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

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

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Rubi [A]  time = 0.20, 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 = {792, 650} \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 650

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 792

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 = 0.07, size = 89, normalized size = 0.65 \begin {gather*} -\frac {2 \sqrt {(d+e x) (c (d-e x)-b e)} \left (2 c \left (d^2 g+2 d e (f+g x)+e^2 f x\right )-b e (2 d g+e (f+3 g x))\right )}{3 e^2 (d+e x)^2 (b e-2 c d)^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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IntegrateAlgebraic [A]  time = 1.06, size = 105, normalized size = 0.77 \begin {gather*} -\frac {2 \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2} \left (-2 b d e g-b e^2 f-3 b e^2 g x+2 c d^2 g+4 c d e f+4 c d e g x+2 c e^2 f x\right )}{3 e^2 (d+e x)^2 (b e-2 c d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 2.44, size = 182, normalized size = 1.33 \begin {gather*} -\frac {2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left ({\left (4 \, c d e - b e^{2}\right )} f + 2 \, {\left (c d^{2} - b d e\right )} g + {\left (2 \, c e^{2} f + {\left (4 \, c d e - 3 \, b e^{2}\right )} g\right )} x\right )}}{3 \, {\left (4 \, c^{2} d^{4} e^{2} - 4 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4} + {\left (4 \, c^{2} d^{2} e^{4} - 4 \, b c d e^{5} + b^{2} e^{6}\right )} x^{2} + 2 \, {\left (4 \, c^{2} d^{3} e^{3} - 4 \, b c d^{2} e^{4} + b^{2} d e^{5}\right )} x\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*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((4*c*d*e - b*e^2)*f + 2*(c*d^2 - b*d*e)*g + (2*c*e^2*f + (4*c
*d*e - 3*b*e^2)*g)*x)/(4*c^2*d^4*e^2 - 4*b*c*d^3*e^3 + b^2*d^2*e^4 + (4*c^2*d^2*e^4 - 4*b*c*d*e^5 + b^2*e^6)*x
^2 + 2*(4*c^2*d^3*e^3 - 4*b*c*d^2*e^4 + b^2*d*e^5)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \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]

sage0*x

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maple [A]  time = 0.05, size = 127, normalized size = 0.93 \begin {gather*} -\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 ) \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}}\, e^{2}} \end {gather*}

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)

[Out]

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