\(\int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx\) [12]

Optimal result

Integrand size = 29, antiderivative size = 194 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx=\frac {(3 B c-5 A d) \sqrt {a x+b x^2}}{3 c^2 d x^2}-\frac {(2 A b c+9 a B c-15 a A d) \sqrt {a x+b x^2}}{3 a c^3 x}-\frac {(B c-A d) \sqrt {a x+b x^2}}{c d x^2 (c+d x)}-\frac {(a d (3 B c-5 A d)-2 b c (B c-2 A d)) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a x+b x^2}}\right )}{c^{7/2} \sqrt {b c-a d}} \] Output:

1/3*(-5*A*d+3*B*c)*(b*x^2+a*x)^(1/2)/c^2/d/x^2-1/3*(-15*A*a*d+2*A*b*c+9*B* 
a*c)*(b*x^2+a*x)^(1/2)/a/c^3/x-(-A*d+B*c)*(b*x^2+a*x)^(1/2)/c/d/x^2/(d*x+c 
)-(a*d*(-5*A*d+3*B*c)-2*b*c*(-2*A*d+B*c))*arctanh((-a*d+b*c)^(1/2)*x/c^(1/ 
2)/(b*x^2+a*x)^(1/2))/c^(7/2)/(-a*d+b*c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx=\frac {\sqrt {x (a+b x)} \left (-\frac {\sqrt {c} \left (2 A b c x (c+d x)+3 a B c x (2 c+3 d x)+a A \left (2 c^2-10 c d x-15 d^2 x^2\right )\right )}{a (c+d x)}-\frac {3 (2 b c (B c-2 A d)+a d (-3 B c+5 A d)) x^{3/2} \arctan \left (\frac {-d \sqrt {x} \sqrt {a+b x}+\sqrt {b} (c+d x)}{\sqrt {c} \sqrt {-b c+a d}}\right )}{\sqrt {-b c+a d} \sqrt {a+b x}}\right )}{3 c^{7/2} x^2} \] Input:

Integrate[((A + B*x)*Sqrt[a*x + b*x^2])/(x^3*(c + d*x)^2),x]
 

Output:

(Sqrt[x*(a + b*x)]*(-((Sqrt[c]*(2*A*b*c*x*(c + d*x) + 3*a*B*c*x*(2*c + 3*d 
*x) + a*A*(2*c^2 - 10*c*d*x - 15*d^2*x^2)))/(a*(c + d*x))) - (3*(2*b*c*(B* 
c - 2*A*d) + a*d*(-3*B*c + 5*A*d))*x^(3/2)*ArcTan[(-(d*Sqrt[x]*Sqrt[a + b* 
x]) + Sqrt[b]*(c + d*x))/(Sqrt[c]*Sqrt[-(b*c) + a*d])])/(Sqrt[-(b*c) + a*d 
]*Sqrt[a + b*x])))/(3*c^(7/2)*x^2)
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(425\) vs. \(2(194)=388\).

Time = 1.44 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2153, 2009}

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.

\(\displaystyle \int \frac {\sqrt {a x+b x^2} (A+B x)}{x^3 (c+d x)^2} \, dx\)

\(\Big \downarrow \) 2153

\(\displaystyle \int \left (\frac {d^2 \sqrt {a x+b x^2} (2 B c-3 A d)}{c^4 (c+d x)}-\frac {d \sqrt {a x+b x^2} (2 B c-3 A d)}{c^4 x}+\frac {d^2 \sqrt {a x+b x^2} (B c-A d)}{c^3 (c+d x)^2}+\frac {\sqrt {a x+b x^2} (B c-2 A d)}{c^3 x^2}+\frac {A \sqrt {a x+b x^2}}{c^2 x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {b c-a d} (2 B c-3 A d) \text {arctanh}\left (\frac {x (2 b c-a d)+a c}{2 \sqrt {c} \sqrt {a x+b x^2} \sqrt {b c-a d}}\right )}{c^{7/2}}-\frac {(2 b c-a d) (B c-A d) \text {arctanh}\left (\frac {x (2 b c-a d)+a c}{2 \sqrt {c} \sqrt {a x+b x^2} \sqrt {b c-a d}}\right )}{2 c^{7/2} \sqrt {b c-a d}}-\frac {a d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (2 B c-3 A d)}{\sqrt {b} c^4}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (2 b c-a d) (2 B c-3 A d)}{\sqrt {b} c^4}+\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (B c-2 A d)}{c^3}+\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (B c-A d)}{c^3}-\frac {2 \sqrt {a x+b x^2} (B c-2 A d)}{c^3 x}-\frac {d \sqrt {a x+b x^2} (B c-A d)}{c^3 (c+d x)}-\frac {2 A \left (a x+b x^2\right )^{3/2}}{3 a c^2 x^3}\)

Input:

Int[((A + B*x)*Sqrt[a*x + b*x^2])/(x^3*(c + d*x)^2),x]
 

Output:

(-2*(B*c - 2*A*d)*Sqrt[a*x + b*x^2])/(c^3*x) - (d*(B*c - A*d)*Sqrt[a*x + b 
*x^2])/(c^3*(c + d*x)) - (2*A*(a*x + b*x^2)^(3/2))/(3*a*c^2*x^3) - (a*d*(2 
*B*c - 3*A*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(Sqrt[b]*c^4) - ((2* 
b*c - a*d)*(2*B*c - 3*A*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(Sqrt[b 
]*c^4) + (2*Sqrt[b]*(B*c - 2*A*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/ 
c^3 + (2*Sqrt[b]*(B*c - A*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/c^3 + 
 (Sqrt[b*c - a*d]*(2*B*c - 3*A*d)*ArcTanh[(a*c + (2*b*c - a*d)*x)/(2*Sqrt[ 
c]*Sqrt[b*c - a*d]*Sqrt[a*x + b*x^2])])/c^(7/2) - ((2*b*c - a*d)*(B*c - A* 
d)*ArcTanh[(a*c + (2*b*c - a*d)*x)/(2*Sqrt[c]*Sqrt[b*c - a*d]*Sqrt[a*x + b 
*x^2])])/(2*c^(7/2)*Sqrt[b*c - a*d])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2153

Int[(Px_)*((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b 
_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e* 
x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, n, p}, x] && PolyQ[Px, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ 
[m] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] && IGtQ[n, 0])
 

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(\frac {-2 \sqrt {x \left (b x +a \right )}\, \left (\left (\left (3 B x +A \right ) a +A b x \right ) c^{2}-5 d x \left (\left (-\frac {9 B x}{10}+A \right ) a -\frac {A b x}{5}\right ) c -\frac {15 A a \,d^{2} x^{2}}{2}\right ) \sqrt {c \left (a d -b c \right )}-15 \left (\frac {2 B b \,c^{2}}{5}-\frac {4 d \left (A b +\frac {3 B a}{4}\right ) c}{5}+A a \,d^{2}\right ) \left (d x +c \right ) x^{2} a \arctan \left (\frac {\sqrt {x \left (b x +a \right )}\, c}{x \sqrt {c \left (a d -b c \right )}}\right )}{3 \sqrt {c \left (a d -b c \right )}\, c^{3} x^{2} \left (d x +c \right ) a}\)	\(166\)
risch	\(-\frac {2 \left (b x +a \right ) \left (-6 A a d x +A b c x +3 B a c x +A a c \right )}{3 a \,c^{3} \sqrt {x \left (b x +a \right )}\, x}+\frac {\frac {c \left (A a \,d^{2}-A b c d -B a c d +B b \,c^{2}\right ) \left (\frac {d^{2} \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{c \left (a d -b c \right ) \left (x +\frac {c}{d}\right )}-\frac {\left (a d -2 b c \right ) d \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{2 c \left (a d -b c \right ) \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}\right )}{d^{2}}-\frac {\left (2 A a d -A b c -a B c \right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{\sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}}{c^{3}}\)	\(452\)
default	\(\text {Expression too large to display}\)	\(1054\)

Input:

int((B*x+A)*(b*x^2+a*x)^(1/2)/x^3/(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

1/3*(-2*(x*(b*x+a))^(1/2)*(((3*B*x+A)*a+A*b*x)*c^2-5*d*x*((-9/10*B*x+A)*a- 
1/5*A*b*x)*c-15/2*A*a*d^2*x^2)*(c*(a*d-b*c))^(1/2)-15*(2/5*B*b*c^2-4/5*d*( 
A*b+3/4*B*a)*c+A*a*d^2)*(d*x+c)*x^2*a*arctan((x*(b*x+a))^(1/2)/x*c/(c*(a*d 
-b*c))^(1/2)))/(c*(a*d-b*c))^(1/2)/c^3/x^2/(d*x+c)/a
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 644, normalized size of antiderivative = 3.32 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx=\left [\frac {3 \, {\left ({\left (2 \, B a b c^{2} d + 5 \, A a^{2} d^{3} - {\left (3 \, B a^{2} + 4 \, A a b\right )} c d^{2}\right )} x^{3} + {\left (2 \, B a b c^{3} + 5 \, A a^{2} c d^{2} - {\left (3 \, B a^{2} + 4 \, A a b\right )} c^{2} d\right )} x^{2}\right )} \sqrt {b c^{2} - a c d} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x + 2 \, \sqrt {b c^{2} - a c d} \sqrt {b x^{2} + a x}}{d x + c}\right ) - 2 \, {\left (2 \, A a b c^{4} - 2 \, A a^{2} c^{3} d + {\left (15 \, A a^{2} c d^{3} + {\left (9 \, B a b + 2 \, A b^{2}\right )} c^{3} d - {\left (9 \, B a^{2} + 17 \, A a b\right )} c^{2} d^{2}\right )} x^{2} + 2 \, {\left (5 \, A a^{2} c^{2} d^{2} + {\left (3 \, B a b + A b^{2}\right )} c^{4} - 3 \, {\left (B a^{2} + 2 \, A a b\right )} c^{3} d\right )} x\right )} \sqrt {b x^{2} + a x}}{6 \, {\left ({\left (a b c^{5} d - a^{2} c^{4} d^{2}\right )} x^{3} + {\left (a b c^{6} - a^{2} c^{5} d\right )} x^{2}\right )}}, -\frac {3 \, {\left ({\left (2 \, B a b c^{2} d + 5 \, A a^{2} d^{3} - {\left (3 \, B a^{2} + 4 \, A a b\right )} c d^{2}\right )} x^{3} + {\left (2 \, B a b c^{3} + 5 \, A a^{2} c d^{2} - {\left (3 \, B a^{2} + 4 \, A a b\right )} c^{2} d\right )} x^{2}\right )} \sqrt {-b c^{2} + a c d} \arctan \left (\frac {\sqrt {-b c^{2} + a c d} \sqrt {b x^{2} + a x}}{b c x + a c}\right ) + {\left (2 \, A a b c^{4} - 2 \, A a^{2} c^{3} d + {\left (15 \, A a^{2} c d^{3} + {\left (9 \, B a b + 2 \, A b^{2}\right )} c^{3} d - {\left (9 \, B a^{2} + 17 \, A a b\right )} c^{2} d^{2}\right )} x^{2} + 2 \, {\left (5 \, A a^{2} c^{2} d^{2} + {\left (3 \, B a b + A b^{2}\right )} c^{4} - 3 \, {\left (B a^{2} + 2 \, A a b\right )} c^{3} d\right )} x\right )} \sqrt {b x^{2} + a x}}{3 \, {\left ({\left (a b c^{5} d - a^{2} c^{4} d^{2}\right )} x^{3} + {\left (a b c^{6} - a^{2} c^{5} d\right )} x^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x^3/(d*x+c)^2,x, algorithm="fricas")
 

Output:

[1/6*(3*((2*B*a*b*c^2*d + 5*A*a^2*d^3 - (3*B*a^2 + 4*A*a*b)*c*d^2)*x^3 + ( 
2*B*a*b*c^3 + 5*A*a^2*c*d^2 - (3*B*a^2 + 4*A*a*b)*c^2*d)*x^2)*sqrt(b*c^2 - 
 a*c*d)*log((a*c + (2*b*c - a*d)*x + 2*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a* 
x))/(d*x + c)) - 2*(2*A*a*b*c^4 - 2*A*a^2*c^3*d + (15*A*a^2*c*d^3 + (9*B*a 
*b + 2*A*b^2)*c^3*d - (9*B*a^2 + 17*A*a*b)*c^2*d^2)*x^2 + 2*(5*A*a^2*c^2*d 
^2 + (3*B*a*b + A*b^2)*c^4 - 3*(B*a^2 + 2*A*a*b)*c^3*d)*x)*sqrt(b*x^2 + a* 
x))/((a*b*c^5*d - a^2*c^4*d^2)*x^3 + (a*b*c^6 - a^2*c^5*d)*x^2), -1/3*(3*( 
(2*B*a*b*c^2*d + 5*A*a^2*d^3 - (3*B*a^2 + 4*A*a*b)*c*d^2)*x^3 + (2*B*a*b*c 
^3 + 5*A*a^2*c*d^2 - (3*B*a^2 + 4*A*a*b)*c^2*d)*x^2)*sqrt(-b*c^2 + a*c*d)* 
arctan(sqrt(-b*c^2 + a*c*d)*sqrt(b*x^2 + a*x)/(b*c*x + a*c)) + (2*A*a*b*c^ 
4 - 2*A*a^2*c^3*d + (15*A*a^2*c*d^3 + (9*B*a*b + 2*A*b^2)*c^3*d - (9*B*a^2 
 + 17*A*a*b)*c^2*d^2)*x^2 + 2*(5*A*a^2*c^2*d^2 + (3*B*a*b + A*b^2)*c^4 - 3 
*(B*a^2 + 2*A*a*b)*c^3*d)*x)*sqrt(b*x^2 + a*x))/((a*b*c^5*d - a^2*c^4*d^2) 
*x^3 + (a*b*c^6 - a^2*c^5*d)*x^2)]
 

Sympy [F]

\[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx=\int \frac {\sqrt {x \left (a + b x\right )} \left (A + B x\right )}{x^{3} \left (c + d x\right )^{2}}\, dx \] Input:

integrate((B*x+A)*(b*x**2+a*x)**(1/2)/x**3/(d*x+c)**2,x)
 

Output:

Integral(sqrt(x*(a + b*x))*(A + B*x)/(x**3*(c + d*x)**2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx=\int { \frac {\sqrt {b x^{2} + a x} {\left (B x + A\right )}}{{\left (d x + c\right )}^{2} x^{3}} \,d x } \] Input:

integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x^3/(d*x+c)^2,x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^2 + a*x)*(B*x + A)/((d*x + c)^2*x^3), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1319 vs. \(2 (174) = 348\).

Time = 20.09 (sec) , antiderivative size = 1319, normalized size of antiderivative = 6.80 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x^3/(d*x+c)^2,x, algorithm="giac")
 

Output:

1/10*d^2*(((4*b*c*d - 5*a*d^2)*A*sgn(1/(d*x + c))*sgn(d) - (2*b*c^2 - 3*a* 
c*d)*B*sgn(1/(d*x + c))*sgn(d))*log(abs(-2*sqrt(b*c^2 - a*c*d)*c^2*(sqrt(b 
 - 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a*d/(d*x + c) - a*c*d/(d*x + c)^2 
) + sqrt(b*c^2*d^2 - a*c*d^3)/((d*x + c)*d))^5*abs(d) + 2*b^3*c^3*d - 5*a* 
b^2*c^2*d^2 + 4*a^2*b*c*d^3 - a^3*d^4 - 4*(5*b*c^2 - 4*a*c*d)*sqrt(b*c^2 - 
 a*c*d)*(sqrt(b - 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a*d/(d*x + c) - a* 
c*d/(d*x + c)^2) + sqrt(b*c^2*d^2 - a*c*d^3)/((d*x + c)*d))^3*abs(d) + (10 
*b*c^3*d - 9*a*c^2*d^2)*(sqrt(b - 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a* 
d/(d*x + c) - a*c*d/(d*x + c)^2) + sqrt(b*c^2*d^2 - a*c*d^3)/((d*x + c)*d) 
)^4 - 2*(5*b^2*c^2 - 8*a*b*c*d + 3*a^2*d^2)*sqrt(b*c^2 - a*c*d)*(sqrt(b - 
2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a*d/(d*x + c) - a*c*d/(d*x + c)^2) + 
 sqrt(b*c^2*d^2 - a*c*d^3)/((d*x + c)*d))*abs(d) + 2*(10*b^2*c^3*d - 17*a* 
b*c^2*d^2 + 7*a^2*c*d^3)*(sqrt(b - 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a 
*d/(d*x + c) - a*c*d/(d*x + c)^2) + sqrt(b*c^2*d^2 - a*c*d^3)/((d*x + c)*d 
))^2))/(sqrt(b*c^2 - a*c*d)*c^3*d*abs(d)) + (2*B*b*c^2*d*log(abs(32*b^3*c^ 
3*d - 48*a*b^2*c^2*d^2 + 18*a^2*b*c*d^3 - a^3*d^4 - 32*sqrt(b*c^2 - a*c*d) 
*b^(5/2)*c^2*abs(d) + 32*sqrt(b*c^2 - a*c*d)*a*b^(3/2)*c*d*abs(d) - 6*sqrt 
(b*c^2 - a*c*d)*a^2*sqrt(b)*d^2*abs(d))) - 3*B*a*c*d^2*log(abs(32*b^3*c^3* 
d - 48*a*b^2*c^2*d^2 + 18*a^2*b*c*d^3 - a^3*d^4 - 32*sqrt(b*c^2 - a*c*d)*b 
^(5/2)*c^2*abs(d) + 32*sqrt(b*c^2 - a*c*d)*a*b^(3/2)*c*d*abs(d) - 6*sqr...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx=\int \frac {\sqrt {b\,x^2+a\,x}\,\left (A+B\,x\right )}{x^3\,{\left (c+d\,x\right )}^2} \,d x \] Input:

int(((a*x + b*x^2)^(1/2)*(A + B*x))/(x^3*(c + d*x)^2),x)
 

Output:

int(((a*x + b*x^2)^(1/2)*(A + B*x))/(x^3*(c + d*x)^2), x)
 

Reduce [B] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 980, normalized size of antiderivative = 5.05 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^3 (c+d x)^2} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(b*x^2+a*x)^(1/2)/x^3/(d*x+c)^2,x)
 

Output:

( - 75*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b* 
x) - sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a**2*c*d**2*x**2 - 75*sqr 
t(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt( 
x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a**2*d**3*x**3 + 90*sqrt(c)*sqrt(a* 
d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*s 
qrt(b))/(sqrt(c)*sqrt(b)))*a*b*c**2*d*x**2 + 90*sqrt(c)*sqrt(a*d - b*c)*at 
an((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*sqrt(b))/(sq 
rt(c)*sqrt(b)))*a*b*c*d**2*x**3 - 24*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a* 
d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt( 
b)))*b**2*c**3*x**2 - 24*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - s 
qrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*b**2*c* 
*2*d*x**3 - 75*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqr 
t(a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a**2*c*d**2*x**2 
- 75*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) 
 + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a**2*d**3*x**3 + 90*sqrt(c) 
*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(x)*s 
qrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a*b*c**2*d*x**2 + 90*sqrt(c)*sqrt(a*d - 
 b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(x)*sqrt(d)*sqrt 
(b))/(sqrt(c)*sqrt(b)))*a*b*c*d**2*x**3 - 24*sqrt(c)*sqrt(a*d - b*c)*atan( 
(sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/(sq...