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

Optimal. Leaf size=325 \[ -\frac {\sqrt {a+b x+c x^2} \left (b^2 e^2 g^2-2 c e g x (-b e g-2 c d g+4 c e f)-2 b c e g (2 e f-d g)-8 c^2 (e f-d g)^2\right )}{8 c^2 e^3}+\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (g \left (-4 c e (b d-a e)-b^2 e^2+8 c^2 d^2\right ) (-b e g-2 c d g+4 c e f)-4 c e (2 c d-b e) \left (2 c e f^2-b d g^2\right )\right )}{16 c^{5/2} e^4}+\frac {(e f-d g)^2 \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^4}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 c e} \]

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Rubi [A]  time = 0.71, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1653, 814, 843, 621, 206, 724} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (g \left (-4 c e (b d-a e)-b^2 e^2+8 c^2 d^2\right ) (-b e g-2 c d g+4 c e f)-4 c e (2 c d-b e) \left (2 c e f^2-b d g^2\right )\right )}{16 c^{5/2} e^4}-\frac {\sqrt {a+b x+c x^2} \left (b^2 e^2 g^2-2 c e g x (-b e g-2 c d g+4 c e f)-2 b c e g (2 e f-d g)-8 c^2 (e f-d g)^2\right )}{8 c^2 e^3}+\frac {(e f-d g)^2 \sqrt {a e^2-b d e+c d^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^4}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

\begin {align*} \int \frac {(f+g x)^2 \sqrt {a+b x+c x^2}}{d+e x} \, dx &=\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 c e}+\frac {\int \frac {\left (\frac {3}{2} e \left (2 c e f^2-b d g^2\right )+\frac {3}{2} e g (4 c e f-2 c d g-b e g) x\right ) \sqrt {a+b x+c x^2}}{d+e x} \, dx}{3 c e^2}\\ &=-\frac {\left (b^2 e^2 g^2-8 c^2 (e f-d g)^2-2 b c e g (2 e f-d g)-2 c e g (4 c e f-2 c d g-b e g) x\right ) \sqrt {a+b x+c x^2}}{8 c^2 e^3}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\int \frac {-\frac {3}{4} e \left (d \left (4 b c d-b^2 e-4 a c e\right ) g (4 c e f-2 c d g-b e g)-4 c e (b d-2 a e) \left (2 c e f^2-b d g^2\right )\right )-\frac {3}{4} e \left (\left (8 c^2 d^2-b^2 e^2-4 c e (b d-a e)\right ) g (4 c e f-2 c d g-b e g)-4 c e (2 c d-b e) \left (2 c e f^2-b d g^2\right )\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{12 c^2 e^4}\\ &=-\frac {\left (b^2 e^2 g^2-8 c^2 (e f-d g)^2-2 b c e g (2 e f-d g)-2 c e g (4 c e f-2 c d g-b e g) x\right ) \sqrt {a+b x+c x^2}}{8 c^2 e^3}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 c e}+\frac {\left (\left (c d^2-b d e+a e^2\right ) (e f-d g)^2\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e^4}+\frac {\left (\left (8 c^2 d^2-b^2 e^2-4 c e (b d-a e)\right ) g (4 c e f-2 c d g-b e g)-4 c e (2 c d-b e) \left (2 c e f^2-b d g^2\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^2 e^4}\\ &=-\frac {\left (b^2 e^2 g^2-8 c^2 (e f-d g)^2-2 b c e g (2 e f-d g)-2 c e g (4 c e f-2 c d g-b e g) x\right ) \sqrt {a+b x+c x^2}}{8 c^2 e^3}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\left (2 \left (c d^2-b d e+a e^2\right ) (e f-d g)^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^4}+\frac {\left (\left (8 c^2 d^2-b^2 e^2-4 c e (b d-a e)\right ) g (4 c e f-2 c d g-b e g)-4 c e (2 c d-b e) \left (2 c e f^2-b d g^2\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^2 e^4}\\ &=-\frac {\left (b^2 e^2 g^2-8 c^2 (e f-d g)^2-2 b c e g (2 e f-d g)-2 c e g (4 c e f-2 c d g-b e g) x\right ) \sqrt {a+b x+c x^2}}{8 c^2 e^3}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 c e}+\frac {\left (\left (8 c^2 d^2-b^2 e^2-4 c e (b d-a e)\right ) g (4 c e f-2 c d g-b e g)-4 c e (2 c d-b e) \left (2 c e f^2-b d g^2\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} e^4}+\frac {\sqrt {c d^2-b d e+a e^2} (e f-d g)^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^4}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 372, normalized size = 1.14 \begin {gather*} \frac {-\frac {6 e g \left (b^2-4 a c\right ) (e f-d g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{c^{3/2}}-\frac {3 e^2 g (b g-2 c f) \left (2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{c^{5/2}}+\frac {24 (e f-d g)^2 \left (2 \sqrt {c} \sqrt {e (a e-b d)+c d^2} \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 )+(b e-2 c d) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{\sqrt {c} e}+\frac {12 e g (b+2 c x) \sqrt {a+x (b+c x)} (e f-d g)}{c}+48 \sqrt {a+x (b+c x)} (e f-d g)^2+\frac {16 e^2 g^2 (a+x (b+c x))^{3/2}}{c}}{48 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(48*(e*f - d*g)^2*Sqrt[a + x*(b + c*x)] + (12*e*g*(e*f - d*g)*(b + 2*c*x)*Sqrt[a + x*(b + c*x)])/c + (16*e^2*g
^2*(a + x*(b + c*x))^(3/2))/c - (6*(b^2 - 4*a*c)*e*g*(e*f - d*g)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b
+ c*x)])])/c^(3/2) - (3*e^2*g*(-2*c*f + b*g)*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)] - (b^2 - 4*a*c)*ArcT
anh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/c^(5/2) + (24*(e*f - d*g)^2*((-2*c*d + b*e)*ArcTanh[(b +
2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + 2*Sqrt[c]*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*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)])]))/(Sqrt[c]*e))/(48*e^3)

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IntegrateAlgebraic [A]  time = 2.03, size = 428, normalized size = 1.32 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (8 a c e^2 g^2-3 b^2 e^2 g^2-6 b c d e g^2+12 b c e^2 f g+2 b c e^2 g^2 x+24 c^2 d^2 g^2-48 c^2 d e f g-12 c^2 d e g^2 x+24 c^2 e^2 f^2+24 c^2 e^2 f g x+8 c^2 e^2 g^2 x^2\right )}{24 c^2 e^3}+\frac {\log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right ) \left (4 a b c e^3 g^2+8 a c^2 d e^2 g^2-16 a c^2 e^3 f g-b^3 e^3 g^2-2 b^2 c d e^2 g^2+4 b^2 c e^3 f g-8 b c^2 d^2 e g^2+16 b c^2 d e^2 f g-8 b c^2 e^3 f^2+16 c^3 d^3 g^2-32 c^3 d^2 e f g+16 c^3 d e^2 f^2\right )}{16 c^{5/2} e^4}+\frac {2 \left (d^2 g^2-2 d e f g+e^2 f^2\right ) \sqrt {-a e^2+b d e-c d^2} \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^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(24*c^2*e^2*f^2 - 48*c^2*d*e*f*g + 12*b*c*e^2*f*g + 24*c^2*d^2*g^2 - 6*b*c*d*e*g^2 - 3*
b^2*e^2*g^2 + 8*a*c*e^2*g^2 + 24*c^2*e^2*f*g*x - 12*c^2*d*e*g^2*x + 2*b*c*e^2*g^2*x + 8*c^2*e^2*g^2*x^2))/(24*
c^2*e^3) + (2*Sqrt[-(c*d^2) + b*d*e - a*e^2]*(e^2*f^2 - 2*d*e*f*g + d^2*g^2)*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^4 + ((16*c^3*d*e^2*f^2 - 8*b*c^2*e^3*f^2 - 32*c^3
*d^2*e*f*g + 16*b*c^2*d*e^2*f*g + 4*b^2*c*e^3*f*g - 16*a*c^2*e^3*f*g + 16*c^3*d^3*g^2 - 8*b*c^2*d^2*e*g^2 - 2*
b^2*c*d*e^2*g^2 + 8*a*c^2*d*e^2*g^2 - b^3*e^3*g^2 + 4*a*b*c*e^3*g^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x +
c*x^2]])/(16*c^(5/2)*e^4)

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

[Out]

Timed out

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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)^2*(c*x^2+b*x+a)^(1/2)/(e*x+d),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

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maple [B]  time = 0.02, size = 2602, normalized size = 8.01

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2*(c*x^2+b*x+a)^(1/2)/(e*x+d),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>0)', see `assume?` f
or more details)Is b*e-2*c*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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