\(\int \frac {(a+b x^2)^2}{\sqrt {e x} \sqrt {c+d x}} \, dx\) [826]

Optimal result

Integrand size = 26, antiderivative size = 210 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=-\frac {b c \left (35 b c^2+96 a d^2\right ) \sqrt {e x} \sqrt {c+d x}}{64 d^4 e}+\frac {b \left (35 b c^2+96 a d^2\right ) (e x)^{3/2} \sqrt {c+d x}}{96 d^3 e^2}-\frac {7 b^2 c (e x)^{5/2} \sqrt {c+d x}}{24 d^2 e^3}+\frac {b^2 (e x)^{7/2} \sqrt {c+d x}}{4 d e^4}+\frac {\left (35 b^2 c^4+96 a b c^2 d^2+128 a^2 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{64 d^{9/2} \sqrt {e}} \] Output:

-1/64*b*c*(96*a*d^2+35*b*c^2)*(e*x)^(1/2)*(d*x+c)^(1/2)/d^4/e+1/96*b*(96*a 
*d^2+35*b*c^2)*(e*x)^(3/2)*(d*x+c)^(1/2)/d^3/e^2-7/24*b^2*c*(e*x)^(5/2)*(d 
*x+c)^(1/2)/d^2/e^3+1/4*b^2*(e*x)^(7/2)*(d*x+c)^(1/2)/d/e^4+1/64*(128*a^2* 
d^4+96*a*b*c^2*d^2+35*b^2*c^4)*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 = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.66 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {b \sqrt {d} x \sqrt {c+d x} \left (96 a d^2 (-3 c+2 d x)+b \left (-105 c^3+70 c^2 d x-56 c d^2 x^2+48 d^3 x^3\right )\right )-3 \left (35 b^2 c^4+96 a b c^2 d^2+128 a^2 d^4\right ) \sqrt {x} \log \left (-\sqrt {d} \sqrt {x}+\sqrt {c+d x}\right )}{192 d^{9/2} \sqrt {e x}} \] Input:

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

Output:

(b*Sqrt[d]*x*Sqrt[c + d*x]*(96*a*d^2*(-3*c + 2*d*x) + b*(-105*c^3 + 70*c^2 
*d*x - 56*c*d^2*x^2 + 48*d^3*x^3)) - 3*(35*b^2*c^4 + 96*a*b*c^2*d^2 + 128* 
a^2*d^4)*Sqrt[x]*Log[-(Sqrt[d]*Sqrt[x]) + Sqrt[c + d*x]])/(192*d^(9/2)*Sqr 
t[e*x])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 233, 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 = {521, 27, 2125, 27, 521, 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.

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

\(\Big \downarrow \) 521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2125

\(\displaystyle \frac {\frac {\int \frac {e^7 \left (48 a^2 d^2+b \left (35 b c^2+96 a d^2\right ) x^2\right )}{2 \sqrt {e x} \sqrt {c+d x}}dx}{3 d e^3}-\frac {7 b^2 c e (e x)^{5/2} \sqrt {c+d x}}{3 d}}{8 d e^4}+\frac {b^2 (e x)^{7/2} \sqrt {c+d x}}{4 d e^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {e^4 \int \frac {48 a^2 d^2+b \left (35 b c^2+96 a d^2\right ) x^2}{\sqrt {e x} \sqrt {c+d x}}dx}{6 d}-\frac {7 b^2 c e (e x)^{5/2} \sqrt {c+d x}}{3 d}}{8 d e^4}+\frac {b^2 (e x)^{7/2} \sqrt {c+d x}}{4 d e^4}\)

\(\Big \downarrow \) 521

\(\displaystyle \frac {\frac {e^4 \left (\frac {\int \frac {3 e^2 \left (64 a^2 d^3-b c \left (35 b c^2+96 a d^2\right ) x\right )}{2 \sqrt {e x} \sqrt {c+d x}}dx}{2 d e^2}+\frac {b (e x)^{3/2} \sqrt {c+d x} \left (96 a d^2+35 b c^2\right )}{2 d e^2}\right )}{6 d}-\frac {7 b^2 c e (e x)^{5/2} \sqrt {c+d x}}{3 d}}{8 d e^4}+\frac {b^2 (e x)^{7/2} \sqrt {c+d x}}{4 d e^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {e^4 \left (\frac {3 \int \frac {64 a^2 d^3-b c \left (35 b c^2+96 a d^2\right ) x}{\sqrt {e x} \sqrt {c+d x}}dx}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x} \left (96 a d^2+35 b c^2\right )}{2 d e^2}\right )}{6 d}-\frac {7 b^2 c e (e x)^{5/2} \sqrt {c+d x}}{3 d}}{8 d e^4}+\frac {b^2 (e x)^{7/2} \sqrt {c+d x}}{4 d e^4}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {e^4 \left (\frac {3 \left (\frac {\left (128 a^2 d^4+96 a b c^2 d^2+35 b^2 c^4\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx}{2 d}-\frac {b c \sqrt {e x} \sqrt {c+d x} \left (96 a d^2+35 b c^2\right )}{d e}\right )}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x} \left (96 a d^2+35 b c^2\right )}{2 d e^2}\right )}{6 d}-\frac {7 b^2 c e (e x)^{5/2} \sqrt {c+d x}}{3 d}}{8 d e^4}+\frac {b^2 (e x)^{7/2} \sqrt {c+d x}}{4 d e^4}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\frac {e^4 \left (\frac {3 \left (\frac {\left (128 a^2 d^4+96 a b c^2 d^2+35 b^2 c^4\right ) \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}}{d}-\frac {b c \sqrt {e x} \sqrt {c+d x} \left (96 a d^2+35 b c^2\right )}{d e}\right )}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x} \left (96 a d^2+35 b c^2\right )}{2 d e^2}\right )}{6 d}-\frac {7 b^2 c e (e x)^{5/2} \sqrt {c+d x}}{3 d}}{8 d e^4}+\frac {b^2 (e x)^{7/2} \sqrt {c+d x}}{4 d e^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {e^4 \left (\frac {3 \left (\frac {\left (128 a^2 d^4+96 a b c^2 d^2+35 b^2 c^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{3/2} \sqrt {e}}-\frac {b c \sqrt {e x} \sqrt {c+d x} \left (96 a d^2+35 b c^2\right )}{d e}\right )}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x} \left (96 a d^2+35 b c^2\right )}{2 d e^2}\right )}{6 d}-\frac {7 b^2 c e (e x)^{5/2} \sqrt {c+d x}}{3 d}}{8 d e^4}+\frac {b^2 (e x)^{7/2} \sqrt {c+d x}}{4 d e^4}\)

Input:

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

Output:

(b^2*(e*x)^(7/2)*Sqrt[c + d*x])/(4*d*e^4) + ((-7*b^2*c*e*(e*x)^(5/2)*Sqrt[ 
c + d*x])/(3*d) + (e^4*((b*(35*b*c^2 + 96*a*d^2)*(e*x)^(3/2)*Sqrt[c + d*x] 
)/(2*d*e^2) + (3*(-((b*c*(35*b*c^2 + 96*a*d^2)*Sqrt[e*x]*Sqrt[c + d*x])/(d 
*e)) + ((35*b^2*c^4 + 96*a*b*c^2*d^2 + 128*a^2*d^4)*ArcTanh[(Sqrt[d]*Sqrt[ 
e*x])/(Sqrt[e]*Sqrt[c + d*x])])/(d^(3/2)*Sqrt[e])))/(4*d)))/(6*d))/(8*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 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 = 158, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {b \left (-48 b \,d^{3} x^{3}+56 b c \,d^{2} x^{2}-192 a x \,d^{3}-70 b \,c^{2} d x +288 a \,d^{2} c +105 b \,c^{3}\right ) x \sqrt {d x +c}}{192 d^{4} \sqrt {e x}}+\frac {\left (128 a^{2} d^{4}+96 b \,c^{2} d^{2} a +35 b^{2} c^{4}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) \sqrt {\left (d x +c \right ) e x}}{128 d^{4} \sqrt {d e}\, \sqrt {e x}\, \sqrt {d x +c}}\)	\(158\)
default	\(\frac {\sqrt {d x +c}\, x \left (96 b^{2} d^{3} x^{3} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-112 b^{2} c \,d^{2} x^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+384 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a^{2} d^{4} e +288 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a b \,c^{2} d^{2} e +105 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b^{2} c^{4} e +384 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, a b \,d^{3} x +140 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b^{2} c^{2} d x -576 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, a b c \,d^{2}-210 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b^{2} c^{3}\right )}{384 d^{4} \sqrt {e x}\, \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}}\)	\(313\)

Input:

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

Output:

-1/192*b*(-48*b*d^3*x^3+56*b*c*d^2*x^2-192*a*d^3*x-70*b*c^2*d*x+288*a*c*d^ 
2+105*b*c^3)*x*(d*x+c)^(1/2)/d^4/(e*x)^(1/2)+1/128*(128*a^2*d^4+96*a*b*c^2 
*d^2+35*b^2*c^4)/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)*((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 = 301, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (35 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 128 \, a^{2} d^{4}\right )} \sqrt {d e} \log \left (2 \, d e x + c e + 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) + 2 \, {\left (48 \, b^{2} d^{4} x^{3} - 56 \, b^{2} c d^{3} x^{2} - 105 \, b^{2} c^{3} d - 288 \, a b c d^{3} + 2 \, {\left (35 \, b^{2} c^{2} d^{2} + 96 \, a b d^{4}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{384 \, d^{5} e}, -\frac {3 \, {\left (35 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 128 \, a^{2} d^{4}\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) - {\left (48 \, b^{2} d^{4} x^{3} - 56 \, b^{2} c d^{3} x^{2} - 105 \, b^{2} c^{3} d - 288 \, a b c d^{3} + 2 \, {\left (35 \, b^{2} c^{2} d^{2} + 96 \, a b d^{4}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{192 \, d^{5} e}\right ] \] Input:

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

Output:

[1/384*(3*(35*b^2*c^4 + 96*a*b*c^2*d^2 + 128*a^2*d^4)*sqrt(d*e)*log(2*d*e* 
x + c*e + 2*sqrt(d*e)*sqrt(d*x + c)*sqrt(e*x)) + 2*(48*b^2*d^4*x^3 - 56*b^ 
2*c*d^3*x^2 - 105*b^2*c^3*d - 288*a*b*c*d^3 + 2*(35*b^2*c^2*d^2 + 96*a*b*d 
^4)*x)*sqrt(d*x + c)*sqrt(e*x))/(d^5*e), -1/192*(3*(35*b^2*c^4 + 96*a*b*c^ 
2*d^2 + 128*a^2*d^4)*sqrt(-d*e)*arctan(sqrt(-d*e)*sqrt(d*x + c)*sqrt(e*x)/ 
(d*e*x + c*e)) - (48*b^2*d^4*x^3 - 56*b^2*c*d^3*x^2 - 105*b^2*c^3*d - 288* 
a*b*c*d^3 + 2*(35*b^2*c^2*d^2 + 96*a*b*d^4)*x)*sqrt(d*x + c)*sqrt(e*x))/(d 
^5*e)]
 

Sympy [A] (verification not implemented)

Time = 42.42 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.77 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {2 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{\sqrt {d} \sqrt {e}} - \frac {3 a b c^{\frac {3}{2}} \sqrt {x}}{2 d^{2} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} - \frac {a b \sqrt {c} x^{\frac {3}{2}}}{2 d \sqrt {e} \sqrt {1 + \frac {d x}{c}}} + \frac {3 a b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{2 d^{\frac {5}{2}} \sqrt {e}} + \frac {a b x^{\frac {5}{2}}}{\sqrt {c} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} - \frac {35 b^{2} c^{\frac {7}{2}} \sqrt {x}}{64 d^{4} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} - \frac {35 b^{2} c^{\frac {5}{2}} x^{\frac {3}{2}}}{192 d^{3} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} + \frac {7 b^{2} c^{\frac {3}{2}} x^{\frac {5}{2}}}{96 d^{2} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} - \frac {b^{2} \sqrt {c} x^{\frac {7}{2}}}{24 d \sqrt {e} \sqrt {1 + \frac {d x}{c}}} + \frac {35 b^{2} c^{4} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{64 d^{\frac {9}{2}} \sqrt {e}} + \frac {b^{2} x^{\frac {9}{2}}}{4 \sqrt {c} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} \] Input:

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

Output:

2*a**2*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(sqrt(d)*sqrt(e)) - 3*a*b*c**(3/2)*s 
qrt(x)/(2*d**2*sqrt(e)*sqrt(1 + d*x/c)) - a*b*sqrt(c)*x**(3/2)/(2*d*sqrt(e 
)*sqrt(1 + d*x/c)) + 3*a*b*c**2*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(2*d**(5/2) 
*sqrt(e)) + a*b*x**(5/2)/(sqrt(c)*sqrt(e)*sqrt(1 + d*x/c)) - 35*b**2*c**(7 
/2)*sqrt(x)/(64*d**4*sqrt(e)*sqrt(1 + d*x/c)) - 35*b**2*c**(5/2)*x**(3/2)/ 
(192*d**3*sqrt(e)*sqrt(1 + d*x/c)) + 7*b**2*c**(3/2)*x**(5/2)/(96*d**2*sqr 
t(e)*sqrt(1 + d*x/c)) - b**2*sqrt(c)*x**(7/2)/(24*d*sqrt(e)*sqrt(1 + d*x/c 
)) + 35*b**2*c**4*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(64*d**(9/2)*sqrt(e)) + b 
**2*x**(9/2)/(4*sqrt(c)*sqrt(e)*sqrt(1 + d*x/c))
 

Maxima [F(-2)]

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

integrate((b*x^2+a)^2/(e*x)^(1/2)/(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 = 0.16 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {{\left (\sqrt {{\left (d x + c\right )} d e - c d e} {\left (2 \, {\left (d x + c\right )} {\left (4 \, {\left (d x + c\right )} {\left (\frac {6 \, {\left (d x + c\right )} b^{2}}{d^{5} e} - \frac {25 \, b^{2} c}{d^{5} e}\right )} + \frac {163 \, b^{2} c^{2} d^{19} e^{3} + 96 \, a b d^{21} e^{3}}{d^{24} e^{4}}\right )} - \frac {3 \, {\left (93 \, b^{2} c^{3} d^{19} e^{3} + 160 \, a b c d^{21} e^{3}\right )}}{d^{24} e^{4}}\right )} \sqrt {d x + c} - \frac {3 \, {\left (35 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 128 \, a^{2} d^{4}\right )} \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e} d^{4}}\right )} d}{192 \, {\left | d \right |}} \] Input:

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

Output:

1/192*(sqrt((d*x + c)*d*e - c*d*e)*(2*(d*x + c)*(4*(d*x + c)*(6*(d*x + c)* 
b^2/(d^5*e) - 25*b^2*c/(d^5*e)) + (163*b^2*c^2*d^19*e^3 + 96*a*b*d^21*e^3) 
/(d^24*e^4)) - 3*(93*b^2*c^3*d^19*e^3 + 160*a*b*c*d^21*e^3)/(d^24*e^4))*sq 
rt(d*x + c) - 3*(35*b^2*c^4 + 96*a*b*c^2*d^2 + 128*a^2*d^4)*log(abs(-sqrt( 
d*e)*sqrt(d*x + c) + sqrt((d*x + c)*d*e - c*d*e)))/(sqrt(d*e)*d^4))*d/abs( 
d)
 

Mupad [B] (verification not implemented)

Time = 49.66 (sec) , antiderivative size = 647, normalized size of antiderivative = 3.08 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\sqrt {e\,x}}{\sqrt {e}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )\,\left (128\,a^2\,d^4+96\,a\,b\,c^2\,d^2+35\,b^2\,c^4\right )}{32\,d^{9/2}\,\sqrt {e}}-\frac {\frac {{\left (e\,x\right )}^{11/2}\,\left (\frac {2681\,b^2\,c^4\,d\,e^2}{96}+51\,a\,b\,c^2\,d^3\,e^2\right )}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{11}}-\frac {{\left (e\,x\right )}^{9/2}\,\left (\frac {5053\,b^2\,c^4\,e^3}{96}+31\,a\,b\,c^2\,d^2\,e^3\right )}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^9}-\frac {{\left (e\,x\right )}^{13/2}\,\left (\frac {805\,e\,b^2\,c^4\,d^2}{96}+23\,a\,e\,b\,c^2\,d^4\right )}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{13}}+\frac {{\left (e\,x\right )}^{15/2}\,\left (\frac {35\,b^2\,c^4\,d^3}{32}+3\,a\,b\,c^2\,d^5\right )}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{15}}+\frac {e^7\,\sqrt {e\,x}\,\left (35\,b^2\,c^4+96\,a\,b\,c^2\,d^2\right )}{32\,d^4\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {23\,e^6\,{\left (e\,x\right )}^{3/2}\,\left (35\,b^2\,c^4+96\,a\,b\,c^2\,d^2\right )}{96\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {31\,e^4\,{\left (e\,x\right )}^{7/2}\,\left (163\,b^2\,c^4+96\,a\,b\,c^2\,d^2\right )}{96\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}+\frac {e^5\,{\left (e\,x\right )}^{5/2}\,\left (2681\,b^2\,c^4+4896\,a\,b\,c^2\,d^2\right )}{96\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}}{e^8-\frac {8\,d\,e^8\,x}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {28\,d^2\,e^8\,x^2}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {56\,d^3\,e^8\,x^3}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {70\,d^4\,e^8\,x^4}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}-\frac {56\,d^5\,e^8\,x^5}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{10}}+\frac {28\,d^6\,e^8\,x^6}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{12}}-\frac {8\,d^7\,e^8\,x^7}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{14}}+\frac {d^8\,e^8\,x^8}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{16}}} \] Input:

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

Output:

(atanh((d^(1/2)*(e*x)^(1/2))/(e^(1/2)*((c + d*x)^(1/2) - c^(1/2))))*(128*a 
^2*d^4 + 35*b^2*c^4 + 96*a*b*c^2*d^2))/(32*d^(9/2)*e^(1/2)) - (((e*x)^(11/ 
2)*((2681*b^2*c^4*d*e^2)/96 + 51*a*b*c^2*d^3*e^2))/((c + d*x)^(1/2) - c^(1 
/2))^11 - ((e*x)^(9/2)*((5053*b^2*c^4*e^3)/96 + 31*a*b*c^2*d^2*e^3))/((c + 
 d*x)^(1/2) - c^(1/2))^9 - ((e*x)^(13/2)*((805*b^2*c^4*d^2*e)/96 + 23*a*b* 
c^2*d^4*e))/((c + d*x)^(1/2) - c^(1/2))^13 + ((e*x)^(15/2)*((35*b^2*c^4*d^ 
3)/32 + 3*a*b*c^2*d^5))/((c + d*x)^(1/2) - c^(1/2))^15 + (e^7*(e*x)^(1/2)* 
(35*b^2*c^4 + 96*a*b*c^2*d^2))/(32*d^4*((c + d*x)^(1/2) - c^(1/2))) - (23* 
e^6*(e*x)^(3/2)*(35*b^2*c^4 + 96*a*b*c^2*d^2))/(96*d^3*((c + d*x)^(1/2) - 
c^(1/2))^3) - (31*e^4*(e*x)^(7/2)*(163*b^2*c^4 + 96*a*b*c^2*d^2))/(96*d*(( 
c + d*x)^(1/2) - c^(1/2))^7) + (e^5*(e*x)^(5/2)*(2681*b^2*c^4 + 4896*a*b*c 
^2*d^2))/(96*d^2*((c + d*x)^(1/2) - c^(1/2))^5))/(e^8 - (8*d*e^8*x)/((c + 
d*x)^(1/2) - c^(1/2))^2 + (28*d^2*e^8*x^2)/((c + d*x)^(1/2) - c^(1/2))^4 - 
 (56*d^3*e^8*x^3)/((c + d*x)^(1/2) - c^(1/2))^6 + (70*d^4*e^8*x^4)/((c + d 
*x)^(1/2) - c^(1/2))^8 - (56*d^5*e^8*x^5)/((c + d*x)^(1/2) - c^(1/2))^10 + 
 (28*d^6*e^8*x^6)/((c + d*x)^(1/2) - c^(1/2))^12 - (8*d^7*e^8*x^7)/((c + d 
*x)^(1/2) - c^(1/2))^14 + (d^8*e^8*x^8)/((c + d*x)^(1/2) - c^(1/2))^16)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {\sqrt {e}\, \left (-288 \sqrt {x}\, \sqrt {d x +c}\, a b c \,d^{3}+192 \sqrt {x}\, \sqrt {d x +c}\, a b \,d^{4} x -105 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{3} d +70 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c^{2} d^{2} x -56 \sqrt {x}\, \sqrt {d x +c}\, b^{2} c \,d^{3} x^{2}+48 \sqrt {x}\, \sqrt {d x +c}\, b^{2} d^{4} x^{3}+384 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a^{2} d^{4}+288 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a b \,c^{2} d^{2}+105 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b^{2} c^{4}\right )}{192 d^{5} e} \] Input:

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

Output:

(sqrt(e)*( - 288*sqrt(x)*sqrt(c + d*x)*a*b*c*d**3 + 192*sqrt(x)*sqrt(c + d 
*x)*a*b*d**4*x - 105*sqrt(x)*sqrt(c + d*x)*b**2*c**3*d + 70*sqrt(x)*sqrt(c 
 + d*x)*b**2*c**2*d**2*x - 56*sqrt(x)*sqrt(c + d*x)*b**2*c*d**3*x**2 + 48* 
sqrt(x)*sqrt(c + d*x)*b**2*d**4*x**3 + 384*sqrt(d)*log((sqrt(c + d*x) + sq 
rt(x)*sqrt(d))/sqrt(c))*a**2*d**4 + 288*sqrt(d)*log((sqrt(c + d*x) + sqrt( 
x)*sqrt(d))/sqrt(c))*a*b*c**2*d**2 + 105*sqrt(d)*log((sqrt(c + d*x) + sqrt 
(x)*sqrt(d))/sqrt(c))*b**2*c**4))/(192*d**5*e)