\(\int \frac {c+d x}{x^2 (a+b x^2)^3} \, dx\) [227]

Optimal result

Integrand size = 18, antiderivative size = 112 \[ \int \frac {c+d x}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {c}{a^3 x}+\frac {a d-b c x}{4 a^2 \left (a+b x^2\right )^2}+\frac {4 a d-7 b c x}{8 a^3 \left (a+b x^2\right )}-\frac {15 \sqrt {b} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}+\frac {d \log (x)}{a^3}-\frac {d \log \left (a+b x^2\right )}{2 a^3} \] Output:

-c/a^3/x+1/4*(-b*c*x+a*d)/a^2/(b*x^2+a)^2+1/8*(-7*b*c*x+4*a*d)/a^3/(b*x^2+ 
a)-15/8*b^(1/2)*c*arctan(b^(1/2)*x/a^(1/2))/a^(7/2)+d*ln(x)/a^3-1/2*d*ln(b 
*x^2+a)/a^3
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {c}{a^3 x}+\frac {a d-b c x}{4 a^2 \left (a+b x^2\right )^2}+\frac {4 a d-7 b c x}{8 a^3 \left (a+b x^2\right )}-\frac {15 \sqrt {b} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}+\frac {d \log (x)}{a^3}-\frac {d \log \left (a+b x^2\right )}{2 a^3} \] Input:

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

Output:

-(c/(a^3*x)) + (a*d - b*c*x)/(4*a^2*(a + b*x^2)^2) + (4*a*d - 7*b*c*x)/(8* 
a^3*(a + b*x^2)) - (15*Sqrt[b]*c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)) 
+ (d*Log[x])/a^3 - (d*Log[a + b*x^2])/(2*a^3)
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {532, 25, 2336, 25, 2333, 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.

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

\(\Big \downarrow \) 532

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2336

\(\displaystyle \frac {\frac {4 a d-7 b c x}{2 a^2 \left (a+b x^2\right )}-\frac {\int -\frac {-\frac {7 b c x^2}{a}+8 d x+8 c}{x^2 \left (b x^2+a\right )}dx}{2 a}}{4 a}+\frac {a d-b c x}{4 a^2 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {-\frac {7 b c x^2}{a}+8 d x+8 c}{x^2 \left (b x^2+a\right )}dx}{2 a}+\frac {4 a d-7 b c x}{2 a^2 \left (a+b x^2\right )}}{4 a}+\frac {a d-b c x}{4 a^2 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2333

\(\displaystyle \frac {\frac {\int \left (\frac {8 c}{a x^2}+\frac {8 d}{a x}+\frac {-15 b c-8 b d x}{a \left (b x^2+a\right )}\right )dx}{2 a}+\frac {4 a d-7 b c x}{2 a^2 \left (a+b x^2\right )}}{4 a}+\frac {a d-b c x}{4 a^2 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a d-b c x}{4 a^2 \left (a+b x^2\right )^2}+\frac {\frac {-\frac {15 \sqrt {b} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {4 d \log \left (a+b x^2\right )}{a}-\frac {8 c}{a x}+\frac {8 d \log (x)}{a}}{2 a}+\frac {4 a d-7 b c x}{2 a^2 \left (a+b x^2\right )}}{4 a}\)

Input:

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

Output:

(a*d - b*c*x)/(4*a^2*(a + b*x^2)^2) + ((4*a*d - 7*b*c*x)/(2*a^2*(a + b*x^2 
)) + ((-8*c)/(a*x) - (15*Sqrt[b]*c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(3/2) + 
(8*d*Log[x])/a - (4*d*Log[a + b*x^2])/a)/(2*a))/(4*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 2009

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

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {c}{a^{3} x}+\frac {d \ln \left (x \right )}{a^{3}}-\frac {b \left (\frac {\frac {7 b c \,x^{3}}{8}-\frac {a d \,x^{2}}{2}+\frac {9 a c x}{8}-\frac {3 a^{2} d}{4 b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {d \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {15 c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}\)	\(95\)
risch	\(\frac {-\frac {15 b^{2} c \,x^{4}}{8 a^{3}}+\frac {d b \,x^{3}}{2 a^{2}}-\frac {25 b c \,x^{2}}{8 a^{2}}+\frac {3 d x}{4 a}-\frac {c}{a}}{x \left (b \,x^{2}+a \right )^{2}}+\frac {d \ln \left (x \right )}{a^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{2} a^{7}+16 a^{4} d \textit {\_Z} +64 a \,d^{2}+225 b \,c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (a^{7} \textit {\_R}^{2}+8 a^{4} d \textit {\_R} +150 b \,c^{2}\right ) x +5 a^{4} c \textit {\_R} -80 a c d \right )\right )}{16}\)	\(140\)

Input:

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

Output:

-c/a^3/x+d*ln(x)/a^3-b/a^3*((7/8*b*c*x^3-1/2*a*d*x^2+9/8*a*c*x-3/4*a^2/b*d 
)/(b*x^2+a)^2+1/2*d*ln(b*x^2+a)/b+15/8*c/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2 
)))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 364, normalized size of antiderivative = 3.25 \[ \int \frac {c+d x}{x^2 \left (a+b x^2\right )^3} \, dx=\left [-\frac {30 \, b^{2} c x^{4} - 8 \, a b d x^{3} + 50 \, a b c x^{2} - 12 \, a^{2} d x + 16 \, a^{2} c - 15 \, {\left (b^{2} c x^{5} + 2 \, a b c x^{3} + a^{2} c x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 8 \, {\left (b^{2} d x^{5} + 2 \, a b d x^{3} + a^{2} d x\right )} \log \left (b x^{2} + a\right ) - 16 \, {\left (b^{2} d x^{5} + 2 \, a b d x^{3} + a^{2} d x\right )} \log \left (x\right )}{16 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}, -\frac {15 \, b^{2} c x^{4} - 4 \, a b d x^{3} + 25 \, a b c x^{2} - 6 \, a^{2} d x + 8 \, a^{2} c + 15 \, {\left (b^{2} c x^{5} + 2 \, a b c x^{3} + a^{2} c x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 4 \, {\left (b^{2} d x^{5} + 2 \, a b d x^{3} + a^{2} d x\right )} \log \left (b x^{2} + a\right ) - 8 \, {\left (b^{2} d x^{5} + 2 \, a b d x^{3} + a^{2} d x\right )} \log \left (x\right )}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}\right ] \] Input:

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

Output:

[-1/16*(30*b^2*c*x^4 - 8*a*b*d*x^3 + 50*a*b*c*x^2 - 12*a^2*d*x + 16*a^2*c 
- 15*(b^2*c*x^5 + 2*a*b*c*x^3 + a^2*c*x)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqr 
t(-b/a) - a)/(b*x^2 + a)) + 8*(b^2*d*x^5 + 2*a*b*d*x^3 + a^2*d*x)*log(b*x^ 
2 + a) - 16*(b^2*d*x^5 + 2*a*b*d*x^3 + a^2*d*x)*log(x))/(a^3*b^2*x^5 + 2*a 
^4*b*x^3 + a^5*x), -1/8*(15*b^2*c*x^4 - 4*a*b*d*x^3 + 25*a*b*c*x^2 - 6*a^2 
*d*x + 8*a^2*c + 15*(b^2*c*x^5 + 2*a*b*c*x^3 + a^2*c*x)*sqrt(b/a)*arctan(x 
*sqrt(b/a)) + 4*(b^2*d*x^5 + 2*a*b*d*x^3 + a^2*d*x)*log(b*x^2 + a) - 8*(b^ 
2*d*x^5 + 2*a*b*d*x^3 + a^2*d*x)*log(x))/(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5* 
x)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (105) = 210\).

Time = 1.20 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.82 \[ \int \frac {c+d x}{x^2 \left (a+b x^2\right )^3} \, dx=\left (- \frac {d}{2 a^{3}} - \frac {15 c \sqrt {- a^{7} b}}{16 a^{7}}\right ) \log {\left (x + \frac {2048 a^{8} d \left (- \frac {d}{2 a^{3}} - \frac {15 c \sqrt {- a^{7} b}}{16 a^{7}}\right )^{2} - 1024 a^{5} d^{2} \left (- \frac {d}{2 a^{3}} - \frac {15 c \sqrt {- a^{7} b}}{16 a^{7}}\right ) - 1200 a^{4} b c^{2} \left (- \frac {d}{2 a^{3}} - \frac {15 c \sqrt {- a^{7} b}}{16 a^{7}}\right ) - 1024 a^{2} d^{3} + 1200 a b c^{2} d}{2880 a b c d^{2} + 1125 b^{2} c^{3}} \right )} + \left (- \frac {d}{2 a^{3}} + \frac {15 c \sqrt {- a^{7} b}}{16 a^{7}}\right ) \log {\left (x + \frac {2048 a^{8} d \left (- \frac {d}{2 a^{3}} + \frac {15 c \sqrt {- a^{7} b}}{16 a^{7}}\right )^{2} - 1024 a^{5} d^{2} \left (- \frac {d}{2 a^{3}} + \frac {15 c \sqrt {- a^{7} b}}{16 a^{7}}\right ) - 1200 a^{4} b c^{2} \left (- \frac {d}{2 a^{3}} + \frac {15 c \sqrt {- a^{7} b}}{16 a^{7}}\right ) - 1024 a^{2} d^{3} + 1200 a b c^{2} d}{2880 a b c d^{2} + 1125 b^{2} c^{3}} \right )} + \frac {- 8 a^{2} c + 6 a^{2} d x - 25 a b c x^{2} + 4 a b d x^{3} - 15 b^{2} c x^{4}}{8 a^{5} x + 16 a^{4} b x^{3} + 8 a^{3} b^{2} x^{5}} + \frac {d \log {\left (x \right )}}{a^{3}} \] Input:

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

Output:

(-d/(2*a**3) - 15*c*sqrt(-a**7*b)/(16*a**7))*log(x + (2048*a**8*d*(-d/(2*a 
**3) - 15*c*sqrt(-a**7*b)/(16*a**7))**2 - 1024*a**5*d**2*(-d/(2*a**3) - 15 
*c*sqrt(-a**7*b)/(16*a**7)) - 1200*a**4*b*c**2*(-d/(2*a**3) - 15*c*sqrt(-a 
**7*b)/(16*a**7)) - 1024*a**2*d**3 + 1200*a*b*c**2*d)/(2880*a*b*c*d**2 + 1 
125*b**2*c**3)) + (-d/(2*a**3) + 15*c*sqrt(-a**7*b)/(16*a**7))*log(x + (20 
48*a**8*d*(-d/(2*a**3) + 15*c*sqrt(-a**7*b)/(16*a**7))**2 - 1024*a**5*d**2 
*(-d/(2*a**3) + 15*c*sqrt(-a**7*b)/(16*a**7)) - 1200*a**4*b*c**2*(-d/(2*a* 
*3) + 15*c*sqrt(-a**7*b)/(16*a**7)) - 1024*a**2*d**3 + 1200*a*b*c**2*d)/(2 
880*a*b*c*d**2 + 1125*b**2*c**3)) + (-8*a**2*c + 6*a**2*d*x - 25*a*b*c*x** 
2 + 4*a*b*d*x**3 - 15*b**2*c*x**4)/(8*a**5*x + 16*a**4*b*x**3 + 8*a**3*b** 
2*x**5) + d*log(x)/a**3
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {15 \, b^{2} c x^{4} - 4 \, a b d x^{3} + 25 \, a b c x^{2} - 6 \, a^{2} d x + 8 \, a^{2} c}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} - \frac {15 \, b c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {d \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {d \log \left (x\right )}{a^{3}} \] Input:

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

Output:

-1/8*(15*b^2*c*x^4 - 4*a*b*d*x^3 + 25*a*b*c*x^2 - 6*a^2*d*x + 8*a^2*c)/(a^ 
3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x) - 15/8*b*c*arctan(b*x/sqrt(a*b))/(sqrt(a* 
b)*a^3) - 1/2*d*log(b*x^2 + a)/a^3 + d*log(x)/a^3
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {15 \, b c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {d \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {d \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {15 \, b^{2} c x^{4} - 4 \, a b d x^{3} + 25 \, a b c x^{2} - 6 \, a^{2} d x + 8 \, a^{2} c}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} x} \] Input:

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

Output:

-15/8*b*c*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) - 1/2*d*log(b*x^2 + a)/a^3 
 + d*log(abs(x))/a^3 - 1/8*(15*b^2*c*x^4 - 4*a*b*d*x^3 + 25*a*b*c*x^2 - 6* 
a^2*d*x + 8*a^2*c)/((b*x^2 + a)^2*a^3*x)
 

Mupad [B] (verification not implemented)

Time = 7.09 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.79 \[ \int \frac {c+d x}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {d\,\ln \left (x\right )}{a^3}-\frac {\ln \left (8\,a^4\,d-5\,c\,\sqrt {-a^7\,b}+8\,d\,x\,\sqrt {-a^7\,b}+5\,a^3\,b\,c\,x\right )\,\left (8\,a^4\,d-15\,c\,\sqrt {-a^7\,b}\right )}{16\,a^7}-\frac {\ln \left (8\,a^4\,d+5\,c\,\sqrt {-a^7\,b}-8\,d\,x\,\sqrt {-a^7\,b}+5\,a^3\,b\,c\,x\right )\,\left (8\,a^4\,d+15\,c\,\sqrt {-a^7\,b}\right )}{16\,a^7}-\frac {\frac {c}{a}-\frac {3\,d\,x}{4\,a}+\frac {15\,b^2\,c\,x^4}{8\,a^3}+\frac {25\,b\,c\,x^2}{8\,a^2}-\frac {b\,d\,x^3}{2\,a^2}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5} \] Input:

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

Output:

(d*log(x))/a^3 - (log(8*a^4*d - 5*c*(-a^7*b)^(1/2) + 8*d*x*(-a^7*b)^(1/2) 
+ 5*a^3*b*c*x)*(8*a^4*d - 15*c*(-a^7*b)^(1/2)))/(16*a^7) - (log(8*a^4*d + 
5*c*(-a^7*b)^(1/2) - 8*d*x*(-a^7*b)^(1/2) + 5*a^3*b*c*x)*(8*a^4*d + 15*c*( 
-a^7*b)^(1/2)))/(16*a^7) - (c/a - (3*d*x)/(4*a) + (15*b^2*c*x^4)/(8*a^3) + 
 (25*b*c*x^2)/(8*a^2) - (b*d*x^3)/(2*a^2))/(a^2*x + b^2*x^5 + 2*a*b*x^3)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.04 \[ \int \frac {c+d x}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} c x -30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b c \,x^{3}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c \,x^{5}-4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} d x -8 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b d \,x^{3}-4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} d \,x^{5}+8 \,\mathrm {log}\left (x \right ) a^{3} d x +16 \,\mathrm {log}\left (x \right ) a^{2} b d \,x^{3}+8 \,\mathrm {log}\left (x \right ) a \,b^{2} d \,x^{5}-8 a^{3} c +4 a^{3} d x -25 a^{2} b c \,x^{2}-15 a \,b^{2} c \,x^{4}-2 a \,b^{2} d \,x^{5}}{8 a^{4} x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*c*x - 30*sqrt(b) 
*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*c*x**3 - 15*sqrt(b)*sqrt(a)*ata 
n((b*x)/(sqrt(b)*sqrt(a)))*b**2*c*x**5 - 4*log(a + b*x**2)*a**3*d*x - 8*lo 
g(a + b*x**2)*a**2*b*d*x**3 - 4*log(a + b*x**2)*a*b**2*d*x**5 + 8*log(x)*a 
**3*d*x + 16*log(x)*a**2*b*d*x**3 + 8*log(x)*a*b**2*d*x**5 - 8*a**3*c + 4* 
a**3*d*x - 25*a**2*b*c*x**2 - 15*a*b**2*c*x**4 - 2*a*b**2*d*x**5)/(8*a**4* 
x*(a**2 + 2*a*b*x**2 + b**2*x**4))