\(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx\) [229]

Optimal result

Integrand size = 28, antiderivative size = 323 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx=\frac {4 (b B+5 A c) x^{3/2} \left (b+c x^2\right )}{5 \sqrt {c} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {2 (b B+5 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{5 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}-\frac {4 \sqrt [4]{b} (b B+5 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {b x^2+c x^4}}+\frac {2 \sqrt [4]{b} (b B+5 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 c^{3/4} \sqrt {b x^2+c x^4}} \] Output:

4/5*(5*A*c+B*b)*x^(3/2)*(c*x^2+b)/c^(1/2)/(b^(1/2)+c^(1/2)*x)/(c*x^4+b*x^2 
)^(1/2)+2/5*(5*A*c+B*b)*x^(1/2)*(c*x^4+b*x^2)^(1/2)/b-2*A*(c*x^4+b*x^2)^(3 
/2)/b/x^(7/2)-4/5*b^(1/4)*(5*A*c+B*b)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^ 
(1/2)+c^(1/2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))) 
,1/2*2^(1/2))/c^(3/4)/(c*x^4+b*x^2)^(1/2)+2/5*b^(1/4)*(5*A*c+B*b)*x*(b^(1/ 
2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*ar 
ctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/c^(3/4)/(c*x^4+b*x^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 9.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.30 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (-3 A \left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}}+(b B+5 A c) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )\right )}{3 b x^{3/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(5/2),x]
 

Output:

(2*Sqrt[x^2*(b + c*x^2)]*(-3*A*(b + c*x^2)*Sqrt[1 + (c*x^2)/b] + (b*B + 5* 
A*c)*x^2*Hypergeometric2F1[-1/2, 3/4, 7/4, -((c*x^2)/b)]))/(3*b*x^(3/2)*Sq 
rt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1944, 1426, 1431, 266, 834, 27, 761, 1510}

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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(5 A c+b B) \int \frac {\sqrt {c x^4+b x^2}}{\sqrt {x}}dx}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {(5 A c+b B) \left (\frac {2}{5} b \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {(5 A c+b B) \left (\frac {2 b x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(5 A c+b B) \left (\frac {4 b x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {(5 A c+b B) \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(5 A c+b B) \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(5 A c+b B) \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {(5 A c+b B) \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^{7/2}}\)

Input:

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(5/2),x]
 

Output:

(-2*A*(b*x^2 + c*x^4)^(3/2))/(b*x^(7/2)) + ((b*B + 5*A*c)*((2*Sqrt[x]*Sqrt 
[b*x^2 + c*x^4])/5 + (4*b*x*Sqrt[b + c*x^2]*(-((-((Sqrt[x]*Sqrt[b + c*x^2] 
)/(Sqrt[b] + Sqrt[c]*x)) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2) 
/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1 
/2])/(c^(1/4)*Sqrt[b + c*x^2]))/Sqrt[c]) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)* 
Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt 
[x])/b^(1/4)], 1/2])/(2*c^(3/4)*Sqrt[b + c*x^2])))/(5*Sqrt[b*x^2 + c*x^4]) 
))/b
 

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1426

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 
*(m + 4*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
 

rule 1431

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p)   Int[(d*x)^(m + 2*p)*(b + c 
*x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {2 \left (-B \,x^{2}+5 A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{5 x^{\frac {3}{2}}}+\frac {\left (2 A c +\frac {2 B b}{5}\right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{c \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(232\)
default	\(\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (10 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c -5 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c +2 B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2}-B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2}+B \,c^{2} x^{4}-5 A \,c^{2} x^{2}+x^{2} B b c -5 A b c \right )}{5 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c}\)	\(399\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/5*(-B*x^2+5*A)*(x^2*(c*x^2+b))^(1/2)/x^(3/2)+(2*A*c+2/5*B*b)/c*(-b*c)^( 
1/2)*((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-2*(x-1/c*(-b*c)^(1/2))* 
c/(-b*c)^(1/2))^(1/2)*(-c/(-b*c)^(1/2)*x)^(1/2)/(c*x^3+b*x)^(1/2)*(-2/c*(- 
b*c)^(1/2)*EllipticE(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/ 
2))+1/c*(-b*c)^(1/2)*EllipticF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2) 
,1/2*2^(1/2)))*(x^2*(c*x^2+b))^(1/2)/x^(3/2)/(c*x^2+b)*(x*(c*x^2+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.22 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx=-\frac {2 \, {\left (2 \, {\left (B b + 5 \, A c\right )} \sqrt {c} x^{2} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - \sqrt {c x^{4} + b x^{2}} {\left (B c x^{2} - 5 \, A c\right )} \sqrt {x}\right )}}{5 \, c x^{2}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(5/2),x, algorithm="fricas")
 

Output:

-2/5*(2*(B*b + 5*A*c)*sqrt(c)*x^2*weierstrassZeta(-4*b/c, 0, weierstrassPI 
nverse(-4*b/c, 0, x)) - sqrt(c*x^4 + b*x^2)*(B*c*x^2 - 5*A*c)*sqrt(x))/(c* 
x^2)
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{\frac {5}{2}}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**(5/2),x)
 

Output:

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**(5/2), x)
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {5}{2}}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(5/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(5/2), x)
 

Giac [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {5}{2}}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(5/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2}}{x^{5/2}} \,d x \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(5/2), x)
 

Reduce [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{5/2}} \, dx=\frac {2 \sqrt {c \,x^{2}+b}\, a c +\frac {4 \sqrt {c \,x^{2}+b}\, b^{2}}{5}+\frac {2 \sqrt {c \,x^{2}+b}\, b c \,x^{2}}{5}+2 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{4}+b \,x^{2}}d x \right ) a b c +\frac {2 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{4}+b \,x^{2}}d x \right ) b^{3}}{5}}{\sqrt {x}\, c} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(5/2),x)
 

Output:

(2*(5*sqrt(b + c*x**2)*a*c + 2*sqrt(b + c*x**2)*b**2 + sqrt(b + c*x**2)*b* 
c*x**2 + 5*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x**2 + c*x**4),x)*a*b 
*c + sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x**2 + c*x**4),x)*b**3))/(5 
*sqrt(x)*c)