\(\int \frac {\sqrt {e x} (a+b x^2)^2}{\sqrt {c+d x}} \, dx\) [825]

Optimal result

Integrand size = 26, antiderivative size = 261 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx=\frac {\left (63 b^2 c^4+160 a b c^2 d^2+128 a^2 d^4\right ) \sqrt {e x} \sqrt {c+d x}}{128 d^5}-\frac {b c \left (63 b c^2+160 a d^2\right ) (e x)^{3/2} \sqrt {c+d x}}{192 d^4 e}+\frac {b \left (63 b c^2+160 a d^2\right ) (e x)^{5/2} \sqrt {c+d x}}{240 d^3 e^2}-\frac {9 b^2 c (e x)^{7/2} \sqrt {c+d x}}{40 d^2 e^3}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}-\frac {c \left (63 b^2 c^4+160 a b c^2 d^2+128 a^2 d^4\right ) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{128 d^{11/2}} \] Output:

1/128*(128*a^2*d^4+160*a*b*c^2*d^2+63*b^2*c^4)*(e*x)^(1/2)*(d*x+c)^(1/2)/d 
^5-1/192*b*c*(160*a*d^2+63*b*c^2)*(e*x)^(3/2)*(d*x+c)^(1/2)/d^4/e+1/240*b* 
(160*a*d^2+63*b*c^2)*(e*x)^(5/2)*(d*x+c)^(1/2)/d^3/e^2-9/40*b^2*c*(e*x)^(7 
/2)*(d*x+c)^(1/2)/d^2/e^3+1/5*b^2*(e*x)^(9/2)*(d*x+c)^(1/2)/d/e^4-1/128*c* 
(128*a^2*d^4+160*a*b*c^2*d^2+63*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 = 0.82 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx=\frac {\sqrt {e x} \left (\sqrt {d} \sqrt {x} \sqrt {c+d x} \left (1920 a^2 d^4+160 a b d^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )+b^2 \left (945 c^4-630 c^3 d x+504 c^2 d^2 x^2-432 c d^3 x^3+384 d^4 x^4\right )\right )+30 c \left (63 b^2 c^4+160 a b c^2 d^2+128 a^2 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}-\sqrt {c+d x}}\right )\right )}{1920 d^{11/2} \sqrt {x}} \] Input:

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

Output:

(Sqrt[e*x]*(Sqrt[d]*Sqrt[x]*Sqrt[c + d*x]*(1920*a^2*d^4 + 160*a*b*d^2*(15* 
c^2 - 10*c*d*x + 8*d^2*x^2) + b^2*(945*c^4 - 630*c^3*d*x + 504*c^2*d^2*x^2 
 - 432*c*d^3*x^3 + 384*d^4*x^4)) + 30*c*(63*b^2*c^4 + 160*a*b*c^2*d^2 + 12 
8*a^2*d^4)*ArcTanh[(Sqrt[d]*Sqrt[x])/(Sqrt[c] - Sqrt[c + d*x])]))/(1920*d^ 
(11/2)*Sqrt[x])
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {521, 27, 2125, 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.

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

\(\Big \downarrow \) 521

\(\displaystyle \frac {\int \frac {\sqrt {e x} \left (-9 b^2 c x^3 e^4+20 a b d x^2 e^4+10 a^2 d e^4\right )}{2 \sqrt {c+d x}}dx}{5 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {e x} \left (-9 b^2 c x^3 e^4+20 a b d x^2 e^4+10 a^2 d e^4\right )}{\sqrt {c+d x}}dx}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 2125

\(\displaystyle \frac {\frac {\int \frac {e^7 \sqrt {e x} \left (80 a^2 d^2+b \left (63 b c^2+160 a d^2\right ) x^2\right )}{2 \sqrt {c+d x}}dx}{4 d e^3}-\frac {9 b^2 c e (e x)^{7/2} \sqrt {c+d x}}{4 d}}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {e^4 \int \frac {\sqrt {e x} \left (80 a^2 d^2+b \left (63 b c^2+160 a d^2\right ) x^2\right )}{\sqrt {c+d x}}dx}{8 d}-\frac {9 b^2 c e (e x)^{7/2} \sqrt {c+d x}}{4 d}}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 521

\(\displaystyle \frac {\frac {e^4 \left (\frac {\int \frac {5 e^2 \sqrt {e x} \left (96 a^2 d^3-b c \left (63 b c^2+160 a d^2\right ) x\right )}{2 \sqrt {c+d x}}dx}{3 d e^2}+\frac {b (e x)^{5/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{3 d e^2}\right )}{8 d}-\frac {9 b^2 c e (e x)^{7/2} \sqrt {c+d x}}{4 d}}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {e^4 \left (\frac {5 \int \frac {\sqrt {e x} \left (96 a^2 d^3-b c \left (63 b c^2+160 a d^2\right ) x\right )}{\sqrt {c+d x}}dx}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{3 d e^2}\right )}{8 d}-\frac {9 b^2 c e (e x)^{7/2} \sqrt {c+d x}}{4 d}}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {e^4 \left (\frac {5 \left (\frac {3 \left (128 a^2 d^4+160 a b c^2 d^2+63 b^2 c^4\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x}}dx}{4 d}-\frac {b c (e x)^{3/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{2 d e}\right )}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{3 d e^2}\right )}{8 d}-\frac {9 b^2 c e (e x)^{7/2} \sqrt {c+d x}}{4 d}}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {e^4 \left (\frac {5 \left (\frac {3 \left (128 a^2 d^4+160 a b c^2 d^2+63 b^2 c^4\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}-\frac {b c (e x)^{3/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{2 d e}\right )}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{3 d e^2}\right )}{8 d}-\frac {9 b^2 c e (e x)^{7/2} \sqrt {c+d x}}{4 d}}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\frac {e^4 \left (\frac {5 \left (\frac {3 \left (128 a^2 d^4+160 a b c^2 d^2+63 b^2 c^4\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}}{d}\right )}{4 d}-\frac {b c (e x)^{3/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{2 d e}\right )}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{3 d e^2}\right )}{8 d}-\frac {9 b^2 c e (e x)^{7/2} \sqrt {c+d x}}{4 d}}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {e^4 \left (\frac {5 \left (\frac {3 \left (128 a^2 d^4+160 a b c^2 d^2+63 b^2 c^4\right ) \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{3/2}}\right )}{4 d}-\frac {b c (e x)^{3/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{2 d e}\right )}{6 d}+\frac {b (e x)^{5/2} \sqrt {c+d x} \left (160 a d^2+63 b c^2\right )}{3 d e^2}\right )}{8 d}-\frac {9 b^2 c e (e x)^{7/2} \sqrt {c+d x}}{4 d}}{10 d e^4}+\frac {b^2 (e x)^{9/2} \sqrt {c+d x}}{5 d e^4}\)

Input:

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

Output:

(b^2*(e*x)^(9/2)*Sqrt[c + d*x])/(5*d*e^4) + ((-9*b^2*c*e*(e*x)^(7/2)*Sqrt[ 
c + d*x])/(4*d) + (e^4*((b*(63*b*c^2 + 160*a*d^2)*(e*x)^(5/2)*Sqrt[c + d*x 
])/(3*d*e^2) + (5*(-1/2*(b*c*(63*b*c^2 + 160*a*d^2)*(e*x)^(3/2)*Sqrt[c + d 
*x])/(d*e) + (3*(63*b^2*c^4 + 160*a*b*c^2*d^2 + 128*a^2*d^4)*((Sqrt[e*x]*S 
qrt[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)))/(8*d))/(10*d*e^4)
 

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 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 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 = 0.26 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {\left (384 b^{2} d^{4} x^{4}-432 b^{2} c \,d^{3} x^{3}+1280 a b \,d^{4} x^{2}+504 d^{2} c^{2} x^{2} b^{2}-1600 a b c \,d^{3} x -630 b^{2} c^{3} d x +1920 a^{2} d^{4}+2400 b \,c^{2} d^{2} a +945 b^{2} c^{4}\right ) x \sqrt {d x +c}\, e}{1920 d^{5} \sqrt {e x}}-\frac {c \left (128 a^{2} d^{4}+160 b \,c^{2} d^{2} a +63 b^{2} c^{4}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) e \sqrt {\left (d x +c \right ) e x}}{256 d^{5} \sqrt {d e}\, \sqrt {e x}\, \sqrt {d x +c}}\)	\(205\)
default	\(-\frac {\sqrt {e x}\, \sqrt {d x +c}\, \left (-768 b^{2} d^{4} x^{4} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+864 b^{2} c \,d^{3} x^{3} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-2560 a b \,d^{4} x^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-1008 b^{2} c^{2} d^{2} x^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+1920 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a^{2} c \,d^{4} e +2400 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a b \,c^{3} d^{2} e +945 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b^{2} c^{5} e +3200 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a b c \,d^{3} x +1260 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, b^{2} c^{3} d x -3840 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a^{2} d^{4}-4800 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a b \,c^{2} d^{2}-1890 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, b^{2} c^{4}\right )}{3840 d^{5} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}}\)	\(393\)

Input:

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

Output:

1/1920*(384*b^2*d^4*x^4-432*b^2*c*d^3*x^3+1280*a*b*d^4*x^2+504*b^2*c^2*d^2 
*x^2-1600*a*b*c*d^3*x-630*b^2*c^3*d*x+1920*a^2*d^4+2400*a*b*c^2*d^2+945*b^ 
2*c^4)*x*(d*x+c)^(1/2)/d^5*e/(e*x)^(1/2)-1/256*c*(128*a^2*d^4+160*a*b*c^2* 
d^2+63*b^2*c^4)/d^5*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 = 0.12 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx=\left [\frac {15 \, {\left (63 \, b^{2} c^{5} + 160 \, a b c^{3} d^{2} + 128 \, a^{2} c d^{4}\right )} \sqrt {\frac {e}{d}} \log \left (2 \, d e x - 2 \, \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {e}{d}} + c e\right ) + 2 \, {\left (384 \, b^{2} d^{4} x^{4} - 432 \, b^{2} c d^{3} x^{3} + 945 \, b^{2} c^{4} + 2400 \, a b c^{2} d^{2} + 1920 \, a^{2} d^{4} + 8 \, {\left (63 \, b^{2} c^{2} d^{2} + 160 \, a b d^{4}\right )} x^{2} - 10 \, {\left (63 \, b^{2} c^{3} d + 160 \, a b c d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{3840 \, d^{5}}, \frac {15 \, {\left (63 \, b^{2} c^{5} + 160 \, a b c^{3} d^{2} + 128 \, a^{2} c d^{4}\right )} \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {d x + c} \sqrt {e x} d \sqrt {-\frac {e}{d}}}{d e x + c e}\right ) + {\left (384 \, b^{2} d^{4} x^{4} - 432 \, b^{2} c d^{3} x^{3} + 945 \, b^{2} c^{4} + 2400 \, a b c^{2} d^{2} + 1920 \, a^{2} d^{4} + 8 \, {\left (63 \, b^{2} c^{2} d^{2} + 160 \, a b d^{4}\right )} x^{2} - 10 \, {\left (63 \, b^{2} c^{3} d + 160 \, a b c d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{1920 \, d^{5}}\right ] \] Input:

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

Output:

[1/3840*(15*(63*b^2*c^5 + 160*a*b*c^3*d^2 + 128*a^2*c*d^4)*sqrt(e/d)*log(2 
*d*e*x - 2*sqrt(d*x + c)*sqrt(e*x)*d*sqrt(e/d) + c*e) + 2*(384*b^2*d^4*x^4 
 - 432*b^2*c*d^3*x^3 + 945*b^2*c^4 + 2400*a*b*c^2*d^2 + 1920*a^2*d^4 + 8*( 
63*b^2*c^2*d^2 + 160*a*b*d^4)*x^2 - 10*(63*b^2*c^3*d + 160*a*b*c*d^3)*x)*s 
qrt(d*x + c)*sqrt(e*x))/d^5, 1/1920*(15*(63*b^2*c^5 + 160*a*b*c^3*d^2 + 12 
8*a^2*c*d^4)*sqrt(-e/d)*arctan(sqrt(d*x + c)*sqrt(e*x)*d*sqrt(-e/d)/(d*e*x 
 + c*e)) + (384*b^2*d^4*x^4 - 432*b^2*c*d^3*x^3 + 945*b^2*c^4 + 2400*a*b*c 
^2*d^2 + 1920*a^2*d^4 + 8*(63*b^2*c^2*d^2 + 160*a*b*d^4)*x^2 - 10*(63*b^2* 
c^3*d + 160*a*b*c*d^3)*x)*sqrt(d*x + c)*sqrt(e*x))/d^5]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((e*x)^(1/2)*(b*x^2+a)^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 = 0.20 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx=\frac {\frac {1920 \, {\left (\frac {c d e \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e}} + \sqrt {{\left (d x + c\right )} d e - c d e} \sqrt {d x + c}\right )} a^{2} {\left | d \right |}}{d^{2}} + \frac {160 \, {\left (\frac {15 \, c^{3} d e \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e}} + \sqrt {{\left (d x + c\right )} d e - c d e} {\left (2 \, {\left (4 \, d x - 9 \, c\right )} {\left (d x + c\right )} + 33 \, c^{2}\right )} \sqrt {d x + c}\right )} a b {\left | d \right |}}{d^{4}} + \frac {3 \, {\left (\frac {315 \, c^{5} d e \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e}} + {\left (965 \, c^{4} - 2 \, {\left (745 \, c^{3} - 4 \, {\left (2 \, {\left (8 \, d x - 33 \, c\right )} {\left (d x + c\right )} + 171 \, c^{2}\right )} {\left (d x + c\right )}\right )} {\left (d x + c\right )}\right )} \sqrt {{\left (d x + c\right )} d e - c d e} \sqrt {d x + c}\right )} b^{2} {\left | d \right |}}{d^{6}}}{1920 \, d} \] Input:

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

Output:

1/1920*(1920*(c*d*e*log(abs(-sqrt(d*e)*sqrt(d*x + c) + sqrt((d*x + c)*d*e 
- c*d*e)))/sqrt(d*e) + sqrt((d*x + c)*d*e - c*d*e)*sqrt(d*x + c))*a^2*abs( 
d)/d^2 + 160*(15*c^3*d*e*log(abs(-sqrt(d*e)*sqrt(d*x + c) + sqrt((d*x + c) 
*d*e - c*d*e)))/sqrt(d*e) + sqrt((d*x + c)*d*e - c*d*e)*(2*(4*d*x - 9*c)*( 
d*x + c) + 33*c^2)*sqrt(d*x + c))*a*b*abs(d)/d^4 + 3*(315*c^5*d*e*log(abs( 
-sqrt(d*e)*sqrt(d*x + c) + sqrt((d*x + c)*d*e - c*d*e)))/sqrt(d*e) + (965* 
c^4 - 2*(745*c^3 - 4*(2*(8*d*x - 33*c)*(d*x + c) + 171*c^2)*(d*x + c))*(d* 
x + c))*sqrt((d*x + c)*d*e - c*d*e)*sqrt(d*x + c))*b^2*abs(d)/d^6)/d
 

Mupad [B] (verification not implemented)

Time = 25.54 (sec) , antiderivative size = 925, normalized size of antiderivative = 3.54 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx =\text {Too large to display} \] Input:

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

Output:

(((e*x)^(19/2)*((63*b^2*c^5*d^4*e)/64 + 2*a^2*c*d^8*e + (5*a*b*c^3*d^6*e)/ 
2))/((c + d*x)^(1/2) - c^(1/2))^19 - ((e*x)^(13/2)*(56*a^2*c*d^5*e^4 + (16 
83*b^2*c^5*d*e^4)/16 + 194*a*b*c^3*d^3*e^4))/((c + d*x)^(1/2) - c^(1/2))^1 
3 - ((e*x)^(17/2)*(14*a^2*c*d^7*e^2 + (609*b^2*c^5*d^3*e^2)/64 + (145*a*b* 
c^3*d^5*e^2)/6))/((c + d*x)^(1/2) - c^(1/2))^17 + ((e*x)^(15/2)*(40*a^2*c* 
d^6*e^3 + (3297*b^2*c^5*d^2*e^3)/80 + (314*a*b*c^3*d^4*e^3)/3))/((c + d*x) 
^(1/2) - c^(1/2))^15 + ((e*x)^(11/2)*((5597*b^2*c^5*e^5)/32 + 28*a^2*c*d^4 
*e^5 + 111*a*b*c^3*d^2*e^5))/((c + d*x)^(1/2) - c^(1/2))^11 - ((e*x)^(7/2) 
*(1683*b^2*c^5*e^7 + 896*a^2*c*d^4*e^7 + 3104*a*b*c^3*d^2*e^7))/(16*d^2*(( 
c + d*x)^(1/2) - c^(1/2))^7) + ((e*x)^(9/2)*(5597*b^2*c^5*e^6 + 896*a^2*c* 
d^4*e^6 + 3552*a*b*c^3*d^2*e^6))/(32*d*((c + d*x)^(1/2) - c^(1/2))^9) + (e 
^9*(e*x)^(1/2)*(63*b^2*c^5*e + 128*a^2*c*d^4*e + 160*a*b*c^3*d^2*e))/(64*d 
^5*((c + d*x)^(1/2) - c^(1/2))) - (e^8*(e*x)^(3/2)*(1827*b^2*c^5*e + 2688* 
a^2*c*d^4*e + 4640*a*b*c^3*d^2*e))/(192*d^4*((c + d*x)^(1/2) - c^(1/2))^3) 
 + (e^7*(e*x)^(5/2)*(9891*b^2*c^5*e + 9600*a^2*c*d^4*e + 25120*a*b*c^3*d^2 
*e))/(240*d^3*((c + d*x)^(1/2) - c^(1/2))^5))/(e^10 - (10*d*e^10*x)/((c + 
d*x)^(1/2) - c^(1/2))^2 + (45*d^2*e^10*x^2)/((c + d*x)^(1/2) - c^(1/2))^4 
- (120*d^3*e^10*x^3)/((c + d*x)^(1/2) - c^(1/2))^6 + (210*d^4*e^10*x^4)/(( 
c + d*x)^(1/2) - c^(1/2))^8 - (252*d^5*e^10*x^5)/((c + d*x)^(1/2) - c^(1/2 
))^10 + (210*d^6*e^10*x^6)/((c + d*x)^(1/2) - c^(1/2))^12 - (120*d^7*e^...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx=\frac {\sqrt {e}\, \left (1920 \sqrt {x}\, \sqrt {d x +c}\, a^{2} d^{5}+2400 \sqrt {x}\, \sqrt {d x +c}\, a b \,c^{2} d^{3}-1600 \sqrt {x}\, \sqrt {d x +c}\, a b c \,d^{4} x +1280 \sqrt {x}\, \sqrt {d x +c}\, a b \,d^{5} x^{2}+945 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{4} d -630 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{3} d^{2} x +504 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{2} d^{3} x^{2}-432 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c \,d^{4} x^{3}+384 \sqrt {x}\, \sqrt {d x +c}\, b^{2} d^{5} x^{4}-1920 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a^{2} c \,d^{4}-2400 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a b \,c^{3} d^{2}-945 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b^{2} c^{5}\right )}{1920 d^{6}} \] Input:

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

Output:

(sqrt(e)*(1920*sqrt(x)*sqrt(c + d*x)*a**2*d**5 + 2400*sqrt(x)*sqrt(c + d*x 
)*a*b*c**2*d**3 - 1600*sqrt(x)*sqrt(c + d*x)*a*b*c*d**4*x + 1280*sqrt(x)*s 
qrt(c + d*x)*a*b*d**5*x**2 + 945*sqrt(x)*sqrt(c + d*x)*b**2*c**4*d - 630*s 
qrt(x)*sqrt(c + d*x)*b**2*c**3*d**2*x + 504*sqrt(x)*sqrt(c + d*x)*b**2*c** 
2*d**3*x**2 - 432*sqrt(x)*sqrt(c + d*x)*b**2*c*d**4*x**3 + 384*sqrt(x)*sqr 
t(c + d*x)*b**2*d**5*x**4 - 1920*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt 
(d))/sqrt(c))*a**2*c*d**4 - 2400*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt 
(d))/sqrt(c))*a*b*c**3*d**2 - 945*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqr 
t(d))/sqrt(c))*b**2*c**5))/(1920*d**6)