\(\int \frac {x^4 (a+b x^2)}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\) [32]

Optimal result

Integrand size = 31, antiderivative size = 169 \[ \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {\left (\frac {a}{c^2}+\frac {b}{d^2}\right ) x^5}{\sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 \left (5 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^6}+\frac {\left (5 b c^2+4 a d^2\right ) x^3 \sqrt {-c+d x} \sqrt {c+d x}}{4 c^2 d^4}+\frac {3 c^2 \left (5 b c^2+4 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{4 d^7} \] Output:

-(a/c^2+b/d^2)*x^5/(d*x-c)^(1/2)/(d*x+c)^(1/2)+3/8*(4*a*d^2+5*b*c^2)*x*(d* 
x-c)^(1/2)*(d*x+c)^(1/2)/d^6+1/4*(4*a*d^2+5*b*c^2)*x^3*(d*x-c)^(1/2)*(d*x+ 
c)^(1/2)/c^2/d^4+3/4*c^2*(4*a*d^2+5*b*c^2)*arctanh((d*x-c)^(1/2)/(d*x+c)^( 
1/2))/d^7
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.81 \[ \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {4 a d^3 x \left (-3 c^2+d^2 x^2\right )+b d x \left (-15 c^4+5 c^2 d^2 x^2+2 d^4 x^4\right )+6 c^2 \left (5 b c^2+4 a d^2\right ) \sqrt {-c+d x} \sqrt {c+d x} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {-c+d x}}\right )}{8 d^7 \sqrt {-c+d x} \sqrt {c+d x}} \] Input:

Integrate[(x^4*(a + b*x^2))/((-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]
 

Output:

(4*a*d^3*x*(-3*c^2 + d^2*x^2) + b*d*x*(-15*c^4 + 5*c^2*d^2*x^2 + 2*d^4*x^4 
) + 6*c^2*(5*b*c^2 + 4*a*d^2)*Sqrt[-c + d*x]*Sqrt[c + d*x]*ArcTanh[Sqrt[c 
+ d*x]/Sqrt[-c + d*x]])/(8*d^7*Sqrt[-c + d*x]*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {960, 109, 27, 101, 27, 45, 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.

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

\(\Big \downarrow \) 960

\(\displaystyle \frac {1}{4} \left (4 a+\frac {5 b c^2}{d^2}\right ) \int \frac {x^4}{(d x-c)^{3/2} (c+d x)^{3/2}}dx+\frac {b x^5}{4 d^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 109

\(\displaystyle \frac {1}{4} \left (4 a+\frac {5 b c^2}{d^2}\right ) \left (-\frac {\int -\frac {3 c x^2}{\sqrt {d x-c} \sqrt {c+d x}}dx}{c d^2}-\frac {x^3}{d^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {b x^5}{4 d^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (4 a+\frac {5 b c^2}{d^2}\right ) \left (\frac {3 \int \frac {x^2}{\sqrt {d x-c} \sqrt {c+d x}}dx}{d^2}-\frac {x^3}{d^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {b x^5}{4 d^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 101

\(\displaystyle \frac {1}{4} \left (4 a+\frac {5 b c^2}{d^2}\right ) \left (\frac {3 \left (\frac {\int \frac {c^2}{\sqrt {d x-c} \sqrt {c+d x}}dx}{2 d^2}+\frac {x \sqrt {d x-c} \sqrt {c+d x}}{2 d^2}\right )}{d^2}-\frac {x^3}{d^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {b x^5}{4 d^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (4 a+\frac {5 b c^2}{d^2}\right ) \left (\frac {3 \left (\frac {c^2 \int \frac {1}{\sqrt {d x-c} \sqrt {c+d x}}dx}{2 d^2}+\frac {x \sqrt {d x-c} \sqrt {c+d x}}{2 d^2}\right )}{d^2}-\frac {x^3}{d^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {b x^5}{4 d^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 45

\(\displaystyle \frac {1}{4} \left (4 a+\frac {5 b c^2}{d^2}\right ) \left (\frac {3 \left (\frac {c^2 \int \frac {1}{d-\frac {d (d x-c)}{c+d x}}d\frac {\sqrt {d x-c}}{\sqrt {c+d x}}}{d^2}+\frac {x \sqrt {d x-c} \sqrt {c+d x}}{2 d^2}\right )}{d^2}-\frac {x^3}{d^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {b x^5}{4 d^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4} \left (4 a+\frac {5 b c^2}{d^2}\right ) \left (\frac {3 \left (\frac {c^2 \text {arctanh}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{d^3}+\frac {x \sqrt {d x-c} \sqrt {c+d x}}{2 d^2}\right )}{d^2}-\frac {x^3}{d^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {b x^5}{4 d^2 \sqrt {d x-c} \sqrt {c+d x}}\)

Input:

Int[(x^4*(a + b*x^2))/((-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]
 

Output:

(b*x^5)/(4*d^2*Sqrt[-c + d*x]*Sqrt[c + d*x]) + ((4*a + (5*b*c^2)/d^2)*(-(x 
^3/(d^2*Sqrt[-c + d*x]*Sqrt[c + d*x])) + (3*((x*Sqrt[-c + d*x]*Sqrt[c + d* 
x])/(2*d^2) + (c^2*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]])/d^3))/d^2))/4
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 45

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] && EqQ[b*c + a*d, 0] &&  !GtQ[c, 0]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + 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]
 

rule 960

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.) 
*(x_)^(non2_.))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^( 
m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(b1*b2*e*(m + n 
*(p + 1) + 1))), x] - Simp[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/ 
(b1*b2*(m + n*(p + 1) + 1))   Int[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n 
/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] && EqQ[non2, 
 n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.63

method	result	size
risch	\(-\frac {x \left (2 b \,x^{2} d^{2}+4 a \,d^{2}+7 b \,c^{2}\right ) \left (-d x +c \right ) \sqrt {d x +c}}{8 d^{6} \sqrt {d x -c}}+\frac {c^{2} \left (\frac {12 a \,d^{2} \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right )}{\sqrt {d^{2}}}+\frac {15 b \,c^{2} \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right )}{\sqrt {d^{2}}}-\frac {4 \left (a \,d^{2}+b \,c^{2}\right ) \sqrt {d^{2} \left (x -\frac {c}{d}\right )^{2}+2 c d \left (x -\frac {c}{d}\right )}}{d^{2} \left (x -\frac {c}{d}\right )}-\frac {4 \left (a \,d^{2}+b \,c^{2}\right ) \sqrt {d^{2} \left (x +\frac {c}{d}\right )^{2}-2 c d \left (x +\frac {c}{d}\right )}}{d^{2} \left (x +\frac {c}{d}\right )}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{8 d^{6} \sqrt {d x -c}\, \sqrt {d x +c}}\)	\(275\)
default	\(-\frac {\left (-2 \,\operatorname {csgn}\left (d \right ) b \,d^{5} x^{5} \sqrt {d^{2} x^{2}-c^{2}}-4 \,\operatorname {csgn}\left (d \right ) a \,d^{5} x^{3} \sqrt {d^{2} x^{2}-c^{2}}-5 \,\operatorname {csgn}\left (d \right ) b \,c^{2} d^{3} x^{3} \sqrt {d^{2} x^{2}-c^{2}}-12 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) a \,c^{2} d^{4} x^{2}-15 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) b \,c^{4} d^{2} x^{2}+12 \,\operatorname {csgn}\left (d \right ) d^{3} \sqrt {d^{2} x^{2}-c^{2}}\, a \,c^{2} x +15 \,\operatorname {csgn}\left (d \right ) d \sqrt {d^{2} x^{2}-c^{2}}\, b \,c^{4} x +12 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) a \,c^{4} d^{2}+15 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) b \,c^{6}\right ) \operatorname {csgn}\left (d \right )}{8 \sqrt {d x -c}\, \sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {d x +c}\, d^{7}}\)	\(316\)

Input:

int(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8*x*(2*b*d^2*x^2+4*a*d^2+7*b*c^2)*(-d*x+c)*(d*x+c)^(1/2)/d^6/(d*x-c)^(1 
/2)+1/8*c^2/d^6*(12*a*d^2*ln(d^2*x/(d^2)^(1/2)+(d^2*x^2-c^2)^(1/2))/(d^2)^ 
(1/2)+15*b*c^2*ln(d^2*x/(d^2)^(1/2)+(d^2*x^2-c^2)^(1/2))/(d^2)^(1/2)-4*(a* 
d^2+b*c^2)/d^2/(x-c/d)*(d^2*(x-c/d)^2+2*c*d*(x-c/d))^(1/2)-4*(a*d^2+b*c^2) 
/d^2/(x+c/d)*(d^2*(x+c/d)^2-2*c*d*(x+c/d))^(1/2))*((d*x-c)*(d*x+c))^(1/2)/ 
(d*x-c)^(1/2)/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.12 \[ \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {8 \, b c^{6} + 8 \, a c^{4} d^{2} - 8 \, {\left (b c^{4} d^{2} + a c^{2} d^{4}\right )} x^{2} + {\left (2 \, b d^{5} x^{5} + {\left (5 \, b c^{2} d^{3} + 4 \, a d^{5}\right )} x^{3} - 3 \, {\left (5 \, b c^{4} d + 4 \, a c^{2} d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {d x - c} + 3 \, {\left (5 \, b c^{6} + 4 \, a c^{4} d^{2} - {\left (5 \, b c^{4} d^{2} + 4 \, a c^{2} d^{4}\right )} x^{2}\right )} \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right )}{8 \, {\left (d^{9} x^{2} - c^{2} d^{7}\right )}} \] Input:

integrate(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

1/8*(8*b*c^6 + 8*a*c^4*d^2 - 8*(b*c^4*d^2 + a*c^2*d^4)*x^2 + (2*b*d^5*x^5 
+ (5*b*c^2*d^3 + 4*a*d^5)*x^3 - 3*(5*b*c^4*d + 4*a*c^2*d^3)*x)*sqrt(d*x + 
c)*sqrt(d*x - c) + 3*(5*b*c^6 + 4*a*c^4*d^2 - (5*b*c^4*d^2 + 4*a*c^2*d^4)* 
x^2)*log(-d*x + sqrt(d*x + c)*sqrt(d*x - c)))/(d^9*x^2 - c^2*d^7)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(x**4*(b*x**2+a)/(d*x-c)**(3/2)/(d*x+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.16 \[ \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {b x^{5}}{4 \, \sqrt {d^{2} x^{2} - c^{2}} d^{2}} + \frac {5 \, b c^{2} x^{3}}{8 \, \sqrt {d^{2} x^{2} - c^{2}} d^{4}} + \frac {a x^{3}}{2 \, \sqrt {d^{2} x^{2} - c^{2}} d^{2}} - \frac {15 \, b c^{4} x}{8 \, \sqrt {d^{2} x^{2} - c^{2}} d^{6}} - \frac {3 \, a c^{2} x}{2 \, \sqrt {d^{2} x^{2} - c^{2}} d^{4}} + \frac {15 \, b c^{4} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{8 \, d^{7}} + \frac {3 \, a c^{2} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{2 \, d^{5}} \] Input:

integrate(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

1/4*b*x^5/(sqrt(d^2*x^2 - c^2)*d^2) + 5/8*b*c^2*x^3/(sqrt(d^2*x^2 - c^2)*d 
^4) + 1/2*a*x^3/(sqrt(d^2*x^2 - c^2)*d^2) - 15/8*b*c^4*x/(sqrt(d^2*x^2 - c 
^2)*d^6) - 3/2*a*c^2*x/(sqrt(d^2*x^2 - c^2)*d^4) + 15/8*b*c^4*log(2*d^2*x 
+ 2*sqrt(d^2*x^2 - c^2)*d)/d^7 + 3/2*a*c^2*log(2*d^2*x + 2*sqrt(d^2*x^2 - 
c^2)*d)/d^5
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.27 \[ \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {{\left ({\left ({\left (d x + c\right )} {\left (2 \, {\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{7}} - \frac {5 \, b c}{d^{7}}\right )} + \frac {25 \, b c^{2} d^{35} + 4 \, a d^{37}}{d^{42}}\right )} - \frac {35 \, b c^{3} d^{35} + 12 \, a c d^{37}}{d^{42}}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (7 \, b c^{4} d^{35} + 2 \, a c^{2} d^{37}\right )}}{d^{42}}\right )} \sqrt {d x + c}}{8 \, \sqrt {d x - c}} - \frac {3 \, {\left (5 \, b c^{4} + 4 \, a c^{2} d^{2}\right )} \log \left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}\right )}{8 \, d^{7}} - \frac {2 \, {\left (b c^{5} + a c^{3} d^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, c\right )} d^{7}} \] Input:

integrate(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

1/8*(((d*x + c)*(2*(d*x + c)*((d*x + c)*b/d^7 - 5*b*c/d^7) + (25*b*c^2*d^3 
5 + 4*a*d^37)/d^42) - (35*b*c^3*d^35 + 12*a*c*d^37)/d^42)*(d*x + c) + 2*(7 
*b*c^4*d^35 + 2*a*c^2*d^37)/d^42)*sqrt(d*x + c)/sqrt(d*x - c) - 3/8*(5*b*c 
^4 + 4*a*c^2*d^2)*log((sqrt(d*x + c) - sqrt(d*x - c))^2)/d^7 - 2*(b*c^5 + 
a*c^3*d^2)/(((sqrt(d*x + c) - sqrt(d*x - c))^2 + 2*c)*d^7)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\int \frac {x^4\,\left (b\,x^2+a\right )}{{\left (c+d\,x\right )}^{3/2}\,{\left (d\,x-c\right )}^{3/2}} \,d x \] Input:

int((x^4*(a + b*x^2))/((c + d*x)^(3/2)*(d*x - c)^(3/2)),x)
 

Output:

int((x^4*(a + b*x^2))/((c + d*x)^(3/2)*(d*x - c)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.96 \[ \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {24 \sqrt {d x -c}\, \mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) a \,c^{3} d^{2}+24 \sqrt {d x -c}\, \mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) a \,c^{2} d^{3} x +30 \sqrt {d x -c}\, \mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) b \,c^{5}+30 \sqrt {d x -c}\, \mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) b \,c^{4} d x -9 \sqrt {d x -c}\, a \,c^{3} d^{2}-9 \sqrt {d x -c}\, a \,c^{2} d^{3} x -10 \sqrt {d x -c}\, b \,c^{5}-10 \sqrt {d x -c}\, b \,c^{4} d x -12 \sqrt {d x +c}\, a \,c^{2} d^{3} x +4 \sqrt {d x +c}\, a \,d^{5} x^{3}-15 \sqrt {d x +c}\, b \,c^{4} d x +5 \sqrt {d x +c}\, b \,c^{2} d^{3} x^{3}+2 \sqrt {d x +c}\, b \,d^{5} x^{5}}{8 \sqrt {d x -c}\, d^{7} \left (d x +c \right )} \] Input:

int(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x)
 

Output:

(24*sqrt( - c + d*x)*log((sqrt( - c + d*x) + sqrt(c + d*x))/(sqrt(c)*sqrt( 
2)))*a*c**3*d**2 + 24*sqrt( - c + d*x)*log((sqrt( - c + d*x) + sqrt(c + d* 
x))/(sqrt(c)*sqrt(2)))*a*c**2*d**3*x + 30*sqrt( - c + d*x)*log((sqrt( - c 
+ d*x) + sqrt(c + d*x))/(sqrt(c)*sqrt(2)))*b*c**5 + 30*sqrt( - c + d*x)*lo 
g((sqrt( - c + d*x) + sqrt(c + d*x))/(sqrt(c)*sqrt(2)))*b*c**4*d*x - 9*sqr 
t( - c + d*x)*a*c**3*d**2 - 9*sqrt( - c + d*x)*a*c**2*d**3*x - 10*sqrt( - 
c + d*x)*b*c**5 - 10*sqrt( - c + d*x)*b*c**4*d*x - 12*sqrt(c + d*x)*a*c**2 
*d**3*x + 4*sqrt(c + d*x)*a*d**5*x**3 - 15*sqrt(c + d*x)*b*c**4*d*x + 5*sq 
rt(c + d*x)*b*c**2*d**3*x**3 + 2*sqrt(c + d*x)*b*d**5*x**5)/(8*sqrt( - c + 
 d*x)*d**7*(c + d*x))