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

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

________________________________________________________________________________________

Rubi [A]  time = 0.53, antiderivative size = 357, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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 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]

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 = 1.04, size = 373, normalized size = 1.04 \begin {gather*} \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 {(e f-d g)^3 \log (d+e x)}{e \left (e (a e-b d)+c d^2\right )^{3/2}}+\frac {(d g-e f)^3 \log \left (2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+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}} \end {gather*}

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 6.47, size = 437, normalized size = 1.22 \begin {gather*} -\frac {2 \left (-a^2 b e g^3-2 a^2 c d g^3+6 a^2 c e f g^2+2 a^2 c e g^3 x+a b^2 d g^3-a b^2 e g^3 x-3 a b c d f g^2-3 a b c d g^3 x-3 a b c e f^2 g+3 a b c e f g^2 x+6 a c^2 d f^2 g+6 a c^2 d f g^2 x-2 a c^2 e f^3-6 a c^2 e f^2 g x+b^3 d g^3 x-3 b^2 c d f g^2 x+b^2 c e f^3-b c^2 d f^3+3 b c^2 d f^2 g x+b c^2 e f^3 x-2 c^3 d f^3 x\right )}{c \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {g^3 \log \left (-2 c^{3/2} e \sqrt {a+b x+c x^2}+b c e+2 c^2 e x\right )}{c^{3/2} e}-\frac {2 \left (d^3 g^3-3 d^2 e f g^2+3 d e^2 f^2 g-e^3 f^3\right ) \tan ^{-1}\left (\frac {-e \sqrt {a+b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

________________________________________________________________________________________

maple [B]  time = 0.02, size = 3127, normalized size = 8.76 \begin {gather*} \text {output too large to display} \end {gather*}

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]

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

________________________________________________________________________________________

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((g*x+f)^3/(e*x+d)/(c*x^2+b*x+a)^(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)/e^2)^2>0)', see `
assume?` for more details)Is (b/e-(2*c*d)/e^2)^2    -(4*c       *((-(b*d)/e)        +(c*d^2)/e^2+a))     /e^2
zero or nonzero?

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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]

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

________________________________________________________________________________________