Integrand size = 34, antiderivative size = 371 \[ \int \frac {(a+b x) \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {C (a+b x)^2 \sqrt {c+d x} \sqrt {e+f x}}{3 b d f}-\frac {\sqrt {c+d x} \sqrt {e+f x} \left (8 a^2 C d^2 f^2-6 a b d f (4 B d f-3 C (d e+c f))-b^2 \left (C \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )+6 d f (4 A d f-3 B (d e+c f))\right )+2 b d f (2 a C d f-b (6 B d f-5 C (d e+c f))) x\right )}{24 b d^3 f^3}+\frac {\left (2 a d f \left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right )-b \left (C \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )+2 d f \left (4 A d f (d e+c f)-B \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{8 d^{7/2} f^{7/2}} \]
1/8*(2*a*d*f*(C*(3*c^2*f^2+2*c*d*e*f+3*d^2*e^2)+4*d*f*(2*A*d*f-B*(c*f+d*e) ))-b*(C*(5*c^3*f^3+3*c^2*d*e*f^2+3*c*d^2*e^2*f+5*d^3*e^3)+2*d*f*(4*A*d*f*( c*f+d*e)-B*(3*c^2*f^2+2*c*d*e*f+3*d^2*e^2))))*arctanh(f^(1/2)*(d*x+c)^(1/2 )/d^(1/2)/(f*x+e)^(1/2))/d^(7/2)/f^(7/2)+1/3*C*(b*x+a)^2*(d*x+c)^(1/2)*(f* x+e)^(1/2)/b/d/f-1/24*(8*a^2*C*d^2*f^2-6*a*b*d*f*(4*B*d*f-3*C*(c*f+d*e))-b ^2*(C*(15*c^2*f^2+14*c*d*e*f+15*d^2*e^2)+6*d*f*(4*A*d*f-3*B*(c*f+d*e)))+2* b*d*f*(2*a*C*d*f-b*(6*B*d*f-5*C*(c*f+d*e)))*x)*(d*x+c)^(1/2)*(f*x+e)^(1/2) /b/d^3/f^3
Time = 0.96 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x) \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {\sqrt {c+d x} \sqrt {e+f x} \left (6 a d f (4 B d f+C (-3 d e-3 c f+2 d f x))+b \left (6 d f (4 A d f+B (-3 d e-3 c f+2 d f x))+C \left (15 c^2 f^2+2 c d f (7 e-5 f x)+d^2 \left (15 e^2-10 e f x+8 f^2 x^2\right )\right )\right )\right )}{24 d^3 f^3}-\frac {\left (-2 a d f \left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right )+b \left (C \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )+2 d f \left (4 A d f (d e+c f)-B \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {f} \sqrt {c+d x}}\right )}{8 d^{7/2} f^{7/2}} \]
(Sqrt[c + d*x]*Sqrt[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)))))/(24*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[d]*Sqrt[e + f*x])/(Sqrt[f]*Sqrt[c + d*x])])/(8*d^(7/2)*f ^(7/2))
Time = 0.61 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {2118, 27, 164, 66, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a+b x) \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx\) |
| \(\Big \downarrow \) 2118 |
| \(\displaystyle \frac {\int -\frac {b (a+b x) (4 b c C e+a C d e+a c C f-6 A b d f-(6 b B d f-2 a C d f-5 b C (d e+c f)) x)}{2 \sqrt {c+d x} \sqrt {e+f x}}dx}{3 b^2 d f}+\frac {C (a+b x)^2 \sqrt {c+d x} \sqrt {e+f x}}{3 b d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {C (a+b x)^2 \sqrt {c+d x} \sqrt {e+f x}}{3 b d f}-\frac {\int \frac {(a+b x) (4 b c C e+a C d e+a c C f-6 A b d f-(6 b B d f-2 a C d f-5 b C (d e+c f)) x)}{\sqrt {c+d x} \sqrt {e+f x}}dx}{6 b d f}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {C (a+b x)^2 \sqrt {c+d x} \sqrt {e+f x}}{3 b d f}-\frac {\frac {\sqrt {c+d x} \sqrt {e+f x} \left (8 a^2 C d^2 f^2-2 b d f x (-2 a C d f+6 b B d f-5 b C (c f+d e))-6 a b d f (4 B d f-3 C (c f+d e))-\left (b^2 \left (6 d f (4 A d f-3 B (c f+d e))+C \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )\right )\right )}{4 d^2 f^2}-\frac {3 b \left (2 a d f \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )-b \left (2 d f \left (4 A d f (c f+d e)-B \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+C \left (5 c^3 f^3+3 c^2 d e f^2+3 c d^2 e^2 f+5 d^3 e^3\right )\right )\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}}dx}{8 d^2 f^2}}{6 b d f}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {C (a+b x)^2 \sqrt {c+d x} \sqrt {e+f x}}{3 b d f}-\frac {\frac {\sqrt {c+d x} \sqrt {e+f x} \left (8 a^2 C d^2 f^2-2 b d f x (-2 a C d f+6 b B d f-5 b C (c f+d e))-6 a b d f (4 B d f-3 C (c f+d e))-\left (b^2 \left (6 d f (4 A d f-3 B (c f+d e))+C \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )\right )\right )}{4 d^2 f^2}-\frac {3 b \left (2 a d f \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )-b \left (2 d f \left (4 A d f (c f+d e)-B \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+C \left (5 c^3 f^3+3 c^2 d e f^2+3 c d^2 e^2 f+5 d^3 e^3\right )\right )\right ) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{4 d^2 f^2}}{6 b d f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {C (a+b x)^2 \sqrt {c+d x} \sqrt {e+f x}}{3 b d f}-\frac {\frac {\sqrt {c+d x} \sqrt {e+f x} \left (8 a^2 C d^2 f^2-2 b d f x (-2 a C d f+6 b B d f-5 b C (c f+d e))-6 a b d f (4 B d f-3 C (c f+d e))-\left (b^2 \left (6 d f (4 A d f-3 B (c f+d e))+C \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )\right )\right )}{4 d^2 f^2}-\frac {3 b \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) \left (2 a d f \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )-b \left (2 d f \left (4 A d f (c f+d e)-B \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+C \left (5 c^3 f^3+3 c^2 d e f^2+3 c d^2 e^2 f+5 d^3 e^3\right )\right )\right )}{4 d^{5/2} f^{5/2}}}{6 b d f}\) |
(C*(a + b*x)^2*Sqrt[c + d*x]*Sqrt[e + f*x])/(3*b*d*f) - ((Sqrt[c + d*x]*Sq rt[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))/(4*d^2*f^2 ) - (3*b*(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])])/(4*d^(5/2)* f^(5/2)))/(6*b*d*f)
3.1.55.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> 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] + Simp[(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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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, Expo n[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] + Simp[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]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1198\) vs. \(2(345)=690\).
Time = 1.68 (sec) , antiderivative size = 1199, normalized size of antiderivative = 3.23
1/48*(-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+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^2-24*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*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)*b*d^2*f^2*x-20*C*((d* x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)*b*d^2*e*f*x-36*C*(d*f)^(1/2)*((d*x+c)*(f*x +e))^(1/2)*a*d^2*e*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))*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+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+24*C*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)*a*d^2*f^2*x+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+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+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^2-2 4*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...
Time = 0.49 (sec) , antiderivative size = 720, normalized size of antiderivative = 1.94 \[ \int \frac {(a+b x) \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\left [-\frac {3 \, {\left (5 \, C b d^{3} e^{3} + 3 \, {\left (C b c d^{2} - 2 \, {\left (C a + B b\right )} d^{3}\right )} e^{2} f + {\left (3 \, C b c^{2} d - 4 \, {\left (C a + B b\right )} c d^{2} + 8 \, {\left (B a + A b\right )} d^{3}\right )} e f^{2} + {\left (5 \, C b c^{3} - 16 \, A a d^{3} - 6 \, {\left (C a + B b\right )} c^{2} d + 8 \, {\left (B a + A b\right )} 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 b d^{3} f^{3} x^{2} + 15 \, C b d^{3} e^{2} f + 2 \, {\left (7 \, C b c d^{2} - 9 \, {\left (C a + B b\right )} d^{3}\right )} e f^{2} + 3 \, {\left (5 \, C b c^{2} d - 6 \, {\left (C a + B b\right )} c d^{2} + 8 \, {\left (B a + A b\right )} d^{3}\right )} f^{3} - 2 \, {\left (5 \, C b d^{3} e f^{2} + {\left (5 \, C b c d^{2} - 6 \, {\left (C a + B b\right )} d^{3}\right )} f^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {f x + e}}{96 \, d^{4} f^{4}}, \frac {3 \, {\left (5 \, C b d^{3} e^{3} + 3 \, {\left (C b c d^{2} - 2 \, {\left (C a + B b\right )} d^{3}\right )} e^{2} f + {\left (3 \, C b c^{2} d - 4 \, {\left (C a + B b\right )} c d^{2} + 8 \, {\left (B a + A b\right )} d^{3}\right )} e f^{2} + {\left (5 \, C b c^{3} - 16 \, A a d^{3} - 6 \, {\left (C a + B b\right )} c^{2} d + 8 \, {\left (B a + A b\right )} 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 b d^{3} f^{3} x^{2} + 15 \, C b d^{3} e^{2} f + 2 \, {\left (7 \, C b c d^{2} - 9 \, {\left (C a + B b\right )} d^{3}\right )} e f^{2} + 3 \, {\left (5 \, C b c^{2} d - 6 \, {\left (C a + B b\right )} c d^{2} + 8 \, {\left (B a + A b\right )} d^{3}\right )} f^{3} - 2 \, {\left (5 \, C b d^{3} e f^{2} + {\left (5 \, C b c d^{2} - 6 \, {\left (C a + B b\right )} d^{3}\right )} f^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {f x + e}}{48 \, d^{4} f^{4}}\right ] \]
[-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)*sqr t(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)*s qrt(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)]
\[ \int \frac {(a+b x) \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\int \frac {\left (a + b x\right ) \left (A + B x + C x^{2}\right )}{\sqrt {c + d x} \sqrt {e + f x}}\, dx \]
Exception generated. \[ \int \frac {(a+b x) \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(c*f-d*e>0)', see `assume?` for m ore detail
Time = 0.33 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x) \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {{\left (\sqrt {d^{2} e + {\left (d x + c\right )} d f - c d f} \sqrt {d x + c} {\left (2 \, {\left (d x + c\right )} {\left (\frac {4 \, {\left (d x + c\right )} C b}{d^{4} f} - \frac {5 \, C b d^{12} e f^{3} + 13 \, C b c d^{11} f^{4} - 6 \, C a d^{12} f^{4} - 6 \, B b d^{12} f^{4}}{d^{15} f^{5}}\right )} + \frac {3 \, {\left (5 \, C b d^{13} e^{2} f^{2} + 8 \, C b c d^{12} e f^{3} - 6 \, C a d^{13} e f^{3} - 6 \, B b d^{13} e f^{3} + 11 \, C b c^{2} d^{11} f^{4} - 10 \, C a c d^{12} f^{4} - 10 \, B b c d^{12} f^{4} + 8 \, B a d^{13} f^{4} + 8 \, A b d^{13} f^{4}\right )}}{d^{15} f^{5}}\right )} + \frac {3 \, {\left (5 \, C b d^{3} e^{3} + 3 \, C b c d^{2} e^{2} f - 6 \, C a d^{3} e^{2} f - 6 \, B b d^{3} e^{2} f + 3 \, C b c^{2} d e f^{2} - 4 \, C a c d^{2} e f^{2} - 4 \, B b c d^{2} e f^{2} + 8 \, B a d^{3} e f^{2} + 8 \, A b d^{3} e f^{2} + 5 \, C b c^{3} f^{3} - 6 \, C a c^{2} d f^{3} - 6 \, B b c^{2} d f^{3} + 8 \, B a c d^{2} f^{3} + 8 \, A b c d^{2} f^{3} - 16 \, A a d^{3} f^{3}\right )} \log \left ({\left | -\sqrt {d f} \sqrt {d x + c} + \sqrt {d^{2} e + {\left (d x + c\right )} d f - c d f} \right |}\right )}{\sqrt {d f} d^{3} f^{3}}\right )} d}{24 \, {\left | d \right |}} \]
1/24*(sqrt(d^2*e + (d*x + c)*d*f - c*d*f)*sqrt(d*x + c)*(2*(d*x + c)*(4*(d *x + c)*C*b/(d^4*f) - (5*C*b*d^12*e*f^3 + 13*C*b*c*d^11*f^4 - 6*C*a*d^12*f ^4 - 6*B*b*d^12*f^4)/(d^15*f^5)) + 3*(5*C*b*d^13*e^2*f^2 + 8*C*b*c*d^12*e* f^3 - 6*C*a*d^13*e*f^3 - 6*B*b*d^13*e*f^3 + 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)/(d^15*f^5 )) + 3*(5*C*b*d^3*e^3 + 3*C*b*c*d^2*e^2*f - 6*C*a*d^3*e^2*f - 6*B*b*d^3*e^ 2*f + 3*C*b*c^2*d*e*f^2 - 4*C*a*c*d^2*e*f^2 - 4*B*b*c*d^2*e*f^2 + 8*B*a*d^ 3*e*f^2 + 8*A*b*d^3*e*f^2 + 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)*log(abs(-sqrt(d* f)*sqrt(d*x + c) + sqrt(d^2*e + (d*x + c)*d*f - c*d*f)))/(sqrt(d*f)*d^3*f^ 3))*d/abs(d)
Time = 140.16 (sec) , antiderivative size = 2621, normalized size of antiderivative = 7.06 \[ \int \frac {(a+b x) \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\text {Too large to display} \]
((((c + d*x)^(1/2) - c^(1/2))*(2*A*b*c*f + 2*A*b*d*e))/(f^3*((e + f*x)^(1/ 2) - e^(1/2))) + (((c + d*x)^(1/2) - c^(1/2))^3*(2*A*b*c*f + 2*A*b*d*e))/( d*f^2*((e + f*x)^(1/2) - e^(1/2))^3) - (8*A*b*c^(1/2)*e^(1/2)*((c + d*x)^( 1/2) - c^(1/2))^2)/(f^2*((e + f*x)^(1/2) - e^(1/2))^2))/(((c + d*x)^(1/2) - c^(1/2))^4/((e + f*x)^(1/2) - e^(1/2))^4 + d^2/f^2 - (2*d*((c + d*x)^(1/ 2) - c^(1/2))^2)/(f*((e + f*x)^(1/2) - e^(1/2))^2)) - ((((c + d*x)^(1/2) - c^(1/2))*((3*C*a*d^3*e^2)/2 + (3*C*a*c^2*d*f^2)/2 + C*a*c*d^2*e*f))/(f^6* ((e + f*x)^(1/2) - e^(1/2))) - (((c + d*x)^(1/2) - c^(1/2))^3*((11*C*a*c^2 *f^2)/2 + (11*C*a*d^2*e^2)/2 + 25*C*a*c*d*e*f))/(f^5*((e + f*x)^(1/2) - e^ (1/2))^3) + (((c + d*x)^(1/2) - c^(1/2))^7*((3*C*a*c^2*f^2)/2 + (3*C*a*d^2 *e^2)/2 + C*a*c*d*e*f))/(d^2*f^3*((e + f*x)^(1/2) - e^(1/2))^7) - (((c + d *x)^(1/2) - c^(1/2))^5*((11*C*a*c^2*f^2)/2 + (11*C*a*d^2*e^2)/2 + 25*C*a*c *d*e*f))/(d*f^4*((e + f*x)^(1/2) - e^(1/2))^5) + (c^(1/2)*e^(1/2)*((c + d* x)^(1/2) - c^(1/2))^4*(32*C*a*c*f + 32*C*a*d*e))/(f^4*((e + f*x)^(1/2) - e ^(1/2))^4))/(((c + d*x)^(1/2) - c^(1/2))^8/((e + f*x)^(1/2) - e^(1/2))^8 + d^4/f^4 - (4*d*((c + d*x)^(1/2) - c^(1/2))^6)/(f*((e + f*x)^(1/2) - e^(1/ 2))^6) - (4*d^3*((c + d*x)^(1/2) - c^(1/2))^2)/(f^3*((e + f*x)^(1/2) - e^( 1/2))^2) + (6*d^2*((c + d*x)^(1/2) - c^(1/2))^4)/(f^2*((e + f*x)^(1/2) - e ^(1/2))^4)) - ((((c + d*x)^(1/2) - c^(1/2))^3*((85*C*b*d^4*e^3)/12 + (85*C *b*c^3*d*f^3)/12 + (17*C*b*c*d^3*e^2*f)/4 + (17*C*b*c^2*d^2*e*f^2)/4))/...