\(\int \frac {(a+b x^2)^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx\) [1193]

Optimal result

Integrand size = 26, antiderivative size = 187 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx=\frac {b (b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c d}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 c x^2}-\frac {a^{3/2} (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 c^{3/2}}-\frac {b^{3/2} (b c-5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 d^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx=\frac {a^{3/2} (-5 b c+a d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a+b x^2}}\right )+\frac {\sqrt {c} \left (\frac {\sqrt {d} \sqrt {a+b x^2} \left (-a^2 d+b^2 c x^2\right ) \sqrt {c+d x^2}}{x^2}-b^{3/2} c (b c-5 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )\right )}{d^{3/2}}}{2 c^{3/2}} \] Input:

Integrate[(a + b*x^2)^(5/2)/(x^3*Sqrt[c + d*x^2]),x]
 

Output:

(a^(3/2)*(-5*b*c + a*d)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x^2])/(Sqrt[c]*Sqrt[a 
+ b*x^2])] + (Sqrt[c]*((Sqrt[d]*Sqrt[a + b*x^2]*(-(a^2*d) + b^2*c*x^2)*Sqr 
t[c + d*x^2])/x^2 - b^(3/2)*c*(b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^ 
2])/(Sqrt[d]*Sqrt[a + b*x^2])]))/d^(3/2))/(2*c^(3/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {354, 109, 27, 171, 175, 66, 104, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

Int[(a + b*x^2)^(5/2)/(x^3*Sqrt[c + d*x^2]),x]
 

Output:

(-((a*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(c*x^2)) + ((2*b*(b*c + a*d)*Sqrt 
[a + b*x^2]*Sqrt[c + d*x^2])/d + ((-2*a^(3/2)*d*(5*b*c - a*d)*ArcTanh[(Sqr 
t[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/Sqrt[c] - (2*b^(3/2)*c*( 
b*c - 5*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])]) 
/Sqrt[d])/d)/(2*c))/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 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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 109

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

rule 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegersQ[2*m, 2*n, 2*p]
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(302\) vs. \(2(147)=294\).

Time = 0.64 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.62

method	result	size
risch	\(-\frac {a^{2} \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}{2 c \,x^{2}}-\frac {\left (-\frac {a^{2} \left (a d -5 b c \right ) \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right )}{2 \sqrt {a c}}-2 b^{3} c \left (\frac {\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{2 b d}-\frac {\left (a d +b c \right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{4 b d \sqrt {b d}}\right )-\frac {3 a \,b^{2} c \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{\sqrt {b d}}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}}{2 c \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(303\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \left (-\frac {a^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{2 c \,x^{2}}+\frac {a^{3} \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) d}{4 c \sqrt {a c}}-\frac {5 a^{2} b \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right )}{4 \sqrt {a c}}+\frac {b^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{2 d}+\frac {5 a \,b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{4 \sqrt {b d}}-\frac {b^{3} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c}{4 d \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(347\)
default	\(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}\, \left (5 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d \,x^{2} \sqrt {a c}-\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x^{2} \sqrt {a c}+\ln \left (\frac {a d \,x^{2}+x^{2} b c +2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}+2 a c}{x^{2}}\right ) a^{3} d^{2} x^{2} \sqrt {b d}-5 \ln \left (\frac {a d \,x^{2}+x^{2} b c +2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}+2 a c}{x^{2}}\right ) a^{2} b c d \,x^{2} \sqrt {b d}+2 b^{2} c \,x^{2} \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, \sqrt {a c}-2 a^{2} d \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, \sqrt {a c}\right )}{4 c \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, x^{2} \sqrt {b d}\, \sqrt {a c}\, d}\)	\(374\)

Input:

int((b*x^2+a)^(5/2)/x^3/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2*a^2/c*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/x^2-1/2/c*(-1/2*a^2*(a*d-5*b*c) 
/(a*c)^(1/2)*ln((2*a*c+(a*d+b*c)*x^2+2*(a*c)^(1/2)*(b*d*x^4+(a*d+b*c)*x^2+ 
a*c)^(1/2))/x^2)-2*b^3*c*(1/2/b/d*(b*d*x^4+(a*d+b*c)*x^2+a*c)^(1/2)-1/4*(a 
*d+b*c)/b/d*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))-3*a*b^2*c*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 = 1.12 (sec) , antiderivative size = 1097, normalized size of antiderivative = 5.87 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx =\text {Too large to display} \] Input:

integrate((b*x^2+a)^(5/2)/x^3/(d*x^2+c)^(1/2),x, algorithm="fricas")
 

Output:

[-1/8*((b^2*c^2 - 5*a*b*c*d)*x^2*sqrt(b/d)*log(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^2*x^2 + b*c*d + 
a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b/d)) + (5*a*b*c*d - a^2*d^2)* 
x^2*sqrt(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a* 
b*c^2 + a^2*c*d)*x^2 + 4*(2*a*c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*s 
qrt(d*x^2 + c)*sqrt(a/c))/x^4) - 4*(b^2*c*x^2 - a^2*d)*sqrt(b*x^2 + a)*sqr 
t(d*x^2 + c))/(c*d*x^2), 1/8*(2*(b^2*c^2 - 5*a*b*c*d)*x^2*sqrt(-b/d)*arcta 
n(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*x^4 + a*b*c + (b^2*c + a*b*d)*x^2)) - (5*a*b*c*d - a^2*d^2)*x^2*sqrt 
(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + 
a^2*c*d)*x^2 + 4*(2*a*c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^ 
2 + c)*sqrt(a/c))/x^4) + 4*(b^2*c*x^2 - a^2*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 
+ c))/(c*d*x^2), 1/8*(2*(5*a*b*c*d - a^2*d^2)*x^2*sqrt(-a/c)*arctan(1/2*(( 
b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a/c)/(a*b*d* 
x^4 + a^2*c + (a*b*c + a^2*d)*x^2)) - (b^2*c^2 - 5*a*b*c*d)*x^2*sqrt(b/d)* 
log(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^2*x^2 + b*c*d + a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt 
(b/d)) + 4*(b^2*c*x^2 - a^2*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(c*d*x^2), 
 1/4*((5*a*b*c*d - a^2*d^2)*x^2*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2 
*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a/c)/(a*b*d*x^4 + a^2*c + (...
 

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {5}{2}}}{x^{3} \sqrt {c + d x^{2}}}\, dx \] Input:

integrate((b*x**2+a)**(5/2)/x**3/(d*x**2+c)**(1/2),x)
 

Output:

Integral((a + b*x**2)**(5/2)/(x**3*sqrt(c + d*x**2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x^2+a)^(5/2)/x^3/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (147) = 294\).

Time = 0.18 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.83 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx=\frac {b^{3} {\left (\frac {{\left (b c - 5 \, a d\right )} \log \left ({\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}{\sqrt {b d} d} + \frac {2 \, \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a}}{b d} + \frac {2 \, {\left (5 \, a^{2} b c d - a^{3} d^{2}\right )} \arctan \left (\frac {b^{2} c + a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} \sqrt {b d} b c} - \frac {4 \, {\left (a^{2} b^{3} c^{2} d - 2 \, a^{3} b^{2} c d^{2} + a^{4} b d^{3} - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{2} b c d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{3} d^{2}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4}\right )} \sqrt {b d} c}\right )}}{4 \, {\left | b \right |}} \] Input:

integrate((b*x^2+a)^(5/2)/x^3/(d*x^2+c)^(1/2),x, algorithm="giac")
 

Output:

1/4*b^3*((b*c - 5*a*d)*log((sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^ 
2 + a)*b*d - a*b*d))^2)/(sqrt(b*d)*d) + 2*sqrt(b^2*c + (b*x^2 + a)*b*d - a 
*b*d)*sqrt(b*x^2 + a)/(b*d) + 2*(5*a^2*b*c*d - a^3*d^2)*arctan(1/2*(b^2*c 
+ a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b* 
d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*sqrt(b*d)*b*c) - 4*(a^2*b^3*c^2 
*d - 2*a^3*b^2*c*d^2 + a^4*b*d^3 - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c 
 + (b*x^2 + a)*b*d - a*b*d))^2*a^2*b*c*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sq 
rt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^3*d^2)/((b^4*c^2 - 2*a*b^3*c*d + 
a^2*b^2*d^2 - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d 
- a*b*d))^2*b^2*c - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a 
)*b*d - a*b*d))^2*a*b*d + (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 
 + a)*b*d - a*b*d))^4)*sqrt(b*d)*c))/abs(b)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/2}}{x^3\,\sqrt {d\,x^2+c}} \,d x \] Input:

int((a + b*x^2)^(5/2)/(x^3*(c + d*x^2)^(1/2)),x)
 

Output:

int((a + b*x^2)^(5/2)/(x^3*(c + d*x^2)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.41 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx=\frac {-\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a^{2} c \,d^{2}+\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{2} c^{2} d \,x^{2}-\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c -\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a^{2} d^{3} x^{2}+5 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c -\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a b c \,d^{2} x^{2}+\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a^{2} d^{3} x^{2}-5 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a b c \,d^{2} x^{2}-5 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a b \,c^{2} d \,x^{2}+\sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) b^{2} c^{3} x^{2}}{2 c^{2} d^{2} x^{2}} \] Input:

int((b*x^2+a)^(5/2)/x^3/(d*x^2+c)^(1/2),x)
 

Output:

( - sqrt(c + d*x**2)*sqrt(a + b*x**2)*a**2*c*d**2 + sqrt(c + d*x**2)*sqrt( 
a + b*x**2)*b**2*c**2*d*x**2 - sqrt(c)*sqrt(a)*log(sqrt(a)*sqrt(a + b*x**2 
)*c - sqrt(c)*sqrt(c + d*x**2)*a)*a**2*d**3*x**2 + 5*sqrt(c)*sqrt(a)*log(s 
qrt(a)*sqrt(a + b*x**2)*c - sqrt(c)*sqrt(c + d*x**2)*a)*a*b*c*d**2*x**2 + 
sqrt(c)*sqrt(a)*log(x)*a**2*d**3*x**2 - 5*sqrt(c)*sqrt(a)*log(x)*a*b*c*d** 
2*x**2 - 5*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d + sqrt(d)*sqr 
t(c + d*x**2)*b)*a*b*c**2*d*x**2 + sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + 
 b*x**2)*d + sqrt(d)*sqrt(c + d*x**2)*b)*b**2*c**3*x**2)/(2*c**2*d**2*x**2 
)