Integrand size = 29, antiderivative size = 164 \[ \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx=-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \text {arctanh}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{4 e^{5/2} g^{5/2}} \]
1/4*(c*(3*d^2*g^2+2*d*e*f*g+3*e^2*f^2)+4*e*g*(2*a*e*g-b*(d*g+e*f)))*arctan h(g^(1/2)*(e*x+d)^(1/2)/e^(1/2)/(g*x+f)^(1/2))/e^(5/2)/g^(5/2)+1/2*c*(e*x+ d)^(3/2)*(g*x+f)^(1/2)/e^2/g-1/4*(-4*b*e*g+5*c*d*g+3*c*e*f)*(e*x+d)^(1/2)* (g*x+f)^(1/2)/e^2/g^2
Time = 0.06 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.86 \[ \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx=\frac {\sqrt {d+e x} \sqrt {f+g x} (4 b e g+c (-3 e f-3 d g+2 e g x))}{4 e^2 g^2}+\frac {\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right )}{4 e^{5/2} g^{5/2}} \]
(Sqrt[d + e*x]*Sqrt[f + g*x]*(4*b*e*g + c*(-3*e*f - 3*d*g + 2*e*g*x)))/(4* e^2*g^2) + ((c*(3*e^2*f^2 + 2*d*e*f*g + 3*d^2*g^2) + 4*e*g*(2*a*e*g - b*(e *f + d*g)))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/(Sqrt[g]*Sqrt[d + e*x])])/(4*e ^(5/2)*g^(5/2))
Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1194, 27, 90, 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.
| \(\displaystyle \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx\) |
| \(\Big \downarrow \) 1194 |
| \(\displaystyle \frac {\int \frac {4 a g e^2-(3 c e f+5 c d g-4 b e g) x e-c d (3 e f+d g)}{2 \sqrt {d+e x} \sqrt {f+g x}}dx}{2 e^2 g}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {4 a g e^2-(3 c e f+5 c d g-4 b e g) x e-c d (3 e f+d g)}{\sqrt {d+e x} \sqrt {f+g x}}dx}{4 e^2 g}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {\frac {\left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {f+g x}}dx}{2 g}-\frac {\sqrt {d+e x} \sqrt {f+g x} (-4 b e g+5 c d g+3 c e f)}{g}}{4 e^2 g}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {\frac {\left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right ) \int \frac {1}{e-\frac {g (d+e x)}{f+g x}}d\frac {\sqrt {d+e x}}{\sqrt {f+g x}}}{g}-\frac {\sqrt {d+e x} \sqrt {f+g x} (-4 b e g+5 c d g+3 c e f)}{g}}{4 e^2 g}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right ) \left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )}{\sqrt {e} g^{3/2}}-\frac {\sqrt {d+e x} \sqrt {f+g x} (-4 b e g+5 c d g+3 c e f)}{g}}{4 e^2 g}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}\) |
(c*(d + e*x)^(3/2)*Sqrt[f + g*x])/(2*e^2*g) + (-(((3*c*e*f + 5*c*d*g - 4*b *e*g)*Sqrt[d + e*x]*Sqrt[f + g*x])/g) + ((c*(3*e^2*f^2 + 2*d*e*f*g + 3*d^2 *g^2) + 4*e*g*(2*a*e*g - b*(e*f + d*g)))*ArcTanh[(Sqrt[g]*Sqrt[d + e*x])/( Sqrt[e]*Sqrt[f + g*x])])/(Sqrt[e]*g^(3/2)))/(4*e^2*g)
3.9.37.3.1 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_))*((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. \(424\) vs. \(2(138)=276\).
Time = 0.48 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.59
| method | result | size |
| default | \(\frac {\left (4 \sqrt {e g}\, \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, c e g x +3 \ln \left (\frac {2 e g x +2 \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c \,d^{2} g^{2}+2 \ln \left (\frac {2 e g x +2 \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c d e f g +3 \ln \left (\frac {2 e g x +2 \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c \,e^{2} f^{2}+8 \ln \left (\frac {2 e g x +2 \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) a \,e^{2} g^{2}-4 \ln \left (\frac {2 e g x +2 \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) b d e \,g^{2}-4 \ln \left (\frac {2 e g x +2 \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) b \,e^{2} f g -6 \sqrt {e g}\, \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, c d g -6 \sqrt {e g}\, \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, c e f +8 \sqrt {e g}\, \sqrt {\left (g x +f \right ) \left (e x +d \right )}\, b e g \right ) \sqrt {e x +d}\, \sqrt {g x +f}}{8 \sqrt {e g}\, g^{2} e^{2} \sqrt {\left (g x +f \right ) \left (e x +d \right )}}\) | \(425\) |
1/8*(4*(e*g)^(1/2)*((g*x+f)*(e*x+d))^(1/2)*c*e*g*x+3*ln(1/2*(2*e*g*x+2*((g *x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d^2*g^2+2*ln(1/2* (2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d*e *f*g+3*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g )^(1/2))*c*e^2*f^2+8*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2) +d*g+e*f)/(e*g)^(1/2))*a*e^2*g^2-4*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/ 2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*b*d*e*g^2-4*ln(1/2*(2*e*g*x+2*((g*x+f )*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*b*e^2*f*g-6*(e*g)^(1/2) *((g*x+f)*(e*x+d))^(1/2)*c*d*g-6*(e*g)^(1/2)*((g*x+f)*(e*x+d))^(1/2)*c*e*f +8*(e*g)^(1/2)*((g*x+f)*(e*x+d))^(1/2)*b*e*g)*(e*x+d)^(1/2)*(g*x+f)^(1/2)/ (e*g)^(1/2)/g^2/e^2/((g*x+f)*(e*x+d))^(1/2)
Time = 0.48 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.32 \[ \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx=\left [\frac {{\left (3 \, c e^{2} f^{2} + 2 \, {\left (c d e - 2 \, b e^{2}\right )} f g + {\left (3 \, c d^{2} - 4 \, b d e + 8 \, a e^{2}\right )} g^{2}\right )} \sqrt {e g} \log \left (8 \, e^{2} g^{2} x^{2} + e^{2} f^{2} + 6 \, d e f g + d^{2} g^{2} + 4 \, {\left (2 \, e g x + e f + d g\right )} \sqrt {e g} \sqrt {e x + d} \sqrt {g x + f} + 8 \, {\left (e^{2} f g + d e g^{2}\right )} x\right ) + 4 \, {\left (2 \, c e^{2} g^{2} x - 3 \, c e^{2} f g - {\left (3 \, c d e - 4 \, b e^{2}\right )} g^{2}\right )} \sqrt {e x + d} \sqrt {g x + f}}{16 \, e^{3} g^{3}}, -\frac {{\left (3 \, c e^{2} f^{2} + 2 \, {\left (c d e - 2 \, b e^{2}\right )} f g + {\left (3 \, c d^{2} - 4 \, b d e + 8 \, a e^{2}\right )} g^{2}\right )} \sqrt {-e g} \arctan \left (\frac {{\left (2 \, e g x + e f + d g\right )} \sqrt {-e g} \sqrt {e x + d} \sqrt {g x + f}}{2 \, {\left (e^{2} g^{2} x^{2} + d e f g + {\left (e^{2} f g + d e g^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, c e^{2} g^{2} x - 3 \, c e^{2} f g - {\left (3 \, c d e - 4 \, b e^{2}\right )} g^{2}\right )} \sqrt {e x + d} \sqrt {g x + f}}{8 \, e^{3} g^{3}}\right ] \]
[1/16*((3*c*e^2*f^2 + 2*(c*d*e - 2*b*e^2)*f*g + (3*c*d^2 - 4*b*d*e + 8*a*e ^2)*g^2)*sqrt(e*g)*log(8*e^2*g^2*x^2 + e^2*f^2 + 6*d*e*f*g + d^2*g^2 + 4*( 2*e*g*x + e*f + d*g)*sqrt(e*g)*sqrt(e*x + d)*sqrt(g*x + f) + 8*(e^2*f*g + d*e*g^2)*x) + 4*(2*c*e^2*g^2*x - 3*c*e^2*f*g - (3*c*d*e - 4*b*e^2)*g^2)*sq rt(e*x + d)*sqrt(g*x + f))/(e^3*g^3), -1/8*((3*c*e^2*f^2 + 2*(c*d*e - 2*b* e^2)*f*g + (3*c*d^2 - 4*b*d*e + 8*a*e^2)*g^2)*sqrt(-e*g)*arctan(1/2*(2*e*g *x + e*f + d*g)*sqrt(-e*g)*sqrt(e*x + d)*sqrt(g*x + f)/(e^2*g^2*x^2 + d*e* f*g + (e^2*f*g + d*e*g^2)*x)) - 2*(2*c*e^2*g^2*x - 3*c*e^2*f*g - (3*c*d*e - 4*b*e^2)*g^2)*sqrt(e*x + d)*sqrt(g*x + f))/(e^3*g^3)]
\[ \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx=\int \frac {a + b x + c x^{2}}{\sqrt {d + e x} \sqrt {f + g x}}\, dx \]
Exception generated. \[ \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(d*g-e*f>0)', see `assume?` for m ore detail
Time = 0.31 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.16 \[ \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx=\frac {{\left (\sqrt {e^{2} f + {\left (e x + d\right )} e g - d e g} \sqrt {e x + d} {\left (\frac {2 \, {\left (e x + d\right )} c}{e^{3} g} - \frac {3 \, c e^{6} f g + 5 \, c d e^{5} g^{2} - 4 \, b e^{6} g^{2}}{e^{8} g^{3}}\right )} - \frac {{\left (3 \, c e^{2} f^{2} + 2 \, c d e f g - 4 \, b e^{2} f g + 3 \, c d^{2} g^{2} - 4 \, b d e g^{2} + 8 \, a e^{2} g^{2}\right )} \log \left ({\left | -\sqrt {e g} \sqrt {e x + d} + \sqrt {e^{2} f + {\left (e x + d\right )} e g - d e g} \right |}\right )}{\sqrt {e g} e^{2} g^{2}}\right )} e}{4 \, {\left | e \right |}} \]
1/4*(sqrt(e^2*f + (e*x + d)*e*g - d*e*g)*sqrt(e*x + d)*(2*(e*x + d)*c/(e^3 *g) - (3*c*e^6*f*g + 5*c*d*e^5*g^2 - 4*b*e^6*g^2)/(e^8*g^3)) - (3*c*e^2*f^ 2 + 2*c*d*e*f*g - 4*b*e^2*f*g + 3*c*d^2*g^2 - 4*b*d*e*g^2 + 8*a*e^2*g^2)*l og(abs(-sqrt(e*g)*sqrt(e*x + d) + sqrt(e^2*f + (e*x + d)*e*g - d*e*g)))/(s qrt(e*g)*e^2*g^2))*e/abs(e)
Time = 0.00 (sec) , antiderivative size = 833, normalized size of antiderivative = 5.08 \[ \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx=\frac {\frac {\left (2\,b\,d\,g+2\,b\,e\,f\right )\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{g^3\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}+\frac {\left (2\,b\,d\,g+2\,b\,e\,f\right )\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}{e\,g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^3}-\frac {8\,b\,\sqrt {d}\,\sqrt {f}\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}}{\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}+\frac {e^2}{g^2}-\frac {2\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}}-\frac {\frac {\left (\sqrt {d+e\,x}-\sqrt {d}\right )\,\left (\frac {3\,c\,d^2\,e\,g^2}{2}+c\,d\,e^2\,f\,g+\frac {3\,c\,e^3\,f^2}{2}\right )}{g^6\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}-\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3\,\left (\frac {11\,c\,d^2\,g^2}{2}+25\,c\,d\,e\,f\,g+\frac {11\,c\,e^2\,f^2}{2}\right )}{g^5\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^3}+\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7\,\left (\frac {3\,c\,d^2\,g^2}{2}+c\,d\,e\,f\,g+\frac {3\,c\,e^2\,f^2}{2}\right )}{e^2\,g^3\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^7}-\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5\,\left (\frac {11\,c\,d^2\,g^2}{2}+25\,c\,d\,e\,f\,g+\frac {11\,c\,e^2\,f^2}{2}\right )}{e\,g^4\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^5}+\frac {\sqrt {d}\,\sqrt {f}\,\left (32\,c\,d\,g+32\,c\,e\,f\right )\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{g^4\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}}{\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^8}{{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^8}+\frac {e^4}{g^4}-\frac {4\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}{g\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^6}-\frac {4\,e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g^3\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}+\frac {6\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}}-\frac {4\,a\,\mathrm {atan}\left (\frac {e\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}{\sqrt {-e\,g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )}{\sqrt {-e\,g}}-\frac {2\,b\,\mathrm {atanh}\left (\frac {\sqrt {g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {e}\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}\right )\,\left (d\,g+e\,f\right )}{e^{3/2}\,g^{3/2}}+\frac {c\,\mathrm {atanh}\left (\frac {\sqrt {g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {e}\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}\right )\,\left (3\,d^2\,g^2+2\,d\,e\,f\,g+3\,e^2\,f^2\right )}{2\,e^{5/2}\,g^{5/2}} \]
(((2*b*d*g + 2*b*e*f)*((d + e*x)^(1/2) - d^(1/2)))/(g^3*((f + g*x)^(1/2) - f^(1/2))) + ((2*b*d*g + 2*b*e*f)*((d + e*x)^(1/2) - d^(1/2))^3)/(e*g^2*(( f + g*x)^(1/2) - f^(1/2))^3) - (8*b*d^(1/2)*f^(1/2)*((d + e*x)^(1/2) - d^( 1/2))^2)/(g^2*((f + g*x)^(1/2) - f^(1/2))^2))/(((d + e*x)^(1/2) - d^(1/2)) ^4/((f + g*x)^(1/2) - f^(1/2))^4 + e^2/g^2 - (2*e*((d + e*x)^(1/2) - d^(1/ 2))^2)/(g*((f + g*x)^(1/2) - f^(1/2))^2)) - ((((d + e*x)^(1/2) - d^(1/2))* ((3*c*e^3*f^2)/2 + (3*c*d^2*e*g^2)/2 + c*d*e^2*f*g))/(g^6*((f + g*x)^(1/2) - f^(1/2))) - (((d + e*x)^(1/2) - d^(1/2))^3*((11*c*d^2*g^2)/2 + (11*c*e^ 2*f^2)/2 + 25*c*d*e*f*g))/(g^5*((f + g*x)^(1/2) - f^(1/2))^3) + (((d + e*x )^(1/2) - d^(1/2))^7*((3*c*d^2*g^2)/2 + (3*c*e^2*f^2)/2 + c*d*e*f*g))/(e^2 *g^3*((f + g*x)^(1/2) - f^(1/2))^7) - (((d + e*x)^(1/2) - d^(1/2))^5*((11* c*d^2*g^2)/2 + (11*c*e^2*f^2)/2 + 25*c*d*e*f*g))/(e*g^4*((f + g*x)^(1/2) - f^(1/2))^5) + (d^(1/2)*f^(1/2)*(32*c*d*g + 32*c*e*f)*((d + e*x)^(1/2) - d ^(1/2))^4)/(g^4*((f + g*x)^(1/2) - f^(1/2))^4))/(((d + e*x)^(1/2) - d^(1/2 ))^8/((f + g*x)^(1/2) - f^(1/2))^8 + e^4/g^4 - (4*e*((d + e*x)^(1/2) - d^( 1/2))^6)/(g*((f + g*x)^(1/2) - f^(1/2))^6) - (4*e^3*((d + e*x)^(1/2) - d^( 1/2))^2)/(g^3*((f + g*x)^(1/2) - f^(1/2))^2) + (6*e^2*((d + e*x)^(1/2) - d ^(1/2))^4)/(g^2*((f + g*x)^(1/2) - f^(1/2))^4)) - (4*a*atan((e*((f + g*x)^ (1/2) - f^(1/2)))/((-e*g)^(1/2)*((d + e*x)^(1/2) - d^(1/2)))))/(-e*g)^(1/2 ) - (2*b*atanh((g^(1/2)*((d + e*x)^(1/2) - d^(1/2)))/(e^(1/2)*((f + g*x...