\(\int \sqrt {e x} (c+d x)^{3/2} (a+b x^2) \, dx\) [753]

Optimal result

Integrand size = 24, antiderivative size = 232 \[ \int \sqrt {e x} (c+d x)^{3/2} \left (a+b x^2\right ) \, dx=\frac {c^2 \left (3 b c^2+16 a d^2\right ) \sqrt {e x} \sqrt {c+d x}}{128 d^3}+\frac {7 c \left (3 b c^2+16 a d^2\right ) (e x)^{3/2} \sqrt {c+d x}}{192 d^2 e}+\frac {\left (3 b c^2+16 a d^2\right ) (e x)^{5/2} \sqrt {c+d x}}{48 d e^2}-\frac {b c (e x)^{3/2} (c+d x)^{5/2}}{8 d^2 e}+\frac {b (e x)^{5/2} (c+d x)^{5/2}}{5 d e^2}-\frac {c^3 \left (3 b c^2+16 a d^2\right ) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{128 d^{7/2}} \] Output:

1/128*c^2*(16*a*d^2+3*b*c^2)*(e*x)^(1/2)*(d*x+c)^(1/2)/d^3+7/192*c*(16*a*d 
^2+3*b*c^2)*(e*x)^(3/2)*(d*x+c)^(1/2)/d^2/e+1/48*(16*a*d^2+3*b*c^2)*(e*x)^ 
(5/2)*(d*x+c)^(1/2)/d/e^2-1/8*b*c*(e*x)^(3/2)*(d*x+c)^(5/2)/d^2/e+1/5*b*(e 
*x)^(5/2)*(d*x+c)^(5/2)/d/e^2-1/128*c^3*(16*a*d^2+3*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 = 0.70 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.69 \[ \int \sqrt {e x} (c+d x)^{3/2} \left (a+b x^2\right ) \, dx=\frac {\sqrt {e x} \left (\sqrt {d} \sqrt {x} \sqrt {c+d x} \left (80 a d^2 \left (3 c^2+14 c d x+8 d^2 x^2\right )+3 b \left (15 c^4-10 c^3 d x+8 c^2 d^2 x^2+176 c d^3 x^3+128 d^4 x^4\right )\right )+30 c^3 \left (3 b c^2+16 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}-\sqrt {c+d x}}\right )\right )}{1920 d^{7/2} \sqrt {x}} \] Input:

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

Output:

(Sqrt[e*x]*(Sqrt[d]*Sqrt[x]*Sqrt[c + d*x]*(80*a*d^2*(3*c^2 + 14*c*d*x + 8* 
d^2*x^2) + 3*b*(15*c^4 - 10*c^3*d*x + 8*c^2*d^2*x^2 + 176*c*d^3*x^3 + 128* 
d^4*x^4)) + 30*c^3*(3*b*c^2 + 16*a*d^2)*ArcTanh[(Sqrt[d]*Sqrt[x])/(Sqrt[c] 
 - Sqrt[c + d*x])]))/(1920*d^(7/2)*Sqrt[x])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {521, 27, 90, 60, 60, 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.

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

\(\Big \downarrow \) 521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\left (16 a d^2+3 b c^2\right ) \left (\frac {1}{2} c \int \sqrt {e x} \sqrt {c+d x}dx+\frac {(e x)^{3/2} (c+d x)^{3/2}}{3 e}\right )}{8 d}-\frac {b c (e x)^{3/2} (c+d x)^{5/2}}{4 d e}}{2 d}+\frac {b (e x)^{5/2} (c+d x)^{5/2}}{5 d e^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\left (16 a d^2+3 b c^2\right ) \left (\frac {1}{2} c \left (\frac {1}{4} c \int \frac {\sqrt {e x}}{\sqrt {c+d x}}dx+\frac {(e x)^{3/2} \sqrt {c+d x}}{2 e}\right )+\frac {(e x)^{3/2} (c+d x)^{3/2}}{3 e}\right )}{8 d}-\frac {b c (e x)^{3/2} (c+d x)^{5/2}}{4 d e}}{2 d}+\frac {b (e x)^{5/2} (c+d x)^{5/2}}{5 d e^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\left (16 a d^2+3 b c^2\right ) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx}{2 d}\right )+\frac {(e x)^{3/2} \sqrt {c+d x}}{2 e}\right )+\frac {(e x)^{3/2} (c+d x)^{3/2}}{3 e}\right )}{8 d}-\frac {b c (e x)^{3/2} (c+d x)^{5/2}}{4 d e}}{2 d}+\frac {b (e x)^{5/2} (c+d x)^{5/2}}{5 d e^2}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\frac {\left (16 a d^2+3 b c^2\right ) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}}{d}\right )+\frac {(e x)^{3/2} \sqrt {c+d x}}{2 e}\right )+\frac {(e x)^{3/2} (c+d x)^{3/2}}{3 e}\right )}{8 d}-\frac {b c (e x)^{3/2} (c+d x)^{5/2}}{4 d e}}{2 d}+\frac {b (e x)^{5/2} (c+d x)^{5/2}}{5 d e^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\left (16 a d^2+3 b c^2\right ) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{3/2}}\right )+\frac {(e x)^{3/2} \sqrt {c+d x}}{2 e}\right )+\frac {(e x)^{3/2} (c+d x)^{3/2}}{3 e}\right )}{8 d}-\frac {b c (e x)^{3/2} (c+d x)^{5/2}}{4 d e}}{2 d}+\frac {b (e x)^{5/2} (c+d x)^{5/2}}{5 d e^2}\)

Input:

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

Output:

(b*(e*x)^(5/2)*(c + d*x)^(5/2))/(5*d*e^2) + (-1/4*(b*c*(e*x)^(3/2)*(c + d* 
x)^(5/2))/(d*e) + ((3*b*c^2 + 16*a*d^2)*(((e*x)^(3/2)*(c + d*x)^(3/2))/(3* 
e) + (c*(((e*x)^(3/2)*Sqrt[c + d*x])/(2*e) + (c*((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))/(8*d))/(2*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 = 172, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\left (384 b \,x^{4} d^{4}+528 c b \,d^{3} x^{3}+640 a \,d^{4} x^{2}+24 b \,c^{2} d^{2} x^{2}+1120 a c \,d^{3} x -30 b \,c^{3} d x +240 a \,c^{2} d^{2}+45 b \,c^{4}\right ) x \sqrt {d x +c}\, e}{1920 d^{3} \sqrt {e x}}-\frac {c^{3} \left (16 a \,d^{2}+3 b \,c^{2}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) e \sqrt {\left (d x +c \right ) e x}}{256 d^{3} \sqrt {d e}\, \sqrt {e x}\, \sqrt {d x +c}}\)	\(172\)
default	\(-\frac {\sqrt {e x}\, \sqrt {d x +c}\, \left (-768 b \,d^{4} x^{4} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}-1056 b c \,d^{3} x^{3} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}-1280 a \,d^{4} x^{2} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}-48 b \,c^{2} d^{2} x^{2} \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}+240 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a \,c^{3} d^{2} e +45 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{5} e -2240 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, a c \,d^{3} x +60 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b \,c^{3} d x -480 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, a \,c^{2} d^{2}-90 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b \,c^{4}\right )}{3840 d^{3} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}}\)	\(310\)

Input:

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

Output:

1/1920*(384*b*d^4*x^4+528*b*c*d^3*x^3+640*a*d^4*x^2+24*b*c^2*d^2*x^2+1120* 
a*c*d^3*x-30*b*c^3*d*x+240*a*c^2*d^2+45*b*c^4)*x*(d*x+c)^(1/2)/d^3*e/(e*x) 
^(1/2)-1/256*c^3*(16*a*d^2+3*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 = 0.10 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.31 \[ \int \sqrt {e x} (c+d x)^{3/2} \left (a+b x^2\right ) \, dx=\left [\frac {15 \, {\left (3 \, b c^{5} + 16 \, a c^{3} d^{2}\right )} \sqrt {\frac {e}{d}} \log \left (2 \, d e x - 2 \, \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {e}{d}} + c e\right ) + 2 \, {\left (384 \, b d^{4} x^{4} + 528 \, b c d^{3} x^{3} + 45 \, b c^{4} + 240 \, a c^{2} d^{2} + 8 \, {\left (3 \, b c^{2} d^{2} + 80 \, a d^{4}\right )} x^{2} - 10 \, {\left (3 \, b c^{3} d - 112 \, a c d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{3840 \, d^{3}}, \frac {15 \, {\left (3 \, b c^{5} + 16 \, a c^{3} d^{2}\right )} \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {d x + c} \sqrt {e x} d \sqrt {-\frac {e}{d}}}{d e x + c e}\right ) + {\left (384 \, b d^{4} x^{4} + 528 \, b c d^{3} x^{3} + 45 \, b c^{4} + 240 \, a c^{2} d^{2} + 8 \, {\left (3 \, b c^{2} d^{2} + 80 \, a d^{4}\right )} x^{2} - 10 \, {\left (3 \, b c^{3} d - 112 \, a c d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{1920 \, d^{3}}\right ] \] Input:

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

Output:

[1/3840*(15*(3*b*c^5 + 16*a*c^3*d^2)*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*d^4*x^4 + 528*b*c*d^3*x^3 + 45* 
b*c^4 + 240*a*c^2*d^2 + 8*(3*b*c^2*d^2 + 80*a*d^4)*x^2 - 10*(3*b*c^3*d - 1 
12*a*c*d^3)*x)*sqrt(d*x + c)*sqrt(e*x))/d^3, 1/1920*(15*(3*b*c^5 + 16*a*c^ 
3*d^2)*sqrt(-e/d)*arctan(sqrt(d*x + c)*sqrt(e*x)*d*sqrt(-e/d)/(d*e*x + c*e 
)) + (384*b*d^4*x^4 + 528*b*c*d^3*x^3 + 45*b*c^4 + 240*a*c^2*d^2 + 8*(3*b* 
c^2*d^2 + 80*a*d^4)*x^2 - 10*(3*b*c^3*d - 112*a*c*d^3)*x)*sqrt(d*x + c)*sq 
rt(e*x))/d^3]
 

Sympy [A] (verification not implemented)

Time = 144.27 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.66 \[ \int \sqrt {e x} (c+d x)^{3/2} \left (a+b x^2\right ) \, dx=\frac {a c^{\frac {5}{2}} \sqrt {e} \sqrt {x}}{8 d \sqrt {1 + \frac {d x}{c}}} + \frac {17 a c^{\frac {3}{2}} \sqrt {e} x^{\frac {3}{2}}}{24 \sqrt {1 + \frac {d x}{c}}} + \frac {11 a \sqrt {c} d \sqrt {e} x^{\frac {5}{2}}}{12 \sqrt {1 + \frac {d x}{c}}} - \frac {a c^{3} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{8 d^{\frac {3}{2}}} + \frac {a d^{2} \sqrt {e} x^{\frac {7}{2}}}{3 \sqrt {c} \sqrt {1 + \frac {d x}{c}}} + \frac {3 b c^{\frac {9}{2}} \sqrt {e} \sqrt {x}}{128 d^{3} \sqrt {1 + \frac {d x}{c}}} + \frac {b c^{\frac {7}{2}} \sqrt {e} x^{\frac {3}{2}}}{128 d^{2} \sqrt {1 + \frac {d x}{c}}} - \frac {b c^{\frac {5}{2}} \sqrt {e} x^{\frac {5}{2}}}{320 d \sqrt {1 + \frac {d x}{c}}} + \frac {23 b c^{\frac {3}{2}} \sqrt {e} x^{\frac {7}{2}}}{80 \sqrt {1 + \frac {d x}{c}}} + \frac {19 b \sqrt {c} d \sqrt {e} x^{\frac {9}{2}}}{40 \sqrt {1 + \frac {d x}{c}}} - \frac {3 b c^{5} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{128 d^{\frac {7}{2}}} + \frac {b d^{2} \sqrt {e} x^{\frac {11}{2}}}{5 \sqrt {c} \sqrt {1 + \frac {d x}{c}}} \] Input:

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

Output:

a*c**(5/2)*sqrt(e)*sqrt(x)/(8*d*sqrt(1 + d*x/c)) + 17*a*c**(3/2)*sqrt(e)*x 
**(3/2)/(24*sqrt(1 + d*x/c)) + 11*a*sqrt(c)*d*sqrt(e)*x**(5/2)/(12*sqrt(1 
+ d*x/c)) - a*c**3*sqrt(e)*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(8*d**(3/2)) + a 
*d**2*sqrt(e)*x**(7/2)/(3*sqrt(c)*sqrt(1 + d*x/c)) + 3*b*c**(9/2)*sqrt(e)* 
sqrt(x)/(128*d**3*sqrt(1 + d*x/c)) + b*c**(7/2)*sqrt(e)*x**(3/2)/(128*d**2 
*sqrt(1 + d*x/c)) - b*c**(5/2)*sqrt(e)*x**(5/2)/(320*d*sqrt(1 + d*x/c)) + 
23*b*c**(3/2)*sqrt(e)*x**(7/2)/(80*sqrt(1 + d*x/c)) + 19*b*sqrt(c)*d*sqrt( 
e)*x**(9/2)/(40*sqrt(1 + d*x/c)) - 3*b*c**5*sqrt(e)*asinh(sqrt(d)*sqrt(x)/ 
sqrt(c))/(128*d**(7/2)) + b*d**2*sqrt(e)*x**(11/2)/(5*sqrt(c)*sqrt(1 + d*x 
/c))
 

Maxima [F(-2)]

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

integrate((e*x)^(1/2)*(d*x+c)^(3/2)*(b*x^2+a),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 629 vs. \(2 (188) = 376\).

Time = 0.30 (sec) , antiderivative size = 629, normalized size of antiderivative = 2.71 \[ \int \sqrt {e x} (c+d x)^{3/2} \left (a+b x^2\right ) \, dx =\text {Too large to display} \] Input:

integrate((e*x)^(1/2)*(d*x+c)^(3/2)*(b*x^2+a),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*c^2*ab 
s(d)/d^2 - 20*(105*c^4*e*log(abs(-sqrt(d*e)*sqrt(d*x + c) + sqrt((d*x + c) 
*d*e - c*d*e)))/(sqrt(d*e)*d^2) - sqrt((d*x + c)*d*e - c*d*e)*(2*(d*x + c) 
*(4*(d*x + c)*(6*(d*x + c)/d^3 - 25*c/d^3) + 163*c^2/d^3) - 279*c^3/d^3)*s 
qrt(d*x + c))*b*c*abs(d)/d - 960*(3*c^2*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*d*x - 3*c)*sqrt(d*x + c))*a*c*abs(d)/d^2 + 80*(15*c^3*d*e*log(abs(-sqr 
t(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*c 
^2*abs(d)/d^4 + 80*(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*abs(d)/d^2 + 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*abs(d)/d^4)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.87 \[ \int \sqrt {e x} (c+d x)^{3/2} \left (a+b x^2\right ) \, dx=\frac {\sqrt {e}\, \left (240 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{2} d^{3}+1120 \sqrt {x}\, \sqrt {d x +c}\, a c \,d^{4} x +640 \sqrt {x}\, \sqrt {d x +c}\, a \,d^{5} x^{2}+45 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{4} d -30 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{3} d^{2} x +24 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{2} d^{3} x^{2}+528 \sqrt {x}\, \sqrt {d x +c}\, b c \,d^{4} x^{3}+384 \sqrt {x}\, \sqrt {d x +c}\, b \,d^{5} x^{4}-240 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a \,c^{3} d^{2}-45 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{5}\right )}{1920 d^{4}} \] Input:

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

Output:

(sqrt(e)*(240*sqrt(x)*sqrt(c + d*x)*a*c**2*d**3 + 1120*sqrt(x)*sqrt(c + d* 
x)*a*c*d**4*x + 640*sqrt(x)*sqrt(c + d*x)*a*d**5*x**2 + 45*sqrt(x)*sqrt(c 
+ d*x)*b*c**4*d - 30*sqrt(x)*sqrt(c + d*x)*b*c**3*d**2*x + 24*sqrt(x)*sqrt 
(c + d*x)*b*c**2*d**3*x**2 + 528*sqrt(x)*sqrt(c + d*x)*b*c*d**4*x**3 + 384 
*sqrt(x)*sqrt(c + d*x)*b*d**5*x**4 - 240*sqrt(d)*log((sqrt(c + d*x) + sqrt 
(x)*sqrt(d))/sqrt(c))*a*c**3*d**2 - 45*sqrt(d)*log((sqrt(c + d*x) + sqrt(x 
)*sqrt(d))/sqrt(c))*b*c**5))/(1920*d**4)