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

Optimal. Leaf size=149 \[ \frac {(2 c d-b e) (-3 b e g+2 c d g+4 c e f) \tan ^{-1}\left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{8 c^{5/2} e^2}+\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (3 b e g-4 c (d g+e f)-2 c e g x)}{4 c^2 e^2} \]

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Rubi [A]  time = 0.16, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {779, 621, 204} \begin {gather*} \frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (3 b e g-4 c (d g+e f)-2 c e g x)}{4 c^2 e^2}+\frac {(2 c d-b e) (-3 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^{5/2} e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b*e*g - 4*c*(e*f + d*g) - 2*c*e*g*x)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(4*c^2*e^2) + ((2*c*d - b*
e)*(4*c*e*f + 2*c*d*g - 3*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^(5/2)*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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {(d+e x) (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx &=\frac {(3 b e g-4 c (e f+d g)-2 c e g x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c^2 e^2}+\frac {((2 c d-b e) (4 c e f+2 c d g-3 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^2 e}\\ &=\frac {(3 b e g-4 c (e f+d g)-2 c e g x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c^2 e^2}+\frac {((2 c d-b e) (4 c e f+2 c d g-3 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^2 e}\\ &=\frac {(3 b e g-4 c (e f+d g)-2 c e g x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c^2 e^2}+\frac {(2 c d-b e) (4 c e f+2 c d g-3 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^{5/2} e^2}\\ \end {align*}

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Mathematica [A]  time = 0.79, size = 218, normalized size = 1.46 \begin {gather*} \frac {\sqrt {c} \sqrt {e} (d+e x) \sqrt {e (2 c d-b e)} (b e-c d+c e x) (2 c (2 d g+2 e f+e g x)-3 b e g)+e \sqrt {d+e x} (b e-2 c d)^2 \sqrt {\frac {b e-c d+c e x}{b e-2 c d}} (-3 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 )}{4 c^{5/2} e^{5/2} \sqrt {e (2 c d-b e)} \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 3.64, size = 327, normalized size = 2.19 \begin {gather*} \frac {\left (-3 b^2 e^2 g+8 b c d e 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^{5/2} e^2}+\frac {\sqrt {-c e^2} \left (3 b^2 e^2 g-8 b c d e 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^3 e^3}+\frac {\sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2} (3 b e g-4 c d g-4 c e f-2 c e g x)}{4 c^2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*c*e*f - 4*c*d*g + 3*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^2*e^2) + ((-8*c^2*
d*e*f + 4*b*c*e^2*f - 4*c^2*d^2*g + 8*b*c*d*e*g - 3*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^(5/2)*e^2) + (Sqrt[-(c*e^2)]*(8*c^2*d*e*f - 4*b*c*e^2*f + 4
*c^2*d^2*g - 8*b*c*d*e*g + 3*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^3*e^3)

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fricas [A]  time = 0.63, size = 403, normalized size = 2.70 \begin {gather*} \left [-\frac {{\left (4 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} f + {\left (4 \, c^{2} d^{2} - 8 \, b c d e + 3 \, 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 - 3 \, b c e\right )} g\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}{16 \, c^{3} e^{2}}, -\frac {{\left (4 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} f + {\left (4 \, c^{2} d^{2} - 8 \, b c d e + 3 \, 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 - 3 \, b c e\right )} g\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}{8 \, c^{3} e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((4*(2*c^2*d*e - b*c*e^2)*f + (4*c^2*d^2 - 8*b*c*d*e + 3*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 - 3*b*c*e)*g)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e))/(c^3*e^2
), -1/8*((4*(2*c^2*d*e - b*c*e^2)*f + (4*c^2*d^2 - 8*b*c*d*e + 3*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 - 3*b*c*e)*g)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e))/(c^3*e^2)]

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giac [A]  time = 0.54, size = 179, normalized size = 1.20 \begin {gather*} -\frac {1}{4} \, \sqrt {-c x^{2} e^{2} + c d^{2} - b x e^{2} - b d e} {\left (\frac {2 \, g x e^{\left (-1\right )}}{c} + \frac {{\left (4 \, c d g e + 4 \, c f e^{2} - 3 \, b g e^{2}\right )} e^{\left (-3\right )}}{c^{2}}\right )} + \frac {{\left (4 \, c^{2} d^{2} g + 8 \, c^{2} d f e - 8 \, b c d g e - 4 \, b c f e^{2} + 3 \, b^{2} g e^{2}\right )} \sqrt {-c e^{2}} e^{\left (-3\right )} \log \left ({\left | -2 \, {\left (\sqrt {-c e^{2}} x - \sqrt {-c x^{2} e^{2} + c d^{2} - b x e^{2} - b d e}\right )} c - \sqrt {-c e^{2}} b \right |}\right )}{8 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(-c*x^2*e^2 + c*d^2 - b*x*e^2 - b*d*e)*(2*g*x*e^(-1)/c + (4*c*d*g*e + 4*c*f*e^2 - 3*b*g*e^2)*e^(-3)/c
^2) + 1/8*(4*c^2*d^2*g + 8*c^2*d*f*e - 8*b*c*d*g*e - 4*b*c*f*e^2 + 3*b^2*g*e^2)*sqrt(-c*e^2)*e^(-3)*log(abs(-2
*(sqrt(-c*e^2)*x - sqrt(-c*x^2*e^2 + c*d^2 - b*x*e^2 - b*d*e))*c - sqrt(-c*e^2)*b))/c^3

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maple [B]  time = 0.06, size = 460, normalized size = 3.09 \begin {gather*} \frac {3 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^{2}}-\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 )}{\sqrt {c \,e^{2}}\, c}-\frac {b e f \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}}\, c}+\frac {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 {d f \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 )}{\sqrt {c \,e^{2}}}-\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, g x}{2 c e}+\frac {3 \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, b g}{4 c^{2} e}-\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, d g}{c \,e^{2}}-\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, f}{c e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/e*g*x/c*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)+3/4/e*g*b/c^2*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)+3/8*e
*g*b^2/c^2/(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))-g/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*d+1/2/e*g/(c*e^2)^(1/2)*arcta
n((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/c/e^2*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^
2)^(1/2)*d*g-1/c/e*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*f-1/2*b/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))*e*f+d*f/(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))

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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((e*x+d)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(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(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 )\,\left (d+e\,x\right )}{\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))/(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))/(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 {\left (d + e x\right ) \left (f + g x\right )}{\sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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