\(\int \frac {(a+b x^2)^2}{(e x)^{3/2} (c+d x)^2} \, dx\) [351]

Optimal result

Integrand size = 24, antiderivative size = 160 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)^2} \, dx=-\frac {2 a^2}{c^2 e \sqrt {e x}}-\frac {4 b^2 c \sqrt {e x}}{d^3 e^2}+\frac {2 b^2 (e x)^{3/2}}{3 d^2 e^3}-\frac {\left (b c^2+a d^2\right )^2 \sqrt {e x}}{c^2 d^3 e^2 (c+d x)}+\frac {\left (5 b c^2-3 a d^2\right ) \left (b c^2+a d^2\right ) \arctan \left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {c} \sqrt {e}}\right )}{c^{5/2} d^{7/2} e^{3/2}} \] Output:

-2*a^2/c^2/e/(e*x)^(1/2)-4*b^2*c*(e*x)^(1/2)/d^3/e^2+2/3*b^2*(e*x)^(3/2)/d 
^2/e^3-(a*d^2+b*c^2)^2*(e*x)^(1/2)/c^2/d^3/e^2/(d*x+c)+(-3*a*d^2+5*b*c^2)* 
(a*d^2+b*c^2)*arctan(d^(1/2)*(e*x)^(1/2)/c^(1/2)/e^(1/2))/c^(5/2)/d^(7/2)/ 
e^(3/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)^2} \, dx=\frac {x \left (-\frac {\sqrt {c} \sqrt {d} \left (6 a b c^2 d^2 x+3 a^2 d^3 (2 c+3 d x)+b^2 c^2 x \left (15 c^2+10 c d x-2 d^2 x^2\right )\right )}{c+d x}+3 \left (5 b^2 c^4+2 a b c^2 d^2-3 a^2 d^4\right ) \sqrt {x} \arctan \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}}\right )\right )}{3 c^{5/2} d^{7/2} (e x)^{3/2}} \] Input:

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

Output:

(x*(-((Sqrt[c]*Sqrt[d]*(6*a*b*c^2*d^2*x + 3*a^2*d^3*(2*c + 3*d*x) + b^2*c^ 
2*x*(15*c^2 + 10*c*d*x - 2*d^2*x^2)))/(c + d*x)) + 3*(5*b^2*c^4 + 2*a*b*c^ 
2*d^2 - 3*a^2*d^4)*Sqrt[x]*ArcTan[(Sqrt[d]*Sqrt[x])/Sqrt[c]]))/(3*c^(5/2)* 
d^(7/2)*(e*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {518, 1583, 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.

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

\(\Big \downarrow \) 518

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

\(\Big \downarrow \) 1583

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2333

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {-\frac {2 a^2 d^4 e^4}{\sqrt {e x}}+\frac {\sqrt {d} e^{7/2} \left (5 b c^2-3 a d^2\right ) \left (a d^2+b c^2\right ) \arctan \left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {c} \sqrt {e}}\right )}{\sqrt {c}}-4 b^2 c^3 d e^3 \sqrt {e x}+\frac {2}{3} b^2 c^2 d^2 e^2 (e x)^{3/2}}{2 c^2 d^4 e^2}-\frac {e^2 \sqrt {e x} \left (a d^2+b c^2\right )^2}{2 c^2 d^3 (c e+d e x)}\right )}{e^3}\)

Input:

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

Output:

(2*(-1/2*((b*c^2 + a*d^2)^2*e^2*Sqrt[e*x])/(c^2*d^3*(c*e + d*e*x)) + ((-2* 
a^2*d^4*e^4)/Sqrt[e*x] - 4*b^2*c^3*d*e^3*Sqrt[e*x] + (2*b^2*c^2*d^2*e^2*(e 
*x)^(3/2))/3 + (Sqrt[d]*(5*b*c^2 - 3*a*d^2)*(b*c^2 + a*d^2)*e^(7/2)*ArcTan 
[(Sqrt[d]*Sqrt[e*x])/(Sqrt[c]*Sqrt[e])])/Sqrt[c])/(2*c^2*d^4*e^2)))/e^3
 

Defintions of rubi rules used

rule 25

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

rule 518

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[2/e^(n + 2*p + 1)   Subst[Int[x^(2*m + 1)*(e*c + d*x^2)^ 
n*(a*e^2 + b*x^4)^p, x], x, Sqrt[e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && 
IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
 

rule 1583

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Sym 
bol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^( 
2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^(2*p)*(q + 1))   Int[x^ 
m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 
1)*e^(2*p)*(q + 1)*(a + c*x^4)^p - ((c*d^2 + a*e^2)^p/(e^(m/2)*x^m))*(d + e 
*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, c, d, e}, x] && IGtQ[p, 0] && IL 
tQ[q, -1] && ILtQ[m/2, 0]
 

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]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.86

method	result	size
risch	\(-\frac {2 \left (-b^{2} c^{2} d \,x^{2}+6 c^{3} b^{2} x +3 a^{2} d^{3}\right )}{3 c^{2} \sqrt {e x}\, d^{3} e}-\frac {\left (2 a \,d^{2}+2 b \,c^{2}\right ) \left (\frac {\left (\frac {a \,d^{2}}{2}+\frac {b \,c^{2}}{2}\right ) \sqrt {e x}}{d e x +c e}+\frac {\left (3 a \,d^{2}-5 b \,c^{2}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{2 \sqrt {d e c}}\right )}{c^{2} d^{3} e}\)	\(137\)
pseudoelliptic	\(-\frac {2 \left (\frac {3 \left (a \,d^{2}-\frac {5 b \,c^{2}}{3}\right ) \left (d x +c \right ) \sqrt {e x}\, \left (a \,d^{2}+b \,c^{2}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{2}+\sqrt {d e c}\, \left (\frac {5 b^{2} c^{4} x}{2}+\frac {5 b^{2} c^{3} d \,x^{2}}{3}+b \,d^{2} x \left (-\frac {b \,x^{2}}{3}+a \right ) c^{2}+d^{3} c \,a^{2}+\frac {3 a^{2} d^{4} x}{2}\right )\right )}{e \sqrt {e x}\, \sqrt {d e c}\, d^{3} c^{2} \left (d x +c \right )}\)	\(143\)
derivativedivides	\(\frac {-\frac {2 b^{2} \left (-\frac {d \left (e x \right )^{\frac {3}{2}}}{3}+2 c e \sqrt {e x}\right )}{d^{3}}-\frac {2 e^{2} \left (\frac {\left (\frac {1}{2} a^{2} d^{4}+b \,c^{2} d^{2} a +\frac {1}{2} b^{2} c^{4}\right ) \sqrt {e x}}{d e x +c e}+\frac {\left (3 a^{2} d^{4}-2 b \,c^{2} d^{2} a -5 b^{2} c^{4}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{2 \sqrt {d e c}}\right )}{c^{2} d^{3}}-\frac {2 a^{2} e^{2}}{c^{2} \sqrt {e x}}}{e^{3}}\)	\(152\)
default	\(\frac {-\frac {2 b^{2} \left (-\frac {d \left (e x \right )^{\frac {3}{2}}}{3}+2 c e \sqrt {e x}\right )}{d^{3}}-\frac {2 e^{2} \left (\frac {\left (\frac {1}{2} a^{2} d^{4}+b \,c^{2} d^{2} a +\frac {1}{2} b^{2} c^{4}\right ) \sqrt {e x}}{d e x +c e}+\frac {\left (3 a^{2} d^{4}-2 b \,c^{2} d^{2} a -5 b^{2} c^{4}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{2 \sqrt {d e c}}\right )}{c^{2} d^{3}}-\frac {2 a^{2} e^{2}}{c^{2} \sqrt {e x}}}{e^{3}}\)	\(152\)

Input:

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

Output:

-2/3*(-b^2*c^2*d*x^2+6*b^2*c^3*x+3*a^2*d^3)/c^2/(e*x)^(1/2)/d^3/e-1/c^2/d^ 
3*(2*a*d^2+2*b*c^2)*((1/2*a*d^2+1/2*b*c^2)*(e*x)^(1/2)/(d*e*x+c*e)+1/2*(3* 
a*d^2-5*b*c^2)/(d*e*c)^(1/2)*arctan(d*(e*x)^(1/2)/(d*e*c)^(1/2)))/e
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 416, normalized size of antiderivative = 2.60 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)^2} \, dx=\left [\frac {3 \, \sqrt {-c d e} {\left ({\left (5 \, b^{2} c^{4} d + 2 \, a b c^{2} d^{3} - 3 \, a^{2} d^{5}\right )} x^{2} + {\left (5 \, b^{2} c^{5} + 2 \, a b c^{3} d^{2} - 3 \, a^{2} c d^{4}\right )} x\right )} \log \left (\frac {d e x - c e + 2 \, \sqrt {-c d e} \sqrt {e x}}{d x + c}\right ) + 2 \, {\left (2 \, b^{2} c^{3} d^{3} x^{3} - 10 \, b^{2} c^{4} d^{2} x^{2} - 6 \, a^{2} c^{2} d^{4} - 3 \, {\left (5 \, b^{2} c^{5} d + 2 \, a b c^{3} d^{3} + 3 \, a^{2} c d^{5}\right )} x\right )} \sqrt {e x}}{6 \, {\left (c^{3} d^{5} e^{2} x^{2} + c^{4} d^{4} e^{2} x\right )}}, -\frac {3 \, \sqrt {c d e} {\left ({\left (5 \, b^{2} c^{4} d + 2 \, a b c^{2} d^{3} - 3 \, a^{2} d^{5}\right )} x^{2} + {\left (5 \, b^{2} c^{5} + 2 \, a b c^{3} d^{2} - 3 \, a^{2} c d^{4}\right )} x\right )} \arctan \left (\frac {\sqrt {c d e} \sqrt {e x}}{d e x}\right ) - {\left (2 \, b^{2} c^{3} d^{3} x^{3} - 10 \, b^{2} c^{4} d^{2} x^{2} - 6 \, a^{2} c^{2} d^{4} - 3 \, {\left (5 \, b^{2} c^{5} d + 2 \, a b c^{3} d^{3} + 3 \, a^{2} c d^{5}\right )} x\right )} \sqrt {e x}}{3 \, {\left (c^{3} d^{5} e^{2} x^{2} + c^{4} d^{4} e^{2} x\right )}}\right ] \] Input:

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

Output:

[1/6*(3*sqrt(-c*d*e)*((5*b^2*c^4*d + 2*a*b*c^2*d^3 - 3*a^2*d^5)*x^2 + (5*b 
^2*c^5 + 2*a*b*c^3*d^2 - 3*a^2*c*d^4)*x)*log((d*e*x - c*e + 2*sqrt(-c*d*e) 
*sqrt(e*x))/(d*x + c)) + 2*(2*b^2*c^3*d^3*x^3 - 10*b^2*c^4*d^2*x^2 - 6*a^2 
*c^2*d^4 - 3*(5*b^2*c^5*d + 2*a*b*c^3*d^3 + 3*a^2*c*d^5)*x)*sqrt(e*x))/(c^ 
3*d^5*e^2*x^2 + c^4*d^4*e^2*x), -1/3*(3*sqrt(c*d*e)*((5*b^2*c^4*d + 2*a*b* 
c^2*d^3 - 3*a^2*d^5)*x^2 + (5*b^2*c^5 + 2*a*b*c^3*d^2 - 3*a^2*c*d^4)*x)*ar 
ctan(sqrt(c*d*e)*sqrt(e*x)/(d*e*x)) - (2*b^2*c^3*d^3*x^3 - 10*b^2*c^4*d^2* 
x^2 - 6*a^2*c^2*d^4 - 3*(5*b^2*c^5*d + 2*a*b*c^3*d^3 + 3*a^2*c*d^5)*x)*sqr 
t(e*x))/(c^3*d^5*e^2*x^2 + c^4*d^4*e^2*x)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c)^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(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)^2} \, dx=\frac {\frac {3 \, {\left (5 \, b^{2} c^{4} + 2 \, a b c^{2} d^{2} - 3 \, a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {e x} d}{\sqrt {c d e}}\right )}{\sqrt {c d e} c^{2} d^{3}} - \frac {3 \, {\left (b^{2} c^{4} e x + 2 \, a b c^{2} d^{2} e x + 3 \, a^{2} d^{4} e x + 2 \, a^{2} c d^{3} e\right )}}{{\left (\sqrt {e x} d e x + \sqrt {e x} c e\right )} c^{2} d^{3}} + \frac {2 \, {\left (\sqrt {e x} b^{2} d^{4} e^{5} x - 6 \, \sqrt {e x} b^{2} c d^{3} e^{5}\right )}}{d^{6} e^{6}}}{3 \, e} \] Input:

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

Output:

1/3*(3*(5*b^2*c^4 + 2*a*b*c^2*d^2 - 3*a^2*d^4)*arctan(sqrt(e*x)*d/sqrt(c*d 
*e))/(sqrt(c*d*e)*c^2*d^3) - 3*(b^2*c^4*e*x + 2*a*b*c^2*d^2*e*x + 3*a^2*d^ 
4*e*x + 2*a^2*c*d^3*e)/((sqrt(e*x)*d*e*x + sqrt(e*x)*c*e)*c^2*d^3) + 2*(sq 
rt(e*x)*b^2*d^4*e^5*x - 6*sqrt(e*x)*b^2*c*d^3*e^5)/(d^6*e^6))/e
 

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)^2} \, dx=\frac {2\,b^2\,{\left (e\,x\right )}^{3/2}}{3\,d^2\,e^3}-\frac {\frac {2\,a^2\,d^3\,e}{c}+\frac {e\,x\,\left (3\,a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}{c^2}}{d^4\,e\,{\left (e\,x\right )}^{3/2}+c\,d^3\,e^2\,\sqrt {e\,x}}-\frac {4\,b^2\,c\,\sqrt {e\,x}}{d^3\,e^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e\,x}\,\left (b\,c^2+a\,d^2\right )\,\left (3\,a\,d^2-5\,b\,c^2\right )}{\sqrt {c}\,\sqrt {e}\,\left (-3\,a^2\,d^4+2\,a\,b\,c^2\,d^2+5\,b^2\,c^4\right )}\right )\,\left (b\,c^2+a\,d^2\right )\,\left (3\,a\,d^2-5\,b\,c^2\right )}{c^{5/2}\,d^{7/2}\,e^{3/2}} \] Input:

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

Output:

(2*b^2*(e*x)^(3/2))/(3*d^2*e^3) - ((2*a^2*d^3*e)/c + (e*x*(3*a^2*d^4 + b^2 
*c^4 + 2*a*b*c^2*d^2))/c^2)/(d^4*e*(e*x)^(3/2) + c*d^3*e^2*(e*x)^(1/2)) - 
(4*b^2*c*(e*x)^(1/2))/(d^3*e^2) + (atan((d^(1/2)*(e*x)^(1/2)*(a*d^2 + b*c^ 
2)*(3*a*d^2 - 5*b*c^2))/(c^(1/2)*e^(1/2)*(5*b^2*c^4 - 3*a^2*d^4 + 2*a*b*c^ 
2*d^2)))*(a*d^2 + b*c^2)*(3*a*d^2 - 5*b*c^2))/(c^(5/2)*d^(7/2)*e^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.66 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} (c+d x)^2} \, dx=\frac {\sqrt {e}\, \left (-9 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} c \,d^{4}-9 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} d^{5} x +6 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) a b \,c^{3} d^{2}+6 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) a b \,c^{2} d^{3} x +15 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{5}+15 \sqrt {x}\, \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{4} d x -6 a^{2} c^{2} d^{4}-9 a^{2} c \,d^{5} x -6 a b \,c^{3} d^{3} x -15 b^{2} c^{5} d x -10 b^{2} c^{4} d^{2} x^{2}+2 b^{2} c^{3} d^{3} x^{3}\right )}{3 \sqrt {x}\, c^{3} d^{4} e^{2} \left (d x +c \right )} \] Input:

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

Output:

(sqrt(e)*( - 9*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)*sqrt(c))) 
*a**2*c*d**4 - 9*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)*sqrt(c) 
))*a**2*d**5*x + 6*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)*sqrt( 
c)))*a*b*c**3*d**2 + 6*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)*s 
qrt(c)))*a*b*c**2*d**3*x + 15*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sq 
rt(d)*sqrt(c)))*b**2*c**5 + 15*sqrt(x)*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(s 
qrt(d)*sqrt(c)))*b**2*c**4*d*x - 6*a**2*c**2*d**4 - 9*a**2*c*d**5*x - 6*a* 
b*c**3*d**3*x - 15*b**2*c**5*d*x - 10*b**2*c**4*d**2*x**2 + 2*b**2*c**3*d* 
*3*x**3))/(3*sqrt(x)*c**3*d**4*e**2*(c + d*x))