3.49 \(\int \frac{\sqrt{c+d x} (A+B x+C x^2)}{\sqrt{e+f x}} \, dx\)

Optimal. Leaf size=246 \[ \frac{\sqrt{c+d x} \sqrt{e+f x} \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{8 d^2 f^3}-\frac{(d e-c f) \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right ) \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{8 d^{5/2} f^{7/2}}-\frac{(c+d x)^{3/2} \sqrt{e+f x} (-6 B d f+7 c C f+5 C d e)}{12 d^2 f^2}+\frac{C (c+d x)^{5/2} \sqrt{e+f x}}{3 d^2 f} \]

[Out]

((C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*Sqrt[c + d*x]*Sqrt[e + f*x])/(8*d^2
*f^3) - ((5*C*d*e + 7*c*C*f - 6*B*d*f)*(c + d*x)^(3/2)*Sqrt[e + f*x])/(12*d^2*f^2) + (C*(c + d*x)^(5/2)*Sqrt[e
+ f*x])/(3*d^2*f) - ((d*e - c*f)*(C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*Ar
cTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(8*d^(5/2)*f^(7/2))

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Rubi [A]  time = 0.230315, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {951, 80, 50, 63, 217, 206} \[ \frac{\sqrt{c+d x} \sqrt{e+f x} \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{8 d^2 f^3}-\frac{(d e-c f) \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right ) \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{8 d^{5/2} f^{7/2}}-\frac{(c+d x)^{3/2} \sqrt{e+f x} (-6 B d f+7 c C f+5 C d e)}{12 d^2 f^2}+\frac{C (c+d x)^{5/2} \sqrt{e+f x}}{3 d^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c + d*x]*(A + B*x + C*x^2))/Sqrt[e + f*x],x]

[Out]

((C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*Sqrt[c + d*x]*Sqrt[e + f*x])/(8*d^2
*f^3) - ((5*C*d*e + 7*c*C*f - 6*B*d*f)*(c + d*x)^(3/2)*Sqrt[e + f*x])/(12*d^2*f^2) + (C*(c + d*x)^(5/2)*Sqrt[e
+ f*x])/(3*d^2*f) - ((d*e - c*f)*(C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*Ar
cTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(8*d^(5/2)*f^(7/2))

Rule 951

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

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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])

Rubi steps

\begin{align*} \int \frac{\sqrt{c+d x} \left (A+B x+C x^2\right )}{\sqrt{e+f x}} \, dx &=\frac{C (c+d x)^{5/2} \sqrt{e+f x}}{3 d^2 f}+\frac{\int \frac{\sqrt{c+d x} \left (\frac{1}{2} \left (-5 c C d e-c^2 C f+6 A d^2 f\right )-\frac{1}{2} d (5 C d e+7 c C f-6 B d f) x\right )}{\sqrt{e+f x}} \, dx}{3 d^2 f}\\ &=-\frac{(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt{e+f x}}{12 d^2 f^2}+\frac{C (c+d x)^{5/2} \sqrt{e+f x}}{3 d^2 f}+\frac{\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \int \frac{\sqrt{c+d x}}{\sqrt{e+f x}} \, dx}{8 d^2 f^2}\\ &=\frac{\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt{c+d x} \sqrt{e+f x}}{8 d^2 f^3}-\frac{(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt{e+f x}}{12 d^2 f^2}+\frac{C (c+d x)^{5/2} \sqrt{e+f x}}{3 d^2 f}-\frac{\left ((d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right )\right ) \int \frac{1}{\sqrt{c+d x} \sqrt{e+f x}} \, dx}{16 d^2 f^3}\\ &=\frac{\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt{c+d x} \sqrt{e+f x}}{8 d^2 f^3}-\frac{(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt{e+f x}}{12 d^2 f^2}+\frac{C (c+d x)^{5/2} \sqrt{e+f x}}{3 d^2 f}-\frac{\left ((d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{e-\frac{c f}{d}+\frac{f x^2}{d}}} \, dx,x,\sqrt{c+d x}\right )}{8 d^3 f^3}\\ &=\frac{\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt{c+d x} \sqrt{e+f x}}{8 d^2 f^3}-\frac{(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt{e+f x}}{12 d^2 f^2}+\frac{C (c+d x)^{5/2} \sqrt{e+f x}}{3 d^2 f}-\frac{\left ((d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{f x^2}{d}} \, dx,x,\frac{\sqrt{c+d x}}{\sqrt{e+f x}}\right )}{8 d^3 f^3}\\ &=\frac{\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt{c+d x} \sqrt{e+f x}}{8 d^2 f^3}-\frac{(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt{e+f x}}{12 d^2 f^2}+\frac{C (c+d x)^{5/2} \sqrt{e+f x}}{3 d^2 f}-\frac{(d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right )}{8 d^{5/2} f^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.06982, size = 225, normalized size = 0.91 \[ \frac{-d \sqrt{f} \sqrt{c+d x} (e+f x) \left (C \left (3 c^2 f^2-2 c d f (f x-2 e)+d^2 \left (-15 e^2+10 e f x-8 f^2 x^2\right )\right )-6 d f (4 A d f+B (c f-3 d e+2 d f x))\right )-3 (d e-c f)^{3/2} \sqrt{\frac{d (e+f x)}{d e-c f}} \sinh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d e-c f}}\right ) \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{24 d^3 f^{7/2} \sqrt{e+f x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[c + d*x]*(A + B*x + C*x^2))/Sqrt[e + f*x],x]

[Out]

(-(d*Sqrt[f]*Sqrt[c + d*x]*(e + f*x)*(-6*d*f*(4*A*d*f + B*(-3*d*e + c*f + 2*d*f*x)) + C*(3*c^2*f^2 - 2*c*d*f*(
-2*e + f*x) + d^2*(-15*e^2 + 10*e*f*x - 8*f^2*x^2)))) - 3*(d*e - c*f)^(3/2)*(C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^
2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*Sqrt[(d*(e + f*x))/(d*e - c*f)]*ArcSinh[(Sqrt[f]*Sqrt[c + d*x])/Sqrt[d
*e - c*f]])/(24*d^3*f^(7/2)*Sqrt[e + f*x])

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Maple [B]  time = 0.018, size = 763, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)*(d*x+c)^(1/2)/(f*x+e)^(1/2),x)

[Out]

1/48*(d*x+c)^(1/2)*(f*x+e)^(1/2)*(16*C*x^2*d^2*f^2*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)+24*A*ln(1/2*(2*d*f*x+2*
((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c*d^2*f^3-24*A*ln(1/2*(2*d*f*x+2*((d*x+c)*(f*x+e))^(
1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*d^3*e*f^2-6*B*ln(1/2*(2*d*f*x+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+c*f
+d*e)/(d*f)^(1/2))*c^2*d*f^3-12*B*ln(1/2*(2*d*f*x+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*
c*d^2*e*f^2+18*B*ln(1/2*(2*d*f*x+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*d^3*e^2*f+24*B*(d
*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*x*d^2*f^2+3*C*ln(1/2*(2*d*f*x+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+c*f+d*e)
/(d*f)^(1/2))*c^3*f^3+3*C*ln(1/2*(2*d*f*x+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c^2*d*e*
f^2+9*C*ln(1/2*(2*d*f*x+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c*d^2*e^2*f-15*C*ln(1/2*(2
*d*f*x+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*d^3*e^3+4*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(
1/2)*x*c*d*f^2-20*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*x*d^2*e*f+48*A*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d^2
*f^2+12*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*c*d*f^2-36*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d^2*e*f-6*C*(d*
f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*c^2*f^2-8*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*c*d*e*f+30*C*(d*f)^(1/2)*((d*
x+c)*(f*x+e))^(1/2)*d^2*e^2)/f^3/((d*x+c)*(f*x+e))^(1/2)/d^2/(d*f)^(1/2)

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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((C*x^2+B*x+A)*(d*x+c)^(1/2)/(f*x+e)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.5108, size = 1277, normalized size = 5.19 \begin{align*} \left [-\frac{3 \,{\left (5 \, C d^{3} e^{3} - 3 \,{\left (C c d^{2} + 2 \, B d^{3}\right )} e^{2} f -{\left (C c^{2} d - 4 \, B c d^{2} - 8 \, A d^{3}\right )} e f^{2} -{\left (C c^{3} - 2 \, B c^{2} d + 8 \, A c d^{2}\right )} f^{3}\right )} \sqrt{d f} \log \left (8 \, d^{2} f^{2} x^{2} + d^{2} e^{2} + 6 \, c d e f + c^{2} f^{2} + 4 \,{\left (2 \, d f x + d e + c f\right )} \sqrt{d f} \sqrt{d x + c} \sqrt{f x + e} + 8 \,{\left (d^{2} e f + c d f^{2}\right )} x\right ) - 4 \,{\left (8 \, C d^{3} f^{3} x^{2} + 15 \, C d^{3} e^{2} f - 2 \,{\left (2 \, C c d^{2} + 9 \, B d^{3}\right )} e f^{2} - 3 \,{\left (C c^{2} d - 2 \, B c d^{2} - 8 \, A d^{3}\right )} f^{3} - 2 \,{\left (5 \, C d^{3} e f^{2} -{\left (C c d^{2} + 6 \, B d^{3}\right )} f^{3}\right )} x\right )} \sqrt{d x + c} \sqrt{f x + e}}{96 \, d^{3} f^{4}}, \frac{3 \,{\left (5 \, C d^{3} e^{3} - 3 \,{\left (C c d^{2} + 2 \, B d^{3}\right )} e^{2} f -{\left (C c^{2} d - 4 \, B c d^{2} - 8 \, A d^{3}\right )} e f^{2} -{\left (C c^{3} - 2 \, B c^{2} d + 8 \, A c d^{2}\right )} f^{3}\right )} \sqrt{-d f} \arctan \left (\frac{{\left (2 \, d f x + d e + c f\right )} \sqrt{-d f} \sqrt{d x + c} \sqrt{f x + e}}{2 \,{\left (d^{2} f^{2} x^{2} + c d e f +{\left (d^{2} e f + c d f^{2}\right )} x\right )}}\right ) + 2 \,{\left (8 \, C d^{3} f^{3} x^{2} + 15 \, C d^{3} e^{2} f - 2 \,{\left (2 \, C c d^{2} + 9 \, B d^{3}\right )} e f^{2} - 3 \,{\left (C c^{2} d - 2 \, B c d^{2} - 8 \, A d^{3}\right )} f^{3} - 2 \,{\left (5 \, C d^{3} e f^{2} -{\left (C c d^{2} + 6 \, B d^{3}\right )} f^{3}\right )} x\right )} \sqrt{d x + c} \sqrt{f x + e}}{48 \, d^{3} f^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(d*x+c)^(1/2)/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(5*C*d^3*e^3 - 3*(C*c*d^2 + 2*B*d^3)*e^2*f - (C*c^2*d - 4*B*c*d^2 - 8*A*d^3)*e*f^2 - (C*c^3 - 2*B*c^
2*d + 8*A*c*d^2)*f^3)*sqrt(d*f)*log(8*d^2*f^2*x^2 + d^2*e^2 + 6*c*d*e*f + c^2*f^2 + 4*(2*d*f*x + d*e + c*f)*sq
rt(d*f)*sqrt(d*x + c)*sqrt(f*x + e) + 8*(d^2*e*f + c*d*f^2)*x) - 4*(8*C*d^3*f^3*x^2 + 15*C*d^3*e^2*f - 2*(2*C*
c*d^2 + 9*B*d^3)*e*f^2 - 3*(C*c^2*d - 2*B*c*d^2 - 8*A*d^3)*f^3 - 2*(5*C*d^3*e*f^2 - (C*c*d^2 + 6*B*d^3)*f^3)*x
)*sqrt(d*x + c)*sqrt(f*x + e))/(d^3*f^4), 1/48*(3*(5*C*d^3*e^3 - 3*(C*c*d^2 + 2*B*d^3)*e^2*f - (C*c^2*d - 4*B*
c*d^2 - 8*A*d^3)*e*f^2 - (C*c^3 - 2*B*c^2*d + 8*A*c*d^2)*f^3)*sqrt(-d*f)*arctan(1/2*(2*d*f*x + d*e + c*f)*sqrt
(-d*f)*sqrt(d*x + c)*sqrt(f*x + e)/(d^2*f^2*x^2 + c*d*e*f + (d^2*e*f + c*d*f^2)*x)) + 2*(8*C*d^3*f^3*x^2 + 15*
C*d^3*e^2*f - 2*(2*C*c*d^2 + 9*B*d^3)*e*f^2 - 3*(C*c^2*d - 2*B*c*d^2 - 8*A*d^3)*f^3 - 2*(5*C*d^3*e*f^2 - (C*c*
d^2 + 6*B*d^3)*f^3)*x)*sqrt(d*x + c)*sqrt(f*x + e))/(d^3*f^4)]

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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((C*x**2+B*x+A)*(d*x+c)**(1/2)/(f*x+e)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 2.66155, size = 425, normalized size = 1.73 \begin{align*} \frac{{\left (\sqrt{{\left (d x + c\right )} d f - c d f + d^{2} e} \sqrt{d x + c}{\left (2 \,{\left (d x + c\right )}{\left (\frac{4 \,{\left (d x + c\right )} C}{d^{3} f} - \frac{7 \, C c d^{6} f^{4} - 6 \, B d^{7} f^{4} + 5 \, C d^{7} f^{3} e}{d^{9} f^{5}}\right )} + \frac{3 \,{\left (C c^{2} d^{6} f^{4} - 2 \, B c d^{7} f^{4} + 8 \, A d^{8} f^{4} + 2 \, C c d^{7} f^{3} e - 6 \, B d^{8} f^{3} e + 5 \, C d^{8} f^{2} e^{2}\right )}}{d^{9} f^{5}}\right )} - \frac{3 \,{\left (C c^{3} f^{3} - 2 \, B c^{2} d f^{3} + 8 \, A c d^{2} f^{3} + C c^{2} d f^{2} e - 4 \, B c d^{2} f^{2} e - 8 \, A d^{3} f^{2} e + 3 \, C c d^{2} f e^{2} + 6 \, B d^{3} f e^{2} - 5 \, C d^{3} e^{3}\right )} \log \left ({\left | -\sqrt{d f} \sqrt{d x + c} + \sqrt{{\left (d x + c\right )} d f - c d f + d^{2} e} \right |}\right )}{\sqrt{d f} d^{2} f^{3}}\right )} d}{24 \,{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(d*x+c)^(1/2)/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

1/24*(sqrt((d*x + c)*d*f - c*d*f + d^2*e)*sqrt(d*x + c)*(2*(d*x + c)*(4*(d*x + c)*C/(d^3*f) - (7*C*c*d^6*f^4 -
6*B*d^7*f^4 + 5*C*d^7*f^3*e)/(d^9*f^5)) + 3*(C*c^2*d^6*f^4 - 2*B*c*d^7*f^4 + 8*A*d^8*f^4 + 2*C*c*d^7*f^3*e -
6*B*d^8*f^3*e + 5*C*d^8*f^2*e^2)/(d^9*f^5)) - 3*(C*c^3*f^3 - 2*B*c^2*d*f^3 + 8*A*c*d^2*f^3 + C*c^2*d*f^2*e - 4
*B*c*d^2*f^2*e - 8*A*d^3*f^2*e + 3*C*c*d^2*f*e^2 + 6*B*d^3*f*e^2 - 5*C*d^3*e^3)*log(abs(-sqrt(d*f)*sqrt(d*x +
c) + sqrt((d*x + c)*d*f - c*d*f + d^2*e)))/(sqrt(d*f)*d^2*f^3))*d/abs(d)