\(\int \frac {\sqrt {e x} (a+b x^2)^2}{(c+d x)^{3/2}} \, dx\) [836]

Optimal result

Integrand size = 26, antiderivative size = 257 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{(c+d x)^{3/2}} \, dx=-\frac {\left (9 b^2 c^4+16 a b c^2 d^2+8 a^2 d^4\right ) \sqrt {e x}}{4 d^5 \sqrt {c+d x}}+\frac {b^2 (e x)^{9/2}}{4 d e^4 \sqrt {c+d x}}-\frac {b c \left (171 b c^2+224 a d^2\right ) \sqrt {e x} \sqrt {c+d x}}{64 d^5}+\frac {b \left (45 b c^2+32 a d^2\right ) (e x)^{3/2} \sqrt {c+d x}}{32 d^4 e}-\frac {3 b^2 c (e x)^{3/2} (c+d x)^{3/2}}{8 d^4 e}+\frac {\left (315 b^2 c^4+480 a b c^2 d^2+128 a^2 d^4\right ) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{64 d^{11/2}} \] Output:

-1/4*(8*a^2*d^4+16*a*b*c^2*d^2+9*b^2*c^4)*(e*x)^(1/2)/d^5/(d*x+c)^(1/2)+1/ 
4*b^2*(e*x)^(9/2)/d/e^4/(d*x+c)^(1/2)-1/64*b*c*(224*a*d^2+171*b*c^2)*(e*x) 
^(1/2)*(d*x+c)^(1/2)/d^5+1/32*b*(32*a*d^2+45*b*c^2)*(e*x)^(3/2)*(d*x+c)^(1 
/2)/d^4/e-3/8*b^2*c*(e*x)^(3/2)*(d*x+c)^(3/2)/d^4/e+1/64*(128*a^2*d^4+480* 
a*b*c^2*d^2+315*b^2*c^4)*e^(1/2)*arctanh(d^(1/2)*(e*x)^(1/2)/e^(1/2)/(d*x+ 
c)^(1/2))/d^(11/2)
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {e x} \left (\frac {\sqrt {d} \left (-128 a^2 d^4+32 a b d^2 \left (-15 c^2-5 c d x+2 d^2 x^2\right )+b^2 \left (-315 c^4-105 c^3 d x+42 c^2 d^2 x^2-24 c d^3 x^3+16 d^4 x^4\right )\right )}{\sqrt {c+d x}}+\frac {2 \left (315 b^2 c^4+480 a b c^2 d^2+128 a^2 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{-\sqrt {c}+\sqrt {c+d x}}\right )}{\sqrt {x}}\right )}{64 d^{11/2}} \] Input:

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

Output:

(Sqrt[e*x]*((Sqrt[d]*(-128*a^2*d^4 + 32*a*b*d^2*(-15*c^2 - 5*c*d*x + 2*d^2 
*x^2) + b^2*(-315*c^4 - 105*c^3*d*x + 42*c^2*d^2*x^2 - 24*c*d^3*x^3 + 16*d 
^4*x^4)))/Sqrt[c + d*x] + (2*(315*b^2*c^4 + 480*a*b*c^2*d^2 + 128*a^2*d^4) 
*ArcTanh[(Sqrt[d]*Sqrt[x])/(-Sqrt[c] + Sqrt[c + d*x])])/Sqrt[x]))/(64*d^(1 
1/2))
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {519, 27, 2125, 27, 1194, 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.

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

\(\Big \downarrow \) 519

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2125

\(\displaystyle \frac {2 (e x)^{3/2} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\int \frac {\sqrt {e x} \left (\frac {15 b^2 c^2 x^2 e^3}{d}+\frac {8 \left (3 b^2 c^4+6 a b d^2 c^2+2 a^2 d^4\right ) e^3}{d^3}-8 b c \left (\frac {b c^2}{d^2}+2 a\right ) x e^3\right )}{2 \sqrt {c+d x}}dx}{4 d e^3}-\frac {b^2 c (e x)^{7/2} \sqrt {c+d x}}{4 d^2 e^3}}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 (e x)^{3/2} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\int \frac {\sqrt {e x} \left (\frac {15 b^2 c^2 x^2 e^3}{d}+\frac {8 \left (3 b^2 c^4+6 a b d^2 c^2+2 a^2 d^4\right ) e^3}{d^3}-8 b c \left (\frac {b c^2}{d^2}+2 a\right ) x e^3\right )}{\sqrt {c+d x}}dx}{8 d e^3}-\frac {b^2 c (e x)^{7/2} \sqrt {c+d x}}{4 d^2 e^3}}{c}\)

\(\Big \downarrow \) 1194

\(\displaystyle \frac {2 (e x)^{3/2} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {\int \frac {3 e^5 \sqrt {e x} \left (16 \left (3 b^2 c^4+6 a b d^2 c^2+2 a^2 d^4\right )-b c d^2 \left (\frac {41 b c^2}{d}+32 a d\right ) x\right )}{2 d^2 \sqrt {c+d x}}dx}{3 d e^2}+\frac {5 b^2 c^2 e (e x)^{5/2} \sqrt {c+d x}}{d^2}}{8 d e^3}-\frac {b^2 c (e x)^{7/2} \sqrt {c+d x}}{4 d^2 e^3}}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 (e x)^{3/2} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \int \frac {\sqrt {e x} \left (16 \left (3 b^2 c^4+6 a b d^2 c^2+2 a^2 d^4\right )-b c d \left (41 b c^2+32 a d^2\right ) x\right )}{\sqrt {c+d x}}dx}{2 d^3}+\frac {5 b^2 c^2 e (e x)^{5/2} \sqrt {c+d x}}{d^2}}{8 d e^3}-\frac {b^2 c (e x)^{7/2} \sqrt {c+d x}}{4 d^2 e^3}}{c}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {2 (e x)^{3/2} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \left (\frac {1}{4} \left (128 a^2 d^4+480 a b c^2 d^2+315 b^2 c^4\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x}}dx-\frac {b c (e x)^{3/2} \sqrt {c+d x} \left (32 a d^2+41 b c^2\right )}{2 e}\right )}{2 d^3}+\frac {5 b^2 c^2 e (e x)^{5/2} \sqrt {c+d x}}{d^2}}{8 d e^3}-\frac {b^2 c (e x)^{7/2} \sqrt {c+d x}}{4 d^2 e^3}}{c}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (e x)^{3/2} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \left (\frac {1}{4} \left (128 a^2 d^4+480 a b c^2 d^2+315 b^2 c^4\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx}{2 d}\right )-\frac {b c (e x)^{3/2} \sqrt {c+d x} \left (32 a d^2+41 b c^2\right )}{2 e}\right )}{2 d^3}+\frac {5 b^2 c^2 e (e x)^{5/2} \sqrt {c+d x}}{d^2}}{8 d e^3}-\frac {b^2 c (e x)^{7/2} \sqrt {c+d x}}{4 d^2 e^3}}{c}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {2 (e x)^{3/2} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \left (\frac {1}{4} \left (128 a^2 d^4+480 a b c^2 d^2+315 b^2 c^4\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}}{d}\right )-\frac {b c (e x)^{3/2} \sqrt {c+d x} \left (32 a d^2+41 b c^2\right )}{2 e}\right )}{2 d^3}+\frac {5 b^2 c^2 e (e x)^{5/2} \sqrt {c+d x}}{d^2}}{8 d e^3}-\frac {b^2 c (e x)^{7/2} \sqrt {c+d x}}{4 d^2 e^3}}{c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (e x)^{3/2} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \left (\frac {1}{4} \left (128 a^2 d^4+480 a b c^2 d^2+315 b^2 c^4\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{3/2}}\right )-\frac {b c (e x)^{3/2} \sqrt {c+d x} \left (32 a d^2+41 b c^2\right )}{2 e}\right )}{2 d^3}+\frac {5 b^2 c^2 e (e x)^{5/2} \sqrt {c+d x}}{d^2}}{8 d e^3}-\frac {b^2 c (e x)^{7/2} \sqrt {c+d x}}{4 d^2 e^3}}{c}\)

Input:

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

Output:

(2*(b*c^2 + a*d^2)^2*(e*x)^(3/2))/(c*d^4*e*Sqrt[c + d*x]) - (-1/4*(b^2*c*( 
e*x)^(7/2)*Sqrt[c + d*x])/(d^2*e^3) + ((5*b^2*c^2*e*(e*x)^(5/2)*Sqrt[c + d 
*x])/d^2 + (e^3*(-1/2*(b*c*(41*b*c^2 + 32*a*d^2)*(e*x)^(3/2)*Sqrt[c + d*x] 
)/e + ((315*b^2*c^4 + 480*a*b*c^2*d^2 + 128*a^2*d^4)*((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))/(2*d^3))/(8*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 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 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 = 298, normalized size of antiderivative = 1.16

method	result	size
risch	\(-\frac {b \left (-16 b \,d^{3} x^{3}+40 b c \,d^{2} x^{2}-64 a x \,d^{3}-82 b \,c^{2} d x +224 a \,d^{2} c +187 b \,c^{3}\right ) x \sqrt {d x +c}\, e}{64 d^{5} \sqrt {e x}}+\frac {\left (\frac {128 a^{2} d^{4} \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right )}{\sqrt {d e}}+\frac {315 b^{2} c^{4} \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right )}{\sqrt {d e}}-\frac {256 \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \sqrt {d e \left (x +\frac {c}{d}\right )^{2}-c e \left (x +\frac {c}{d}\right )}}{d e \left (x +\frac {c}{d}\right )}+\frac {480 b \,c^{2} d^{2} a \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right )}{\sqrt {d e}}\right ) e \sqrt {\left (d x +c \right ) e x}}{128 d^{5} \sqrt {e x}\, \sqrt {d x +c}}\)	\(298\)
default	\(\frac {\left (32 b^{2} d^{4} x^{4} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}-48 b^{2} c \,d^{3} x^{3} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}+128 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a^{2} d^{5} e x +480 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a b \,c^{2} d^{3} e x +315 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b^{2} c^{4} d e x +128 a b \,d^{4} x^{2} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}+84 b^{2} c^{2} d^{2} x^{2} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}+128 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a^{2} c \,d^{4} e +480 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a b \,c^{3} d^{2} e +315 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b^{2} c^{5} e -320 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a b c \,d^{3} x -210 b^{2} c^{3} d x \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}-256 a^{2} d^{4} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}-960 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a b \,c^{2} d^{2}-630 b^{2} c^{4} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\right ) \sqrt {e x}}{128 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, d^{5} \sqrt {d x +c}}\)	\(528\)

Input:

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

Output:

-1/64*b*(-16*b*d^3*x^3+40*b*c*d^2*x^2-64*a*d^3*x-82*b*c^2*d*x+224*a*c*d^2+ 
187*b*c^3)*x*(d*x+c)^(1/2)/d^5*e/(e*x)^(1/2)+1/128/d^5*(128*a^2*d^4*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)+315*b^2*c^4*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)-256*(a^2*d 
^4+2*a*b*c^2*d^2+b^2*c^4)/d/e/(x+c/d)*(d*e*(x+c/d)^2-c*e*(x+c/d))^(1/2)+48 
0*b*c^2*d^2*a*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*((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 = 449, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{(c+d x)^{3/2}} \, dx=\left [\frac {{\left (315 \, b^{2} c^{5} + 480 \, a b c^{3} d^{2} + 128 \, a^{2} c d^{4} + {\left (315 \, b^{2} c^{4} d + 480 \, a b c^{2} d^{3} + 128 \, a^{2} d^{5}\right )} x\right )} \sqrt {\frac {e}{d}} \log \left (2 \, d e x + 2 \, \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {e}{d}} + c e\right ) + 2 \, {\left (16 \, b^{2} d^{4} x^{4} - 24 \, b^{2} c d^{3} x^{3} - 315 \, b^{2} c^{4} - 480 \, a b c^{2} d^{2} - 128 \, a^{2} d^{4} + 2 \, {\left (21 \, b^{2} c^{2} d^{2} + 32 \, a b d^{4}\right )} x^{2} - 5 \, {\left (21 \, b^{2} c^{3} d + 32 \, a b c d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{128 \, {\left (d^{6} x + c d^{5}\right )}}, -\frac {{\left (315 \, b^{2} c^{5} + 480 \, a b c^{3} d^{2} + 128 \, a^{2} c d^{4} + {\left (315 \, b^{2} c^{4} d + 480 \, a b c^{2} d^{3} + 128 \, a^{2} d^{5}\right )} x\right )} \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {d x + c} \sqrt {e x} d \sqrt {-\frac {e}{d}}}{d e x + c e}\right ) - {\left (16 \, b^{2} d^{4} x^{4} - 24 \, b^{2} c d^{3} x^{3} - 315 \, b^{2} c^{4} - 480 \, a b c^{2} d^{2} - 128 \, a^{2} d^{4} + 2 \, {\left (21 \, b^{2} c^{2} d^{2} + 32 \, a b d^{4}\right )} x^{2} - 5 \, {\left (21 \, b^{2} c^{3} d + 32 \, a b c d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{64 \, {\left (d^{6} x + c d^{5}\right )}}\right ] \] Input:

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

Output:

[1/128*((315*b^2*c^5 + 480*a*b*c^3*d^2 + 128*a^2*c*d^4 + (315*b^2*c^4*d + 
480*a*b*c^2*d^3 + 128*a^2*d^5)*x)*sqrt(e/d)*log(2*d*e*x + 2*sqrt(d*x + c)* 
sqrt(e*x)*d*sqrt(e/d) + c*e) + 2*(16*b^2*d^4*x^4 - 24*b^2*c*d^3*x^3 - 315* 
b^2*c^4 - 480*a*b*c^2*d^2 - 128*a^2*d^4 + 2*(21*b^2*c^2*d^2 + 32*a*b*d^4)* 
x^2 - 5*(21*b^2*c^3*d + 32*a*b*c*d^3)*x)*sqrt(d*x + c)*sqrt(e*x))/(d^6*x + 
 c*d^5), -1/64*((315*b^2*c^5 + 480*a*b*c^3*d^2 + 128*a^2*c*d^4 + (315*b^2* 
c^4*d + 480*a*b*c^2*d^3 + 128*a^2*d^5)*x)*sqrt(-e/d)*arctan(sqrt(d*x + c)* 
sqrt(e*x)*d*sqrt(-e/d)/(d*e*x + c*e)) - (16*b^2*d^4*x^4 - 24*b^2*c*d^3*x^3 
 - 315*b^2*c^4 - 480*a*b*c^2*d^2 - 128*a^2*d^4 + 2*(21*b^2*c^2*d^2 + 32*a* 
b*d^4)*x^2 - 5*(21*b^2*c^3*d + 32*a*b*c*d^3)*x)*sqrt(d*x + c)*sqrt(e*x))/( 
d^6*x + c*d^5)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((e*x)^(1/2)*(b*x^2+a)^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 = 0.26 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{(c+d x)^{3/2}} \, dx=\frac {1}{64} \, \sqrt {{\left (d x + c\right )} d e - c d e} {\left (2 \, {\left (d x + c\right )} {\left (4 \, {\left (d x + c\right )} {\left (\frac {2 \, {\left (d x + c\right )} b^{2} {\left | d \right |}}{d^{7}} - \frac {11 \, b^{2} c {\left | d \right |}}{d^{7}}\right )} + \frac {105 \, b^{2} c^{2} d^{27} {\left | d \right |} + 32 \, a b d^{29} {\left | d \right |}}{d^{34}}\right )} - \frac {325 \, b^{2} c^{3} d^{27} {\left | d \right |} + 288 \, a b c d^{29} {\left | d \right |}}{d^{34}}\right )} \sqrt {d x + c} - \frac {{\left (315 \, \sqrt {d e} b^{2} c^{4} {\left | d \right |} + 480 \, \sqrt {d e} a b c^{2} d^{2} {\left | d \right |} + 128 \, \sqrt {d e} a^{2} d^{4} {\left | d \right |}\right )} \log \left ({\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2}\right )}{128 \, d^{7}} - \frac {4 \, {\left (\sqrt {d e} b^{2} c^{5} e {\left | d \right |} + 2 \, \sqrt {d e} a b c^{3} d^{2} e {\left | d \right |} + \sqrt {d e} a^{2} c d^{4} e {\left | d \right |}\right )}}{{\left (c d e + {\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2}\right )} d^{6}} \] Input:

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

Output:

1/64*sqrt((d*x + c)*d*e - c*d*e)*(2*(d*x + c)*(4*(d*x + c)*(2*(d*x + c)*b^ 
2*abs(d)/d^7 - 11*b^2*c*abs(d)/d^7) + (105*b^2*c^2*d^27*abs(d) + 32*a*b*d^ 
29*abs(d))/d^34) - (325*b^2*c^3*d^27*abs(d) + 288*a*b*c*d^29*abs(d))/d^34) 
*sqrt(d*x + c) - 1/128*(315*sqrt(d*e)*b^2*c^4*abs(d) + 480*sqrt(d*e)*a*b*c 
^2*d^2*abs(d) + 128*sqrt(d*e)*a^2*d^4*abs(d))*log((sqrt(d*e)*sqrt(d*x + c) 
 - sqrt((d*x + c)*d*e - c*d*e))^2)/d^7 - 4*(sqrt(d*e)*b^2*c^5*e*abs(d) + 2 
*sqrt(d*e)*a*b*c^3*d^2*e*abs(d) + sqrt(d*e)*a^2*c*d^4*e*abs(d))/((c*d*e + 
(sqrt(d*e)*sqrt(d*x + c) - sqrt((d*x + c)*d*e - c*d*e))^2)*d^6)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {e}\, \left (128 \sqrt {d}\, \sqrt {d x +c}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a^{2} d^{4}+480 \sqrt {d}\, \sqrt {d x +c}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a b \,c^{2} d^{2}+315 \sqrt {d}\, \sqrt {d x +c}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b^{2} c^{4}-128 \sqrt {d}\, \sqrt {d x +c}\, a^{2} d^{4}-320 \sqrt {d}\, \sqrt {d x +c}\, a b \,c^{2} d^{2}-189 \sqrt {d}\, \sqrt {d x +c}\, b^{2} c^{4}-128 \sqrt {x}\, a^{2} d^{5}-480 \sqrt {x}\, a b \,c^{2} d^{3}-160 \sqrt {x}\, a b c \,d^{4} x +64 \sqrt {x}\, a b \,d^{5} x^{2}-315 \sqrt {x}\, b^{2} c^{4} d -105 \sqrt {x}\, b^{2} c^{3} d^{2} x +42 \sqrt {x}\, b^{2} c^{2} d^{3} x^{2}-24 \sqrt {x}\, b^{2} c \,d^{4} x^{3}+16 \sqrt {x}\, b^{2} d^{5} x^{4}\right )}{64 \sqrt {d x +c}\, d^{6}} \] Input:

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

Output:

(sqrt(e)*(128*sqrt(d)*sqrt(c + d*x)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/ 
sqrt(c))*a**2*d**4 + 480*sqrt(d)*sqrt(c + d*x)*log((sqrt(c + d*x) + sqrt(x 
)*sqrt(d))/sqrt(c))*a*b*c**2*d**2 + 315*sqrt(d)*sqrt(c + d*x)*log((sqrt(c 
+ d*x) + sqrt(x)*sqrt(d))/sqrt(c))*b**2*c**4 - 128*sqrt(d)*sqrt(c + d*x)*a 
**2*d**4 - 320*sqrt(d)*sqrt(c + d*x)*a*b*c**2*d**2 - 189*sqrt(d)*sqrt(c + 
d*x)*b**2*c**4 - 128*sqrt(x)*a**2*d**5 - 480*sqrt(x)*a*b*c**2*d**3 - 160*s 
qrt(x)*a*b*c*d**4*x + 64*sqrt(x)*a*b*d**5*x**2 - 315*sqrt(x)*b**2*c**4*d - 
 105*sqrt(x)*b**2*c**3*d**2*x + 42*sqrt(x)*b**2*c**2*d**3*x**2 - 24*sqrt(x 
)*b**2*c*d**4*x**3 + 16*sqrt(x)*b**2*d**5*x**4))/(64*sqrt(c + d*x)*d**6)