3.20.89 \(\int \frac {(d+e x) (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=129 \[ \frac {2 (d+e x) (-b e g+c d g+c e f)}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {g \tan ^{-1}\left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{c^{3/2} e^2} \]

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Rubi [A]  time = 0.17, antiderivative size = 148, normalized size of antiderivative = 1.15, 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 = {777, 621, 204} \begin {gather*} \frac {2 (e x (2 c d-b e)+d (2 c d-b e)) (-b e g+c d g+c e f)}{c e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {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 )}{c^{3/2} e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*e*f + c*d*g - b*e*g)*(d*(2*c*d - b*e) + e*(2*c*d - b*e)*x))/(c*e^2*(2*c*d - b*e)^2*Sqrt[d*(c*d - b*e) -
b*e^2*x - c*e^2*x^2]) - (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])])/(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 777

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

Rubi steps

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

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Mathematica [A]  time = 0.44, size = 173, normalized size = 1.34 \begin {gather*} -\frac {2 \left (\sqrt {c} \sqrt {e} (d+e x) (-b e g+c d g+c e f)+g \sqrt {d+e x} \sqrt {e (2 c d-b e)} (b e-2 c d) \sqrt {\frac {b e-c d+c e x}{b e-2 c d}} \sin ^{-1}\left (\frac {\sqrt {c} \sqrt {e} \sqrt {d+e x}}{\sqrt {e (2 c d-b e)}}\right )\right )}{c^{3/2} e^{5/2} (b e-2 c d) \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 1.70, size = 267, normalized size = 2.07 \begin {gather*} -\frac {g \sqrt {-c e^2} \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 )}{2 c^2 e^3}-\frac {g \tan ^{-1}\left (\frac {2 \sqrt {c} x \sqrt {-c e^2}}{b e}-\frac {2 \sqrt {c} \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2}}{b e}\right )}{c^{3/2} e^2}+\frac {2 \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2} (-b e g+c d g+c e f)}{c e^2 (b e-2 c d) (b e-c d+c e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 483, normalized size = 3.74 \begin {gather*} \left [\frac {{\left ({\left (2 \, c^{2} d e - b c e^{2}\right )} g x - {\left (2 \, c^{2} d^{2} - 3 \, b c d e + 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 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (c^{2} e f + {\left (c^{2} d - b c e\right )} g\right )}}{2 \, {\left (2 \, c^{4} d^{2} e^{2} - 3 \, b c^{3} d e^{3} + b^{2} c^{2} e^{4} - {\left (2 \, c^{4} d e^{3} - b c^{3} e^{4}\right )} x\right )}}, -\frac {{\left ({\left (2 \, c^{2} d e - b c e^{2}\right )} g x - {\left (2 \, c^{2} d^{2} - 3 \, b c d e + 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 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (c^{2} e f + {\left (c^{2} d - b c e\right )} g\right )}}{2 \, c^{4} d^{2} e^{2} - 3 \, b c^{3} d e^{3} + b^{2} c^{2} e^{4} - {\left (2 \, c^{4} d e^{3} - b c^{3} e^{4}\right )} x}\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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.56, size = 282, normalized size = 2.19 \begin {gather*} -\frac {\sqrt {-c e^{2}} g 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 )}{c^{2}} - \frac {2 \, \sqrt {-c x^{2} e^{2} + c d^{2} - b x e^{2} - b d e} {\left (\frac {{\left (2 \, c^{2} d^{2} g e^{2} + 2 \, c^{2} d f e^{3} - 3 \, b c d g e^{3} - b c f e^{4} + b^{2} g e^{4}\right )} x}{4 \, c^{3} d^{2} e^{3} - 4 \, b c^{2} d e^{4} + b^{2} c e^{5}} + \frac {2 \, c^{2} d^{3} g e + 2 \, c^{2} d^{2} f e^{2} - 3 \, b c d^{2} g e^{2} - b c d f e^{3} + b^{2} d g e^{3}}{4 \, c^{3} d^{2} e^{3} - 4 \, b c^{2} d e^{4} + b^{2} c e^{5}}\right )}}{c x^{2} e^{2} - c d^{2} + b x e^{2} + b d e} \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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 710, normalized size = 5.50 \begin {gather*} -\frac {b^{2} e^{3} g x}{\left (-b^{2} e^{4}+4 b c d \,e^{3}-4 d^{2} e^{2} c^{2}\right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, c}+\frac {2 b d \,e^{2} g x}{\left (-b^{2} e^{4}+4 b c d \,e^{3}-4 d^{2} e^{2} c^{2}\right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}+\frac {2 b \,e^{3} f x}{\left (-b^{2} e^{4}+4 b c d \,e^{3}-4 d^{2} e^{2} c^{2}\right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}-\frac {b^{3} e^{3} g}{2 \left (-b^{2} e^{4}+4 b c d \,e^{3}-4 d^{2} e^{2} c^{2}\right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, c^{2}}+\frac {b^{2} d \,e^{2} g}{\left (-b^{2} e^{4}+4 b c d \,e^{3}-4 d^{2} e^{2} c^{2}\right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, c}+\frac {b^{2} e^{3} f}{\left (-b^{2} e^{4}+4 b c d \,e^{3}-4 d^{2} e^{2} c^{2}\right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, c}+\frac {2 \left (-2 c \,e^{2} x -b \,e^{2}\right ) d f}{\left (-b^{2} e^{4}-4 \left (-b d e +c \,d^{2}\right ) c \,e^{2}\right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}+\frac {g x}{\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, c e}-\frac {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 e}-\frac {b g}{2 \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, c^{2} e}+\frac {d g}{\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, c \,e^{2}}+\frac {f}{\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, 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)^(3/2),x)

[Out]

1/e*g*x/c/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)-1/2/e*g*b/c^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)-e^3*g*b^
2/c/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*x-1/2*e^3*g*b^3/c^2/(-b^2*e^4+
4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)-1/e*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))+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+2*b/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^
2)^(1/2)*e^2*x*d*g+2*b/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*e^3*x*f+b^2
/c/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*e^2*d*g+b^2/c/(-b^2*e^4+4*b*c*d
*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*e^3*f+2*d*f*(-2*c*e^2*x-b*e^2)/(-b^2*e^4-4*(-b*d*e+
c*d^2)*c*e^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((e*x+d)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2),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 = 4.04, size = 344, normalized size = 2.67 \begin {gather*} \frac {4\,c\,d^3\,g+2\,b\,d\,e^2\,f-4\,b\,d^2\,e\,g-2\,b\,d\,e^2\,g\,x+4\,c\,d\,e^2\,f\,x}{\left (b^2\,e^4+4\,c\,e^2\,\left (c\,d^2-b\,d\,e\right )\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}+\frac {e\,g\,\ln \left (b\,e^2-2\,\sqrt {-c\,e^2}\,\sqrt {-\left (d+e\,x\right )\,\left (b\,e-c\,d+c\,e\,x\right )}+2\,c\,e^2\,x\right )}{{\left (-c\,e^2\right )}^{3/2}}-\frac {e\,f\,\left (-4\,c\,d^2+4\,b\,d\,e+2\,b\,x\,e^2\right )}{\left (b^2\,e^4+4\,c\,e^2\,\left (c\,d^2-b\,d\,e\right )\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}+\frac {g\,\left (x\,\left (\frac {b^2\,e^4}{2}+c\,e^2\,\left (c\,d^2-b\,d\,e\right )\right )-\frac {b\,e^2\,\left (c\,d^2-b\,d\,e\right )}{2}\right )}{c\,e\,\left (\frac {b^2\,e^4}{4}+c\,e^2\,\left (c\,d^2-b\,d\,e\right )\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,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)^(3/2),x)

[Out]

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

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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 )}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}}}\, 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)**(3/2),x)

[Out]

Integral((d + e*x)*(f + g*x)/(-(d + e*x)*(b*e - c*d + c*e*x))**(3/2), x)

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