Integrand size = 29, antiderivative size = 177 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx=-\frac {2 (b f g+b e h-2 a f h) \sqrt {c+d x}}{b^3 \sqrt {a+b x}}+\frac {f h \sqrt {a+b x} \sqrt {c+d x}}{b^3}-\frac {2 (b e-a f) (b g-a h) (c+d x)^{3/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {(b c f h-5 a d f h+2 b d (f g+e h)) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2} \sqrt {d}} \] Output:
-2*(-2*a*f*h+b*e*h+b*f*g)*(d*x+c)^(1/2)/b^3/(b*x+a)^(1/2)+f*h*(b*x+a)^(1/2 )*(d*x+c)^(1/2)/b^3-2/3*(-a*f+b*e)*(-a*h+b*g)*(d*x+c)^(3/2)/b^2/(-a*d+b*c) /(b*x+a)^(3/2)+(b*c*f*h-5*a*d*f*h+2*b*d*(e*h+f*g))*arctanh(d^(1/2)*(b*x+a) ^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)/d^(1/2)
Time = 0.94 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx=\frac {\sqrt {c+d x} \left (-15 a^3 d f h-b^3 (2 d e g x-3 c f x (-2 g+h x)+2 c e (g+3 h x))+a^2 b (13 c f h+d (6 f g+6 e h-20 f h x))+a b^2 (-2 c (2 f g+2 e h-9 f h x)+d x (8 f g+8 e h-3 f h x))\right )}{3 b^3 (b c-a d) (a+b x)^{3/2}}+\frac {(b c f h-5 a d f h+2 b d (f g+e h)) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{7/2} \sqrt {d}} \] Input:
Integrate[(Sqrt[c + d*x]*(e + f*x)*(g + h*x))/(a + b*x)^(5/2),x]
Output:
(Sqrt[c + d*x]*(-15*a^3*d*f*h - b^3*(2*d*e*g*x - 3*c*f*x*(-2*g + h*x) + 2* c*e*(g + 3*h*x)) + a^2*b*(13*c*f*h + d*(6*f*g + 6*e*h - 20*f*h*x)) + a*b^2 *(-2*c*(2*f*g + 2*e*h - 9*f*h*x) + d*x*(8*f*g + 8*e*h - 3*f*h*x))))/(3*b^3 *(b*c - a*d)*(a + b*x)^(3/2)) + ((b*c*f*h - 5*a*d*f*h + 2*b*d*(f*g + e*h)) *ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(b^(7/2)*Sqrt[d ])
Time = 0.30 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {160, 57, 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.
\(\displaystyle \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 160 |
\(\displaystyle \frac {(-5 a d f h+b c f h+2 b d (e h+f g)) \int \frac {\sqrt {c+d x}}{(a+b x)^{3/2}}dx}{2 b^2 d}-\frac {(c+d x)^{3/2} \left (5 a^2 d f h-2 a b \left (\frac {3 c f h}{2}+d e h+d f g\right )-3 b f h x (b c-a d)+2 b^2 d e g\right )}{3 b^2 d (a+b x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {(-5 a d f h+b c f h+2 b d (e h+f g)) \left (\frac {d \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{b}-\frac {2 \sqrt {c+d x}}{b \sqrt {a+b x}}\right )}{2 b^2 d}-\frac {(c+d x)^{3/2} \left (5 a^2 d f h-2 a b \left (\frac {3 c f h}{2}+d e h+d f g\right )-3 b f h x (b c-a d)+2 b^2 d e g\right )}{3 b^2 d (a+b x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {(-5 a d f h+b c f h+2 b d (e h+f g)) \left (\frac {2 d \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}-\frac {2 \sqrt {c+d x}}{b \sqrt {a+b x}}\right )}{2 b^2 d}-\frac {(c+d x)^{3/2} \left (5 a^2 d f h-2 a b \left (\frac {3 c f h}{2}+d e h+d f g\right )-3 b f h x (b c-a d)+2 b^2 d e g\right )}{3 b^2 d (a+b x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\left (\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}-\frac {2 \sqrt {c+d x}}{b \sqrt {a+b x}}\right ) (-5 a d f h+b c f h+2 b d (e h+f g))}{2 b^2 d}-\frac {(c+d x)^{3/2} \left (5 a^2 d f h-2 a b \left (\frac {3 c f h}{2}+d e h+d f g\right )-3 b f h x (b c-a d)+2 b^2 d e g\right )}{3 b^2 d (a+b x)^{3/2} (b c-a d)}\) |
Input:
Int[(Sqrt[c + d*x]*(e + f*x)*(g + h*x))/(a + b*x)^(5/2),x]
Output:
-1/3*((c + d*x)^(3/2)*(2*b^2*d*e*g + 5*a^2*d*f*h - 2*a*b*(d*f*g + d*e*h + (3*c*f*h)/2) - 3*b*(b*c - a*d)*f*h*x))/(b^2*d*(b*c - a*d)*(a + b*x)^(3/2)) + ((b*c*f*h - 5*a*d*f*h + 2*b*d*(f*g + e*h))*((-2*Sqrt[c + d*x])/(b*Sqrt[ a + b*x]) + (2*Sqrt[d]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d *x])])/b^(3/2)))/(2*b^2*d)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) )*((g_.) + (h_.)*(x_)), x_] :> Simp[(b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d*(b*c - a*d)*(m + 1))), x] + Simp[(a*d*f*h*m + b*(d* (f*g + e*h) - c*f*h*(m + 2)))/(b^2*d) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1587\) vs. \(2(151)=302\).
Time = 0.34 (sec) , antiderivative size = 1588, normalized size of antiderivative = 8.97
Input:
int((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/6*(-12*b^3*c*e*h*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-12*b^3*c*f*g*x*( (b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-4*b^3*d*e*g*x*((b*x+a)*(d*x+c))^(1/2)*( d*b)^(1/2)+16*a*b^2*d*f*g*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-18*ln(1/2* (2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^3 *c*d*f*h*x^2-36*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+ b*c)/(d*b)^(1/2))*a^2*b^2*c*d*f*h*x+12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c)) ^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^3*c*d*e*h*x+12*ln(1/2*(2*b*d* x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^3*c*d*f* g*x-30*a^3*d*f*h*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-4*b^3*c*e*g*((b*x+a)* (d*x+c))^(1/2)*(d*b)^(1/2)-6*a*b^2*d*f*h*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b) ^(1/2)-40*a^2*b*d*f*h*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+6*ln(1/2*(2*b* d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^4*c*d*f* g*x^2+30*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d *b)^(1/2))*a^3*b*d^2*f*h*x-12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d *b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2*d^2*e*h*x-12*ln(1/2*(2*b*d*x+2*((b *x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2*d^2*f*g*x+6 *ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2 ))*a*b^3*c^2*f*h*x-18*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2 )+a*d+b*c)/(d*b)^(1/2))*a^3*b*c*d*f*h+6*b^3*c*f*h*x^2*((b*x+a)*(d*x+c))^(1 /2)*(d*b)^(1/2)-8*a*b^2*c*f*g*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+15*ln...
Leaf count of result is larger than twice the leaf count of optimal. 582 vs. \(2 (151) = 302\).
Time = 9.91 (sec) , antiderivative size = 1178, normalized size of antiderivative = 6.66 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(5/2),x, algorithm="fricas ")
Output:
[-1/12*(3*(2*(a^2*b^2*c*d - a^3*b*d^2)*f*g + (2*(b^4*c*d - a*b^3*d^2)*f*g + (2*(b^4*c*d - a*b^3*d^2)*e + (b^4*c^2 - 6*a*b^3*c*d + 5*a^2*b^2*d^2)*f)* h)*x^2 + (2*(a^2*b^2*c*d - a^3*b*d^2)*e + (a^2*b^2*c^2 - 6*a^3*b*c*d + 5*a ^4*d^2)*f)*h + 2*(2*(a*b^3*c*d - a^2*b^2*d^2)*f*g + (2*(a*b^3*c*d - a^2*b^ 2*d^2)*e + (a*b^3*c^2 - 6*a^2*b^2*c*d + 5*a^3*b*d^2)*f)*h)*x)*sqrt(b*d)*lo g(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)* sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(3*(b ^4*c*d - a*b^3*d^2)*f*h*x^2 - 2*(b^4*c*d*e + (2*a*b^3*c*d - 3*a^2*b^2*d^2) *f)*g - (2*(2*a*b^3*c*d - 3*a^2*b^2*d^2)*e - (13*a^2*b^2*c*d - 15*a^3*b*d^ 2)*f)*h - 2*((b^4*d^2*e + (3*b^4*c*d - 4*a*b^3*d^2)*f)*g + ((3*b^4*c*d - 4 *a*b^3*d^2)*e - (9*a*b^3*c*d - 10*a^2*b^2*d^2)*f)*h)*x)*sqrt(b*x + a)*sqrt (d*x + c))/(a^2*b^5*c*d - a^3*b^4*d^2 + (b^7*c*d - a*b^6*d^2)*x^2 + 2*(a*b ^6*c*d - a^2*b^5*d^2)*x), -1/6*(3*(2*(a^2*b^2*c*d - a^3*b*d^2)*f*g + (2*(b ^4*c*d - a*b^3*d^2)*f*g + (2*(b^4*c*d - a*b^3*d^2)*e + (b^4*c^2 - 6*a*b^3* c*d + 5*a^2*b^2*d^2)*f)*h)*x^2 + (2*(a^2*b^2*c*d - a^3*b*d^2)*e + (a^2*b^2 *c^2 - 6*a^3*b*c*d + 5*a^4*d^2)*f)*h + 2*(2*(a*b^3*c*d - a^2*b^2*d^2)*f*g + (2*(a*b^3*c*d - a^2*b^2*d^2)*e + (a*b^3*c^2 - 6*a^2*b^2*c*d + 5*a^3*b*d^ 2)*f)*h)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b* x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 2* (3*(b^4*c*d - a*b^3*d^2)*f*h*x^2 - 2*(b^4*c*d*e + (2*a*b^3*c*d - 3*a^2*...
\[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx=\int \frac {\sqrt {c + d x} \left (e + f x\right ) \left (g + h x\right )}{\left (a + b x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((d*x+c)**(1/2)*(f*x+e)*(h*x+g)/(b*x+a)**(5/2),x)
Output:
Integral(sqrt(c + d*x)*(e + f*x)*(g + h*x)/(a + b*x)**(5/2), x)
Exception generated. \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(5/2),x, algorithm="maxima ")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 1238 vs. \(2 (151) = 302\).
Time = 0.37 (sec) , antiderivative size = 1238, normalized size of antiderivative = 6.99 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(5/2),x, algorithm="giac")
Output:
sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*f*h*abs(b)/b^5 - 1/2*(2* b*d*f*g*abs(b) + 2*b*d*e*h*abs(b) + b*c*f*h*abs(b) - 5*a*d*f*h*abs(b))*log ((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt( b*d)*b^4) - 4/3*(b^6*c^2*d^2*e*g*abs(b) - 2*a*b^5*c*d^3*e*g*abs(b) + a^2*b ^4*d^4*e*g*abs(b) + 3*b^6*c^3*d*f*g*abs(b) - 10*a*b^5*c^2*d^2*f*g*abs(b) + 11*a^2*b^4*c*d^3*f*g*abs(b) - 4*a^3*b^3*d^4*f*g*abs(b) + 3*b^6*c^3*d*e*h* abs(b) - 10*a*b^5*c^2*d^2*e*h*abs(b) + 11*a^2*b^4*c*d^3*e*h*abs(b) - 4*a^3 *b^3*d^4*e*h*abs(b) - 6*a*b^5*c^3*d*f*h*abs(b) + 19*a^2*b^4*c^2*d^2*f*h*ab s(b) - 20*a^3*b^3*c*d^3*f*h*abs(b) + 7*a^4*b^2*d^4*f*h*abs(b) - 6*(sqrt(b* d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^4*c^2*d*f*g*ab s(b) + 12*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^ 2*a*b^3*c*d^2*f*g*abs(b) - 6*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^2*d^3*f*g*abs(b) - 6*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^4*c^2*d*e*h*abs(b) + 12*(sqrt(b* d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*c*d^2*e*h* abs(b) - 6*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) ^2*a^2*b^2*d^3*e*h*abs(b) + 12*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b* x + a)*b*d - a*b*d))^2*a*b^3*c^2*d*f*h*abs(b) - 24*(sqrt(b*d)*sqrt(b*x + a ) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^2*c*d^2*f*h*abs(b) + 12*( sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b*...
Timed out. \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx=\int \frac {\left (e+f\,x\right )\,\left (g+h\,x\right )\,\sqrt {c+d\,x}}{{\left (a+b\,x\right )}^{5/2}} \,d x \] Input:
int(((e + f*x)*(g + h*x)*(c + d*x)^(1/2))/(a + b*x)^(5/2),x)
Output:
int(((e + f*x)*(g + h*x)*(c + d*x)^(1/2))/(a + b*x)^(5/2), x)
Time = 0.28 (sec) , antiderivative size = 1308, normalized size of antiderivative = 7.39 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{5/2}} \, dx =\text {Too large to display} \] Input:
int((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(5/2),x)
Output:
( - 30*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)* sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*d**2*f*h + 36*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c) )*a**2*b*c*d*f*h + 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b*d**2*e*h + 12*sqrt(d )*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x) )/sqrt(a*d - b*c))*a**2*b*d**2*f*g - 30*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log( (sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b*d* *2*f*h*x - 6*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sq rt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c**2*f*h - 12*sqrt(d)*sqrt(b) *sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a* d - b*c))*a*b**2*c*d*e*h - 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*s qrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c*d*f*g + 36 *sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c*d*f*h*x + 12*sqrt(d)*sqrt(b)*sqrt(a + b *x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a *b**2*d**2*e*h*x + 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*d**2*f*g*x - 6*sqrt( d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x ))/sqrt(a*d - b*c))*b**3*c**2*f*h*x - 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*...