Integrand size = 22, antiderivative size = 157 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x} \, dx=2 a^2 c \sqrt {c+d x}+\frac {2}{3} a^2 (c+d x)^{3/2}-\frac {2 b c \left (b c^2+2 a d^2\right ) (c+d x)^{5/2}}{5 d^4}+\frac {2 b \left (3 b c^2+2 a d^2\right ) (c+d x)^{7/2}}{7 d^4}-\frac {2 b^2 c (c+d x)^{9/2}}{3 d^4}+\frac {2 b^2 (c+d x)^{11/2}}{11 d^4}-2 a^2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Output:
2*a^2*c*(d*x+c)^(1/2)+2/3*a^2*(d*x+c)^(3/2)-2/5*b*c*(2*a*d^2+b*c^2)*(d*x+c )^(5/2)/d^4+2/7*b*(2*a*d^2+3*b*c^2)*(d*x+c)^(7/2)/d^4-2/3*b^2*c*(d*x+c)^(9 /2)/d^4+2/11*b^2*(d*x+c)^(11/2)/d^4-2*a^2*c^(3/2)*arctanh((d*x+c)^(1/2)/c^ (1/2))
Time = 0.22 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.78 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x} \, dx=\frac {2 \sqrt {c+d x} \left (385 a^2 d^4 (4 c+d x)+66 a b d^2 (c+d x)^2 (-2 c+5 d x)-b^2 (c+d x)^2 \left (16 c^3-40 c^2 d x+70 c d^2 x^2-105 d^3 x^3\right )\right )}{1155 d^4}-2 a^2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Input:
Integrate[((c + d*x)^(3/2)*(a + b*x^2)^2)/x,x]
Output:
(2*Sqrt[c + d*x]*(385*a^2*d^4*(4*c + d*x) + 66*a*b*d^2*(c + d*x)^2*(-2*c + 5*d*x) - b^2*(c + d*x)^2*(16*c^3 - 40*c^2*d*x + 70*c*d^2*x^2 - 105*d^3*x^ 3)))/(1155*d^4) - 2*a^2*c^(3/2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]
Time = 0.50 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {517, 25, 1584, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^2 (c+d x)^{3/2}}{x} \, dx\) |
| \(\Big \downarrow \) 517 |
| \(\displaystyle \frac {2 \int \frac {(c+d x)^2 \left (b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2\right )^2}{d x}d\sqrt {c+d x}}{d^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 \int -\frac {(c+d x)^2 \left (b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2\right )^2}{d x}d\sqrt {c+d x}}{d^4}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle -\frac {2 \int \left (-b^2 (c+d x)^5+3 b^2 c (c+d x)^4-b \left (3 b c^2+2 a d^2\right ) (c+d x)^3+b c \left (b c^2+2 a d^2\right ) (c+d x)^2-a^2 d^4 (c+d x)-a^2 c d^4-\frac {a^2 c^2 d^3}{x}\right )d\sqrt {c+d x}}{d^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (-a^2 c^{3/2} d^4 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\frac {1}{3} a^2 d^4 (c+d x)^{3/2}+a^2 c d^4 \sqrt {c+d x}+\frac {1}{7} b (c+d x)^{7/2} \left (2 a d^2+3 b c^2\right )-\frac {1}{5} b c (c+d x)^{5/2} \left (2 a d^2+b c^2\right )+\frac {1}{11} b^2 (c+d x)^{11/2}-\frac {1}{3} b^2 c (c+d x)^{9/2}\right )}{d^4}\) |
Input:
Int[((c + d*x)^(3/2)*(a + b*x^2)^2)/x,x]
Output:
(2*(a^2*c*d^4*Sqrt[c + d*x] + (a^2*d^4*(c + d*x)^(3/2))/3 - (b*c*(b*c^2 + 2*a*d^2)*(c + d*x)^(5/2))/5 + (b*(3*b*c^2 + 2*a*d^2)*(c + d*x)^(7/2))/7 - (b^2*c*(c + d*x)^(9/2))/3 + (b^2*(c + d*x)^(11/2))/11 - a^2*c^(3/2)*d^4*Ar cTanh[Sqrt[c + d*x]/Sqrt[c]]))/d^4
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.91
| method | result | size |
| pseudoelliptic | \(\frac {-2310 a^{2} c^{\frac {3}{2}} d^{4} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+3080 \left (\left (\frac {3}{44} b^{2} x^{5}+\frac {3}{14} a b \,x^{3}+\frac {1}{4} a^{2} x \right ) d^{5}+c \left (\frac {1}{11} b^{2} x^{4}+\frac {12}{35} a b \,x^{2}+a^{2}\right ) d^{4}+\frac {3 x b \,c^{2} \left (\frac {5 b \,x^{2}}{66}+a \right ) d^{3}}{70}-\frac {3 b \left (\frac {b \,x^{2}}{22}+a \right ) c^{3} d^{2}}{35}+\frac {2 b^{2} c^{4} d x}{385}-\frac {4 c^{5} b^{2}}{385}\right ) \sqrt {d x +c}}{1155 d^{4}}\) | \(143\) |
| derivativedivides | \(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {11}{2}}}{11}-\frac {2 b^{2} c \left (d x +c \right )^{\frac {9}{2}}}{3}+\frac {4 a b \,d^{2} \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {6 b^{2} c^{2} \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {4 a b c \,d^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 b^{2} c^{3} \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} d^{4} \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a^{2} c \,d^{4} \sqrt {d x +c}-2 a^{2} c^{\frac {3}{2}} d^{4} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d^{4}}\) | \(144\) |
| default | \(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {11}{2}}}{11}-\frac {2 b^{2} c \left (d x +c \right )^{\frac {9}{2}}}{3}+\frac {4 a b \,d^{2} \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {6 b^{2} c^{2} \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {4 a b c \,d^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 b^{2} c^{3} \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} d^{4} \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a^{2} c \,d^{4} \sqrt {d x +c}-2 a^{2} c^{\frac {3}{2}} d^{4} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d^{4}}\) | \(144\) |
Input:
int((d*x+c)^(3/2)*(b*x^2+a)^2/x,x,method=_RETURNVERBOSE)
Output:
1/1155*(-2310*a^2*c^(3/2)*d^4*arctanh((d*x+c)^(1/2)/c^(1/2))+3080*((3/44*b ^2*x^5+3/14*a*b*x^3+1/4*a^2*x)*d^5+c*(1/11*b^2*x^4+12/35*a*b*x^2+a^2)*d^4+ 3/70*x*b*c^2*(5/66*b*x^2+a)*d^3-3/35*b*(1/22*b*x^2+a)*c^3*d^2+2/385*b^2*c^ 4*d*x-4/385*c^5*b^2)*(d*x+c)^(1/2))/d^4
Time = 0.11 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.22 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x} \, dx=\left [\frac {1155 \, a^{2} c^{\frac {3}{2}} d^{4} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (105 \, b^{2} d^{5} x^{5} + 140 \, b^{2} c d^{4} x^{4} - 16 \, b^{2} c^{5} - 132 \, a b c^{3} d^{2} + 1540 \, a^{2} c d^{4} + 5 \, {\left (b^{2} c^{2} d^{3} + 66 \, a b d^{5}\right )} x^{3} - 6 \, {\left (b^{2} c^{3} d^{2} - 88 \, a b c d^{4}\right )} x^{2} + {\left (8 \, b^{2} c^{4} d + 66 \, a b c^{2} d^{3} + 385 \, a^{2} d^{5}\right )} x\right )} \sqrt {d x + c}}{1155 \, d^{4}}, \frac {2 \, {\left (1155 \, a^{2} \sqrt {-c} c d^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (105 \, b^{2} d^{5} x^{5} + 140 \, b^{2} c d^{4} x^{4} - 16 \, b^{2} c^{5} - 132 \, a b c^{3} d^{2} + 1540 \, a^{2} c d^{4} + 5 \, {\left (b^{2} c^{2} d^{3} + 66 \, a b d^{5}\right )} x^{3} - 6 \, {\left (b^{2} c^{3} d^{2} - 88 \, a b c d^{4}\right )} x^{2} + {\left (8 \, b^{2} c^{4} d + 66 \, a b c^{2} d^{3} + 385 \, a^{2} d^{5}\right )} x\right )} \sqrt {d x + c}\right )}}{1155 \, d^{4}}\right ] \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)^2/x,x, algorithm="fricas")
Output:
[1/1155*(1155*a^2*c^(3/2)*d^4*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(105*b^2*d^5*x^5 + 140*b^2*c*d^4*x^4 - 16*b^2*c^5 - 132*a*b*c^3*d^2 + 1540*a^2*c*d^4 + 5*(b^2*c^2*d^3 + 66*a*b*d^5)*x^3 - 6*(b^2*c^3*d^2 - 88*a *b*c*d^4)*x^2 + (8*b^2*c^4*d + 66*a*b*c^2*d^3 + 385*a^2*d^5)*x)*sqrt(d*x + c))/d^4, 2/1155*(1155*a^2*sqrt(-c)*c*d^4*arctan(sqrt(-c)/sqrt(d*x + c)) + (105*b^2*d^5*x^5 + 140*b^2*c*d^4*x^4 - 16*b^2*c^5 - 132*a*b*c^3*d^2 + 154 0*a^2*c*d^4 + 5*(b^2*c^2*d^3 + 66*a*b*d^5)*x^3 - 6*(b^2*c^3*d^2 - 88*a*b*c *d^4)*x^2 + (8*b^2*c^4*d + 66*a*b*c^2*d^3 + 385*a^2*d^5)*x)*sqrt(d*x + c)) /d^4]
Time = 3.96 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.25 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x} \, dx=\begin {cases} \frac {2 a^{2} c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a^{2} c \sqrt {c + d x} + \frac {2 a^{2} \left (c + d x\right )^{\frac {3}{2}}}{3} - \frac {2 b^{2} c \left (c + d x\right )^{\frac {9}{2}}}{3 d^{4}} + \frac {2 b^{2} \left (c + d x\right )^{\frac {11}{2}}}{11 d^{4}} + \frac {2 \left (c + d x\right )^{\frac {7}{2}} \cdot \left (2 a b d^{2} + 3 b^{2} c^{2}\right )}{7 d^{4}} + \frac {2 \left (c + d x\right )^{\frac {5}{2}} \left (- 2 a b c d^{2} - b^{2} c^{3}\right )}{5 d^{4}} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\frac {a^{2} \log {\left (x^{2} \right )}}{2} + a b x^{2} + \frac {b^{2} x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)**(3/2)*(b*x**2+a)**2/x,x)
Output:
Piecewise((2*a**2*c**2*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + 2*a**2*c*sq rt(c + d*x) + 2*a**2*(c + d*x)**(3/2)/3 - 2*b**2*c*(c + d*x)**(9/2)/(3*d** 4) + 2*b**2*(c + d*x)**(11/2)/(11*d**4) + 2*(c + d*x)**(7/2)*(2*a*b*d**2 + 3*b**2*c**2)/(7*d**4) + 2*(c + d*x)**(5/2)*(-2*a*b*c*d**2 - b**2*c**3)/(5 *d**4), Ne(d, 0)), (c**(3/2)*(a**2*log(x**2)/2 + a*b*x**2 + b**2*x**4/4), True))
Time = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.94 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x} \, dx=a^{2} c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right ) + \frac {2 \, {\left (105 \, {\left (d x + c\right )}^{\frac {11}{2}} b^{2} - 385 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{2} c + 385 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} d^{4} + 1155 \, \sqrt {d x + c} a^{2} c d^{4} + 165 \, {\left (3 \, b^{2} c^{2} + 2 \, a b d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}} - 231 \, {\left (b^{2} c^{3} + 2 \, a b c d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}}\right )}}{1155 \, d^{4}} \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)^2/x,x, algorithm="maxima")
Output:
a^2*c^(3/2)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c))) + 2/1 155*(105*(d*x + c)^(11/2)*b^2 - 385*(d*x + c)^(9/2)*b^2*c + 385*(d*x + c)^ (3/2)*a^2*d^4 + 1155*sqrt(d*x + c)*a^2*c*d^4 + 165*(3*b^2*c^2 + 2*a*b*d^2) *(d*x + c)^(7/2) - 231*(b^2*c^3 + 2*a*b*c*d^2)*(d*x + c)^(5/2))/d^4
Time = 0.13 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x} \, dx=\frac {2 \, a^{2} c^{2} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {2 \, {\left (105 \, {\left (d x + c\right )}^{\frac {11}{2}} b^{2} d^{40} - 385 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{2} c d^{40} + 495 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{2} c^{2} d^{40} - 231 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} c^{3} d^{40} + 330 \, {\left (d x + c\right )}^{\frac {7}{2}} a b d^{42} - 462 \, {\left (d x + c\right )}^{\frac {5}{2}} a b c d^{42} + 385 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} d^{44} + 1155 \, \sqrt {d x + c} a^{2} c d^{44}\right )}}{1155 \, d^{44}} \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)^2/x,x, algorithm="giac")
Output:
2*a^2*c^2*arctan(sqrt(d*x + c)/sqrt(-c))/sqrt(-c) + 2/1155*(105*(d*x + c)^ (11/2)*b^2*d^40 - 385*(d*x + c)^(9/2)*b^2*c*d^40 + 495*(d*x + c)^(7/2)*b^2 *c^2*d^40 - 231*(d*x + c)^(5/2)*b^2*c^3*d^40 + 330*(d*x + c)^(7/2)*a*b*d^4 2 - 462*(d*x + c)^(5/2)*a*b*c*d^42 + 385*(d*x + c)^(3/2)*a^2*d^44 + 1155*s qrt(d*x + c)*a^2*c*d^44)/d^44
Time = 8.29 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.14 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x} \, dx=\left (\frac {12\,b^2\,c^2+4\,a\,b\,d^2}{7\,d^4}-\frac {6\,b^2\,c^2}{7\,d^4}\right )\,{\left (c+d\,x\right )}^{7/2}+\left (\frac {c\,\left (\frac {12\,b^2\,c^2+4\,a\,b\,d^2}{d^4}-\frac {6\,b^2\,c^2}{d^4}\right )}{5}-\frac {8\,b^2\,c^3+8\,a\,b\,c\,d^2}{5\,d^4}\right )\,{\left (c+d\,x\right )}^{5/2}+\left (\frac {c\,\left (c\,\left (\frac {12\,b^2\,c^2+4\,a\,b\,d^2}{d^4}-\frac {6\,b^2\,c^2}{d^4}\right )-\frac {8\,b^2\,c^3+8\,a\,b\,c\,d^2}{d^4}\right )}{3}+\frac {2\,{\left (b\,c^2+a\,d^2\right )}^2}{3\,d^4}\right )\,{\left (c+d\,x\right )}^{3/2}+\frac {2\,b^2\,{\left (c+d\,x\right )}^{11/2}}{11\,d^4}+c\,\left (c\,\left (c\,\left (\frac {12\,b^2\,c^2+4\,a\,b\,d^2}{d^4}-\frac {6\,b^2\,c^2}{d^4}\right )-\frac {8\,b^2\,c^3+8\,a\,b\,c\,d^2}{d^4}\right )+\frac {2\,{\left (b\,c^2+a\,d^2\right )}^2}{d^4}\right )\,\sqrt {c+d\,x}-\frac {2\,b^2\,c\,{\left (c+d\,x\right )}^{9/2}}{3\,d^4}+a^2\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,2{}\mathrm {i} \] Input:
int(((a + b*x^2)^2*(c + d*x)^(3/2))/x,x)
Output:
((12*b^2*c^2 + 4*a*b*d^2)/(7*d^4) - (6*b^2*c^2)/(7*d^4))*(c + d*x)^(7/2) + ((c*((12*b^2*c^2 + 4*a*b*d^2)/d^4 - (6*b^2*c^2)/d^4))/5 - (8*b^2*c^3 + 8* a*b*c*d^2)/(5*d^4))*(c + d*x)^(5/2) + ((c*(c*((12*b^2*c^2 + 4*a*b*d^2)/d^4 - (6*b^2*c^2)/d^4) - (8*b^2*c^3 + 8*a*b*c*d^2)/d^4))/3 + (2*(a*d^2 + b*c^ 2)^2)/(3*d^4))*(c + d*x)^(3/2) + (2*b^2*(c + d*x)^(11/2))/(11*d^4) + c*(c* (c*((12*b^2*c^2 + 4*a*b*d^2)/d^4 - (6*b^2*c^2)/d^4) - (8*b^2*c^3 + 8*a*b*c *d^2)/d^4) + (2*(a*d^2 + b*c^2)^2)/d^4)*(c + d*x)^(1/2) + a^2*c^(3/2)*atan (((c + d*x)^(1/2)*1i)/c^(1/2))*2i - (2*b^2*c*(c + d*x)^(9/2))/(3*d^4)
Time = 0.19 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.60 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x} \, dx=\frac {3080 \sqrt {d x +c}\, a^{2} c \,d^{4}+770 \sqrt {d x +c}\, a^{2} d^{5} x -264 \sqrt {d x +c}\, a b \,c^{3} d^{2}+132 \sqrt {d x +c}\, a b \,c^{2} d^{3} x +1056 \sqrt {d x +c}\, a b c \,d^{4} x^{2}+660 \sqrt {d x +c}\, a b \,d^{5} x^{3}-32 \sqrt {d x +c}\, b^{2} c^{5}+16 \sqrt {d x +c}\, b^{2} c^{4} d x -12 \sqrt {d x +c}\, b^{2} c^{3} d^{2} x^{2}+10 \sqrt {d x +c}\, b^{2} c^{2} d^{3} x^{3}+280 \sqrt {d x +c}\, b^{2} c \,d^{4} x^{4}+210 \sqrt {d x +c}\, b^{2} d^{5} x^{5}+1155 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} c \,d^{4}-1155 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} c \,d^{4}}{1155 d^{4}} \] Input:
int((d*x+c)^(3/2)*(b*x^2+a)^2/x,x)
Output:
(3080*sqrt(c + d*x)*a**2*c*d**4 + 770*sqrt(c + d*x)*a**2*d**5*x - 264*sqrt (c + d*x)*a*b*c**3*d**2 + 132*sqrt(c + d*x)*a*b*c**2*d**3*x + 1056*sqrt(c + d*x)*a*b*c*d**4*x**2 + 660*sqrt(c + d*x)*a*b*d**5*x**3 - 32*sqrt(c + d*x )*b**2*c**5 + 16*sqrt(c + d*x)*b**2*c**4*d*x - 12*sqrt(c + d*x)*b**2*c**3* d**2*x**2 + 10*sqrt(c + d*x)*b**2*c**2*d**3*x**3 + 280*sqrt(c + d*x)*b**2* c*d**4*x**4 + 210*sqrt(c + d*x)*b**2*d**5*x**5 + 1155*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a**2*c*d**4 - 1155*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a **2*c*d**4)/(1155*d**4)