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

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

[Out]

(c*x^2+b*x+a)^(3/2)/(-d*g+e*f)/(g*x+f)-1/16*(-b*e+2*c*d)*(8*c^2*d^2-b^2*e^2-4*c*e*(-3*a*e+2*b*d))*arctanh(1/2*
(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)/e^2/(-d*g+e*f)^2+1/16*e*(-b*g+2*c*f)*(8*c^2*f^2-b^2*g^2-4*c*g*(
-3*a*g+2*b*f))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)/g^3/(-d*g+e*f)^2+(a*e^2-b*d*e+c*d^2)
^(3/2)*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/e^2/(-d*g+e*f)^2-
e*(a*g^2-b*f*g+c*f^2)^(3/2)*arctanh(1/2*(b*f-2*a*g+(-b*g+2*c*f)*x)/(a*g^2-b*f*g+c*f^2)^(1/2)/(c*x^2+b*x+a)^(1/
2))/g^3/(-d*g+e*f)^2-3/8*(8*c^2*f^2+b^2*g^2-4*c*g*(-a*g+2*b*f))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1
/2))/g^3/(-d*g+e*f)/c^(1/2)+3/2*(-b*g+2*c*f)*arctanh(1/2*(b*f-2*a*g+(-b*g+2*c*f)*x)/(a*g^2-b*f*g+c*f^2)^(1/2)/
(c*x^2+b*x+a)^(1/2))*(a*g^2-b*f*g+c*f^2)^(1/2)/g^3/(-d*g+e*f)+1/8*(8*c^2*d^2+b^2*e^2-2*c*e*(-4*a*e+5*b*d)-2*c*
e*(-b*e+2*c*d)*x)*(c*x^2+b*x+a)^(1/2)/c/e/(-d*g+e*f)^2+3/4*(-2*c*g*x-3*b*g+4*c*f)*(c*x^2+b*x+a)^(1/2)/g^2/(-d*
g+e*f)-1/8*e*(8*c^2*f^2+b^2*g^2-2*c*g*(-4*a*g+5*b*f)-2*c*g*(-b*g+2*c*f)*x)*(c*x^2+b*x+a)^(1/2)/c/g^2/(-d*g+e*f
)^2

________________________________________________________________________________________

Rubi [A]  time = 1.38, antiderivative size = 787, normalized size of antiderivative = 1.00, 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 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 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 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]

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 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])

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 = 1.40, 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))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.02, size = 7959, normalized size = 10.11 \[ \text {output too large to display} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{{\left (e x + d\right )} {\left (g x + f\right )}^{2}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (f+g\,x\right )}^2\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right ) \left (f + g x\right )^{2}}\, dx \]

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]

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

________________________________________________________________________________________