3.5.70 \(\int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx\) [470]

3.5.70.1 Optimal result
3.5.70.2 Mathematica [A] (verified)
3.5.70.3 Rubi [A] (verified)
3.5.70.4 Maple [A] (verified)
3.5.70.5 Fricas [A] (verification not implemented)
3.5.70.6 Sympy [F]
3.5.70.7 Maxima [F(-2)]
3.5.70.8 Giac [A] (verification not implemented)
3.5.70.9 Mupad [B] (verification not implemented)

3.5.70.1 Optimal result

Integrand size = 18, antiderivative size = 178 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx=\frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x}}{b^4}+\frac {(2 b c-7 a d) (c+d x)^{3/2}}{3 b^3}+\frac {(2 b c-7 a d) (c+d x)^{5/2}}{5 b^2 (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (b c-a d) (a+b x)}-\frac {(2 b c-7 a d) (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{9/2}} \]

output
1/3*(-7*a*d+2*b*c)*(d*x+c)^(3/2)/b^3+1/5*(-7*a*d+2*b*c)*(d*x+c)^(5/2)/b^2/ 
(-a*d+b*c)+a*(d*x+c)^(7/2)/b/(-a*d+b*c)/(b*x+a)-(-7*a*d+2*b*c)*(-a*d+b*c)^ 
(3/2)*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(9/2)+(-7*a*d+2*b* 
c)*(-a*d+b*c)*(d*x+c)^(1/2)/b^4
 
3.5.70.2 Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.94 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx=\frac {\sqrt {c+d x} \left (105 a^3 d^2+10 a^2 b d (-17 c+7 d x)+a b^2 \left (61 c^2-118 c d x-14 d^2 x^2\right )+2 b^3 x \left (23 c^2+11 c d x+3 d^2 x^2\right )\right )}{15 b^4 (a+b x)}-\frac {\sqrt {-b c+a d} \left (2 b^2 c^2-9 a b c d+7 a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{9/2}} \]

input
Integrate[(x*(c + d*x)^(5/2))/(a + b*x)^2,x]
 
output
(Sqrt[c + d*x]*(105*a^3*d^2 + 10*a^2*b*d*(-17*c + 7*d*x) + a*b^2*(61*c^2 - 
 118*c*d*x - 14*d^2*x^2) + 2*b^3*x*(23*c^2 + 11*c*d*x + 3*d^2*x^2)))/(15*b 
^4*(a + b*x)) - (Sqrt[-(b*c) + a*d]*(2*b^2*c^2 - 9*a*b*c*d + 7*a^2*d^2)*Ar 
cTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(9/2)
 
3.5.70.3 Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {87, 60, 60, 60, 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.

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {(2 b c-7 a d) \int \frac {(c+d x)^{5/2}}{a+b x}dx}{2 b (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (a+b x) (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(2 b c-7 a d) \left (\frac {(b c-a d) \int \frac {(c+d x)^{3/2}}{a+b x}dx}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\right )}{2 b (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (a+b x) (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(2 b c-7 a d) \left (\frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {\sqrt {c+d x}}{a+b x}dx}{b}+\frac {2 (c+d x)^{3/2}}{3 b}\right )}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\right )}{2 b (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (a+b x) (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(2 b c-7 a d) \left (\frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{b}+\frac {2 \sqrt {c+d x}}{b}\right )}{b}+\frac {2 (c+d x)^{3/2}}{3 b}\right )}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\right )}{2 b (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (a+b x) (b c-a d)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(2 b c-7 a d) \left (\frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (b c-a d) \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b d}+\frac {2 \sqrt {c+d x}}{b}\right )}{b}+\frac {2 (c+d x)^{3/2}}{3 b}\right )}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\right )}{2 b (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (a+b x) (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(2 b c-7 a d) \left (\frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {2 \sqrt {c+d x}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (c+d x)^{3/2}}{3 b}\right )}{b}+\frac {2 (c+d x)^{5/2}}{5 b}\right )}{2 b (b c-a d)}+\frac {a (c+d x)^{7/2}}{b (a+b x) (b c-a d)}\)

input
Int[(x*(c + d*x)^(5/2))/(a + b*x)^2,x]
 
output
(a*(c + d*x)^(7/2))/(b*(b*c - a*d)*(a + b*x)) + ((2*b*c - 7*a*d)*((2*(c + 
d*x)^(5/2))/(5*b) + ((b*c - a*d)*((2*(c + d*x)^(3/2))/(3*b) + ((b*c - a*d) 
*((2*Sqrt[c + d*x])/b - (2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x]) 
/Sqrt[b*c - a*d]])/b^(3/2)))/b))/b))/(2*b*(b*c - a*d))
 

3.5.70.3.1 Defintions of rubi rules used

rule 60
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]
 

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 87
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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]
 
3.5.70.4 Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.90

method result size
pseudoelliptic \(-\frac {7 \left (\left (a d -b c \right )^{2} \left (a d -\frac {2 b c}{7}\right ) \left (b x +a \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\sqrt {d x +c}\, \sqrt {\left (a d -b c \right ) b}\, \left (\frac {46 x \left (\frac {3}{23} d^{2} x^{2}+\frac {11}{23} c d x +c^{2}\right ) b^{3}}{105}+\frac {61 \left (-\frac {14}{61} d^{2} x^{2}-\frac {118}{61} c d x +c^{2}\right ) a \,b^{2}}{105}-\frac {34 d \,a^{2} \left (-\frac {7 d x}{17}+c \right ) b}{21}+a^{3} d^{2}\right )\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{4} \left (b x +a \right )}\) \(160\)
risch \(\frac {2 \left (3 d^{2} x^{2} b^{2}-10 x a b \,d^{2}+11 x \,b^{2} c d +45 a^{2} d^{2}-70 a b c d +23 b^{2} c^{2}\right ) \sqrt {d x +c}}{15 b^{4}}-\frac {\left (2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) \left (-\frac {a d \sqrt {d x +c}}{2 \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {\left (7 a d -2 b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{4}}\) \(166\)
derivativedivides \(\frac {\frac {2 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {4 a b d \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+6 \sqrt {d x +c}\, a^{2} d^{2}-8 \sqrt {d x +c}\, a b c d +2 b^{2} c^{2} \sqrt {d x +c}}{b^{4}}-\frac {2 \left (\frac {\left (-\frac {1}{2} a^{3} d^{3}+a^{2} b c \,d^{2}-\frac {1}{2} a \,b^{2} c^{2} d \right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (7 a^{3} d^{3}-16 a^{2} b c \,d^{2}+11 a \,b^{2} c^{2} d -2 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{4}}\) \(219\)
default \(\frac {\frac {2 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {4 a b d \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+6 \sqrt {d x +c}\, a^{2} d^{2}-8 \sqrt {d x +c}\, a b c d +2 b^{2} c^{2} \sqrt {d x +c}}{b^{4}}-\frac {2 \left (\frac {\left (-\frac {1}{2} a^{3} d^{3}+a^{2} b c \,d^{2}-\frac {1}{2} a \,b^{2} c^{2} d \right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (7 a^{3} d^{3}-16 a^{2} b c \,d^{2}+11 a \,b^{2} c^{2} d -2 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{4}}\) \(219\)

input
int(x*(d*x+c)^(5/2)/(b*x+a)^2,x,method=_RETURNVERBOSE)
 
output
-7*((a*d-b*c)^2*(a*d-2/7*b*c)*(b*x+a)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b) 
^(1/2))-(d*x+c)^(1/2)*((a*d-b*c)*b)^(1/2)*(46/105*x*(3/23*d^2*x^2+11/23*c* 
d*x+c^2)*b^3+61/105*(-14/61*d^2*x^2-118/61*c*d*x+c^2)*a*b^2-34/21*d*a^2*(- 
7/17*d*x+c)*b+a^3*d^2))/((a*d-b*c)*b)^(1/2)/b^4/(b*x+a)
 
3.5.70.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.53 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx=\left [\frac {15 \, {\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (6 \, b^{3} d^{2} x^{3} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{2} + 2 \, {\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x\right )} \sqrt {d x + c}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (6 \, b^{3} d^{2} x^{3} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{2} + 2 \, {\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x\right )} \sqrt {d x + c}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]

input
integrate(x*(d*x+c)^(5/2)/(b*x+a)^2,x, algorithm="fricas")
 
output
[1/30*(15*(2*a*b^2*c^2 - 9*a^2*b*c*d + 7*a^3*d^2 + (2*b^3*c^2 - 9*a*b^2*c* 
d + 7*a^2*b*d^2)*x)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt( 
d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(6*b^3*d^2*x^3 + 61*a*b^2*c 
^2 - 170*a^2*b*c*d + 105*a^3*d^2 + 2*(11*b^3*c*d - 7*a*b^2*d^2)*x^2 + 2*(2 
3*b^3*c^2 - 59*a*b^2*c*d + 35*a^2*b*d^2)*x)*sqrt(d*x + c))/(b^5*x + a*b^4) 
, -1/15*(15*(2*a*b^2*c^2 - 9*a^2*b*c*d + 7*a^3*d^2 + (2*b^3*c^2 - 9*a*b^2* 
c*d + 7*a^2*b*d^2)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-( 
b*c - a*d)/b)/(b*c - a*d)) - (6*b^3*d^2*x^3 + 61*a*b^2*c^2 - 170*a^2*b*c*d 
 + 105*a^3*d^2 + 2*(11*b^3*c*d - 7*a*b^2*d^2)*x^2 + 2*(23*b^3*c^2 - 59*a*b 
^2*c*d + 35*a^2*b*d^2)*x)*sqrt(d*x + c))/(b^5*x + a*b^4)]
 
3.5.70.6 Sympy [F]

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

input
integrate(x*(d*x+c)**(5/2)/(b*x+a)**2,x)
 
output
Integral(x*(c + d*x)**(5/2)/(a + b*x)**2, x)
 
3.5.70.7 Maxima [F(-2)]

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

input
integrate(x*(d*x+c)^(5/2)/(b*x+a)^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
 
3.5.70.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.35 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx=\frac {{\left (2 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 16 \, a^{2} b c d^{2} - 7 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{4}} + \frac {\sqrt {d x + c} a b^{2} c^{2} d - 2 \, \sqrt {d x + c} a^{2} b c d^{2} + \sqrt {d x + c} a^{3} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{8} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{8} c + 15 \, \sqrt {d x + c} b^{8} c^{2} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{7} d - 60 \, \sqrt {d x + c} a b^{7} c d + 45 \, \sqrt {d x + c} a^{2} b^{6} d^{2}\right )}}{15 \, b^{10}} \]

input
integrate(x*(d*x+c)^(5/2)/(b*x+a)^2,x, algorithm="giac")
 
output
(2*b^3*c^3 - 11*a*b^2*c^2*d + 16*a^2*b*c*d^2 - 7*a^3*d^3)*arctan(sqrt(d*x 
+ c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^4) + (sqrt(d*x + c)*a 
*b^2*c^2*d - 2*sqrt(d*x + c)*a^2*b*c*d^2 + sqrt(d*x + c)*a^3*d^3)/(((d*x + 
 c)*b - b*c + a*d)*b^4) + 2/15*(3*(d*x + c)^(5/2)*b^8 + 5*(d*x + c)^(3/2)* 
b^8*c + 15*sqrt(d*x + c)*b^8*c^2 - 10*(d*x + c)^(3/2)*a*b^7*d - 60*sqrt(d* 
x + c)*a*b^7*c*d + 45*sqrt(d*x + c)*a^2*b^6*d^2)/b^10
 
3.5.70.9 Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.48 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx=\frac {2\,{\left (c+d\,x\right )}^{5/2}}{5\,b^2}-\left (\frac {2\,{\left (a\,d-b\,c\right )}^2}{b^4}+\frac {\left (2\,b^2\,c-2\,a\,b\,d\right )\,\left (\frac {2\,c}{b^2}-\frac {2\,\left (2\,b^2\,c-2\,a\,b\,d\right )}{b^4}\right )}{b^2}\right )\,\sqrt {c+d\,x}-\left (\frac {2\,c}{3\,b^2}-\frac {2\,\left (2\,b^2\,c-2\,a\,b\,d\right )}{3\,b^4}\right )\,{\left (c+d\,x\right )}^{3/2}+\frac {\sqrt {c+d\,x}\,\left (a^3\,d^3-2\,a^2\,b\,c\,d^2+a\,b^2\,c^2\,d\right )}{b^5\,\left (c+d\,x\right )-b^5\,c+a\,b^4\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (7\,a\,d-2\,b\,c\right )\,\sqrt {c+d\,x}}{7\,a^3\,d^3-16\,a^2\,b\,c\,d^2+11\,a\,b^2\,c^2\,d-2\,b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (7\,a\,d-2\,b\,c\right )}{b^{9/2}} \]

input
int((x*(c + d*x)^(5/2))/(a + b*x)^2,x)
 
output
(2*(c + d*x)^(5/2))/(5*b^2) - ((2*(a*d - b*c)^2)/b^4 + ((2*b^2*c - 2*a*b*d 
)*((2*c)/b^2 - (2*(2*b^2*c - 2*a*b*d))/b^4))/b^2)*(c + d*x)^(1/2) - ((2*c) 
/(3*b^2) - (2*(2*b^2*c - 2*a*b*d))/(3*b^4))*(c + d*x)^(3/2) + ((c + d*x)^( 
1/2)*(a^3*d^3 + a*b^2*c^2*d - 2*a^2*b*c*d^2))/(b^5*(c + d*x) - b^5*c + a*b 
^4*d) - (atan((b^(1/2)*(a*d - b*c)^(3/2)*(7*a*d - 2*b*c)*(c + d*x)^(1/2))/ 
(7*a^3*d^3 - 2*b^3*c^3 + 11*a*b^2*c^2*d - 16*a^2*b*c*d^2))*(a*d - b*c)^(3/ 
2)*(7*a*d - 2*b*c))/b^(9/2)