\(\int \frac {(a+b x^2)^2}{(e x)^{3/2} \sqrt {c+d x}} \, dx\) [827]

Optimal result

Integrand size = 26, antiderivative size = 184 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} \sqrt {c+d x}} \, dx=-\frac {2 a^2 \sqrt {c+d x}}{c e \sqrt {e x}}+\frac {b \left (5 b c^2+16 a d^2\right ) \sqrt {e x} \sqrt {c+d x}}{8 d^3 e^2}-\frac {5 b^2 c (e x)^{3/2} \sqrt {c+d x}}{12 d^2 e^3}+\frac {b^2 (e x)^{5/2} \sqrt {c+d x}}{3 d e^4}-\frac {b c \left (5 b c^2+16 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{8 d^{7/2} e^{3/2}} \] Output:

-2*a^2*(d*x+c)^(1/2)/c/e/(e*x)^(1/2)+1/8*b*(16*a*d^2+5*b*c^2)*(e*x)^(1/2)* 
(d*x+c)^(1/2)/d^3/e^2-5/12*b^2*c*(e*x)^(3/2)*(d*x+c)^(1/2)/d^2/e^3+1/3*b^2 
*(e*x)^(5/2)*(d*x+c)^(1/2)/d/e^4-1/8*b*c*(16*a*d^2+5*b*c^2)*arctanh(d^(1/2 
)*(e*x)^(1/2)/e^(1/2)/(d*x+c)^(1/2))/d^(7/2)/e^(3/2)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} \sqrt {c+d x}} \, dx=\frac {x \left (\sqrt {d} \sqrt {c+d x} \left (-48 a^2 d^3+48 a b c d^2 x+b^2 c x \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )+6 b c^2 \left (5 b c^2+16 a d^2\right ) \sqrt {x} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}-\sqrt {c+d x}}\right )\right )}{24 c d^{7/2} (e x)^{3/2}} \] Input:

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

Output:

(x*(Sqrt[d]*Sqrt[c + d*x]*(-48*a^2*d^3 + 48*a*b*c*d^2*x + b^2*c*x*(15*c^2 
- 10*c*d*x + 8*d^2*x^2)) + 6*b*c^2*(5*b*c^2 + 16*a*d^2)*Sqrt[x]*ArcTanh[(S 
qrt[d]*Sqrt[x])/(Sqrt[c] - Sqrt[c + d*x])]))/(24*c*d^(7/2)*(e*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {520, 9, 27, 521, 27, 90, 60, 65, 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.

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

\(\Big \downarrow \) 520

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 521

\(\displaystyle \frac {b \left (\frac {\int \frac {e^2 \sqrt {e x} (12 a d-5 b c x)}{2 \sqrt {c+d x}}dx}{3 d e^2}+\frac {b (e x)^{5/2} \sqrt {c+d x}}{3 d e^2}\right )}{e^2}-\frac {2 a^2 \sqrt {c+d x}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \left (\frac {\int \frac {\sqrt {e x} (12 a d-5 b c x)}{\sqrt {c+d x}}dx}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x}}{3 d e^2}\right )}{e^2}-\frac {2 a^2 \sqrt {c+d x}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {b \left (\frac {\frac {3 \left (16 a d^2+5 b c^2\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x}}dx}{4 d}-\frac {5 b c (e x)^{3/2} \sqrt {c+d x}}{2 d e}}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x}}{3 d e^2}\right )}{e^2}-\frac {2 a^2 \sqrt {c+d x}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {b \left (\frac {\frac {3 \left (16 a d^2+5 b c^2\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}-\frac {5 b c (e x)^{3/2} \sqrt {c+d x}}{2 d e}}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x}}{3 d e^2}\right )}{e^2}-\frac {2 a^2 \sqrt {c+d x}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {b \left (\frac {\frac {3 \left (16 a d^2+5 b c^2\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}}{d}\right )}{4 d}-\frac {5 b c (e x)^{3/2} \sqrt {c+d x}}{2 d e}}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x}}{3 d e^2}\right )}{e^2}-\frac {2 a^2 \sqrt {c+d x}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {b \left (\frac {\frac {3 \left (16 a d^2+5 b c^2\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{3/2}}\right )}{4 d}-\frac {5 b c (e x)^{3/2} \sqrt {c+d x}}{2 d e}}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x}}{3 d e^2}\right )}{e^2}-\frac {2 a^2 \sqrt {c+d x}}{c e \sqrt {e x}}\)

Input:

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

Output:

(-2*a^2*Sqrt[c + d*x])/(c*e*Sqrt[e*x]) + (b*((b*(e*x)^(5/2)*Sqrt[c + d*x]) 
/(3*d*e^2) + ((-5*b*c*(e*x)^(3/2)*Sqrt[c + d*x])/(2*d*e) + (3*(5*b*c^2 + 1 
6*a*d^2)*((Sqrt[e*x]*Sqrt[c + d*x])/d - (c*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[e 
*x])/(Sqrt[e]*Sqrt[c + d*x])])/d^(3/2)))/(4*d))/(6*d)))/e^2
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

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 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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 90

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

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 520

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, e*x, x], R = Pol 
ynomialRemainder[(a + b*x^2)^p, e*x, x]}, Simp[R*(e*x)^(m + 1)*((c + d*x)^( 
n + 1)/((m + 1)*(e*c))), x] + Simp[1/((m + 1)*(e*c))   Int[(e*x)^(m + 1)*(c 
 + d*x)^n*ExpandToSum[(m + 1)*(e*c)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[m, -1] &&  !IntegerQ[n]
 

rule 521

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.85

method	result	size
risch	\(-\frac {\sqrt {d x +c}\, \left (-8 b^{2} c \,x^{3} d^{2}+10 b^{2} c^{2} d \,x^{2}-48 a b c \,d^{2} x -15 c^{3} b^{2} x +48 a^{2} d^{3}\right )}{24 c \,d^{3} e \sqrt {e x}}-\frac {c b \left (16 a \,d^{2}+5 b \,c^{2}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) \sqrt {\left (d x +c \right ) e x}}{16 d^{3} \sqrt {d e}\, e \sqrt {e x}\, \sqrt {d x +c}}\)	\(156\)
default	\(-\frac {\sqrt {d x +c}\, \left (-16 b^{2} c \,d^{2} x^{3} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+48 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a b \,c^{2} d^{2} e x +15 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b^{2} c^{4} e x +20 x^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, b^{2} c^{2} d -96 a b c \,d^{2} x \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-30 b^{2} c^{3} x \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+96 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a^{2} d^{3}\right )}{48 e c \,d^{3} \sqrt {e x}\, \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}}\)	\(255\)

Input:

int((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(d*x+c)^(1/2)*(-8*b^2*c*d^2*x^3+10*b^2*c^2*d*x^2-48*a*b*c*d^2*x-15*b 
^2*c^3*x+48*a^2*d^3)/c/d^3/e/(e*x)^(1/2)-1/16*c*b*(16*a*d^2+5*b*c^2)/d^3*l 
n((1/2*c*e+d*e*x)/(d*e)^(1/2)+(d*e*x^2+c*e*x)^(1/2))/(d*e)^(1/2)/e*((d*x+c 
)*e*x)^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.54 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (5 \, b^{2} c^{4} + 16 \, a b c^{2} d^{2}\right )} \sqrt {d e} x \log \left (2 \, d e x + c e - 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) + 2 \, {\left (8 \, b^{2} c d^{3} x^{3} - 10 \, b^{2} c^{2} d^{2} x^{2} - 48 \, a^{2} d^{4} + 3 \, {\left (5 \, b^{2} c^{3} d + 16 \, a b c d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{48 \, c d^{4} e^{2} x}, \frac {3 \, {\left (5 \, b^{2} c^{4} + 16 \, a b c^{2} d^{2}\right )} \sqrt {-d e} x \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) + {\left (8 \, b^{2} c d^{3} x^{3} - 10 \, b^{2} c^{2} d^{2} x^{2} - 48 \, a^{2} d^{4} + 3 \, {\left (5 \, b^{2} c^{3} d + 16 \, a b c d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{24 \, c d^{4} e^{2} x}\right ] \] Input:

integrate((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/48*(3*(5*b^2*c^4 + 16*a*b*c^2*d^2)*sqrt(d*e)*x*log(2*d*e*x + c*e - 2*sq 
rt(d*e)*sqrt(d*x + c)*sqrt(e*x)) + 2*(8*b^2*c*d^3*x^3 - 10*b^2*c^2*d^2*x^2 
 - 48*a^2*d^4 + 3*(5*b^2*c^3*d + 16*a*b*c*d^3)*x)*sqrt(d*x + c)*sqrt(e*x)) 
/(c*d^4*e^2*x), 1/24*(3*(5*b^2*c^4 + 16*a*b*c^2*d^2)*sqrt(-d*e)*x*arctan(s 
qrt(-d*e)*sqrt(d*x + c)*sqrt(e*x)/(d*e*x + c*e)) + (8*b^2*c*d^3*x^3 - 10*b 
^2*c^2*d^2*x^2 - 48*a^2*d^4 + 3*(5*b^2*c^3*d + 16*a*b*c*d^3)*x)*sqrt(d*x + 
 c)*sqrt(e*x))/(c*d^4*e^2*x)]
 

Sympy [A] (verification not implemented)

Time = 10.66 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} \sqrt {c+d x}} \, dx=- \frac {2 a^{2} \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{c e^{\frac {3}{2}}} + \frac {2 a b \sqrt {c} \sqrt {x} \sqrt {1 + \frac {d x}{c}}}{d e^{\frac {3}{2}}} - \frac {2 a b c \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{d^{\frac {3}{2}} e^{\frac {3}{2}}} + \frac {5 b^{2} c^{\frac {5}{2}} \sqrt {x}}{8 d^{3} e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {5 b^{2} c^{\frac {3}{2}} x^{\frac {3}{2}}}{24 d^{2} e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} - \frac {b^{2} \sqrt {c} x^{\frac {5}{2}}}{12 d e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} - \frac {5 b^{2} c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{8 d^{\frac {7}{2}} e^{\frac {3}{2}}} + \frac {b^{2} x^{\frac {7}{2}}}{3 \sqrt {c} e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} \] Input:

integrate((b*x**2+a)**2/(e*x)**(3/2)/(d*x+c)**(1/2),x)
 

Output:

-2*a**2*sqrt(d)*sqrt(c/(d*x) + 1)/(c*e**(3/2)) + 2*a*b*sqrt(c)*sqrt(x)*sqr 
t(1 + d*x/c)/(d*e**(3/2)) - 2*a*b*c*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(d**(3/ 
2)*e**(3/2)) + 5*b**2*c**(5/2)*sqrt(x)/(8*d**3*e**(3/2)*sqrt(1 + d*x/c)) + 
 5*b**2*c**(3/2)*x**(3/2)/(24*d**2*e**(3/2)*sqrt(1 + d*x/c)) - b**2*sqrt(c 
)*x**(5/2)/(12*d*e**(3/2)*sqrt(1 + d*x/c)) - 5*b**2*c**3*asinh(sqrt(d)*sqr 
t(x)/sqrt(c))/(8*d**(7/2)*e**(3/2)) + b**2*x**(7/2)/(3*sqrt(c)*e**(3/2)*sq 
rt(1 + d*x/c))
 

Maxima [F(-2)]

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

integrate((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c)^(1/2),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 = 0.16 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} \sqrt {c+d x}} \, dx=\frac {{\left (\frac {{\left ({\left (d x + c\right )} {\left (2 \, {\left (d x + c\right )} {\left (\frac {4 \, {\left (d x + c\right )} b^{2}}{d^{4}} - \frac {17 \, b^{2} c}{d^{4}}\right )} + \frac {59 \, b^{2} c^{3} d^{16} + 48 \, a b c d^{18}}{c d^{20}}\right )} - \frac {3 \, {\left (11 \, b^{2} c^{4} d^{16} + 16 \, a b c^{2} d^{18} + 16 \, a^{2} d^{20}\right )}}{c d^{20}}\right )} \sqrt {d x + c}}{\sqrt {{\left (d x + c\right )} d e - c d e}} + \frac {3 \, {\left (5 \, b^{2} c^{3} + 16 \, a b c d^{2}\right )} \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e} d^{4}}\right )} d^{2}}{24 \, e {\left | d \right |}} \] Input:

integrate((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

1/24*(((d*x + c)*(2*(d*x + c)*(4*(d*x + c)*b^2/d^4 - 17*b^2*c/d^4) + (59*b 
^2*c^3*d^16 + 48*a*b*c*d^18)/(c*d^20)) - 3*(11*b^2*c^4*d^16 + 16*a*b*c^2*d 
^18 + 16*a^2*d^20)/(c*d^20))*sqrt(d*x + c)/sqrt((d*x + c)*d*e - c*d*e) + 3 
*(5*b^2*c^3 + 16*a*b*c*d^2)*log(abs(-sqrt(d*e)*sqrt(d*x + c) + sqrt((d*x + 
 c)*d*e - c*d*e)))/(sqrt(d*e)*d^4))*d^2/(e*abs(d))
 

Mupad [F(-1)]

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

int((a + b*x^2)^2/((e*x)^(3/2)*(c + d*x)^(1/2)),x)
 

Output:

int((a + b*x^2)^2/((e*x)^(3/2)*(c + d*x)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{3/2} \sqrt {c+d x}} \, dx=\frac {\sqrt {e}\, \left (-384 \sqrt {x}\, \sqrt {d x +c}\, a^{2} d^{4}+384 \sqrt {x}\, \sqrt {d x +c}\, a b c \,d^{3} x +120 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{3} d x -80 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{2} d^{2} x^{2}+64 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c \,d^{3} x^{3}-384 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a b \,c^{2} d^{2} x -120 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b^{2} c^{4} x -384 \sqrt {d}\, a^{2} d^{4} x -96 \sqrt {d}\, a b \,c^{2} d^{2} x -45 \sqrt {d}\, b^{2} c^{4} x \right )}{192 c \,d^{4} e^{2} x} \] Input:

int((b*x^2+a)^2/(e*x)^(3/2)/(d*x+c)^(1/2),x)
 

Output:

(sqrt(e)*( - 384*sqrt(x)*sqrt(c + d*x)*a**2*d**4 + 384*sqrt(x)*sqrt(c + d* 
x)*a*b*c*d**3*x + 120*sqrt(x)*sqrt(c + d*x)*b**2*c**3*d*x - 80*sqrt(x)*sqr 
t(c + d*x)*b**2*c**2*d**2*x**2 + 64*sqrt(x)*sqrt(c + d*x)*b**2*c*d**3*x**3 
 - 384*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*a*b*c**2*d** 
2*x - 120*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*b**2*c**4 
*x - 384*sqrt(d)*a**2*d**4*x - 96*sqrt(d)*a*b*c**2*d**2*x - 45*sqrt(d)*b** 
2*c**4*x))/(192*c*d**4*e**2*x)