\(\int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx\) [1167]

Optimal result

Integrand size = 26, antiderivative size = 206 \[ \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {a+b x^2} \sqrt {c+d x^2}}{16 b^2 d^3}-\frac {(5 b c+7 a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 b^2 d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b^2 d}-\frac {(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{16 b^{5/2} d^{7/2}} \] Output:

1/16*(a^2*d^2+2*a*b*c*d+5*b^2*c^2)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b^2/d^3 
-1/24*(7*a*d+5*b*c)*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/b^2/d^2+1/6*(b*x^2+a)^ 
(5/2)*(d*x^2+c)^(1/2)/b^2/d-1/16*(-a*d+b*c)*(a^2*d^2+2*a*b*c*d+5*b^2*c^2)* 
arctanh(d^(1/2)*(b*x^2+a)^(1/2)/b^(1/2)/(d*x^2+c)^(1/2))/b^(5/2)/d^(7/2)
 

Mathematica [A] (verified)

Time = 1.48 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.91 \[ \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\frac {-b \sqrt {d} \sqrt {a+b x^2} \left (c+d x^2\right ) \left (3 a^2 d^2-2 a b d \left (-2 c+d x^2\right )+b^2 \left (-15 c^2+10 c d x^2-8 d^2 x^4\right )\right )-3 (b c-a d)^{3/2} \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{48 b^3 d^{7/2} \sqrt {c+d x^2}} \] Input:

Integrate[(x^5*Sqrt[a + b*x^2])/Sqrt[c + d*x^2],x]
 

Output:

(-(b*Sqrt[d]*Sqrt[a + b*x^2]*(c + d*x^2)*(3*a^2*d^2 - 2*a*b*d*(-2*c + d*x^ 
2) + b^2*(-15*c^2 + 10*c*d*x^2 - 8*d^2*x^4))) - 3*(b*c - a*d)^(3/2)*(5*b^2 
*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqr 
t[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(48*b^3*d^(7/2)*Sqrt[c + d*x^2])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {354, 101, 27, 90, 60, 66, 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.

\(\displaystyle \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {x^4 \sqrt {b x^2+a}}{\sqrt {d x^2+c}}dx^2\)

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

Int[(x^5*Sqrt[a + b*x^2])/Sqrt[c + d*x^2],x]
 

Output:

((x^2*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(3*b*d) - (((5*b*c + 3*a*d)*(a + 
b*x^2)^(3/2)*Sqrt[c + d*x^2])/(2*b*d) - (3*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^ 
2)*((Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sq 
rt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(Sqrt[b]*d^(3/2))))/(4*b*d))/(6 
*b*d))/2
 

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 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 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 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.02

method	result	size
risch	\(-\frac {\left (-8 b^{2} d^{2} x^{4}-2 x^{2} a b \,d^{2}+10 x^{2} b^{2} c d +3 a^{2} d^{2}+4 a b c d -15 b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}{48 b^{2} d^{3}}+\frac {\left (a^{3} d^{3}+a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}}{32 b^{2} d^{3} \sqrt {b d}\, \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(211\)
default	\(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}\, \left (16 b^{2} d^{2} x^{4} \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+4 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, a b \,d^{2} x^{2}-20 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, b^{2} c d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}+9 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d -15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}-6 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, a^{2} d^{2}-8 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, a b c d +30 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, b^{2} c^{2}\right )}{96 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, d^{3} b^{2} \sqrt {b d}}\)	\(455\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \left (\frac {a \,x^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{24 b d}-\frac {\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a^{2}}{16 b^{2} d}-\frac {a \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c}{12 b \,d^{2}}+\frac {\ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a^{3}}{32 b^{2} \sqrt {b d}}+\frac {\ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a^{2} c}{32 b d \sqrt {b d}}+\frac {3 a \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c^{2}}{32 d^{2} \sqrt {b d}}+\frac {x^{4} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{6 d}-\frac {5 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, x^{2} c}{24 d^{2}}+\frac {5 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c^{2}}{16 d^{3}}-\frac {5 b \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c^{3}}{32 d^{3} \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(476\)

Input:

int(x^5*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/48*(-8*b^2*d^2*x^4-2*a*b*d^2*x^2+10*b^2*c*d*x^2+3*a^2*d^2+4*a*b*c*d-15* 
b^2*c^2)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b^2/d^3+1/32*(a^3*d^3+a^2*b*c*d^2 
+3*a*b^2*c^2*d-5*b^3*c^3)/b^2/d^3*ln((1/2*a*d+1/2*b*c+b*d*x^2)/(b*d)^(1/2) 
+(b*d*x^4+(a*d+b*c)*x^2+a*c)^(1/2))/(b*d)^(1/2)*((b*x^2+a)*(d*x^2+c))^(1/2 
)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 442, normalized size of antiderivative = 2.15 \[ \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\left [-\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{192 \, b^{3} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{96 \, b^{3} d^{4}}\right ] \] Input:

integrate(x^5*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")
 

Output:

[-1/192*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*l 
og(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x 
^2 + 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d)) 
- 4*(8*b^3*d^3*x^4 + 15*b^3*c^2*d - 4*a*b^2*c*d^2 - 3*a^2*b*d^3 - 2*(5*b^3 
*c*d^2 - a*b^2*d^3)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^3*d^4), 1/96* 
(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(-b*d)*arctan(1 
/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2 
*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2)) + 2*(8*b^3*d^3*x^4 + 15*b^3 
*c^2*d - 4*a*b^2*c*d^2 - 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - a*b^2*d^3)*x^2)*sq 
rt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^3*d^4)]
 

Sympy [F]

\[ \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\int \frac {x^{5} \sqrt {a + b x^{2}}}{\sqrt {c + d x^{2}}}\, dx \] Input:

integrate(x**5*(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2),x)
 

Output:

Integral(x**5*sqrt(a + b*x**2)/sqrt(c + d*x**2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^5*(b*x^2+a)^(1/2)/(d*x^2+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(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.10 \[ \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (2 \, {\left (b x^{2} + a\right )} {\left (\frac {4 \, {\left (b x^{2} + a\right )}}{b^{3} d} - \frac {5 \, b^{7} c d^{3} + 7 \, a b^{6} d^{4}}{b^{9} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{8} c^{2} d^{2} + 2 \, a b^{7} c d^{3} + a^{2} b^{6} d^{4}\right )}}{b^{9} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{2} d^{3}}\right )} b}{48 \, {\left | b \right |}} \] Input:

integrate(x^5*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")
 

Output:

1/48*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a) 
*(4*(b*x^2 + a)/(b^3*d) - (5*b^7*c*d^3 + 7*a*b^6*d^4)/(b^9*d^5)) + 3*(5*b^ 
8*c^2*d^2 + 2*a*b^7*c*d^3 + a^2*b^6*d^4)/(b^9*d^5)) + 3*(5*b^3*c^3 - 3*a*b 
^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*log(abs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqr 
t(b^2*c + (b*x^2 + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3))*b/abs(b)
 

Mupad [B] (verification not implemented)

Time = 52.31 (sec) , antiderivative size = 993, normalized size of antiderivative = 4.82 \[ \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx =\text {Too large to display} \] Input:

int((x^5*(a + b*x^2)^(1/2))/(c + d*x^2)^(1/2),x)
 

Output:

(atanh((d^(1/2)*((a + b*x^2)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x^2)^(1/2) 
 - c^(1/2))))*(a*d - b*c)*(a^2*d^2 + 5*b^2*c^2 + 2*a*b*c*d))/(8*b^(5/2)*d^ 
(7/2)) - ((((a + b*x^2)^(1/2) - a^(1/2))*((a^3*b^3*d^3)/8 - (5*b^6*c^3)/8 
+ (a^2*b^4*c*d^2)/8 + (3*a*b^5*c^2*d)/8))/(d^9*((c + d*x^2)^(1/2) - c^(1/2 
))) - (((a + b*x^2)^(1/2) - a^(1/2))^5*((33*b^4*c^3)/4 + (19*a^3*b*d^3)/4 
+ (275*a^2*b^2*c*d^2)/4 + (313*a*b^3*c^2*d)/4))/(d^7*((c + d*x^2)^(1/2) - 
c^(1/2))^5) - (((a + b*x^2)^(1/2) - a^(1/2))^7*((19*a^3*d^3)/4 + (33*b^3*c 
^3)/4 + (313*a*b^2*c^2*d)/4 + (275*a^2*b*c*d^2)/4))/(d^6*((c + d*x^2)^(1/2 
) - c^(1/2))^7) - (((a + b*x^2)^(1/2) - a^(1/2))^3*((17*a^3*b^2*d^3)/24 - 
(85*b^5*c^3)/24 + (91*a^2*b^3*c*d^2)/8 + (17*a*b^4*c^2*d)/8))/(d^8*((c + d 
*x^2)^(1/2) - c^(1/2))^3) + (((a + b*x^2)^(1/2) - a^(1/2))^11*((a^3*d^3)/8 
 - (5*b^3*c^3)/8 + (3*a*b^2*c^2*d)/8 + (a^2*b*c*d^2)/8))/(b^2*d^4*((c + d* 
x^2)^(1/2) - c^(1/2))^11) - (((a + b*x^2)^(1/2) - a^(1/2))^9*((17*a^3*d^3) 
/24 - (85*b^3*c^3)/24 + (17*a*b^2*c^2*d)/8 + (91*a^2*b*c*d^2)/8))/(b*d^5*( 
(c + d*x^2)^(1/2) - c^(1/2))^9) + (a^(1/2)*c^(1/2)*((a + b*x^2)^(1/2) - a^ 
(1/2))^8*(16*a^2*d + 48*a*b*c))/(d^4*((c + d*x^2)^(1/2) - c^(1/2))^8) + (a 
^(1/2)*c^(1/2)*((a + b*x^2)^(1/2) - a^(1/2))^4*(16*a^2*b^2*d + 48*a*b^3*c) 
)/(d^6*((c + d*x^2)^(1/2) - c^(1/2))^4) + (a^(1/2)*c^(1/2)*((a + b*x^2)^(1 
/2) - a^(1/2))^6*(64*b^3*c^2 + 32*a^2*b*d^2 + (352*a*b^2*c*d)/3))/(d^6*((c 
 + d*x^2)^(1/2) - c^(1/2))^6))/(((a + b*x^2)^(1/2) - a^(1/2))^12/((c + ...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.59 \[ \int \frac {x^5 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\frac {-3 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{3}-4 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,d^{2}+2 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{3} x^{2}+15 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} c^{2} d -10 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} c \,d^{2} x^{2}+8 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} d^{3} x^{4}-3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{3} d^{3}-3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{2} b c \,d^{2}-9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a \,b^{2} c^{2} d +15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) b^{3} c^{3}}{48 b^{3} d^{4}} \] Input:

int(x^5*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x)
 

Output:

( - 3*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a**2*b*d**3 - 4*sqrt(c + d*x**2)*s 
qrt(a + b*x**2)*a*b**2*c*d**2 + 2*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*b**2 
*d**3*x**2 + 15*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**3*c**2*d - 10*sqrt(c 
+ d*x**2)*sqrt(a + b*x**2)*b**3*c*d**2*x**2 + 8*sqrt(c + d*x**2)*sqrt(a + 
b*x**2)*b**3*d**3*x**4 - 3*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2) 
*d + sqrt(d)*sqrt(c + d*x**2)*b)*a**3*d**3 - 3*sqrt(d)*sqrt(b)*log( - sqrt 
(b)*sqrt(a + b*x**2)*d + sqrt(d)*sqrt(c + d*x**2)*b)*a**2*b*c*d**2 - 9*sqr 
t(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d + sqrt(d)*sqrt(c + d*x**2)* 
b)*a*b**2*c**2*d + 15*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d + 
sqrt(d)*sqrt(c + d*x**2)*b)*b**3*c**3)/(48*b**3*d**4)