\(\int x^{3/2} \sqrt {a x+b x^3+c x^5} \, dx\) [52]

Optimal result

Integrand size = 24, antiderivative size = 381 \[ \int x^{3/2} \sqrt {a x+b x^3+c x^5} \, dx=-\frac {2 \left (b^2-3 a c\right ) x^{3/2} \left (a+b x^2+c x^4\right )}{15 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}}+\frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}+\frac {2 \sqrt [4]{a} \left (b^2-3 a c\right ) \sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{15 c^{7/4} \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt [4]{a} \left (\sqrt {a} b+\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}\right ) \sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{30 c^{5/4} \sqrt {a x+b x^3+c x^5}} \] Output:

-2/15*(-3*a*c+b^2)*x^(3/2)*(c*x^4+b*x^2+a)/c^(3/2)/(a^(1/2)+c^(1/2)*x^2)/( 
c*x^5+b*x^3+a*x)^(1/2)+1/15*x^(1/2)*(3*c*x^2+b)*(c*x^5+b*x^3+a*x)^(1/2)/c+ 
2/15*a^(1/4)*(-3*a*c+b^2)*x^(1/2)*(a^(1/2)+c^(1/2)*x^2)*((c*x^4+b*x^2+a)/( 
a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1 
/2*(2-b/a^(1/2)/c^(1/2))^(1/2))/c^(7/4)/(c*x^5+b*x^3+a*x)^(1/2)-1/30*a^(1/ 
4)*(a^(1/2)*b+2*(-3*a*c+b^2)/c^(1/2))*x^(1/2)*(a^(1/2)+c^(1/2)*x^2)*((c*x^ 
4+b*x^2+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4) 
*x/a^(1/4)),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))/c^(5/4)/(c*x^5+b*x^3+a*x)^(1/ 
2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.07 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.28 \[ \int x^{3/2} \sqrt {a x+b x^3+c x^5} \, dx=\frac {\sqrt {x} \left (2 c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )-i \left (b^2-3 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+i \left (-b^3+4 a b c+b^2 \sqrt {b^2-4 a c}-3 a c \sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{30 c^2 \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {x \left (a+b x^2+c x^4\right )}} \] Input:

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

Output:

(Sqrt[x]*(2*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(b + 3*c*x^2)*(a + b*x^2 + 
 c*x^4) - I*(b^2 - 3*a*c)*(-b + Sqrt[b^2 - 4*a*c])*Sqrt[(b + Sqrt[b^2 - 4* 
a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4*a*c] + 
 4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + 
 Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])] 
+ I*(-b^3 + 4*a*b*c + b^2*Sqrt[b^2 - 4*a*c] - 3*a*c*Sqrt[b^2 - 4*a*c])*Sqr 
t[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2 
*Sqrt[b^2 - 4*a*c] + 4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[I*ArcSinh 
[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - 
Sqrt[b^2 - 4*a*c])]))/(30*c^2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[x*(a + 
b*x^2 + c*x^4)])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1966, 25, 2000, 1511, 27, 1416, 1509}

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.

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

\(\Big \downarrow \) 1966

\(\displaystyle \frac {\int -\frac {\sqrt {x} \left (2 \left (b^2-3 a c\right ) x^2+a b\right )}{\sqrt {c x^5+b x^3+a x}}dx}{15 c}+\frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\int \frac {\sqrt {x} \left (2 \left (b^2-3 a c\right ) x^2+a b\right )}{\sqrt {c x^5+b x^3+a x}}dx}{15 c}\)

\(\Big \downarrow \) 2000

\(\displaystyle \frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \int \frac {2 \left (b^2-3 a c\right ) x^2+a b}{\sqrt {c x^4+b x^2+a}}dx}{15 c \sqrt {a x+b x^3+c x^5}}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\sqrt {a} \left (\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}+\sqrt {a} b\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx-\frac {2 \sqrt {a} \left (b^2-3 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{15 c \sqrt {a x+b x^3+c x^5}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\sqrt {a} \left (\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}+\sqrt {a} b\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx-\frac {2 \left (b^2-3 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{15 c \sqrt {a x+b x^3+c x^5}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {\sqrt [4]{a} \left (\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}+\sqrt {a} b\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {2 \left (b^2-3 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{15 c \sqrt {a x+b x^3+c x^5}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {\sqrt [4]{a} \left (\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}+\sqrt {a} b\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {2 \left (b^2-3 a c\right ) \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {x \sqrt {a+b x^2+c x^4}}{\sqrt {a}+\sqrt {c} x^2}\right )}{\sqrt {c}}\right )}{15 c \sqrt {a x+b x^3+c x^5}}\)

Input:

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

Output:

(Sqrt[x]*(b + 3*c*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(15*c) - (Sqrt[x]*Sqrt[a 
 + b*x^2 + c*x^4]*((-2*(b^2 - 3*a*c)*(-((x*Sqrt[a + b*x^2 + c*x^4])/(Sqrt[ 
a] + Sqrt[c]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c* 
x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], ( 
2 - b/(Sqrt[a]*Sqrt[c]))/4])/(c^(1/4)*Sqrt[a + b*x^2 + c*x^4])))/Sqrt[c] + 
 (a^(1/4)*(Sqrt[a]*b + (2*(b^2 - 3*a*c))/Sqrt[c])*(Sqrt[a] + Sqrt[c]*x^2)* 
Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^ 
(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*c^(1/4)*Sqrt[a + b*x^2 
 + c*x^4])))/(15*c*Sqrt[a*x + b*x^3 + c*x^5])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509

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

rule 1511

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + b*x^2 + c*x^ 
4], x], x] - Simp[e/q   Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; 
NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos 
Q[c/a]
 

rule 1966

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ 
), x_Symbol] :> Simp[x^(m - n + q + 1)*(b*(n - q)*p + c*(m + p*q + (n - q)* 
(2*p - 1) + 1)*x^(n - q))*((a*x^q + b*x^n + c*x^(2*n - q))^p/(c*(m + p*(2*n 
 - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1))), x] + Simp[(n - q)*(p/(c*(m 
+ p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1)))   Int[x^(m - (n - 2* 
q))*Simp[(-a)*b*(m + p*q - n + q + 1) + (2*a*c*(m + p*q + (n - q)*(2*p - 1) 
 + 1) - b^2*(m + p*q + (n - q)*(p - 1) + 1))*x^(n - q), x]*(a*x^q + b*x^n + 
 c*x^(2*n - q))^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] & 
& PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[ 
p, 0] && RationalQ[m, q] && GtQ[m + p*q + 1, n - q] && NeQ[m + p*(2*n - q) 
+ 1, 0] && NeQ[m + p*q + (n - q)*(2*p - 1) + 1, 0]
 

rule 2000

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(j_.)))/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x 
_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Simp[x^(q/2)*(Sqrt[a + b*x^(n - 
q) + c*x^(2*(n - q))]/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)])   Int[x^(m - q/2 
)*((A + B*x^(n - q))/Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), x], x] /; Fr 
eeQ[{a, b, c, A, B, m, n, q}, x] && EqQ[j, n - q] && EqQ[r, 2*n - q] && Pos 
Q[n - q] && (EqQ[m, 1/2] || EqQ[m, -2^(-1)]) && EqQ[n, 3] && EqQ[q, 1]
 

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.19

method	result	size
risch	\(\frac {x^{\frac {3}{2}} \left (3 c \,x^{2}+b \right ) \left (c \,x^{4}+b \,x^{2}+a \right )}{15 c \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}-\frac {\left (\frac {\left (6 a c -2 b^{2}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}+\frac {a b \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {x}}{15 c \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\)	\(455\)
default	\(\text {Expression too large to display}\)	\(1042\)

Input:

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

Output:

1/15*x^(3/2)*(3*c*x^2+b)/c*(c*x^4+b*x^2+a)/(x*(c*x^4+b*x^2+a))^(1/2)-1/15/ 
c*(1/2*(6*a*c-2*b^2)*a*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b+ 
(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/ 
(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/2*x*2^(1/2)*((-b 
+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2 
))-EllipticE(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*( 
b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2)))+1/4*a*b*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2) 
)/a)^(1/2)*(4-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^ 
(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2*x*2^(1/2)*((-b+(-4 
*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2)))* 
(c*x^4+b*x^2+a)^(1/2)*x^(1/2)/(x*(c*x^4+b*x^2+a))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 368, normalized size of antiderivative = 0.97 \[ \int x^{3/2} \sqrt {a x+b x^3+c x^5} \, dx=-\frac {2 \, \sqrt {\frac {1}{2}} {\left ({\left (b^{2} c - 3 \, a c^{2}\right )} x^{2} \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - {\left (b^{3} - 3 \, a b c\right )} x^{2}\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} E(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) - \sqrt {\frac {1}{2}} {\left ({\left (2 \, b^{2} c - {\left (6 \, a + b\right )} c^{2}\right )} x^{2} \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - {\left (2 \, b^{3} - {\left (6 \, a b - b^{2}\right )} c\right )} x^{2}\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} F(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) - 2 \, {\left (3 \, c^{3} x^{4} + b c^{2} x^{2} - 2 \, b^{2} c + 6 \, a c^{2}\right )} \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}}{30 \, c^{3} x^{2}} \] Input:

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

Output:

-1/30*(2*sqrt(1/2)*((b^2*c - 3*a*c^2)*x^2*sqrt((b^2 - 4*a*c)/c^2) - (b^3 - 
 3*a*b*c)*x^2)*sqrt(c)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)*elliptic_e( 
arcsin(sqrt(1/2)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)/x), 1/2*(b*c*sqrt 
((b^2 - 4*a*c)/c^2) + b^2 - 2*a*c)/(a*c)) - sqrt(1/2)*((2*b^2*c - (6*a + b 
)*c^2)*x^2*sqrt((b^2 - 4*a*c)/c^2) - (2*b^3 - (6*a*b - b^2)*c)*x^2)*sqrt(c 
)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)*elliptic_f(arcsin(sqrt(1/2)*sqrt 
((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)/x), 1/2*(b*c*sqrt((b^2 - 4*a*c)/c^2) + 
 b^2 - 2*a*c)/(a*c)) - 2*(3*c^3*x^4 + b*c^2*x^2 - 2*b^2*c + 6*a*c^2)*sqrt( 
c*x^5 + b*x^3 + a*x)*sqrt(x))/(c^3*x^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(sqrt(c*x^5 + b*x^3 + a*x)*x^(3/2), x)
 

Giac [F]

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

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

Output:

integrate(sqrt(c*x^5 + b*x^3 + a*x)*x^(3/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^{3/2} \sqrt {a x+b x^3+c x^5} \, dx=\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b x +3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c \,x^{3}-\left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a b +6 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a c -2 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) b^{2}}{15 c} \] Input:

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

Output:

(sqrt(a + b*x**2 + c*x**4)*b*x + 3*sqrt(a + b*x**2 + c*x**4)*c*x**3 - int( 
sqrt(a + b*x**2 + c*x**4)/(a + b*x**2 + c*x**4),x)*a*b + 6*int((sqrt(a + b 
*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*a*c - 2*int((sqrt(a + b*x** 
2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*b**2)/(15*c)