3.23.13 \(\int \frac {f+g x}{(d+e x)^2 \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\) [2213]

Optimal. Leaf size=137 \[ -\frac {2 (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 e^2 (2 c d-b e) (d+e x)^2}-\frac {2 (2 c e f+4 c d g-3 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 e^2 (2 c d-b e)^2 (d+e x)} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {806, 664} \begin {gather*} -\frac {2 (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 e^2 (d+e x)^2 (2 c d-b e)}-\frac {2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-3 b e g+4 c d g+2 c e f)}{3 e^2 (d+e x) (2 c d-b e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 664

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

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(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] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {f+g x}{(d+e x)^2 \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx &=-\frac {2 (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 e^2 (2 c d-b e) (d+e x)^2}+\frac {(2 c e f+4 c d g-3 b e g) \int \frac {1}{(d+e x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{3 e (2 c d-b e)}\\ &=-\frac {2 (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 e^2 (2 c d-b e) (d+e x)^2}-\frac {2 (2 c e f+4 c d g-3 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 e^2 (2 c d-b e)^2 (d+e x)}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 89, normalized size = 0.65 \begin {gather*} -\frac {2 \sqrt {(d+e x) (-b e+c (d-e x))} \left (2 c \left (d^2 g+e^2 f x+2 d e (f+g x)\right )-b e (2 d g+e (f+3 g x))\right )}{3 e^2 (-2 c d+b e)^2 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.05, size = 212, normalized size = 1.55

method	result	size
trager	\(\frac {2 \left (3 b \,e^{2} g x -4 c d e g x -2 c \,e^{2} f x +2 b d e g +b \,e^{2} f -2 c \,d^{2} g -4 c d e f \right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}{3 \left (b^{2} e^{2}-4 b c d e +4 c^{2} d^{2}\right ) \left (e x +d \right )^{2} e^{2}}\)	\(115\)
gosper	\(-\frac {2 \left (c e x +b e -c d \right ) \left (3 b \,e^{2} g x -4 c d e g x -2 c \,e^{2} f x +2 b d e g +b \,e^{2} f -2 c \,d^{2} g -4 c d e f \right )}{3 \left (e x +d \right ) e^{2} \left (b^{2} e^{2}-4 b c d e +4 c^{2} d^{2}\right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}\)	\(127\)
default	\(-\frac {2 g \sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}{e^{2} \left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}+\frac {\left (-d g +e f \right ) \left (-\frac {2 \sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}{3 \left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )^{2}}-\frac {4 c \,e^{2} \sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}{3 \left (-b \,e^{2}+2 c d e \right )^{2} \left (x +\frac {d}{e}\right )}\right )}{e^{3}}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*g/e^2/(-b*e^2+2*c*d*e)/(x+d/e)*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)+(-d*g+e*f)/e^3*(-2/3/(-b*e
^2+2*c*d*e)/(x+d/e)^2*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)-4/3*c*e^2/(-b*e^2+2*c*d*e)^2/(x+d/e)*(
-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(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((g*x+f)/(e*x+d)^2/(-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(2*c*d-%e*b>0)', see `assume?`
for more det

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Fricas [A]
time = 6.31, size = 171, normalized size = 1.25 \begin {gather*} -\frac {2 \, {\left (2 \, c d^{2} g - {\left (b f - {\left (2 \, c f - 3 \, b g\right )} x\right )} e^{2} + 2 \, {\left (2 \, c d g x + 2 \, c d f - b d g\right )} e\right )} \sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}}}{3 \, {\left (4 \, c^{2} d^{4} e^{2} + b^{2} x^{2} e^{6} - 2 \, {\left (2 \, b c d x^{2} - b^{2} d x\right )} e^{5} + {\left (4 \, c^{2} d^{2} x^{2} - 8 \, b c d^{2} x + b^{2} d^{2}\right )} e^{4} + 4 \, {\left (2 \, c^{2} d^{3} x - b c d^{3}\right )} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(2*c*d^2*g - (b*f - (2*c*f - 3*b*g)*x)*e^2 + 2*(2*c*d*g*x + 2*c*d*f - b*d*g)*e)*sqrt(c*d^2 - b*d*e - (c*x
^2 + b*x)*e^2)/(4*c^2*d^4*e^2 + b^2*x^2*e^6 - 2*(2*b*c*d*x^2 - b^2*d*x)*e^5 + (4*c^2*d^2*x^2 - 8*b*c*d^2*x + b
^2*d^2)*e^4 + 4*(2*c^2*d^3*x - b*c*d^3)*e^3)

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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 )} \left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)**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))*(d + e*x)**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (132) = 264\).
time = 1.08, size = 312, normalized size = 2.28 \begin {gather*} \frac {2}{3} \, {\left (\frac {{\left (4 \, \sqrt {-c} c d g + 2 \, \sqrt {-c} c f e - 3 \, b \sqrt {-c} g e\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{4 \, c^{2} d^{2} e - 4 \, b c d e^{2} + b^{2} e^{3}} + \frac {\frac {{\left (3 \, c \sqrt {-c + \frac {2 \, c d}{x e + d} - \frac {b e}{x e + d}} + {\left (-c + \frac {2 \, c d}{x e + d} - \frac {b e}{x e + d}\right )}^{\frac {3}{2}}\right )} d g e}{2 \, c d e - b e^{2}} - \frac {{\left (3 \, c \sqrt {-c + \frac {2 \, c d}{x e + d} - \frac {b e}{x e + d}} + {\left (-c + \frac {2 \, c d}{x e + d} - \frac {b e}{x e + d}\right )}^{\frac {3}{2}}\right )} f e^{2}}{2 \, c d e - b e^{2}} - 3 \, \sqrt {-c + \frac {2 \, c d}{x e + d} - \frac {b e}{x e + d}} g}{2 \, c d e \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - b e^{2} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.79, size = 101, normalized size = 0.74 \begin {gather*} -\frac {2\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}\,\left (2\,c\,d^2\,g-b\,e^2\,f-3\,b\,e^2\,g\,x+2\,c\,e^2\,f\,x-2\,b\,d\,e\,g+4\,c\,d\,e\,f+4\,c\,d\,e\,g\,x\right )}{3\,e^2\,{\left (b\,e-2\,c\,d\right )}^2\,{\left (d+e\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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