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

Optimal. Leaf size=787 \[ \frac{\sqrt{a+b x+c x^2} \left (-2 c e (5 b d-4 a e)+b^2 e^2-2 c e x (2 c d-b e)+8 c^2 d^2\right )}{8 c e (e f-d g)^2}-\frac{(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2} e^2 (e f-d g)^2}-\frac{e \sqrt{a+b x+c x^2} \left (-2 c g (5 b f-4 a g)+b^2 g^2-2 c g x (2 c f-b g)+8 c^2 f^2\right )}{8 c g^2 (e f-d g)^2}+\frac{e (2 c f-b g) \left (-4 c g (2 b f-3 a g)-b^2 g^2+8 c^2 f^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2} g^3 (e f-d g)^2}-\frac{3 \left (-4 c g (2 b f-a g)+b^2 g^2+8 c^2 f^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{c} g^3 (e f-d g)}+\frac{\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e^2 (e f-d g)^2}-\frac{e \left (a g^2-b f g+c f^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{g^3 (e f-d g)^2}+\frac{3 (2 c f-b g) \sqrt{a g^2-b f g+c f^2} \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{2 g^3 (e f-d g)}+\frac{3 \sqrt{a+b x+c x^2} (-3 b g+4 c f-2 c g x)}{4 g^2 (e f-d g)}+\frac{\left (a+b x+c x^2\right )^{3/2}}{(f+g x) (e f-d g)} \]

[Out]

((8*c^2*d^2 + b^2*e^2 - 2*c*e*(5*b*d - 4*a*e) - 2*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(8*c*e*(e*f - d*
g)^2) + (3*(4*c*f - 3*b*g - 2*c*g*x)*Sqrt[a + b*x + c*x^2])/(4*g^2*(e*f - d*g)) - (e*(8*c^2*f^2 + b^2*g^2 - 2*
c*g*(5*b*f - 4*a*g) - 2*c*g*(2*c*f - b*g)*x)*Sqrt[a + b*x + c*x^2])/(8*c*g^2*(e*f - d*g)^2) + (a + b*x + c*x^2
)^(3/2)/((e*f - d*g)*(f + g*x)) - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 - 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b + 2*
c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(3/2)*e^2*(e*f - d*g)^2) + (e*(2*c*f - b*g)*(8*c^2*f^2 - b^2*g^
2 - 4*c*g*(2*b*f - 3*a*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(3/2)*g^3*(e*f - d*g)
^2) - (3*(8*c^2*f^2 + b^2*g^2 - 4*c*g*(2*b*f - a*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(
8*Sqrt[c]*g^3*(e*f - d*g)) + ((c*d^2 - b*d*e + a*e^2)^(3/2)*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])])/(e^2*(e*f - d*g)^2) + (3*(2*c*f - b*g)*Sqrt[c*f^2 - b*f*g + a*g^
2]*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])])/(2*g^3*(e*f
- d*g)) - (e*(c*f^2 - b*f*g + a*g^2)^(3/2)*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])])/(g^3*(e*f - d*g)^2)

________________________________________________________________________________________

Rubi [A]  time = 1.3761, antiderivative size = 787, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {960, 734, 814, 843, 621, 206, 724, 732} \[ \frac{\sqrt{a+b x+c x^2} \left (-2 c e (5 b d-4 a e)+b^2 e^2-2 c e x (2 c d-b e)+8 c^2 d^2\right )}{8 c e (e f-d g)^2}-\frac{(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2} e^2 (e f-d g)^2}-\frac{e \sqrt{a+b x+c x^2} \left (-2 c g (5 b f-4 a g)+b^2 g^2-2 c g x (2 c f-b g)+8 c^2 f^2\right )}{8 c g^2 (e f-d g)^2}+\frac{e (2 c f-b g) \left (-4 c g (2 b f-3 a g)-b^2 g^2+8 c^2 f^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2} g^3 (e f-d g)^2}-\frac{3 \left (-4 c g (2 b f-a g)+b^2 g^2+8 c^2 f^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{c} g^3 (e f-d g)}+\frac{\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e^2 (e f-d g)^2}-\frac{e \left (a g^2-b f g+c f^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{g^3 (e f-d g)^2}+\frac{3 (2 c f-b g) \sqrt{a g^2-b f g+c f^2} \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{2 g^3 (e f-d g)}+\frac{3 \sqrt{a+b x+c x^2} (-3 b g+4 c f-2 c g x)}{4 g^2 (e f-d g)}+\frac{\left (a+b x+c x^2\right )^{3/2}}{(f+g x) (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*c^2*d^2 + b^2*e^2 - 2*c*e*(5*b*d - 4*a*e) - 2*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(8*c*e*(e*f - d*
g)^2) + (3*(4*c*f - 3*b*g - 2*c*g*x)*Sqrt[a + b*x + c*x^2])/(4*g^2*(e*f - d*g)) - (e*(8*c^2*f^2 + b^2*g^2 - 2*
c*g*(5*b*f - 4*a*g) - 2*c*g*(2*c*f - b*g)*x)*Sqrt[a + b*x + c*x^2])/(8*c*g^2*(e*f - d*g)^2) + (a + b*x + c*x^2
)^(3/2)/((e*f - d*g)*(f + g*x)) - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 - 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b + 2*
c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(3/2)*e^2*(e*f - d*g)^2) + (e*(2*c*f - b*g)*(8*c^2*f^2 - b^2*g^
2 - 4*c*g*(2*b*f - 3*a*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(3/2)*g^3*(e*f - d*g)
^2) - (3*(8*c^2*f^2 + b^2*g^2 - 4*c*g*(2*b*f - a*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(
8*Sqrt[c]*g^3*(e*f - d*g)) + ((c*d^2 - b*d*e + a*e^2)^(3/2)*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])])/(e^2*(e*f - d*g)^2) + (3*(2*c*f - b*g)*Sqrt[c*f^2 - b*f*g + a*g^
2]*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])])/(2*g^3*(e*f
- d*g)) - (e*(c*f^2 - b*f*g + a*g^2)^(3/2)*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])])/(g^3*(e*f - d*g)^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 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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]

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(d+e x) (f+g x)^2} \, dx &=\int \left (\frac{e^2 \left (a+b x+c x^2\right )^{3/2}}{(e f-d g)^2 (d+e x)}-\frac{g \left (a+b x+c x^2\right )^{3/2}}{(e f-d g) (f+g x)^2}-\frac{e g \left (a+b x+c x^2\right )^{3/2}}{(e f-d g)^2 (f+g x)}\right ) \, dx\\ &=\frac{e^2 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx}{(e f-d g)^2}-\frac{(e g) \int \frac{\left (a+b x+c x^2\right )^{3/2}}{f+g x} \, dx}{(e f-d g)^2}-\frac{g \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(f+g x)^2} \, dx}{e f-d g}\\ &=\frac{\left (a+b x+c x^2\right )^{3/2}}{(e f-d g) (f+g x)}-\frac{e \int \frac{(b d-2 a e+(2 c d-b e) x) \sqrt{a+b x+c x^2}}{d+e x} \, dx}{2 (e f-d g)^2}+\frac{e \int \frac{(b f-2 a g+(2 c f-b g) x) \sqrt{a+b x+c x^2}}{f+g x} \, dx}{2 (e f-d g)^2}-\frac{3 \int \frac{(b+2 c x) \sqrt{a+b x+c x^2}}{f+g x} \, dx}{2 (e f-d g)}\\ &=\frac{\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{8 c e (e f-d g)^2}+\frac{3 (4 c f-3 b g-2 c g x) \sqrt{a+b x+c x^2}}{4 g^2 (e f-d g)}-\frac{e \left (8 c^2 f^2+b^2 g^2-2 c g (5 b f-4 a g)-2 c g (2 c f-b g) x\right ) \sqrt{a+b x+c x^2}}{8 c g^2 (e f-d g)^2}+\frac{\left (a+b x+c x^2\right )^{3/2}}{(e f-d g) (f+g x)}+\frac{\int \frac{\frac{1}{2} \left (4 c e (b d-2 a e)^2-d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )-\frac{1}{2} (2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{8 c e (e f-d g)^2}-\frac{e \int \frac{\frac{1}{2} \left (4 c g (b f-2 a g)^2-f (2 c f-b g) \left (4 b c f-b^2 g-4 a c g\right )\right )-\frac{1}{2} (2 c f-b g) \left (8 c^2 f^2-b^2 g^2-4 c g (2 b f-3 a g)\right ) x}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{8 c g^2 (e f-d g)^2}+\frac{3 \int \frac{c \left (3 b^2 f g+4 a c f g-4 b \left (c f^2+a g^2\right )\right )-c \left (8 c^2 f^2+b^2 g^2-4 c g (2 b f-a g)\right ) x}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{8 c g^2 (e f-d g)}\\ &=\frac{\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{8 c e (e f-d g)^2}+\frac{3 (4 c f-3 b g-2 c g x) \sqrt{a+b x+c x^2}}{4 g^2 (e f-d g)}-\frac{e \left (8 c^2 f^2+b^2 g^2-2 c g (5 b f-4 a g)-2 c g (2 c f-b g) x\right ) \sqrt{a+b x+c x^2}}{8 c g^2 (e f-d g)^2}+\frac{\left (a+b x+c x^2\right )^{3/2}}{(e f-d g) (f+g x)}+\frac{\left (c d^2-b d e+a e^2\right )^2 \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{e^2 (e f-d g)^2}-\frac{\left ((2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{16 c e^2 (e f-d g)^2}+\frac{\left (3 (2 c f-b g) \left (c f^2-b f g+a g^2\right )\right ) \int \frac{1}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{2 g^3 (e f-d g)}-\frac{\left (e \left (c f^2-b f g+a g^2\right )^2\right ) \int \frac{1}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{g^3 (e f-d g)^2}+\frac{\left (e (2 c f-b g) \left (8 c^2 f^2-b^2 g^2-4 c g (2 b f-3 a g)\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{16 c g^3 (e f-d g)^2}-\frac{\left (3 \left (8 c^2 f^2+b^2 g^2-4 c g (2 b f-a g)\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{8 g^3 (e f-d g)}\\ &=\frac{\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{8 c e (e f-d g)^2}+\frac{3 (4 c f-3 b g-2 c g x) \sqrt{a+b x+c x^2}}{4 g^2 (e f-d g)}-\frac{e \left (8 c^2 f^2+b^2 g^2-2 c g (5 b f-4 a g)-2 c g (2 c f-b g) x\right ) \sqrt{a+b x+c x^2}}{8 c g^2 (e f-d g)^2}+\frac{\left (a+b x+c x^2\right )^{3/2}}{(e f-d g) (f+g x)}-\frac{\left (2 \left (c d^2-b d e+a e^2\right )^2\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^2 (e f-d g)^2}-\frac{\left ((2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right )\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 )}{8 c e^2 (e f-d g)^2}-\frac{\left (3 (2 c f-b g) \left (c f^2-b f g+a g^2\right )\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 )}{g^3 (e f-d g)}+\frac{\left (2 e \left (c f^2-b f g+a g^2\right )^2\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 )}{g^3 (e f-d g)^2}+\frac{\left (e (2 c f-b g) \left (8 c^2 f^2-b^2 g^2-4 c g (2 b f-3 a g)\right )\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 )}{8 c g^3 (e f-d g)^2}-\frac{\left (3 \left (8 c^2 f^2+b^2 g^2-4 c g (2 b f-a g)\right )\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 )}{4 g^3 (e f-d g)}\\ &=\frac{\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{8 c e (e f-d g)^2}+\frac{3 (4 c f-3 b g-2 c g x) \sqrt{a+b x+c x^2}}{4 g^2 (e f-d g)}-\frac{e \left (8 c^2 f^2+b^2 g^2-2 c g (5 b f-4 a g)-2 c g (2 c f-b g) x\right ) \sqrt{a+b x+c x^2}}{8 c g^2 (e f-d g)^2}+\frac{\left (a+b x+c x^2\right )^{3/2}}{(e f-d g) (f+g x)}-\frac{(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2} e^2 (e f-d g)^2}+\frac{e (2 c f-b g) \left (8 c^2 f^2-b^2 g^2-4 c g (2 b f-3 a g)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2} g^3 (e f-d g)^2}-\frac{3 \left (8 c^2 f^2+b^2 g^2-4 c g (2 b f-a g)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{c} g^3 (e f-d g)}+\frac{\left (c d^2-b d e+a e^2\right )^{3/2} \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^2 (e f-d g)^2}+\frac{3 (2 c f-b g) \sqrt{c f^2-b f g+a g^2} \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 )}{2 g^3 (e f-d g)}-\frac{e \left (c f^2-b f g+a g^2\right )^{3/2} \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 )}{g^3 (e f-d g)^2}\\ \end{align*}

Mathematica [A]  time = 1.5675, size = 357, normalized size = 0.45 \[ \frac{-2 g^3 (f+g x) \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )+e \left (2 g \sqrt{a+x (b+c x)} (d g-e f) (e g (b f-a g)+c d g (f+g x)-c e f (2 f+g x))-e (f+g x) \sqrt{g (a g-b f)+c f^2} (g (-2 a e g+3 b d g-b e f)+2 c f (2 e f-3 d g)) \tanh ^{-1}\left (\frac{2 a g-b f+b g x-2 c f x}{2 \sqrt{a+x (b+c x)} \sqrt{g (a g-b f)+c f^2}}\right )\right )-\sqrt{c} (f+g x) (e f-d g)^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) (-3 b e g+2 c d g+4 c e f)}{2 e^2 g^3 (f+g x) (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.408, size = 7959, normalized size = 10.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Exception raised: TypeError