Integrand size = 29, antiderivative size = 330 \[ \int \sqrt {c+d x} \sqrt {e+f x} \left (A+B x+C x^2\right ) \, dx=\frac {(d e-c f) \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{64 d^3 f^3}+\frac {\left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) (c+d x)^{3/2} \sqrt {e+f x}}{32 d^3 f^2}-\frac {(5 C d e+11 c C f-8 B d f) (c+d x)^{3/2} (e+f x)^{3/2}}{24 d^2 f^2}+\frac {C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac {(d e-c f)^2 \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{64 d^{7/2} f^{7/2}} \]
-1/24*(-8*B*d*f+11*C*c*f+5*C*d*e)*(d*x+c)^(3/2)*(f*x+e)^(3/2)/d^2/f^2+1/4* C*(d*x+c)^(5/2)*(f*x+e)^(3/2)/d^2/f-1/64*(-c*f+d*e)^2*(C*(5*c^2*f^2+6*c*d* e*f+5*d^2*e^2)+8*d*f*(2*A*d*f-B*(c*f+d*e)))*arctanh(f^(1/2)*(d*x+c)^(1/2)/ d^(1/2)/(f*x+e)^(1/2))/d^(7/2)/f^(7/2)+1/32*(C*(5*c^2*f^2+6*c*d*e*f+5*d^2* e^2)+8*d*f*(2*A*d*f-B*(c*f+d*e)))*(d*x+c)^(3/2)*(f*x+e)^(1/2)/d^3/f^2+1/64 *(-c*f+d*e)*(C*(5*c^2*f^2+6*c*d*e*f+5*d^2*e^2)+8*d*f*(2*A*d*f-B*(c*f+d*e)) )*(d*x+c)^(1/2)*(f*x+e)^(1/2)/d^3/f^3
Time = 0.87 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.86 \[ \int \sqrt {c+d x} \sqrt {e+f x} \left (A+B x+C x^2\right ) \, dx=\frac {\sqrt {c+d x} \sqrt {e+f x} \left (C \left (15 c^3 f^3-c^2 d f^2 (7 e+10 f x)+c d^2 f \left (-7 e^2+4 e f x+8 f^2 x^2\right )+d^3 \left (15 e^3-10 e^2 f x+8 e f^2 x^2+48 f^3 x^3\right )\right )+8 d f \left (6 A d f (c f+d (e+2 f x))+B \left (-3 c^2 f^2+2 c d f (e+f x)+d^2 \left (-3 e^2+2 e f x+8 f^2 x^2\right )\right )\right )\right )}{192 d^3 f^3}-\frac {(d e-c f)^2 \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {f} \sqrt {c+d x}}\right )}{64 d^{7/2} f^{7/2}} \]
(Sqrt[c + d*x]*Sqrt[e + f*x]*(C*(15*c^3*f^3 - c^2*d*f^2*(7*e + 10*f*x) + c *d^2*f*(-7*e^2 + 4*e*f*x + 8*f^2*x^2) + d^3*(15*e^3 - 10*e^2*f*x + 8*e*f^2 *x^2 + 48*f^3*x^3)) + 8*d*f*(6*A*d*f*(c*f + d*(e + 2*f*x)) + B*(-3*c^2*f^2 + 2*c*d*f*(e + f*x) + d^2*(-3*e^2 + 2*e*f*x + 8*f^2*x^2)))))/(192*d^3*f^3 ) - ((d*e - c*f)^2*(C*(5*d^2*e^2 + 6*c*d*e*f + 5*c^2*f^2) + 8*d*f*(2*A*d*f - B*(d*e + c*f)))*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/(Sqrt[f]*Sqrt[c + d*x]) ])/(64*d^(7/2)*f^(7/2))
Time = 0.34 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.75, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {1194, 27, 90, 60, 60, 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.
| \(\displaystyle \int \sqrt {c+d x} \sqrt {e+f x} \left (A+B x+C x^2\right ) \, dx\) |
| \(\Big \downarrow \) 1194 |
| \(\displaystyle \frac {\int -\frac {1}{2} \sqrt {c+d x} \sqrt {e+f x} \left (3 C f c^2+5 C d e c-8 A d^2 f+d (5 C d e+11 c C f-8 B d f) x\right )dx}{4 d^2 f}+\frac {C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac {\int \sqrt {c+d x} \sqrt {e+f x} \left (3 C f c^2+5 C d e c-8 A d^2 f+d (5 C d e+11 c C f-8 B d f) x\right )dx}{8 d^2 f}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac {\frac {(c+d x)^{3/2} (e+f x)^{3/2} (-8 B d f+11 c C f+5 C d e)}{3 f}-\frac {\left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right ) \int \sqrt {c+d x} \sqrt {e+f x}dx}{2 f}}{8 d^2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac {\frac {(c+d x)^{3/2} (e+f x)^{3/2} (-8 B d f+11 c C f+5 C d e)}{3 f}-\frac {\left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right ) \left (\frac {(d e-c f) \int \frac {\sqrt {c+d x}}{\sqrt {e+f x}}dx}{4 d}+\frac {(c+d x)^{3/2} \sqrt {e+f x}}{2 d}\right )}{2 f}}{8 d^2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac {\frac {(c+d x)^{3/2} (e+f x)^{3/2} (-8 B d f+11 c C f+5 C d e)}{3 f}-\frac {\left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right ) \left (\frac {(d e-c f) \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}}dx}{2 f}\right )}{4 d}+\frac {(c+d x)^{3/2} \sqrt {e+f x}}{2 d}\right )}{2 f}}{8 d^2 f}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac {\frac {(c+d x)^{3/2} (e+f x)^{3/2} (-8 B d f+11 c C f+5 C d e)}{3 f}-\frac {\left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right ) \left (\frac {(d e-c f) \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{f}\right )}{4 d}+\frac {(c+d x)^{3/2} \sqrt {e+f x}}{2 d}\right )}{2 f}}{8 d^2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac {\frac {(c+d x)^{3/2} (e+f x)^{3/2} (-8 B d f+11 c C f+5 C d e)}{3 f}-\frac {\left (\frac {(d e-c f) \left (\frac {\sqrt {c+d x} \sqrt {e+f x}}{f}-\frac {(d e-c f) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{\sqrt {d} f^{3/2}}\right )}{4 d}+\frac {(c+d x)^{3/2} \sqrt {e+f x}}{2 d}\right ) \left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right )}{2 f}}{8 d^2 f}\) |
(C*(c + d*x)^(5/2)*(e + f*x)^(3/2))/(4*d^2*f) - (((5*C*d*e + 11*c*C*f - 8* B*d*f)*(c + d*x)^(3/2)*(e + f*x)^(3/2))/(3*f) - ((C*(5*d^2*e^2 + 6*c*d*e*f + 5*c^2*f^2) + 8*d*f*(2*A*d*f - B*(d*e + c*f)))*(((c + d*x)^(3/2)*Sqrt[e + f*x])/(2*d) + ((d*e - c*f)*((Sqrt[c + d*x]*Sqrt[e + f*x])/f - ((d*e - c* f)*ArcTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(Sqrt[d]*f^(3 /2))))/(4*d)))/(2*f))/(8*d^2*f)
3.1.43.3.1 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[((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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(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]
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[((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] + Simp[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] && IGtQ [p, 0] && !IntegerQ[m] && !IntegerQ[n] && NeQ[m + n + 2*p + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1206\) vs. \(2(292)=584\).
Time = 1.66 (sec) , antiderivative size = 1207, normalized size of antiderivative = 3.66
-1/384*(d*x+c)^(1/2)*(f*x+e)^(1/2)*(-6*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^2*e^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))*c^4*f^4+ 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^4*e^4-12*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))*c^3*d*e*f^3-16*C*c*d^2*f^3*x^2*((d*x+c)*(f*x+e))^(1/ 2)*(d*f)^(1/2)-16*C*d^3*e*f^2*x^2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)-32*B *(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d^3*e*f^2*x+20*C*(d*f)^(1/2)*((d*x+c) *(f*x+e))^(1/2)*c^2*d*f^3*x-96*A*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*c*d^2 *f^3-96*A*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d^3*e*f^2+48*B*(d*f)^(1/2)*( (d*x+c)*(f*x+e))^(1/2)*c^2*d*f^3+48*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)* d^3*e^2*f-96*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^3*e*f^3-192*A*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d^3 *f^3*x-8*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*c*d^2*e*f^2*x+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))*c^2*d^ 2*f^4+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))*d^4*e^2*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))*c^3*d*f^4-24*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))*d^4*e^3*f-30*C*(d*f)^(1/2 )*((d*x+c)*(f*x+e))^(1/2)*c^3*f^3-30*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1...
Time = 0.32 (sec) , antiderivative size = 840, normalized size of antiderivative = 2.55 \[ \int \sqrt {c+d x} \sqrt {e+f x} \left (A+B x+C x^2\right ) \, dx=\left [\frac {3 \, {\left (5 \, C d^{4} e^{4} - 4 \, {\left (C c d^{3} + 2 \, B d^{4}\right )} e^{3} f - 2 \, {\left (C c^{2} d^{2} - 4 \, B c d^{3} - 8 \, A d^{4}\right )} e^{2} f^{2} - 4 \, {\left (C c^{3} d - 2 \, B c^{2} d^{2} + 8 \, A c d^{3}\right )} e f^{3} + {\left (5 \, C c^{4} - 8 \, B c^{3} d + 16 \, A c^{2} d^{2}\right )} f^{4}\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 (48 \, C d^{4} f^{4} x^{3} + 15 \, C d^{4} e^{3} f - {\left (7 \, C c d^{3} + 24 \, B d^{4}\right )} e^{2} f^{2} - {\left (7 \, C c^{2} d^{2} - 16 \, B c d^{3} - 48 \, A d^{4}\right )} e f^{3} + 3 \, {\left (5 \, C c^{3} d - 8 \, B c^{2} d^{2} + 16 \, A c d^{3}\right )} f^{4} + 8 \, {\left (C d^{4} e f^{3} + {\left (C c d^{3} + 8 \, B d^{4}\right )} f^{4}\right )} x^{2} - 2 \, {\left (5 \, C d^{4} e^{2} f^{2} - 2 \, {\left (C c d^{3} + 4 \, B d^{4}\right )} e f^{3} + {\left (5 \, C c^{2} d^{2} - 8 \, B c d^{3} - 48 \, A d^{4}\right )} f^{4}\right )} x\right )} \sqrt {d x + c} \sqrt {f x + e}}{768 \, d^{4} f^{4}}, \frac {3 \, {\left (5 \, C d^{4} e^{4} - 4 \, {\left (C c d^{3} + 2 \, B d^{4}\right )} e^{3} f - 2 \, {\left (C c^{2} d^{2} - 4 \, B c d^{3} - 8 \, A d^{4}\right )} e^{2} f^{2} - 4 \, {\left (C c^{3} d - 2 \, B c^{2} d^{2} + 8 \, A c d^{3}\right )} e f^{3} + {\left (5 \, C c^{4} - 8 \, B c^{3} d + 16 \, A c^{2} d^{2}\right )} f^{4}\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 (48 \, C d^{4} f^{4} x^{3} + 15 \, C d^{4} e^{3} f - {\left (7 \, C c d^{3} + 24 \, B d^{4}\right )} e^{2} f^{2} - {\left (7 \, C c^{2} d^{2} - 16 \, B c d^{3} - 48 \, A d^{4}\right )} e f^{3} + 3 \, {\left (5 \, C c^{3} d - 8 \, B c^{2} d^{2} + 16 \, A c d^{3}\right )} f^{4} + 8 \, {\left (C d^{4} e f^{3} + {\left (C c d^{3} + 8 \, B d^{4}\right )} f^{4}\right )} x^{2} - 2 \, {\left (5 \, C d^{4} e^{2} f^{2} - 2 \, {\left (C c d^{3} + 4 \, B d^{4}\right )} e f^{3} + {\left (5 \, C c^{2} d^{2} - 8 \, B c d^{3} - 48 \, A d^{4}\right )} f^{4}\right )} x\right )} \sqrt {d x + c} \sqrt {f x + e}}{384 \, d^{4} f^{4}}\right ] \]
[1/768*(3*(5*C*d^4*e^4 - 4*(C*c*d^3 + 2*B*d^4)*e^3*f - 2*(C*c^2*d^2 - 4*B* c*d^3 - 8*A*d^4)*e^2*f^2 - 4*(C*c^3*d - 2*B*c^2*d^2 + 8*A*c*d^3)*e*f^3 + ( 5*C*c^4 - 8*B*c^3*d + 16*A*c^2*d^2)*f^4)*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*(48*C*d^4*f^4*x^3 + 15*C*d ^4*e^3*f - (7*C*c*d^3 + 24*B*d^4)*e^2*f^2 - (7*C*c^2*d^2 - 16*B*c*d^3 - 48 *A*d^4)*e*f^3 + 3*(5*C*c^3*d - 8*B*c^2*d^2 + 16*A*c*d^3)*f^4 + 8*(C*d^4*e* f^3 + (C*c*d^3 + 8*B*d^4)*f^4)*x^2 - 2*(5*C*d^4*e^2*f^2 - 2*(C*c*d^3 + 4*B *d^4)*e*f^3 + (5*C*c^2*d^2 - 8*B*c*d^3 - 48*A*d^4)*f^4)*x)*sqrt(d*x + c)*s qrt(f*x + e))/(d^4*f^4), 1/384*(3*(5*C*d^4*e^4 - 4*(C*c*d^3 + 2*B*d^4)*e^3 *f - 2*(C*c^2*d^2 - 4*B*c*d^3 - 8*A*d^4)*e^2*f^2 - 4*(C*c^3*d - 2*B*c^2*d^ 2 + 8*A*c*d^3)*e*f^3 + (5*C*c^4 - 8*B*c^3*d + 16*A*c^2*d^2)*f^4)*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*(48*C*d^4*f^4*x^3 + 1 5*C*d^4*e^3*f - (7*C*c*d^3 + 24*B*d^4)*e^2*f^2 - (7*C*c^2*d^2 - 16*B*c*d^3 - 48*A*d^4)*e*f^3 + 3*(5*C*c^3*d - 8*B*c^2*d^2 + 16*A*c*d^3)*f^4 + 8*(C*d ^4*e*f^3 + (C*c*d^3 + 8*B*d^4)*f^4)*x^2 - 2*(5*C*d^4*e^2*f^2 - 2*(C*c*d^3 + 4*B*d^4)*e*f^3 + (5*C*c^2*d^2 - 8*B*c*d^3 - 48*A*d^4)*f^4)*x)*sqrt(d*x + c)*sqrt(f*x + e))/(d^4*f^4)]
\[ \int \sqrt {c+d x} \sqrt {e+f x} \left (A+B x+C x^2\right ) \, dx=\int \sqrt {c + d x} \sqrt {e + f x} \left (A + B x + C x^{2}\right )\, dx \]
Exception generated. \[ \int \sqrt {c+d x} \sqrt {e+f x} \left (A+B x+C x^2\right ) \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 1073 vs. \(2 (292) = 584\).
Time = 0.43 (sec) , antiderivative size = 1073, normalized size of antiderivative = 3.25 \[ \int \sqrt {c+d x} \sqrt {e+f x} \left (A+B x+C x^2\right ) \, dx=\text {Too large to display} \]
-1/192*(192*((d^2*e - c*d*f)*log(abs(-sqrt(d*f)*sqrt(d*x + c) + sqrt(d^2*e + (d*x + c)*d*f - c*d*f)))/sqrt(d*f) - sqrt(d^2*e + (d*x + c)*d*f - c*d*f )*sqrt(d*x + c))*A*c*abs(d)/d^2 - 8*(sqrt(d^2*e + (d*x + c)*d*f - c*d*f)*s qrt(d*x + c)*(2*(d*x + c)*(4*(d*x + c)/d^2 + (d^6*e*f^3 - 13*c*d^5*f^4)/(d ^7*f^4)) - 3*(d^7*e^2*f^2 + 2*c*d^6*e*f^3 - 11*c^2*d^5*f^4)/(d^7*f^4)) - 3 *(d^3*e^3 + c*d^2*e^2*f + 3*c^2*d*e*f^2 - 5*c^3*f^3)*log(abs(-sqrt(d*f)*sq rt(d*x + c) + sqrt(d^2*e + (d*x + c)*d*f - c*d*f)))/(sqrt(d*f)*d*f^2))*C*c *abs(d)/d^2 - 8*(sqrt(d^2*e + (d*x + c)*d*f - c*d*f)*sqrt(d*x + c)*(2*(d*x + c)*(4*(d*x + c)/d^2 + (d^6*e*f^3 - 13*c*d^5*f^4)/(d^7*f^4)) - 3*(d^7*e^ 2*f^2 + 2*c*d^6*e*f^3 - 11*c^2*d^5*f^4)/(d^7*f^4)) - 3*(d^3*e^3 + c*d^2*e^ 2*f + 3*c^2*d*e*f^2 - 5*c^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*f^2))*B*abs(d)/d - (sqrt(d^2* e + (d*x + c)*d*f - c*d*f)*(2*(d*x + c)*(4*(d*x + c)*(6*(d*x + c)/d^3 + (d ^12*e*f^5 - 25*c*d^11*f^6)/(d^14*f^6)) - (5*d^13*e^2*f^4 + 14*c*d^12*e*f^5 - 163*c^2*d^11*f^6)/(d^14*f^6)) + 3*(5*d^14*e^3*f^3 + 9*c*d^13*e^2*f^4 + 15*c^2*d^12*e*f^5 - 93*c^3*d^11*f^6)/(d^14*f^6))*sqrt(d*x + c) + 3*(5*d^4* e^4 + 4*c*d^3*e^3*f + 6*c^2*d^2*e^2*f^2 + 20*c^3*d*e*f^3 - 35*c^4*f^4)*log (abs(-sqrt(d*f)*sqrt(d*x + c) + sqrt(d^2*e + (d*x + c)*d*f - c*d*f)))/(sqr t(d*f)*d^2*f^3))*C*abs(d)/d - 48*(sqrt(d^2*e + (d*x + c)*d*f - c*d*f)*(2*d *x + 2*c + (d*e*f - 5*c*f^2)/f^2)*sqrt(d*x + c) + (d^3*e^2 + 2*c*d^2*e*...
Timed out. \[ \int \sqrt {c+d x} \sqrt {e+f x} \left (A+B x+C x^2\right ) \, dx=\text {Hanged} \]