\(\int \frac {(a+b x^2)^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx\) [828]

Optimal result

Integrand size = 26, antiderivative size = 175 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx=-\frac {2 a^2 \sqrt {c+d x}}{3 c e (e x)^{3/2}}+\frac {4 a^2 d \sqrt {c+d x}}{3 c^2 e^2 \sqrt {e x}}-\frac {3 b^2 c \sqrt {e x} \sqrt {c+d x}}{4 d^2 e^3}+\frac {b^2 (e x)^{3/2} \sqrt {c+d x}}{2 d e^4}+\frac {b \left (3 b c^2+16 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{4 d^{5/2} e^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.67 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx=\frac {x \left (\sqrt {d} \sqrt {c+d x} \left (-8 a^2 d^2 (c-2 d x)+3 b^2 c^2 x^2 (-3 c+2 d x)\right )-3 b c^2 \left (3 b c^2+16 a d^2\right ) x^{3/2} \log \left (-\sqrt {d} \sqrt {x}+\sqrt {c+d x}\right )\right )}{12 c^2 d^{5/2} (e x)^{5/2}} \] Input:

Integrate[(a + b*x^2)^2/((e*x)^(5/2)*Sqrt[c + d*x]),x]
 

Output:

(x*(Sqrt[d]*Sqrt[c + d*x]*(-8*a^2*d^2*(c - 2*d*x) + 3*b^2*c^2*x^2*(-3*c + 
2*d*x)) - 3*b*c^2*(3*b*c^2 + 16*a*d^2)*x^(3/2)*Log[-(Sqrt[d]*Sqrt[x]) + Sq 
rt[c + d*x]]))/(12*c^2*d^(5/2)*(e*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {520, 27, 2124, 27, 521, 27, 90, 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.

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

\(\Big \downarrow \) 520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2124

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 521

\(\displaystyle -\frac {-\frac {3 b c \left (\frac {\int \frac {e^2 (8 a d-3 b c x)}{2 \sqrt {e x} \sqrt {c+d x}}dx}{2 d e^2}+\frac {b (e x)^{3/2} \sqrt {c+d x}}{2 d e^2}\right )}{e}-\frac {4 a^2 d \sqrt {c+d x}}{c e \sqrt {e x}}}{3 c e}-\frac {2 a^2 \sqrt {c+d x}}{3 c e (e x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {3 b c \left (\frac {\int \frac {8 a d-3 b c x}{\sqrt {e x} \sqrt {c+d x}}dx}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x}}{2 d e^2}\right )}{e}-\frac {4 a^2 d \sqrt {c+d x}}{c e \sqrt {e x}}}{3 c e}-\frac {2 a^2 \sqrt {c+d x}}{3 c e (e x)^{3/2}}\)

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 221

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

Input:

Int[(a + b*x^2)^2/((e*x)^(5/2)*Sqrt[c + d*x]),x]
 

Output:

(-2*a^2*Sqrt[c + d*x])/(3*c*e*(e*x)^(3/2)) - ((-4*a^2*d*Sqrt[c + d*x])/(c* 
e*Sqrt[e*x]) - (3*b*c*((b*(e*x)^(3/2)*Sqrt[c + d*x])/(2*d*e^2) + ((-3*b*c* 
Sqrt[e*x]*Sqrt[c + d*x])/(d*e) + ((3*b*c^2 + 16*a*d^2)*ArcTanh[(Sqrt[d]*Sq 
rt[e*x])/(Sqrt[e]*Sqrt[c + d*x])])/(d^(3/2)*Sqrt[e]))/(4*d)))/e)/(3*c*e)
 

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 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]
 

rule 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.85

method	result	size
risch	\(-\frac {\sqrt {d x +c}\, \left (-6 b^{2} c^{2} d \,x^{3}+9 b^{2} c^{3} x^{2}-16 a^{2} d^{3} x +8 a^{2} c \,d^{2}\right )}{12 d^{2} c^{2} x \,e^{2} \sqrt {e x}}+\frac {b \left (16 a \,d^{2}+3 b \,c^{2}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) \sqrt {\left (d x +c \right ) e x}}{8 d^{2} \sqrt {d e}\, e^{2} \sqrt {e x}\, \sqrt {d x +c}}\)	\(149\)
default	\(\frac {\sqrt {d x +c}\, \left (48 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a b \,c^{2} d^{2} e \,x^{2}+9 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b^{2} c^{4} e \,x^{2}+12 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, b^{2} c^{2} d \,x^{3}-18 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, b^{2} c^{3} x^{2}+32 a^{2} d^{3} x \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-16 a^{2} c \,d^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\right )}{24 e^{2} x \,c^{2} \sqrt {e x}\, \sqrt {\left (d x +c \right ) e x}\, d^{2} \sqrt {d e}}\)	\(238\)

Input:

int((b*x^2+a)^2/(e*x)^(5/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/12*(d*x+c)^(1/2)*(-6*b^2*c^2*d*x^3+9*b^2*c^3*x^2-16*a^2*d^3*x+8*a^2*c*d 
^2)/d^2/c^2/x/e^2/(e*x)^(1/2)+1/8*b*(16*a*d^2+3*b*c^2)/d^2*ln((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^2*((d*x+c)*e*x)^(1/2) 
/(e*x)^(1/2)/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (3 \, b^{2} c^{4} + 16 \, a b c^{2} d^{2}\right )} \sqrt {d e} x^{2} \log \left (2 \, d e x + c e + 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) + 2 \, {\left (6 \, b^{2} c^{2} d^{2} x^{3} - 9 \, b^{2} c^{3} d x^{2} + 16 \, a^{2} d^{4} x - 8 \, a^{2} c d^{3}\right )} \sqrt {d x + c} \sqrt {e x}}{24 \, c^{2} d^{3} e^{3} x^{2}}, -\frac {3 \, {\left (3 \, b^{2} c^{4} + 16 \, a b c^{2} d^{2}\right )} \sqrt {-d e} x^{2} \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) - {\left (6 \, b^{2} c^{2} d^{2} x^{3} - 9 \, b^{2} c^{3} d x^{2} + 16 \, a^{2} d^{4} x - 8 \, a^{2} c d^{3}\right )} \sqrt {d x + c} \sqrt {e x}}{12 \, c^{2} d^{3} e^{3} x^{2}}\right ] \] Input:

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/24*(3*(3*b^2*c^4 + 16*a*b*c^2*d^2)*sqrt(d*e)*x^2*log(2*d*e*x + c*e + 2* 
sqrt(d*e)*sqrt(d*x + c)*sqrt(e*x)) + 2*(6*b^2*c^2*d^2*x^3 - 9*b^2*c^3*d*x^ 
2 + 16*a^2*d^4*x - 8*a^2*c*d^3)*sqrt(d*x + c)*sqrt(e*x))/(c^2*d^3*e^3*x^2) 
, -1/12*(3*(3*b^2*c^4 + 16*a*b*c^2*d^2)*sqrt(-d*e)*x^2*arctan(sqrt(-d*e)*s 
qrt(d*x + c)*sqrt(e*x)/(d*e*x + c*e)) - (6*b^2*c^2*d^2*x^3 - 9*b^2*c^3*d*x 
^2 + 16*a^2*d^4*x - 8*a^2*c*d^3)*sqrt(d*x + c)*sqrt(e*x))/(c^2*d^3*e^3*x^2 
)]
 

Sympy [A] (verification not implemented)

Time = 8.97 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx=- \frac {2 a^{2} \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{3 c e^{\frac {5}{2}} x} + \frac {4 a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}}{3 c^{2} e^{\frac {5}{2}}} + \frac {4 a b \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{\sqrt {d} e^{\frac {5}{2}}} - \frac {3 b^{2} c^{\frac {3}{2}} \sqrt {x}}{4 d^{2} e^{\frac {5}{2}} \sqrt {1 + \frac {d x}{c}}} - \frac {b^{2} \sqrt {c} x^{\frac {3}{2}}}{4 d e^{\frac {5}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {3 b^{2} c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{4 d^{\frac {5}{2}} e^{\frac {5}{2}}} + \frac {b^{2} x^{\frac {5}{2}}}{2 \sqrt {c} e^{\frac {5}{2}} \sqrt {1 + \frac {d x}{c}}} \] Input:

integrate((b*x**2+a)**2/(e*x)**(5/2)/(d*x+c)**(1/2),x)
 

Output:

-2*a**2*sqrt(d)*sqrt(c/(d*x) + 1)/(3*c*e**(5/2)*x) + 4*a**2*d**(3/2)*sqrt( 
c/(d*x) + 1)/(3*c**2*e**(5/2)) + 4*a*b*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(sqr 
t(d)*e**(5/2)) - 3*b**2*c**(3/2)*sqrt(x)/(4*d**2*e**(5/2)*sqrt(1 + d*x/c)) 
 - b**2*sqrt(c)*x**(3/2)/(4*d*e**(5/2)*sqrt(1 + d*x/c)) + 3*b**2*c**2*asin 
h(sqrt(d)*sqrt(x)/sqrt(c))/(4*d**(5/2)*e**(5/2)) + b**2*x**(5/2)/(2*sqrt(c 
)*e**(5/2)*sqrt(1 + d*x/c))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x^2+a)^2/(e*x)^(5/2)/(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 = 211, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx=\frac {d^{3} {\left (\frac {{\left ({\left (3 \, {\left (d x + c\right )} {\left (\frac {2 \, {\left (d x + c\right )} b^{2} e}{d^{3}} - \frac {9 \, b^{2} c e}{d^{3}}\right )} + \frac {4 \, {\left (9 \, b^{2} c^{4} d^{13} e^{2} + 4 \, a^{2} d^{17} e^{2}\right )}}{c^{2} d^{16} e}\right )} {\left (d x + c\right )} - \frac {3 \, {\left (5 \, b^{2} c^{5} d^{13} e^{2} + 8 \, a^{2} c d^{17} e^{2}\right )}}{c^{2} d^{16} e}\right )} \sqrt {d x + c}}{{\left ({\left (d x + c\right )} d e - c d e\right )}^{\frac {3}{2}}} - \frac {3 \, {\left (3 \, b^{2} c^{2} + 16 \, a b d^{2}\right )} \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e} d^{4}}\right )}}{12 \, e^{2} {\left | d \right |}} \] Input:

integrate((b*x^2+a)^2/(e*x)^(5/2)/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

1/12*d^3*(((3*(d*x + c)*(2*(d*x + c)*b^2*e/d^3 - 9*b^2*c*e/d^3) + 4*(9*b^2 
*c^4*d^13*e^2 + 4*a^2*d^17*e^2)/(c^2*d^16*e))*(d*x + c) - 3*(5*b^2*c^5*d^1 
3*e^2 + 8*a^2*c*d^17*e^2)/(c^2*d^16*e))*sqrt(d*x + c)/((d*x + c)*d*e - c*d 
*e)^(3/2) - 3*(3*b^2*c^2 + 16*a*b*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))/(e^2*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2}{{\left (e\,x\right )}^{5/2}\,\sqrt {c+d\,x}} \,d x \] Input:

int((a + b*x^2)^2/((e*x)^(5/2)*(c + d*x)^(1/2)),x)
 

Output:

int((a + b*x^2)^2/((e*x)^(5/2)*(c + d*x)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{5/2} \sqrt {c+d x}} \, dx=\frac {\sqrt {e}\, \left (-64 \sqrt {x}\, \sqrt {d x +c}\, a^{2} c \,d^{3}+128 \sqrt {x}\, \sqrt {d x +c}\, a^{2} d^{4} x -72 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{3} d \,x^{2}+48 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{2} d^{2} x^{3}+384 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a b \,c^{2} d^{2} x^{2}+72 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b^{2} c^{4} x^{2}-128 \sqrt {d}\, a^{2} d^{4} x^{2}-17 \sqrt {d}\, b^{2} c^{4} x^{2}\right )}{96 c^{2} d^{3} e^{3} x^{2}} \] Input:

int((b*x^2+a)^2/(e*x)^(5/2)/(d*x+c)^(1/2),x)
 

Output:

(sqrt(e)*( - 64*sqrt(x)*sqrt(c + d*x)*a**2*c*d**3 + 128*sqrt(x)*sqrt(c + d 
*x)*a**2*d**4*x - 72*sqrt(x)*sqrt(c + d*x)*b**2*c**3*d*x**2 + 48*sqrt(x)*s 
qrt(c + d*x)*b**2*c**2*d**2*x**3 + 384*sqrt(d)*log((sqrt(c + d*x) + sqrt(x 
)*sqrt(d))/sqrt(c))*a*b*c**2*d**2*x**2 + 72*sqrt(d)*log((sqrt(c + d*x) + s 
qrt(x)*sqrt(d))/sqrt(c))*b**2*c**4*x**2 - 128*sqrt(d)*a**2*d**4*x**2 - 17* 
sqrt(d)*b**2*c**4*x**2))/(96*c**2*d**3*e**3*x**2)