArcTan[(Sqrt[c]*(-2*Sqrt[-(c*e^2)]*x + 2*Sqrt[c*d^2 - b*d*e - b*e^2*
x - c*e^2*x^2]))/(b*e)])/(8*c^(3/2)*e^2) - (Sqrt[-(c*e^2)]*(-8*c^2*d*e*f + 4*b*c*e^2*f + 4*c^2*d^2*g - b^2*e^2
*g)*Log[4*c^2*d^2 - 4*b*c*d*e + b^2*e^2 - 4*b*c*e^2*x - 8*c^2*e^2*x^2 - 8*c*Sqrt[-(c*e^2)]*x*Sqrt[c*d^2 - b*d*
e - b*e^2*x - c*e^2*x^2]])/(16*c^2*e^3)

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fricas [A]  time = 0.47, size = 395, normalized size = 2.06 \begin {gather*} \left [-\frac {{\left (4 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} f - {\left (4 \, c^{2} d^{2} - b^{2} e^{2}\right )} g\right )} \sqrt {-c} \log \left (8 \, c^{2} e^{2} x^{2} + 8 \, b c e^{2} x - 4 \, c^{2} d^{2} + 4 \, b c d e + b^{2} e^{2} - 4 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {-c}\right ) - 4 \, {\left (2 \, c^{2} e g x + 4 \, c^{2} e f - {\left (4 \, c^{2} d - b c e\right )} g\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}{16 \, c^{2} e^{2}}, -\frac {{\left (4 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} f - {\left (4 \, c^{2} d^{2} - b^{2} e^{2}\right )} g\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {c}}{2 \, {\left (c^{2} e^{2} x^{2} + b c e^{2} x - c^{2} d^{2} + b c d e\right )}}\right ) - 2 \, {\left (2 \, c^{2} e g x + 4 \, c^{2} e f - {\left (4 \, c^{2} d - b c e\right )} g\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}{8 \, c^{2} e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((4*(2*c^2*d*e - b*c*e^2)*f - (4*c^2*d^2 - b^2*e^2)*g)*sqrt(-c)*log(8*c^2*e^2*x^2 + 8*b*c*e^2*x - 4*c^2
*d^2 + 4*b*c*d*e + b^2*e^2 - 4*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(-c)) - 4*(2*c^2
*e*g*x + 4*c^2*e*f - (4*c^2*d - b*c*e)*g)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e))/(c^2*e^2), -1/8*((4*(2*c
^2*d*e - b*c*e^2)*f - (4*c^2*d^2 - b^2*e^2)*g)*sqrt(c)*arctan(1/2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(
2*c*e*x + b*e)*sqrt(c)/(c^2*e^2*x^2 + b*c*e^2*x - c^2*d^2 + b*c*d*e)) - 2*(2*c^2*e*g*x + 4*c^2*e*f - (4*c^2*d
- b*c*e)*g)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e))/(c^2*e^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.06, size = 697, normalized size = 3.63 \begin {gather*} \frac {b^{2} e g \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {b}{2 c}\right )}{\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}\right )}{8 \sqrt {c \,e^{2}}\, c}-\frac {b d g \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {b}{2 c}\right )}{\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}\right )}{2 \sqrt {c \,e^{2}}}+\frac {b d g \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {d}{e}-\frac {-b \,e^{2}+2 c d e}{2 c \,e^{2}}\right )}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} c \,e^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {c \,e^{2}}}-\frac {b e f \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {d}{e}-\frac {-b \,e^{2}+2 c d e}{2 c \,e^{2}}\right )}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} c \,e^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {c \,e^{2}}}+\frac {c \,d^{2} g \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {b}{2 c}\right )}{\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}\right )}{2 \sqrt {c \,e^{2}}\, e}-\frac {c \,d^{2} g \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {d}{e}-\frac {-b \,e^{2}+2 c d e}{2 c \,e^{2}}\right )}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} c \,e^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {c \,e^{2}}\, e}+\frac {c d f \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {d}{e}-\frac {-b \,e^{2}+2 c d e}{2 c \,e^{2}}\right )}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} c \,e^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {c \,e^{2}}}+\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, g x}{2 e}+\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, b g}{4 c e}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} c \,e^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}\, d g}{e^{2}}+\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} c \,e^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}\, f}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*g/e*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*x+1/4*g/e/c*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*b+1/8*g*e/c/
(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2))*b^2-1/2*g/(c*e^2)^(1/2)
*arctan((c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2))*b*d+1/2*g/e*c/(c*e^2)^(1/2)*arctan((
c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2))*d^2-1/e^2*(-c*(x+d/e)^2*e^2+(-b*e^2+2*c*d*e)*
(x+d/e))^(1/2)*d*g+1/e*(-c*(x+d/e)^2*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)*f+1/2/(c*e^2)^(1/2)*arctan((c*e^2)^(1
/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-c*(x+d/e)^2*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2))*b*d*g-1/2*e/(c*e^2)^
(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-c*(x+d/e)^2*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2
))*b*f-1/e/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-c*(x+d/e)^2*e^2+(-b*e^2+2*c
*d*e)*(x+d/e))^(1/2))*c*d^2*g+1/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-c*(x+d
/e)^2*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2))*c*d*f

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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((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/(e*x+d),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(b*e-2*c*d>0)', see `assume?` f
or more details)Is b*e-2*c*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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