3.883 \(\int \frac{1}{(d+e x) (f+g x) (a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=352 \[ -\frac{2 e \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (e f-d g) \left (a e^2-b d e+c d^2\right )}+\frac{2 g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac{e^3 \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{(e f-d g) \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{g^3 \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{(e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.443355, antiderivative size = 352, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {960, 740, 12, 724, 206} \[ -\frac{2 e \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (e f-d g) \left (a e^2-b d e+c d^2\right )}+\frac{2 g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac{e^3 \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{(e f-d g) \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{g^3 \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{(e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(f + g*x)*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

Rule 960

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 740

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 724

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 1.61699, size = 317, normalized size = 0.9 \[ \frac{-\frac{2 e \left (-2 c (a e+c d x)+b^2 e+b c (e x-d)\right )}{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \left (e (b d-a e)-c d^2\right )}+\frac{2 g \left (-2 c (a g+c f x)+b^2 g+b c (g x-f)\right )}{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \left (g (b f-a g)-c f^2\right )}+\frac{e^3 \tanh ^{-1}\left (\frac{-2 a e+b (d-e x)+2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )}{\left (e (a e-b d)+c d^2\right )^{3/2}}-\frac{g^3 \tanh ^{-1}\left (\frac{-2 a g+b (f-g x)+2 c f x}{2 \sqrt{a+x (b+c x)} \sqrt{g (a g-b f)+c f^2}}\right )}{\left (g (a g-b f)+c f^2\right )^{3/2}}}{e f-d g} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(f + g*x)*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.453, size = 1343, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(g*x+f)/(c*x^2+b*x+a)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}{\left (e x + d\right )}{\left (g x + f\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(g*x+f)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x + a)^(3/2)*(e*x + d)*(g*x + f)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(g*x+f)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x\right ) \left (f + g x\right ) \left (a + b x + c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(g*x+f)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(1/((d + e*x)*(f + g*x)*(a + b*x + c*x**2)**(3/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(g*x+f)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError