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

Optimal. Leaf size=371 \[ -\frac{\sqrt{c+d x} \sqrt{e+f x} \left (8 a^2 C d^2 f^2+2 b d f x (2 a C d f-b (6 B d f-5 C (c f+d e)))-6 a b d f (4 B d f-3 C (c f+d e))+b^2 \left (-\left (6 d f (4 A d f-3 B (c f+d e))+C \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )\right )\right )}{24 b d^3 f^3}+\frac{\tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right ) \left (2 a d f \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )-b \left (2 d f \left (4 A d f (c f+d e)-B \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+C \left (3 c^2 d e f^2+5 c^3 f^3+3 c d^2 e^2 f+5 d^3 e^3\right )\right )\right )}{8 d^{7/2} f^{7/2}}+\frac{C (a+b x)^2 \sqrt{c+d x} \sqrt{e+f x}}{3 b d f} \]

[Out]

(C*(a + b*x)^2*Sqrt[c + d*x]*Sqrt[e + f*x])/(3*b*d*f) - (Sqrt[c + d*x]*Sqrt[e + f*x]*(8*a^2*C*d^2*f^2 - 6*a*b*
d*f*(4*B*d*f - 3*C*(d*e + c*f)) - b^2*(C*(15*d^2*e^2 + 14*c*d*e*f + 15*c^2*f^2) + 6*d*f*(4*A*d*f - 3*B*(d*e +
c*f))) + 2*b*d*f*(2*a*C*d*f - b*(6*B*d*f - 5*C*(d*e + c*f)))*x))/(24*b*d^3*f^3) + ((2*a*d*f*(C*(3*d^2*e^2 + 2*
c*d*e*f + 3*c^2*f^2) + 4*d*f*(2*A*d*f - B*(d*e + c*f))) - b*(C*(5*d^3*e^3 + 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 + 5*
c^3*f^3) + 2*d*f*(4*A*d*f*(d*e + c*f) - B*(3*d^2*e^2 + 2*c*d*e*f + 3*c^2*f^2))))*ArcTanh[(Sqrt[f]*Sqrt[c + d*x
])/(Sqrt[d]*Sqrt[e + f*x])])/(8*d^(7/2)*f^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.5094, antiderivative size = 369, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {1615, 147, 63, 217, 206} \[ -\frac{\sqrt{c+d x} \sqrt{e+f x} \left (8 a^2 C d^2 f^2-2 b d f x (-2 a C d f+6 b B d f-5 b C (c f+d e))-6 a b d f (4 B d f-3 C (c f+d e))+b^2 \left (-\left (6 d f (4 A d f-3 B (c f+d e))+C \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )\right )\right )}{24 b d^3 f^3}+\frac{\tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right ) \left (2 a d f \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )-b \left (2 d f \left (4 A d f (c f+d e)-B \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+C \left (3 c^2 d e f^2+5 c^3 f^3+3 c d^2 e^2 f+5 d^3 e^3\right )\right )\right )}{8 d^{7/2} f^{7/2}}+\frac{C (a+b x)^2 \sqrt{c+d x} \sqrt{e+f x}}{3 b d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(a + b*x)^2*Sqrt[c + d*x]*Sqrt[e + f*x])/(3*b*d*f) - (Sqrt[c + d*x]*Sqrt[e + f*x]*(8*a^2*C*d^2*f^2 - 6*a*b*
d*f*(4*B*d*f - 3*C*(d*e + c*f)) - b^2*(C*(15*d^2*e^2 + 14*c*d*e*f + 15*c^2*f^2) + 6*d*f*(4*A*d*f - 3*B*(d*e +
c*f))) - 2*b*d*f*(6*b*B*d*f - 2*a*C*d*f - 5*b*C*(d*e + c*f))*x))/(24*b*d^3*f^3) + ((2*a*d*f*(C*(3*d^2*e^2 + 2*
c*d*e*f + 3*c^2*f^2) + 4*d*f*(2*A*d*f - B*(d*e + c*f))) - b*(C*(5*d^3*e^3 + 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 + 5*
c^3*f^3) + 2*d*f*(4*A*d*f*(d*e + c*f) - B*(3*d^2*e^2 + 2*c*d*e*f + 3*c^2*f^2))))*ArcTanh[(Sqrt[f]*Sqrt[c + d*x
])/(Sqrt[d]*Sqrt[e + f*x])])/(8*d^(7/2)*f^(7/2))

Rule 1615

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[
{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[(k*(a + b*x)^(m + q - 1)*(c + d*x)^(n + 1)*(e + f*x)^
(p + 1))/(d*f*b^(q - 1)*(m + n + p + q + 1)), x] + Dist[1/(d*f*b^q*(m + n + p + q + 1)), Int[(a + b*x)^m*(c +
d*x)^n*(e + f*x)^p*ExpandToSum[d*f*b^q*(m + n + p + q + 1)*Px - d*f*k*(m + n + p + q + 1)*(a + b*x)^q + k*(a +
b*x)^(q - 2)*(a^2*d*f*(m + n + p + q + 1) - b*(b*c*e*(m + q - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*
(2*(m + q) + n + p) - b*(d*e*(m + q + n) + c*f*(m + q + p)))*x), x], x], x] /; NeQ[m + n + p + q + 1, 0]] /; F
reeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && IntegersQ[2*m, 2*n, 2*p]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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{(a+b x) \left (A+B x+C x^2\right )}{\sqrt{c+d x} \sqrt{e+f x}} \, dx &=\frac{C (a+b x)^2 \sqrt{c+d x} \sqrt{e+f x}}{3 b d f}+\frac{\int \frac{(a+b x) \left (-\frac{1}{2} b (4 b c C e+a C d e+a c C f-6 A b d f)+\frac{1}{2} b (6 b B d f-2 a C d f-5 b C (d e+c f)) x\right )}{\sqrt{c+d x} \sqrt{e+f x}} \, dx}{3 b^2 d f}\\ &=\frac{C (a+b x)^2 \sqrt{c+d x} \sqrt{e+f x}}{3 b d f}-\frac{\sqrt{c+d x} \sqrt{e+f x} \left (8 a^2 C d^2 f^2-6 a b d f (4 B d f-3 C (d e+c f))-b^2 \left (C \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )+6 d f (4 A d f-3 B (d e+c f))\right )-2 b d f (6 b B d f-2 a C d f-5 b C (d e+c f)) x\right )}{24 b d^3 f^3}+\frac{\left (2 a d f \left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right )-b \left (C \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )+2 d f \left (4 A d f (d e+c f)-B \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )\right )\right )\right ) \int \frac{1}{\sqrt{c+d x} \sqrt{e+f x}} \, dx}{16 d^3 f^3}\\ &=\frac{C (a+b x)^2 \sqrt{c+d x} \sqrt{e+f x}}{3 b d f}-\frac{\sqrt{c+d x} \sqrt{e+f x} \left (8 a^2 C d^2 f^2-6 a b d f (4 B d f-3 C (d e+c f))-b^2 \left (C \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )+6 d f (4 A d f-3 B (d e+c f))\right )-2 b d f (6 b B d f-2 a C d f-5 b C (d e+c f)) x\right )}{24 b d^3 f^3}+\frac{\left (2 a d f \left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right )-b \left (C \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )+2 d f \left (4 A d f (d e+c f)-B \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )\right )\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^4 f^3}\\ &=\frac{C (a+b x)^2 \sqrt{c+d x} \sqrt{e+f x}}{3 b d f}-\frac{\sqrt{c+d x} \sqrt{e+f x} \left (8 a^2 C d^2 f^2-6 a b d f (4 B d f-3 C (d e+c f))-b^2 \left (C \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )+6 d f (4 A d f-3 B (d e+c f))\right )-2 b d f (6 b B d f-2 a C d f-5 b C (d e+c f)) x\right )}{24 b d^3 f^3}+\frac{\left (2 a d f \left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right )-b \left (C \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )+2 d f \left (4 A d f (d e+c f)-B \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )\right )\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^4 f^3}\\ &=\frac{C (a+b x)^2 \sqrt{c+d x} \sqrt{e+f x}}{3 b d f}-\frac{\sqrt{c+d x} \sqrt{e+f x} \left (8 a^2 C d^2 f^2-6 a b d f (4 B d f-3 C (d e+c f))-b^2 \left (C \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )+6 d f (4 A d f-3 B (d e+c f))\right )-2 b d f (6 b B d f-2 a C d f-5 b C (d e+c f)) x\right )}{24 b d^3 f^3}+\frac{\left (2 a d f \left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right )-b \left (C \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )+2 d f \left (4 A d f (d e+c f)-B \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right )}{8 d^{7/2} f^{7/2}}\\ \end{align*}

Mathematica [A]  time = 2.01816, size = 379, normalized size = 1.02 \[ \frac{\sqrt{e+f x} \left (3 \sqrt{d e-c f} \sinh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d e-c f}}\right ) \left (b \left (2 d f \left (4 A d f (c f+d e)-B \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+C \left (3 c^2 d e f^2+5 c^3 f^3+3 c d^2 e^2 f+5 d^3 e^3\right )\right )-2 a d f \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )\right )-\frac{d \sqrt{f} \sqrt{c+d x} (e+f x) \left (6 a d f (4 B d f+C (-3 c f-3 d e+2 d f x))+b \left (6 d f (4 A d f+B (-3 c f-3 d e+2 d f x))+C \left (15 c^2 f^2+2 c d f (7 e-5 f x)+d^2 \left (15 e^2-10 e f x+8 f^2 x^2\right )\right )\right )\right )}{\sqrt{\frac{d (e+f x)}{d e-c f}}}\right )}{24 d^3 f^{7/2} (c f-d e) \sqrt{\frac{d (e+f x)}{d e-c f}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.03, size = 1199, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(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))*b*d^3*e^2*f+18*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))*a*c^2*d*f^3+18*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))*a*d^3*e^2*f+48*A*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b*d^
2*f^2+48*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*a*d^2*f^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))*b*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))*b*d^3*e*f^2-24*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))*a*c*d^
2*f^3-24*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))*a*d^3*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))*b*c^2*d*f^3+30*C*(d*f)^(1/2)*((d*x+c)*(f
*x+e))^(1/2)*b*c^2*f^2+30*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b*d^2*e^2+28*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(
1/2)*b*c*d*e*f-20*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*x*b*c*d*f^2-20*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*x
*b*d^2*e*f+48*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))*a*d^3*f^3-36*B*(d*
f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b*c*d*f^2-36*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b*d^2*e*f-36*C*(d*f)^(1/2)
*((d*x+c)*(f*x+e))^(1/2)*a*c*d*f^2-36*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*a*d^2*e*f+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))*b*c*d^2*e*f^2+12*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))*a*c*d^2*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))*b*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))*b*c*d^2*e^2*f+24*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*x*b*d^2*f^2+24*C*(d*f)^(1/2)*((d*x+c)*(f*x+e)
)^(1/2)*x*a*d^2*f^2-15*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))*b*c^3*f^3
-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))*b*d^3*e^3+16*C*x^2*b*d^2*f^2
*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))*(d*x+c)^(1/2)*(f*x+e)^(1/2)/f^3/d^3/(d*f)^(1/2)/((d*x+c)*(f*x+e))^(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((b*x+a)*(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 = 4.98089, size = 1631, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(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*b*d^3*e^3 + 3*(C*b*c*d^2 - 2*(C*a + B*b)*d^3)*e^2*f + (3*C*b*c^2*d - 4*(C*a + B*b)*c*d^2 + 8*(B
*a + A*b)*d^3)*e*f^2 + (5*C*b*c^3 - 16*A*a*d^3 - 6*(C*a + B*b)*c^2*d + 8*(B*a + A*b)*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)*sqrt(d*f)*sqrt(d*x + c)*sqrt(f*x + e)
+ 8*(d^2*e*f + c*d*f^2)*x) - 4*(8*C*b*d^3*f^3*x^2 + 15*C*b*d^3*e^2*f + 2*(7*C*b*c*d^2 - 9*(C*a + B*b)*d^3)*e*
f^2 + 3*(5*C*b*c^2*d - 6*(C*a + B*b)*c*d^2 + 8*(B*a + A*b)*d^3)*f^3 - 2*(5*C*b*d^3*e*f^2 + (5*C*b*c*d^2 - 6*(C
*a + B*b)*d^3)*f^3)*x)*sqrt(d*x + c)*sqrt(f*x + e))/(d^4*f^4), 1/48*(3*(5*C*b*d^3*e^3 + 3*(C*b*c*d^2 - 2*(C*a
+ B*b)*d^3)*e^2*f + (3*C*b*c^2*d - 4*(C*a + B*b)*c*d^2 + 8*(B*a + A*b)*d^3)*e*f^2 + (5*C*b*c^3 - 16*A*a*d^3 -
6*(C*a + B*b)*c^2*d + 8*(B*a + A*b)*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*b*d^3*f^3*x^2 + 15*C*b*d^3*e^2*
f + 2*(7*C*b*c*d^2 - 9*(C*a + B*b)*d^3)*e*f^2 + 3*(5*C*b*c^2*d - 6*(C*a + B*b)*c*d^2 + 8*(B*a + A*b)*d^3)*f^3
- 2*(5*C*b*d^3*e*f^2 + (5*C*b*c*d^2 - 6*(C*a + B*b)*d^3)*f^3)*x)*sqrt(d*x + c)*sqrt(f*x + e))/(d^4*f^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right ) \left (A + B x + C x^{2}\right )}{\sqrt{c + d x} \sqrt{e + f x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 3.43927, size = 603, normalized size = 1.63 \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 b}{d^{4} f} - \frac{13 \, C b c d^{11} f^{4} - 6 \, C a d^{12} f^{4} - 6 \, B b d^{12} f^{4} + 5 \, C b d^{12} f^{3} e}{d^{15} f^{5}}\right )} + \frac{3 \,{\left (11 \, C b c^{2} d^{11} f^{4} - 10 \, C a c d^{12} f^{4} - 10 \, B b c d^{12} f^{4} + 8 \, B a d^{13} f^{4} + 8 \, A b d^{13} f^{4} + 8 \, C b c d^{12} f^{3} e - 6 \, C a d^{13} f^{3} e - 6 \, B b d^{13} f^{3} e + 5 \, C b d^{13} f^{2} e^{2}\right )}}{d^{15} f^{5}}\right )} + \frac{3 \,{\left (5 \, C b c^{3} f^{3} - 6 \, C a c^{2} d f^{3} - 6 \, B b c^{2} d f^{3} + 8 \, B a c d^{2} f^{3} + 8 \, A b c d^{2} f^{3} - 16 \, A a d^{3} f^{3} + 3 \, C b c^{2} d f^{2} e - 4 \, C a c d^{2} f^{2} e - 4 \, B b c d^{2} f^{2} e + 8 \, B a d^{3} f^{2} e + 8 \, A b d^{3} f^{2} e + 3 \, C b c d^{2} f e^{2} - 6 \, C a d^{3} f e^{2} - 6 \, B b d^{3} f e^{2} + 5 \, C b 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^{3} f^{3}}\right )} d}{24 \,{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(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*b/(d^4*f) - (13*C*b*c*d^11
*f^4 - 6*C*a*d^12*f^4 - 6*B*b*d^12*f^4 + 5*C*b*d^12*f^3*e)/(d^15*f^5)) + 3*(11*C*b*c^2*d^11*f^4 - 10*C*a*c*d^1
2*f^4 - 10*B*b*c*d^12*f^4 + 8*B*a*d^13*f^4 + 8*A*b*d^13*f^4 + 8*C*b*c*d^12*f^3*e - 6*C*a*d^13*f^3*e - 6*B*b*d^
13*f^3*e + 5*C*b*d^13*f^2*e^2)/(d^15*f^5)) + 3*(5*C*b*c^3*f^3 - 6*C*a*c^2*d*f^3 - 6*B*b*c^2*d*f^3 + 8*B*a*c*d^
2*f^3 + 8*A*b*c*d^2*f^3 - 16*A*a*d^3*f^3 + 3*C*b*c^2*d*f^2*e - 4*C*a*c*d^2*f^2*e - 4*B*b*c*d^2*f^2*e + 8*B*a*d
^3*f^2*e + 8*A*b*d^3*f^2*e + 3*C*b*c*d^2*f*e^2 - 6*C*a*d^3*f*e^2 - 6*B*b*d^3*f*e^2 + 5*C*b*d^3*e^3)*log(abs(-s
qrt(d*f)*sqrt(d*x + c) + sqrt((d*x + c)*d*f - c*d*f + d^2*e)))/(sqrt(d*f)*d^3*f^3))*d/abs(d)