\(\int \frac {(a+b x^2)^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx\) [837]

Optimal result

Integrand size = 26, antiderivative size = 197 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {2 \left (b c^2+a d^2\right )^2 \sqrt {e x}}{c d^4 e \sqrt {c+d x}}+\frac {b \left (19 b c^2+16 a d^2\right ) \sqrt {e x} \sqrt {c+d x}}{8 d^4 e}-\frac {11 b^2 c (e x)^{3/2} \sqrt {c+d x}}{12 d^3 e^2}+\frac {b^2 (e x)^{5/2} \sqrt {c+d x}}{3 d^2 e^3}-\frac {b c \left (35 b c^2+48 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{8 d^{9/2} \sqrt {e}} \] Output:

2*(a*d^2+b*c^2)^2*(e*x)^(1/2)/c/d^4/e/(d*x+c)^(1/2)+1/8*b*(16*a*d^2+19*b*c 
^2)*(e*x)^(1/2)*(d*x+c)^(1/2)/d^4/e-11/12*b^2*c*(e*x)^(3/2)*(d*x+c)^(1/2)/ 
d^3/e^2+1/3*b^2*(e*x)^(5/2)*(d*x+c)^(1/2)/d^2/e^3-1/8*b*c*(48*a*d^2+35*b*c 
^2)*arctanh(d^(1/2)*(e*x)^(1/2)/e^(1/2)/(d*x+c)^(1/2))/d^(9/2)/e^(1/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {\sqrt {x} \left (\frac {\sqrt {d} \sqrt {x} \left (48 a^2 d^4+48 a b c d^2 (3 c+d x)+b^2 c \left (105 c^3+35 c^2 d x-14 c d^2 x^2+8 d^3 x^3\right )\right )}{\sqrt {c+d x}}+3 b c^2 \left (35 b c^2+48 a d^2\right ) \log \left (-\sqrt {d} \sqrt {x}+\sqrt {c+d x}\right )\right )}{24 c d^{9/2} \sqrt {e x}} \] Input:

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

Output:

(Sqrt[x]*((Sqrt[d]*Sqrt[x]*(48*a^2*d^4 + 48*a*b*c*d^2*(3*c + d*x) + b^2*c* 
(105*c^3 + 35*c^2*d*x - 14*c*d^2*x^2 + 8*d^3*x^3)))/Sqrt[c + d*x] + 3*b*c^ 
2*(35*b*c^2 + 48*a*d^2)*Log[-(Sqrt[d]*Sqrt[x]) + Sqrt[c + d*x]]))/(24*c*d^ 
(9/2)*Sqrt[e*x])
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {519, 27, 2125, 27, 1194, 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.

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

\(\Big \downarrow \) 519

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2125

\(\displaystyle \frac {2 \sqrt {e x} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\int \frac {\frac {11 b^2 c^2 x^2 e^3}{d}+\frac {6 b c^2 \left (b c^2+2 a d^2\right ) e^3}{d^3}-6 b c \left (\frac {b c^2}{d^2}+2 a\right ) x e^3}{2 \sqrt {e x} \sqrt {c+d x}}dx}{3 d e^3}-\frac {b^2 c (e x)^{5/2} \sqrt {c+d x}}{3 d^2 e^3}}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \sqrt {e x} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\int \frac {\frac {11 b^2 c^2 x^2 e^3}{d}+\frac {6 b c^2 \left (b c^2+2 a d^2\right ) e^3}{d^3}-6 b c \left (\frac {b c^2}{d^2}+2 a\right ) x e^3}{\sqrt {e x} \sqrt {c+d x}}dx}{6 d e^3}-\frac {b^2 c (e x)^{5/2} \sqrt {c+d x}}{3 d^2 e^3}}{c}\)

\(\Big \downarrow \) 1194

\(\displaystyle \frac {2 \sqrt {e x} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {\int \frac {3 b c e^5 \left (8 c \left (b c^2+2 a d^2\right )-d^2 \left (\frac {19 b c^2}{d}+16 a d\right ) x\right )}{2 d^2 \sqrt {e x} \sqrt {c+d x}}dx}{2 d e^2}+\frac {11 b^2 c^2 e (e x)^{3/2} \sqrt {c+d x}}{2 d^2}}{6 d e^3}-\frac {b^2 c (e x)^{5/2} \sqrt {c+d x}}{3 d^2 e^3}}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \sqrt {e x} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {3 b c e^3 \int \frac {8 c \left (b c^2+2 a d^2\right )-d \left (19 b c^2+16 a d^2\right ) x}{\sqrt {e x} \sqrt {c+d x}}dx}{4 d^3}+\frac {11 b^2 c^2 e (e x)^{3/2} \sqrt {c+d x}}{2 d^2}}{6 d e^3}-\frac {b^2 c (e x)^{5/2} \sqrt {c+d x}}{3 d^2 e^3}}{c}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {2 \sqrt {e x} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {3 b c e^3 \left (\frac {1}{2} c \left (48 a d^2+35 b c^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx-\frac {\sqrt {e x} \sqrt {c+d x} \left (16 a d^2+19 b c^2\right )}{e}\right )}{4 d^3}+\frac {11 b^2 c^2 e (e x)^{3/2} \sqrt {c+d x}}{2 d^2}}{6 d e^3}-\frac {b^2 c (e x)^{5/2} \sqrt {c+d x}}{3 d^2 e^3}}{c}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {2 \sqrt {e x} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {3 b c e^3 \left (c \left (48 a d^2+35 b c^2\right ) \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}-\frac {\sqrt {e x} \sqrt {c+d x} \left (16 a d^2+19 b c^2\right )}{e}\right )}{4 d^3}+\frac {11 b^2 c^2 e (e x)^{3/2} \sqrt {c+d x}}{2 d^2}}{6 d e^3}-\frac {b^2 c (e x)^{5/2} \sqrt {c+d x}}{3 d^2 e^3}}{c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \sqrt {e x} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {3 b c e^3 \left (\frac {c \left (48 a d^2+35 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{\sqrt {d} \sqrt {e}}-\frac {\sqrt {e x} \sqrt {c+d x} \left (16 a d^2+19 b c^2\right )}{e}\right )}{4 d^3}+\frac {11 b^2 c^2 e (e x)^{3/2} \sqrt {c+d x}}{2 d^2}}{6 d e^3}-\frac {b^2 c (e x)^{5/2} \sqrt {c+d x}}{3 d^2 e^3}}{c}\)

Input:

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

Output:

(2*(b*c^2 + a*d^2)^2*Sqrt[e*x])/(c*d^4*e*Sqrt[c + d*x]) - (-1/3*(b^2*c*(e* 
x)^(5/2)*Sqrt[c + d*x])/(d^2*e^3) + ((11*b^2*c^2*e*(e*x)^(3/2)*Sqrt[c + d* 
x])/(2*d^2) + (3*b*c*e^3*(-(((19*b*c^2 + 16*a*d^2)*Sqrt[e*x]*Sqrt[c + d*x] 
)/e) + (c*(35*b*c^2 + 48*a*d^2)*ArcTanh[(Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[ 
c + d*x])])/(Sqrt[d]*Sqrt[e])))/(4*d^3))/(6*d*e^3))/c
 

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 519

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

rule 1194

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

rule 2125

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> With[{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[k*(a + b*x 
)^(m + q)*((c + d*x)^(n + 1)/(d*b^q*(m + n + q + 1))), x] + Simp[1/(d*b^q*( 
m + n + q + 1))   Int[(a + b*x)^m*(c + d*x)^n*ExpandToSum[d*b^q*(m + n + q 
+ 1)*Px - d*k*(m + n + q + 1)*(a + b*x)^q - k*(b*c - a*d)*(m + q)*(a + b*x) 
^(q - 1), x], x], x] /; NeQ[m + n + q + 1, 0]] /; FreeQ[{a, b, c, d, m, n}, 
 x] && PolyQ[Px, x]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.20

method	result	size
risch	\(\frac {b \left (8 b \,x^{2} d^{2}-22 b c d x +48 a \,d^{2}+57 b \,c^{2}\right ) x \sqrt {d x +c}}{24 d^{4} \sqrt {e x}}-\frac {\left (\frac {2 \left (-16 a^{2} d^{4}-32 b \,c^{2} d^{2} a -16 b^{2} c^{4}\right ) \sqrt {d e \left (x +\frac {c}{d}\right )^{2}-c e \left (x +\frac {c}{d}\right )}}{d c e \left (x +\frac {c}{d}\right )}+\frac {35 c^{3} b^{2} \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right )}{\sqrt {d e}}+\frac {48 a b c \,d^{2} \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right )}{\sqrt {d e}}\right ) \sqrt {\left (d x +c \right ) e x}}{16 d^{4} \sqrt {e x}\, \sqrt {d x +c}}\)	\(236\)
default	\(-\frac {\left (-16 b^{2} c \,d^{3} x^{3} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}+144 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a b \,c^{2} d^{3} e x +105 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b^{2} c^{4} d e x +28 b^{2} c^{2} d^{2} x^{2} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}+144 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a b \,c^{3} d^{2} e +105 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b^{2} c^{5} e -96 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a b c \,d^{3} x -70 b^{2} c^{3} d x \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}-96 a^{2} d^{4} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}-288 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a b \,c^{2} d^{2}-210 b^{2} c^{4} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\right ) x}{48 c \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, \sqrt {e x}\, d^{4} \sqrt {d x +c}}\)	\(393\)

Input:

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

Output:

1/24*b*(8*b*d^2*x^2-22*b*c*d*x+48*a*d^2+57*b*c^2)*x*(d*x+c)^(1/2)/d^4/(e*x 
)^(1/2)-1/16/d^4*(2*(-16*a^2*d^4-32*a*b*c^2*d^2-16*b^2*c^4)/d/c/e/(x+c/d)* 
(d*e*(x+c/d)^2-c*e*(x+c/d))^(1/2)+35*c^3*b^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)+48*a*b*c*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))*((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 = 378, normalized size of antiderivative = 1.92 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\left [\frac {3 \, {\left (35 \, b^{2} c^{5} + 48 \, a b c^{3} d^{2} + {\left (35 \, b^{2} c^{4} d + 48 \, a b c^{2} d^{3}\right )} x\right )} \sqrt {d e} \log \left (2 \, d e x + c e - 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) + 2 \, {\left (8 \, b^{2} c d^{4} x^{3} - 14 \, b^{2} c^{2} d^{3} x^{2} + 105 \, b^{2} c^{4} d + 144 \, a b c^{2} d^{3} + 48 \, a^{2} d^{5} + {\left (35 \, b^{2} c^{3} d^{2} + 48 \, a b c d^{4}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{48 \, {\left (c d^{6} e x + c^{2} d^{5} e\right )}}, \frac {3 \, {\left (35 \, b^{2} c^{5} + 48 \, a b c^{3} d^{2} + {\left (35 \, b^{2} c^{4} d + 48 \, a b c^{2} d^{3}\right )} x\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) + {\left (8 \, b^{2} c d^{4} x^{3} - 14 \, b^{2} c^{2} d^{3} x^{2} + 105 \, b^{2} c^{4} d + 144 \, a b c^{2} d^{3} + 48 \, a^{2} d^{5} + {\left (35 \, b^{2} c^{3} d^{2} + 48 \, a b c d^{4}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{24 \, {\left (c d^{6} e x + c^{2} d^{5} e\right )}}\right ] \] Input:

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

Output:

[1/48*(3*(35*b^2*c^5 + 48*a*b*c^3*d^2 + (35*b^2*c^4*d + 48*a*b*c^2*d^3)*x) 
*sqrt(d*e)*log(2*d*e*x + c*e - 2*sqrt(d*e)*sqrt(d*x + c)*sqrt(e*x)) + 2*(8 
*b^2*c*d^4*x^3 - 14*b^2*c^2*d^3*x^2 + 105*b^2*c^4*d + 144*a*b*c^2*d^3 + 48 
*a^2*d^5 + (35*b^2*c^3*d^2 + 48*a*b*c*d^4)*x)*sqrt(d*x + c)*sqrt(e*x))/(c* 
d^6*e*x + c^2*d^5*e), 1/24*(3*(35*b^2*c^5 + 48*a*b*c^3*d^2 + (35*b^2*c^4*d 
 + 48*a*b*c^2*d^3)*x)*sqrt(-d*e)*arctan(sqrt(-d*e)*sqrt(d*x + c)*sqrt(e*x) 
/(d*e*x + c*e)) + (8*b^2*c*d^4*x^3 - 14*b^2*c^2*d^3*x^2 + 105*b^2*c^4*d + 
144*a*b*c^2*d^3 + 48*a^2*d^5 + (35*b^2*c^3*d^2 + 48*a*b*c*d^4)*x)*sqrt(d*x 
 + c)*sqrt(e*x))/(c*d^6*e*x + c^2*d^5*e)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x^2+a)^2/(e*x)^(1/2)/(d*x+c)^(3/2),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 = 0.18 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {1}{24} \, \sqrt {{\left (d x + c\right )} d e - c d e} \sqrt {d x + c} {\left (2 \, {\left (d x + c\right )} {\left (\frac {4 \, {\left (d x + c\right )} b^{2} {\left | d \right |}}{d^{6} e} - \frac {19 \, b^{2} c {\left | d \right |}}{d^{6} e}\right )} + \frac {3 \, {\left (29 \, b^{2} c^{2} d^{13} e^{2} {\left | d \right |} + 16 \, a b d^{15} e^{2} {\left | d \right |}\right )}}{d^{19} e^{3}}\right )} + \frac {{\left (35 \, b^{2} c^{3} + 48 \, a b c d^{2}\right )} \log \left ({\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2}\right )}{16 \, \sqrt {d e} d^{3} {\left | d \right |}} + \frac {4 \, {\left (b^{2} c^{4} e + 2 \, a b c^{2} d^{2} e + a^{2} d^{4} e\right )}}{{\left (c d e + {\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2}\right )} \sqrt {d e} d^{2} {\left | d \right |}} \] Input:

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

Output:

1/24*sqrt((d*x + c)*d*e - c*d*e)*sqrt(d*x + c)*(2*(d*x + c)*(4*(d*x + c)*b 
^2*abs(d)/(d^6*e) - 19*b^2*c*abs(d)/(d^6*e)) + 3*(29*b^2*c^2*d^13*e^2*abs( 
d) + 16*a*b*d^15*e^2*abs(d))/(d^19*e^3)) + 1/16*(35*b^2*c^3 + 48*a*b*c*d^2 
)*log((sqrt(d*e)*sqrt(d*x + c) - sqrt((d*x + c)*d*e - c*d*e))^2)/(sqrt(d*e 
)*d^3*abs(d)) + 4*(b^2*c^4*e + 2*a*b*c^2*d^2*e + a^2*d^4*e)/((c*d*e + (sqr 
t(d*e)*sqrt(d*x + c) - sqrt((d*x + c)*d*e - c*d*e))^2)*sqrt(d*e)*d^2*abs(d 
))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {\sqrt {e}\, \left (-1152 \sqrt {d}\, \sqrt {d x +c}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a b \,c^{2} d^{2}-840 \sqrt {d}\, \sqrt {d x +c}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b^{2} c^{4}+384 \sqrt {d}\, \sqrt {d x +c}\, a^{2} d^{4}+864 \sqrt {d}\, \sqrt {d x +c}\, a b \,c^{2} d^{2}+525 \sqrt {d}\, \sqrt {d x +c}\, b^{2} c^{4}+384 \sqrt {x}\, a^{2} d^{5}+1152 \sqrt {x}\, a b \,c^{2} d^{3}+384 \sqrt {x}\, a b c \,d^{4} x +840 \sqrt {x}\, b^{2} c^{4} d +280 \sqrt {x}\, b^{2} c^{3} d^{2} x -112 \sqrt {x}\, b^{2} c^{2} d^{3} x^{2}+64 \sqrt {x}\, b^{2} c \,d^{4} x^{3}\right )}{192 \sqrt {d x +c}\, c \,d^{5} e} \] Input:

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

Output:

(sqrt(e)*( - 1152*sqrt(d)*sqrt(c + d*x)*log((sqrt(c + d*x) + sqrt(x)*sqrt( 
d))/sqrt(c))*a*b*c**2*d**2 - 840*sqrt(d)*sqrt(c + d*x)*log((sqrt(c + d*x) 
+ sqrt(x)*sqrt(d))/sqrt(c))*b**2*c**4 + 384*sqrt(d)*sqrt(c + d*x)*a**2*d** 
4 + 864*sqrt(d)*sqrt(c + d*x)*a*b*c**2*d**2 + 525*sqrt(d)*sqrt(c + d*x)*b* 
*2*c**4 + 384*sqrt(x)*a**2*d**5 + 1152*sqrt(x)*a*b*c**2*d**3 + 384*sqrt(x) 
*a*b*c*d**4*x + 840*sqrt(x)*b**2*c**4*d + 280*sqrt(x)*b**2*c**3*d**2*x - 1 
12*sqrt(x)*b**2*c**2*d**3*x**2 + 64*sqrt(x)*b**2*c*d**4*x**3))/(192*sqrt(c 
 + d*x)*c*d**5*e)