\(\int \frac {\sqrt {e x} (a+b x^2)}{\sqrt {c+d x}} \, dx\) [764]

Optimal result

Integrand size = 24, antiderivative size = 148 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {\left (5 b c^2+8 a d^2\right ) \sqrt {e x} \sqrt {c+d x}}{8 d^3}-\frac {5 b c (e x)^{3/2} \sqrt {c+d x}}{12 d^2 e}+\frac {b (e x)^{5/2} \sqrt {c+d x}}{3 d e^2}-\frac {c \left (5 b c^2+8 a d^2\right ) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{8 d^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {\sqrt {e x} \left (\sqrt {d} \sqrt {x} \sqrt {c+d x} \left (24 a d^2+b \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )+6 c \left (5 b c^2+8 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}-\sqrt {c+d x}}\right )\right )}{24 d^{7/2} \sqrt {x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.

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

\(\Big \downarrow \) 521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(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 + 8*a*d^2)*((Sqrt[e*x]*Sqrt[c + d*x])/d - (c*Sqr 
t[e]*ArcTanh[(Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[c + d*x])])/d^(3/2)))/(4*d) 
)/(6*d)
 

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]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {\left (8 b \,x^{2} d^{2}-10 b c d x +24 a \,d^{2}+15 b \,c^{2}\right ) x \sqrt {d x +c}\, e}{24 d^{3} \sqrt {e x}}-\frac {c \left (8 a \,d^{2}+5 b \,c^{2}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) e \sqrt {\left (d x +c \right ) e x}}{16 d^{3} \sqrt {d e}\, \sqrt {e x}\, \sqrt {d x +c}}\)	\(126\)
default	\(-\frac {\sqrt {e x}\, \sqrt {d x +c}\, \left (-16 b \,d^{2} x^{2} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}+24 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a c \,d^{2} e +15 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{3} e +20 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b c d x -48 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, a \,d^{2}-30 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b \,c^{2}\right )}{48 d^{3} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}}\)	\(204\)

Input:

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

Output:

1/24*(8*b*d^2*x^2-10*b*c*d*x+24*a*d^2+15*b*c^2)*x*(d*x+c)^(1/2)/d^3*e/(e*x 
)^(1/2)-1/16*c*(8*a*d^2+5*b*c^2)/d^3*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 = 208, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (5 \, b c^{3} + 8 \, a c d^{2}\right )} \sqrt {\frac {e}{d}} \log \left (2 \, d e x - 2 \, \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {e}{d}} + c e\right ) + 2 \, {\left (8 \, b d^{2} x^{2} - 10 \, b c d x + 15 \, b c^{2} + 24 \, a d^{2}\right )} \sqrt {d x + c} \sqrt {e x}}{48 \, d^{3}}, \frac {3 \, {\left (5 \, b c^{3} + 8 \, a c d^{2}\right )} \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {d x + c} \sqrt {e x} d \sqrt {-\frac {e}{d}}}{d e x + c e}\right ) + {\left (8 \, b d^{2} x^{2} - 10 \, b c d x + 15 \, b c^{2} + 24 \, a d^{2}\right )} \sqrt {d x + c} \sqrt {e x}}{24 \, d^{3}}\right ] \] Input:

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

Output:

[1/48*(3*(5*b*c^3 + 8*a*c*d^2)*sqrt(e/d)*log(2*d*e*x - 2*sqrt(d*x + c)*sqr 
t(e*x)*d*sqrt(e/d) + c*e) + 2*(8*b*d^2*x^2 - 10*b*c*d*x + 15*b*c^2 + 24*a* 
d^2)*sqrt(d*x + c)*sqrt(e*x))/d^3, 1/24*(3*(5*b*c^3 + 8*a*c*d^2)*sqrt(-e/d 
)*arctan(sqrt(d*x + c)*sqrt(e*x)*d*sqrt(-e/d)/(d*e*x + c*e)) + (8*b*d^2*x^ 
2 - 10*b*c*d*x + 15*b*c^2 + 24*a*d^2)*sqrt(d*x + c)*sqrt(e*x))/d^3]
 

Sympy [A] (verification not implemented)

Time = 9.05 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {a \sqrt {c} \sqrt {e} \sqrt {x} \sqrt {1 + \frac {d x}{c}}}{d} - \frac {a c \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{d^{\frac {3}{2}}} + \frac {5 b c^{\frac {5}{2}} \sqrt {e} \sqrt {x}}{8 d^{3} \sqrt {1 + \frac {d x}{c}}} + \frac {5 b c^{\frac {3}{2}} \sqrt {e} x^{\frac {3}{2}}}{24 d^{2} \sqrt {1 + \frac {d x}{c}}} - \frac {b \sqrt {c} \sqrt {e} x^{\frac {5}{2}}}{12 d \sqrt {1 + \frac {d x}{c}}} - \frac {5 b c^{3} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{8 d^{\frac {7}{2}}} + \frac {b \sqrt {e} x^{\frac {7}{2}}}{3 \sqrt {c} \sqrt {1 + \frac {d x}{c}}} \] Input:

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

Output:

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

Maxima [F(-2)]

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

integrate((e*x)^(1/2)*(b*x^2+a)/(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 = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {\frac {24 \, {\left (\frac {c d e \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e}} + \sqrt {{\left (d x + c\right )} d e - c d e} \sqrt {d x + c}\right )} a {\left | d \right |}}{d^{2}} + \frac {{\left (\frac {15 \, c^{3} d e \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e}} + \sqrt {{\left (d x + c\right )} d e - c d e} {\left (2 \, {\left (4 \, d x - 9 \, c\right )} {\left (d x + c\right )} + 33 \, c^{2}\right )} \sqrt {d x + c}\right )} b {\left | d \right |}}{d^{4}}}{24 \, d} \] Input:

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

Output:

1/24*(24*(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*abs(d)/d^2 
 + (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))*b*abs(d)/d^4)/d
 

Mupad [B] (verification not implemented)

Time = 19.47 (sec) , antiderivative size = 552, normalized size of antiderivative = 3.73 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {\frac {33\,b\,c^3\,e^3\,{\left (e\,x\right )}^{7/2}}{2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}-\frac {85\,b\,c^3\,d\,e^2\,{\left (e\,x\right )}^{9/2}}{12\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^9}+\frac {5\,b\,c^3\,d^2\,e\,{\left (e\,x\right )}^{11/2}}{4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{11}}+\frac {5\,b\,c^3\,e^6\,\sqrt {e\,x}}{4\,d^3\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {85\,b\,c^3\,e^5\,{\left (e\,x\right )}^{3/2}}{12\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}+\frac {33\,b\,c^3\,e^4\,{\left (e\,x\right )}^{5/2}}{2\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}}{e^6-\frac {6\,d\,e^6\,x}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {15\,d^2\,e^6\,x^2}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {20\,d^3\,e^6\,x^3}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {15\,d^4\,e^6\,x^4}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}-\frac {6\,d^5\,e^6\,x^5}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{10}}+\frac {d^6\,e^6\,x^6}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{12}}}+\frac {\frac {a\,\left (d\,{\left (e\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2}+e\,\sqrt {e\,x}\,{\left (c+d\,x\right )}^{5/2}-c\,d\,{\left (e\,x\right )}^{3/2}\,\sqrt {c+d\,x}-2\,c\,e\,\sqrt {e\,x}\,{\left (c+d\,x\right )}^{3/2}+c^2\,e\,\sqrt {e\,x}\,\sqrt {c+d\,x}\right )}{2}+\frac {a\,\sqrt {c}\,d^2\,x\,{\left (e\,x\right )}^{3/2}}{2}}{d^3\,e\,x^2}-\frac {a\,\left (4\,c\,d^{3/2}\,e^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {d}\,\sqrt {e\,x}}{\sqrt {e}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )+\sqrt {c}\,d^2\,e\,\sqrt {e\,x}\right )}{2\,d^3\,e}-\frac {5\,b\,c^3\,\sqrt {e}\,\mathrm {atanh}\left (\frac {\sqrt {d}\,\sqrt {e\,x}}{\sqrt {e}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )}{4\,d^{7/2}} \] Input:

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

Output:

((33*b*c^3*e^3*(e*x)^(7/2))/(2*((c + d*x)^(1/2) - c^(1/2))^7) - (85*b*c^3* 
d*e^2*(e*x)^(9/2))/(12*((c + d*x)^(1/2) - c^(1/2))^9) + (5*b*c^3*d^2*e*(e* 
x)^(11/2))/(4*((c + d*x)^(1/2) - c^(1/2))^11) + (5*b*c^3*e^6*(e*x)^(1/2))/ 
(4*d^3*((c + d*x)^(1/2) - c^(1/2))) - (85*b*c^3*e^5*(e*x)^(3/2))/(12*d^2*( 
(c + d*x)^(1/2) - c^(1/2))^3) + (33*b*c^3*e^4*(e*x)^(5/2))/(2*d*((c + d*x) 
^(1/2) - c^(1/2))^5))/(e^6 - (6*d*e^6*x)/((c + d*x)^(1/2) - c^(1/2))^2 + ( 
15*d^2*e^6*x^2)/((c + d*x)^(1/2) - c^(1/2))^4 - (20*d^3*e^6*x^3)/((c + d*x 
)^(1/2) - c^(1/2))^6 + (15*d^4*e^6*x^4)/((c + d*x)^(1/2) - c^(1/2))^8 - (6 
*d^5*e^6*x^5)/((c + d*x)^(1/2) - c^(1/2))^10 + (d^6*e^6*x^6)/((c + d*x)^(1 
/2) - c^(1/2))^12) + ((a*(d*(e*x)^(3/2)*(c + d*x)^(3/2) + e*(e*x)^(1/2)*(c 
 + d*x)^(5/2) - c*d*(e*x)^(3/2)*(c + d*x)^(1/2) - 2*c*e*(e*x)^(1/2)*(c + d 
*x)^(3/2) + c^2*e*(e*x)^(1/2)*(c + d*x)^(1/2)))/2 + (a*c^(1/2)*d^2*x*(e*x) 
^(3/2))/2)/(d^3*e*x^2) - (a*(4*c*d^(3/2)*e^(3/2)*atanh((d^(1/2)*(e*x)^(1/2 
))/(e^(1/2)*((c + d*x)^(1/2) - c^(1/2)))) + c^(1/2)*d^2*e*(e*x)^(1/2)))/(2 
*d^3*e) - (5*b*c^3*e^(1/2)*atanh((d^(1/2)*(e*x)^(1/2))/(e^(1/2)*((c + d*x) 
^(1/2) - c^(1/2)))))/(4*d^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {\sqrt {e}\, \left (24 \sqrt {x}\, \sqrt {d x +c}\, a \,d^{3}+15 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{2} d -10 \sqrt {x}\, \sqrt {d x +c}\, b c \,d^{2} x +8 \sqrt {x}\, \sqrt {d x +c}\, b \,d^{3} x^{2}-24 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a c \,d^{2}-15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{3}\right )}{24 d^{4}} \] Input:

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

Output:

(sqrt(e)*(24*sqrt(x)*sqrt(c + d*x)*a*d**3 + 15*sqrt(x)*sqrt(c + d*x)*b*c** 
2*d - 10*sqrt(x)*sqrt(c + d*x)*b*c*d**2*x + 8*sqrt(x)*sqrt(c + d*x)*b*d**3 
*x**2 - 24*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*a*c*d**2 
 - 15*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*b*c**3))/(24* 
d**4)