\(\int \frac {(a x+b x^3+c x^5)^{3/2}}{x^{3/2}} \, dx\) [59]

Optimal result

Integrand size = 24, antiderivative size = 425 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{x^{3/2}} \, dx=-\frac {2 b \left (b^2-8 a c\right ) x^{3/2} \left (a+b x^2+c x^4\right )}{35 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}}+\frac {\sqrt {x} \left (b^2+10 a c+3 b c x^2\right ) \sqrt {a x+b x^3+c x^5}}{35 c}+\frac {\left (a x+b x^3+c x^5\right )^{3/2}}{7 \sqrt {x}}+\frac {2 \sqrt [4]{a} b \left (b^2-8 a c\right ) \sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{35 c^{7/4} \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt [4]{a} \left (\sqrt {a} \left (b^2-20 a c\right )+\frac {2 b \left (b^2-8 a c\right )}{\sqrt {c}}\right ) \sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{70 c^{5/4} \sqrt {a x+b x^3+c x^5}} \] Output:

-2/35*b*(-8*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)+1/35*x^(1/2)*(3*b*c*x^2+10*a*c+b^2)*(c*x^5+b*x^3+ 
a*x)^(1/2)/c+1/7*(c*x^5+b*x^3+a*x)^(3/2)/x^(1/2)+2/35*a^(1/4)*b*(-8*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)-1/70*a^(1/4)*(a^(1/2)*(-20*a*c+b^ 
2)+2*b*(-8*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 = 9.70 (sec) , antiderivative size = 540, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{x^{3/2}} \, dx=\frac {\sqrt {x} \left (2 c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x \left (15 a^2 c+a \left (b^2+23 b c x^2+20 c^2 x^4\right )+x^2 \left (b^3+9 b^2 c x^2+13 b c^2 x^4+5 c^3 x^6\right )\right )-i b \left (b^2-8 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+i \left (-b^4+9 a b^2 c-20 a^2 c^2+b^3 \sqrt {b^2-4 a c}-8 a b c \sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{70 c^2 \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {x \left (a+b x^2+c x^4\right )}} \] Input:

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

Output:

(Sqrt[x]*(2*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(15*a^2*c + a*(b^2 + 23*b* 
c*x^2 + 20*c^2*x^4) + x^2*(b^3 + 9*b^2*c*x^2 + 13*b*c^2*x^4 + 5*c^3*x^6)) 
- I*b*(b^2 - 8*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^4 + 9*a*b^2*c - 20*a^2*c^2 + b^3*Sqrt[b^2 - 4*a*c] - 8*a*b*c*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])]*Sq 
rt[(2*b - 2*Sqrt[b^2 - 4*a*c] + 4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Elliptic 
F[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])]))/(70*c^2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*S 
qrt[x*(a + b*x^2 + c*x^4)])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1968, 1992, 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.

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

\(\Big \downarrow \) 1968

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

\(\Big \downarrow \) 1992

\(\displaystyle \frac {3}{7} \left (\frac {\int -\frac {\sqrt {x} \left (2 b \left (b^2-8 a c\right ) x^2+a \left (b^2-20 a c\right )\right )}{\sqrt {c x^5+b x^3+a x}}dx}{15 c}+\frac {\sqrt {x} \sqrt {a x+b x^3+c x^5} \left (10 a c+b^2+3 b c x^2\right )}{15 c}\right )+\frac {\left (a x+b x^3+c x^5\right )^{3/2}}{7 \sqrt {x}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3}{7} \left (\frac {\sqrt {x} \left (10 a c+b^2+3 b c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\int \frac {\sqrt {x} \left (2 b \left (b^2-8 a c\right ) x^2+a \left (b^2-20 a c\right )\right )}{\sqrt {c x^5+b x^3+a x}}dx}{15 c}\right )+\frac {\left (a x+b x^3+c x^5\right )^{3/2}}{7 \sqrt {x}}\)

\(\Big \downarrow \) 2000

\(\displaystyle \frac {3}{7} \left (\frac {\sqrt {x} \left (10 a c+b^2+3 b c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \int \frac {2 b \left (b^2-8 a c\right ) x^2+a \left (b^2-20 a c\right )}{\sqrt {c x^4+b x^2+a}}dx}{15 c \sqrt {a x+b x^3+c x^5}}\right )+\frac {\left (a x+b x^3+c x^5\right )^{3/2}}{7 \sqrt {x}}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {3}{7} \left (\frac {\sqrt {x} \left (10 a c+b^2+3 b c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\sqrt {a} \left (\sqrt {a} \left (b^2-20 a c\right )+\frac {2 b \left (b^2-8 a c\right )}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx-\frac {2 \sqrt {a} b \left (b^2-8 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{15 c \sqrt {a x+b x^3+c x^5}}\right )+\frac {\left (a x+b x^3+c x^5\right )^{3/2}}{7 \sqrt {x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{7} \left (\frac {\sqrt {x} \left (10 a c+b^2+3 b c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\sqrt {a} \left (\sqrt {a} \left (b^2-20 a c\right )+\frac {2 b \left (b^2-8 a c\right )}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx-\frac {2 b \left (b^2-8 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{15 c \sqrt {a x+b x^3+c x^5}}\right )+\frac {\left (a x+b x^3+c x^5\right )^{3/2}}{7 \sqrt {x}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {3}{7} \left (\frac {\sqrt {x} \left (10 a c+b^2+3 b c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {\sqrt [4]{a} \left (\sqrt {a} \left (b^2-20 a c\right )+\frac {2 b \left (b^2-8 a c\right )}{\sqrt {c}}\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {2 b \left (b^2-8 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{15 c \sqrt {a x+b x^3+c x^5}}\right )+\frac {\left (a x+b x^3+c x^5\right )^{3/2}}{7 \sqrt {x}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {3}{7} \left (\frac {\sqrt {x} \left (10 a c+b^2+3 b c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {\sqrt [4]{a} \left (\sqrt {a} \left (b^2-20 a c\right )+\frac {2 b \left (b^2-8 a c\right )}{\sqrt {c}}\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {2 b \left (b^2-8 a c\right ) \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {x \sqrt {a+b x^2+c x^4}}{\sqrt {a}+\sqrt {c} x^2}\right )}{\sqrt {c}}\right )}{15 c \sqrt {a x+b x^3+c x^5}}\right )+\frac {\left (a x+b x^3+c x^5\right )^{3/2}}{7 \sqrt {x}}\)

Input:

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

Output:

(a*x + b*x^3 + c*x^5)^(3/2)/(7*Sqrt[x]) + (3*((Sqrt[x]*(b^2 + 10*a*c + 3*b 
*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*(b^2 - 8*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)*(Sq 
rt[a]*(b^2 - 20*a*c) + (2*b*(b^2 - 8*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])))/7
 

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 1968

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ 
), x_Symbol] :> Simp[x^(m + 1)*((a*x^q + b*x^n + c*x^(2*n - q))^p/(m + p*(2 
*n - q) + 1)), x] + Simp[(n - q)*(p/(m + p*(2*n - q) + 1))   Int[x^(m + q)* 
(2*a + b*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] /; Free 
Q[{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]
 

rule 1992

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_ 
.)*((A_) + (B_.)*(x_)^(r_.)), x_Symbol] :> Simp[x^(m + 1)*(b*B*(n - q)*p + 
A*c*(m + p*q + (n - q)*(2*p + 1) + 1) + B*c*(m + p*q + 2*(n - q)*p + 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 + q)*Simp[2*a*A*c*(m + p*q 
+ (n - q)*(2*p + 1) + 1) - a*b*B*(m + p*q + 1) + (2*a*B*c*(m + p*q + 2*(n - 
 q)*p + 1) + A*b*c*(m + p*q + (n - q)*(2*p + 1) + 1) - b^2*B*(m + p*q + (n 
- q)*p + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] 
/; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - q] && EqQ[j, 2*n - q] &&  !Integ 
erQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] 
 && GtQ[m + p*q, -(n - q) - 1] && 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 = 2.72 (sec) , antiderivative size = 620, normalized size of antiderivative = 1.46

method	result	size
risch	\(\frac {x^{\frac {3}{2}} \left (5 c^{2} x^{4}+8 b c \,x^{2}+15 a c +b^{2}\right ) \left (c \,x^{4}+b \,x^{2}+a \right )}{35 c \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}+\frac {\left (-\frac {b \left (8 a c -b^{2}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}+\frac {5 a^{2} c \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b^{2} a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {x}}{35 c \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\)	\(620\)
default	\(\text {Expression too large to display}\)	\(1394\)

Input:

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

Output:

1/35*x^(3/2)*(5*c^2*x^4+8*b*c*x^2+15*a*c+b^2)/c*(c*x^4+b*x^2+a)/(x*(c*x^4+ 
b*x^2+a))^(1/2)+1/35/c*(-b*(8*a*c-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)))+5*a^2*c*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))-1/4*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)*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 = 0.08 (sec) , antiderivative size = 409, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{x^{3/2}} \, dx=-\frac {2 \, \sqrt {\frac {1}{2}} {\left ({\left (b^{3} c - 8 \, a b c^{2}\right )} x^{2} \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - {\left (b^{4} - 8 \, a b^{2} c\right )} x^{2}\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} E(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) - \sqrt {\frac {1}{2}} {\left ({\left (2 \, b^{3} c + 20 \, a c^{3} - {\left (16 \, a b + b^{2}\right )} c^{2}\right )} x^{2} \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - {\left (2 \, b^{4} - 20 \, a b c^{2} - {\left (16 \, a b^{2} - b^{3}\right )} c\right )} x^{2}\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} F(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) - 2 \, {\left (5 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} - 2 \, b^{3} c + 16 \, a b c^{2} + {\left (b^{2} c^{2} + 15 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}}{70 \, c^{3} x^{2}} \] Input:

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

Output:

-1/70*(2*sqrt(1/2)*((b^3*c - 8*a*b*c^2)*x^2*sqrt((b^2 - 4*a*c)/c^2) - (b^4 
 - 8*a*b^2*c)*x^2)*sqrt(c)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)*ellipti 
c_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^3*c + 20*a 
*c^3 - (16*a*b + b^2)*c^2)*x^2*sqrt((b^2 - 4*a*c)/c^2) - (2*b^4 - 20*a*b*c 
^2 - (16*a*b^2 - b^3)*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*(5*c^4*x^6 + 
8*b*c^3*x^4 - 2*b^3*c + 16*a*b*c^2 + (b^2*c^2 + 15*a*c^3)*x^2)*sqrt(c*x^5 
+ b*x^3 + a*x)*sqrt(x))/(c^3*x^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{x^{3/2}} \, dx=\frac {15 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a c x +\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} x +8 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b c \,x^{3}+5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2} x^{5}+20 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a^{2} c -\left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a \,b^{2}+16 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a b c -2 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) b^{3}}{35 c} \] Input:

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

Output:

(15*sqrt(a + b*x**2 + c*x**4)*a*c*x + sqrt(a + b*x**2 + c*x**4)*b**2*x + 8 
*sqrt(a + b*x**2 + c*x**4)*b*c*x**3 + 5*sqrt(a + b*x**2 + c*x**4)*c**2*x** 
5 + 20*int(sqrt(a + b*x**2 + c*x**4)/(a + b*x**2 + c*x**4),x)*a**2*c - int 
(sqrt(a + b*x**2 + c*x**4)/(a + b*x**2 + c*x**4),x)*a*b**2 + 16*int((sqrt( 
a + b*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*a*b*c - 2*int((sqrt(a 
+ b*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*b**3)/(35*c)