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

Optimal. Leaf size=357 \[ \frac{2 \left (-x \left (c g^2 \left (-2 a^2 e g+3 a b d g-3 a b e f+3 b^2 d f\right )-b^2 g^3 (b d-a e)+c^2 f (6 a g (e f-d g)-b f (3 d g+e f))+2 c^3 d f^3\right )-b \left (a^2 e g^3+3 a c f g (d g+e f)+c^2 d f^3\right )+b^2 \left (a d g^3+c e f^3\right )-2 a c \left (c f^2 (e f-3 d g)-a g^2 (3 e f-d g)\right )\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac{g^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2} e}+\frac{(e f-d g)^3 \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e \left (a e^2-b d e+c d^2\right )^{3/2}} \]

[Out]

(2*(b^2*(c*e*f^3 + a*d*g^3) - 2*a*c*(c*f^2*(e*f - 3*d*g) - a*g^2*(3*e*f - d*g)) - b*(c^2*d*f^3 + a^2*e*g^3 + 3
*a*c*f*g*(e*f + d*g)) - (2*c^3*d*f^3 - b^2*(b*d - a*e)*g^3 + c*g^2*(3*b^2*d*f - 3*a*b*e*f + 3*a*b*d*g - 2*a^2*
e*g) + c^2*f*(6*a*g*(e*f - d*g) - b*f*(e*f + 3*d*g)))*x))/(c*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Sqrt[a + b*
x + c*x^2]) + (g^3*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(c^(3/2)*e) + ((e*f - d*g)^3*ArcTan
h[(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])])/(e*(c*d^2 - b*d*e +
a*e^2)^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.52805, antiderivative size = 357, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {1646, 843, 621, 206, 724} \[ \frac{2 \left (-x \left (c g^2 \left (-2 a^2 e g+3 a b d g-3 a b e f+3 b^2 d f\right )-b^2 g^3 (b d-a e)+c^2 f (6 a g (e f-d g)-b f (3 d g+e f))+2 c^3 d f^3\right )-b \left (a^2 e g^3+3 a c f g (d g+e f)+c^2 d f^3\right )+b^2 \left (a d g^3+c e f^3\right )-2 a c \left (c f^2 (e f-3 d g)-a g^2 (3 e f-d g)\right )\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac{g^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2} e}+\frac{(e f-d g)^3 \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e \left (a e^2-b d e+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b^2*(c*e*f^3 + a*d*g^3) - 2*a*c*(c*f^2*(e*f - 3*d*g) - a*g^2*(3*e*f - d*g)) - b*(c^2*d*f^3 + a^2*e*g^3 + 3
*a*c*f*g*(e*f + d*g)) - (2*c^3*d*f^3 - b^2*(b*d - a*e)*g^3 + c*g^2*(3*b^2*d*f - 3*a*b*e*f + 3*a*b*d*g - 2*a^2*
e*g) + c^2*f*(6*a*g*(e*f - d*g) - b*f*(e*f + 3*d*g)))*x))/(c*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Sqrt[a + b*
x + c*x^2]) + (g^3*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(c^(3/2)*e) + ((e*f - d*g)^3*ArcTan
h[(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])])/(e*(c*d^2 - b*d*e +
a*e^2)^(3/2))

Rule 1646

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2
*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 843

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

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]

Rubi steps

\begin{align*} \int \frac{(f+g x)^3}{(d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac{2 \left (b^2 \left (c e f^3+a d g^3\right )-2 a c \left (c f^2 (e f-3 d g)-a g^2 (3 e f-d g)\right )-b \left (c^2 d f^3+a^2 e g^3+3 a c f g (e f+d g)\right )-\left (2 c^3 d f^3-b^2 (b d-a e) g^3+c g^2 \left (3 b^2 d f-3 a b e f+3 a b d g-2 a^2 e g\right )+c^2 f (6 a g (e f-d g)-b f (e f+3 d g))\right ) x\right )}{c \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{a+b x+c x^2}}-\frac{2 \int \frac{\frac{\left (b^2-4 a c\right ) \left (d (b d-a e) g^3-c f \left (e^2 f^2-3 d e f g+3 d^2 g^2\right )\right )}{2 c \left (c d^2-b d e+a e^2\right )}-\frac{\left (b^2-4 a c\right ) g^3 x}{2 c}}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=\frac{2 \left (b^2 \left (c e f^3+a d g^3\right )-2 a c \left (c f^2 (e f-3 d g)-a g^2 (3 e f-d g)\right )-b \left (c^2 d f^3+a^2 e g^3+3 a c f g (e f+d g)\right )-\left (2 c^3 d f^3-b^2 (b d-a e) g^3+c g^2 \left (3 b^2 d f-3 a b e f+3 a b d g-2 a^2 e g\right )+c^2 f (6 a g (e f-d g)-b f (e f+3 d g))\right ) x\right )}{c \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{a+b x+c x^2}}+\frac{g^3 \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{c e}+\frac{(e f-d g)^3 \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{e \left (c d^2-b d e+a e^2\right )}\\ &=\frac{2 \left (b^2 \left (c e f^3+a d g^3\right )-2 a c \left (c f^2 (e f-3 d g)-a g^2 (3 e f-d g)\right )-b \left (c^2 d f^3+a^2 e g^3+3 a c f g (e f+d g)\right )-\left (2 c^3 d f^3-b^2 (b d-a e) g^3+c g^2 \left (3 b^2 d f-3 a b e f+3 a b d g-2 a^2 e g\right )+c^2 f (6 a g (e f-d g)-b f (e f+3 d g))\right ) x\right )}{c \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{a+b x+c x^2}}+\frac{\left (2 g^3\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{c e}-\frac{\left (2 (e f-d g)^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 )}{e \left (c d^2-b d e+a e^2\right )}\\ &=\frac{2 \left (b^2 \left (c e f^3+a d g^3\right )-2 a c \left (c f^2 (e f-3 d g)-a g^2 (3 e f-d g)\right )-b \left (c^2 d f^3+a^2 e g^3+3 a c f g (e f+d g)\right )-\left (2 c^3 d f^3-b^2 (b d-a e) g^3+c g^2 \left (3 b^2 d f-3 a b e f+3 a b d g-2 a^2 e g\right )+c^2 f (6 a g (e f-d g)-b f (e f+3 d g))\right ) x\right )}{c \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{a+b x+c x^2}}+\frac{g^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2} e}+\frac{(e f-d g)^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 )}{e \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.17417, size = 373, normalized size = 1.04 \[ \frac{2 \left (b \left (a^2 e g^3+3 a c g (d g (f+g x)+e f (f-g x))+c^2 f^2 (d (f-3 g x)-e f x)\right )+2 c \left (a^2 g^2 (d g-e (3 f+g x))+a c f (e f (f+3 g x)-3 d g (f+g x))+c^2 d f^3 x\right )+b^2 \left (a g^3 (e x-d)+c \left (3 d f g^2 x-e f^3\right )\right )-b^3 d g^3 x\right )}{c \left (4 a c-b^2\right ) \sqrt{a+x (b+c x)} \left (e (a e-b d)+c d^2\right )}+\frac{g^3 \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )}{c^{3/2} e}+\frac{(d g-e f)^3 \log \left (2 \sqrt{a+x (b+c x)} \sqrt{a e^2-b d e+c d^2}+2 a e-b d+b e x-2 c d x\right )}{e \left (e (a e-b d)+c d^2\right )^{3/2}}+\frac{(e f-d g)^3 \log (d+e x)}{e \left (e (a e-b d)+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.317, size = 3127, normalized size = 8.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*g^3/e^2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d-6*g^2/e*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*f-6/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^2*f^2*g+2/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*d^3*g^3+12
/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*c^2*d
^3*f*g^2-12/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^2*f^2*g+6/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*c*d^3*f*g^2-6/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*b*c*d^2*f*g^2+e/(a*e^2-b*d*e+c*d^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)*f^3+4/(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*f^3+1/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*d^3*g^3+3/(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*d*f^2*g+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*c*d*f^3-3/e/(a*e^2-b*d*e+c*d^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))*d^2*f*g^2+g^3/e*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+6/(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*d*f^2*g-3*g^2/e/c/(c*x^2+b*x+a)^(1
/2)*f+1/2*g^3/e*b/c^2/(c*x^2+b*x+a)^(1/2)-2*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*b*c*f^3-4/e^3/(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^4*g^3-3/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^2*d^2*f*g^2-2/e^3/(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^4*g^3-12*g^2/e^2*d*f/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)
*x*c+g^3/e/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-e/(a*e^2-b*d*e+c*d^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))*f^3+g^3/e^2/c/(c*x^2+b*x+a)^(1/2)*d-g^3/e*x/c/(c*x^
2+b*x+a)^(1/2)-1/e^2/(a*e^2-b*d*e+c*d^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^3
*g^3-3/(a*e^2-b*d*e+c*d^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*f^2*g+g^3/e^2*b
^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d+12*g/e*f^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*c-3*g^2/e*b^2/c/(4*a*c-b^2
)/(c*x^2+b*x+a)^(1/2)*f+4*g^3/e^3*d^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*c-6*g^2/e^2*d*f/(4*a*c-b^2)/(c*x^2+b*x
+a)^(1/2)*b+6*g/e*f^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*b+1/2*g^3/e*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+2*g^
3/e^3*d^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*b+3/(a*e^2-b*d*e+c*d^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))*d*f^2*g+3/e/(a*e^2-b*d*e+c*d^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^2*f*g^2-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^2*f^3+1/e^2/(a*e^2-b*d*e+c*d^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))*d^3*g^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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

[Out]

Timed out

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

[Out]

Exception raised: TypeError