\(\int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx\) [271]

Optimal result

Integrand size = 17, antiderivative size = 124 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=-\frac {5 d}{3 (b c-a d)^2 (c+d x)^{3/2}}-\frac {1}{(b c-a d) (a+b x) (c+d x)^{3/2}}-\frac {5 b d}{(b c-a d)^3 \sqrt {c+d x}}+\frac {5 b^{3/2} d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {2 a^2 d^2-2 a b d (7 c+5 d x)-b^2 \left (3 c^2+20 c d x+15 d^2 x^2\right )}{3 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}+\frac {5 b^{3/2} d \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{7/2}} \] Input:

Integrate[1/((a + b*x)^2*(c + d*x)^(5/2)),x]
 

Output:

(2*a^2*d^2 - 2*a*b*d*(7*c + 5*d*x) - b^2*(3*c^2 + 20*c*d*x + 15*d^2*x^2))/ 
(3*(b*c - a*d)^3*(a + b*x)*(c + d*x)^(3/2)) + (5*b^(3/2)*d*ArcTan[(Sqrt[b] 
*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(7/2)
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {52, 61, 61, 73, 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.

\(\displaystyle \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {5 d \int \frac {1}{(a+b x) (c+d x)^{5/2}}dx}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {5 d \left (\frac {b \int \frac {1}{(a+b x) (c+d x)^{3/2}}dx}{b c-a d}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {5 d \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{b c-a d}+\frac {2}{\sqrt {c+d x} (b c-a d)}\right )}{b c-a d}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {5 d \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{d (b c-a d)}+\frac {2}{\sqrt {c+d x} (b c-a d)}\right )}{b c-a d}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {5 d \left (\frac {b \left (\frac {2}{\sqrt {c+d x} (b c-a d)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}\right )}{b c-a d}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}\)

Input:

Int[1/((a + b*x)^2*(c + d*x)^(5/2)),x]
 

Output:

-(1/((b*c - a*d)*(a + b*x)*(c + d*x)^(3/2))) - (5*d*(2/(3*(b*c - a*d)*(c + 
 d*x)^(3/2)) + (b*(2/((b*c - a*d)*Sqrt[c + d*x]) - (2*Sqrt[b]*ArcTanh[(Sqr 
t[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(3/2)))/(b*c - a*d)))/(2 
*(b*c - a*d))
 

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(2 d \left (-\frac {1}{3 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {3}{2}}}+\frac {2 b}{\left (a d -b c \right )^{3} \sqrt {x d +c}}+\frac {b^{2} \left (\frac {\sqrt {x d +c}}{2 \left (x d +c \right ) b +2 a d -2 b c}+\frac {5 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}\right )\)	\(121\)
default	\(2 d \left (-\frac {1}{3 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {3}{2}}}+\frac {2 b}{\left (a d -b c \right )^{3} \sqrt {x d +c}}+\frac {b^{2} \left (\frac {\sqrt {x d +c}}{2 \left (x d +c \right ) b +2 a d -2 b c}+\frac {5 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}\right )\)	\(121\)
pseudoelliptic	\(d \left (\frac {\sqrt {x d +c}\, b^{2}}{d \left (b x +a \right ) \left (a d -b c \right )^{3}}+\frac {5 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right ) b^{2}}{\sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right )^{3}}-\frac {2}{3 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {3}{2}}}+\frac {4 b}{\left (a d -b c \right )^{3} \sqrt {x d +c}}\right )\)	\(123\)

Input:

int(1/(b*x+a)^2/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2*d*(-1/3/(a*d-b*c)^2/(d*x+c)^(3/2)+2/(a*d-b*c)^3*b/(d*x+c)^(1/2)+1/(a*d-b 
*c)^3*b^2*(1/2*(d*x+c)^(1/2)/((d*x+c)*b+a*d-b*c)+5/2/((a*d-b*c)*b)^(1/2)*a 
rctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (106) = 212\).

Time = 0.14 (sec) , antiderivative size = 762, normalized size of antiderivative = 6.15 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\left [-\frac {15 \, {\left (b^{2} d^{3} x^{3} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x\right )} \sqrt {d x + c}}{6 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}}, -\frac {15 \, {\left (b^{2} d^{3} x^{3} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}\right ) + {\left (15 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}}\right ] \] Input:

integrate(1/(b*x+a)^2/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

[-1/6*(15*(b^2*d^3*x^3 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^2 + (b^2*c^ 
2*d + 2*a*b*c*d^2)*x)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d - 2*(b* 
c - a*d)*sqrt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) + 2*(15*b^2*d^2*x^2 
 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x)*sqrt(d 
*x + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b 
^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c 
^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + 
(b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4 
)*x), -1/3*(15*(b^2*d^3*x^3 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^2 + (b 
^2*c^2*d + 2*a*b*c*d^2)*x)*sqrt(-b/(b*c - a*d))*arctan(sqrt(d*x + c)*sqrt( 
-b/(b*c - a*d))) + (15*b^2*d^2*x^2 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 
10*(2*b^2*c*d + a*b*d^2)*x)*sqrt(d*x + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 
3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2 
*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d 
^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d 
^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)]
 

Sympy [F]

\[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\int \frac {1}{\left (a + b x\right )^{2} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(b*x+a)**2/(d*x+c)**(5/2),x)
 

Output:

Integral(1/((a + b*x)**2*(c + d*x)**(5/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(b*x+a)^2/(d*x+c)^(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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (106) = 212\).

Time = 0.13 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.74 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=-\frac {5 \, b^{2} d \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {\sqrt {d x + c} b^{2} d}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}} - \frac {2 \, {\left (6 \, {\left (d x + c\right )} b d + b c d - a d^{2}\right )}}{3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} \] Input:

integrate(1/(b*x+a)^2/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

-5*b^2*d*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^2* 
c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b^2*c + a*b*d)) - sqrt(d*x + c)*b^2 
*d/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x + c)*b - b*c 
 + a*d)) - 2/3*(6*(d*x + c)*b*d + b*c*d - a*d^2)/((b^3*c^3 - 3*a*b^2*c^2*d 
 + 3*a^2*b*c*d^2 - a^3*d^3)*(d*x + c)^(3/2))
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {\frac {10\,b\,d\,\left (c+d\,x\right )}{3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,d}{3\,\left (a\,d-b\,c\right )}+\frac {5\,b^2\,d\,{\left (c+d\,x\right )}^2}{{\left (a\,d-b\,c\right )}^3}}{b\,{\left (c+d\,x\right )}^{5/2}+\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}+\frac {5\,b^{3/2}\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{{\left (a\,d-b\,c\right )}^{7/2}} \] Input:

int(1/((a + b*x)^2*(c + d*x)^(5/2)),x)
 

Output:

((10*b*d*(c + d*x))/(3*(a*d - b*c)^2) - (2*d)/(3*(a*d - b*c)) + (5*b^2*d*( 
c + d*x)^2)/(a*d - b*c)^3)/(b*(c + d*x)^(5/2) + (a*d - b*c)*(c + d*x)^(3/2 
)) + (5*b^(3/2)*d*atan((b^(1/2)*(c + d*x)^(1/2)*(a^3*d^3 - b^3*c^3 + 3*a*b 
^2*c^2*d - 3*a^2*b*c*d^2))/(a*d - b*c)^(7/2)))/(a*d - b*c)^(7/2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 493, normalized size of antiderivative = 3.98 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a b c d +15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a b \,d^{2} x +15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{2} c d x +15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{2} d^{2} x^{2}-2 a^{3} d^{3}+16 a^{2} b c \,d^{2}+10 a^{2} b \,d^{3} x -11 a \,b^{2} c^{2} d +10 a \,b^{2} c \,d^{2} x +15 a \,b^{2} d^{3} x^{2}-3 b^{3} c^{3}-20 b^{3} c^{2} d x -15 b^{3} c \,d^{2} x^{2}}{3 \sqrt {d x +c}\, \left (a^{4} b \,d^{5} x^{2}-4 a^{3} b^{2} c \,d^{4} x^{2}+6 a^{2} b^{3} c^{2} d^{3} x^{2}-4 a \,b^{4} c^{3} d^{2} x^{2}+b^{5} c^{4} d \,x^{2}+a^{5} d^{5} x -3 a^{4} b c \,d^{4} x +2 a^{3} b^{2} c^{2} d^{3} x +2 a^{2} b^{3} c^{3} d^{2} x -3 a \,b^{4} c^{4} d x +b^{5} c^{5} x +a^{5} c \,d^{4}-4 a^{4} b \,c^{2} d^{3}+6 a^{3} b^{2} c^{3} d^{2}-4 a^{2} b^{3} c^{4} d +a \,b^{4} c^{5}\right )} \] Input:

int(1/(b*x+a)^2/(d*x+c)^(5/2),x)
 

Output:

(15*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)* 
sqrt(a*d - b*c)))*a*b*c*d + 15*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan( 
(sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b*d**2*x + 15*sqrt(b)*sqrt( 
c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c))) 
*b**2*c*d*x + 15*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x) 
*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**2*d**2*x**2 - 2*a**3*d**3 + 16*a**2*b*c* 
d**2 + 10*a**2*b*d**3*x - 11*a*b**2*c**2*d + 10*a*b**2*c*d**2*x + 15*a*b** 
2*d**3*x**2 - 3*b**3*c**3 - 20*b**3*c**2*d*x - 15*b**3*c*d**2*x**2)/(3*sqr 
t(c + d*x)*(a**5*c*d**4 + a**5*d**5*x - 4*a**4*b*c**2*d**3 - 3*a**4*b*c*d* 
*4*x + a**4*b*d**5*x**2 + 6*a**3*b**2*c**3*d**2 + 2*a**3*b**2*c**2*d**3*x 
- 4*a**3*b**2*c*d**4*x**2 - 4*a**2*b**3*c**4*d + 2*a**2*b**3*c**3*d**2*x + 
 6*a**2*b**3*c**2*d**3*x**2 + a*b**4*c**5 - 3*a*b**4*c**4*d*x - 4*a*b**4*c 
**3*d**2*x**2 + b**5*c**5*x + b**5*c**4*d*x**2))