\(\int \frac {(a x+b x^3+c x^5)^{3/2}}{\sqrt {x}} \, dx\) [111]

Optimal result

Integrand size = 24, antiderivative size = 177 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=-\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{128 c^2 \sqrt {x}}+\frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac {3 \left (b^2-4 a c\right )^2 \sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{256 c^{5/2} \sqrt {a x+b x^3+c x^5}} \]

[Out]

1/16*(2*c*x^2+b)*(c*x^5+b*x^3+a*x)^(3/2)/c/x^(3/2)+3/256*(-4*a*c+b^2)^2*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4
+b*x^2+a)^(1/2))*x^(1/2)*(c*x^4+b*x^2+a)^(1/2)/c^(5/2)/(c*x^5+b*x^3+a*x)^(1/2)-3/128*(-4*a*c+b^2)*(2*c*x^2+b)*
(c*x^5+b*x^3+a*x)^(1/2)/c^2/x^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1932, 1928, 1121, 635, 212} \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\frac {3 \sqrt {x} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{256 c^{5/2} \sqrt {a x+b x^3+c x^5}}-\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{128 c^2 \sqrt {x}}+\frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}} \]

[In]

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

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(128*c^2*Sqrt[x]) + ((b + 2*c*x^2)*(a*x + b*x^3 + c
*x^5)^(3/2))/(16*c*x^(3/2)) + (3*(b^2 - 4*a*c)^2*Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*ArcTanh[(b + 2*c*x^2)/(2*Sqrt
[c]*Sqrt[a + b*x^2 + c*x^4])])/(256*c^(5/2)*Sqrt[a*x + b*x^3 + c*x^5])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1121

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rule 1928

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[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)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

Rule 1932

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[x^(m - n + q +
1)*(b + 2*c*x^(n - q))*((a*x^q + b*x^n + c*x^(2*n - q))^p/(2*c*(n - q)*(2*p + 1))), x] - Dist[p*((b^2 - 4*a*c
)/(2*c*(2*p + 1))), Int[x^(m + q)*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && Eq
Q[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] && EqQ[m + p*q + 1, n - q]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx}{16 c} \\ & = -\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{128 c^2 \sqrt {x}}+\frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx}{128 c^2} \\ & = -\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{128 c^2 \sqrt {x}}+\frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac {\left (3 \left (b^2-4 a c\right )^2 \sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx}{128 c^2 \sqrt {a x+b x^3+c x^5}} \\ & = -\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{128 c^2 \sqrt {x}}+\frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac {\left (3 \left (b^2-4 a c\right )^2 \sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{256 c^2 \sqrt {a x+b x^3+c x^5}} \\ & = -\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{128 c^2 \sqrt {x}}+\frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac {\left (3 \left (b^2-4 a c\right )^2 \sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^2}{\sqrt {a+b x^2+c x^4}}\right )}{128 c^2 \sqrt {a x+b x^3+c x^5}} \\ & = -\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{128 c^2 \sqrt {x}}+\frac {\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac {3 \left (b^2-4 a c\right )^2 \sqrt {x} \sqrt {a+b x^2+c x^4} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{256 c^{5/2} \sqrt {a x+b x^3+c x^5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\frac {\left (x \left (a+b x^2+c x^4\right )\right )^{3/2} \left (\frac {\sqrt {c} \left (b+2 c x^2\right ) \left (-3 b^2+8 b c x^2+4 c \left (5 a+2 c x^4\right )\right )}{a+b x^2+c x^4}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2+c x^4}}\right )}{\left (a+b x^2+c x^4\right )^{3/2}}\right )}{128 c^{5/2} x^{3/2}} \]

[In]

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

[Out]

((x*(a + b*x^2 + c*x^4))^(3/2)*((Sqrt[c]*(b + 2*c*x^2)*(-3*b^2 + 8*b*c*x^2 + 4*c*(5*a + 2*c*x^4)))/(a + b*x^2
+ c*x^4) + (3*(b^2 - 4*a*c)^2*ArcTanh[(Sqrt[c]*x^2)/(-Sqrt[a] + Sqrt[a + b*x^2 + c*x^4])])/(a + b*x^2 + c*x^4)
^(3/2)))/(128*c^(5/2)*x^(3/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {\left (16 c^{3} x^{6}+24 b \,c^{2} x^{4}+40 a \,c^{2} x^{2}+2 b^{2} c \,x^{2}+20 a b c -3 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}+a \right ) \sqrt {x}}{128 c^{2} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}+\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {x}}{256 c^{\frac {5}{2}} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\)	\(170\)
default	\(\frac {\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \left (32 c^{\frac {7}{2}} x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}+48 b \,c^{\frac {5}{2}} x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}+80 a \,c^{\frac {5}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+4 b^{2} c^{\frac {3}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+48 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) a^{2} c^{2}-24 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) a \,b^{2} c +3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) b^{4}+40 a b \,c^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}-6 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}\right )}{256 c^{\frac {5}{2}} \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\)	\(295\)

[In]

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

[Out]

1/128*(16*c^3*x^6+24*b*c^2*x^4+40*a*c^2*x^2+2*b^2*c*x^2+20*a*b*c-3*b^3)*(c*x^4+b*x^2+a)/c^2*x^(1/2)/(x*(c*x^4+
b*x^2+a))^(1/2)+3/256*(16*a^2*c^2-8*a*b^2*c+b^4)/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(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)

none

Time = 0.27 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.88 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{5} + 8 \, b c x^{3} + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {c} \sqrt {x} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}}{512 \, c^{3} x}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {-c} \sqrt {x}}{2 \, {\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}}{256 \, c^{3} x}\right ] \]

[In]

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

[Out]

[1/512*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*x*log(-(8*c^2*x^5 + 8*b*c*x^3 + 4*sqrt(c*x^5 + b*x^3 + a*x)*(
2*c*x^2 + b)*sqrt(c)*sqrt(x) + (b^2 + 4*a*c)*x)/x) + 4*(16*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(
b^2*c^2 + 20*a*c^3)*x^2)*sqrt(c*x^5 + b*x^3 + a*x)*sqrt(x))/(c^3*x), -1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*
sqrt(-c)*x*arctan(1/2*sqrt(c*x^5 + b*x^3 + a*x)*(2*c*x^2 + b)*sqrt(-c)*sqrt(x)/(c^2*x^5 + b*c*x^3 + a*c*x)) -
2*(16*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*x^2)*sqrt(c*x^5 + b*x^3 + a*x)*sq
rt(x))/(c^3*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (149) = 298\).

Time = 0.45 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.81 \[ \int \frac {\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt {x}} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}} - \frac {b^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 4 \, a c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 2 \, \sqrt {a} b \sqrt {c}}{c^{\frac {3}{2}}}\right )} a + \frac {1}{96} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}} + \frac {3 \, b^{3} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 12 \, a b c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 6 \, \sqrt {a} b^{2} \sqrt {c} - 16 \, a^{\frac {3}{2}} c^{\frac {3}{2}}}{c^{\frac {5}{2}}}\right )} b + \frac {1}{768} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {7}{2}}} - \frac {15 \, b^{4} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 72 \, a b^{2} c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 48 \, a^{2} c^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 30 \, \sqrt {a} b^{3} \sqrt {c} - 104 \, a^{\frac {3}{2}} b c^{\frac {3}{2}}}{c^{\frac {7}{2}}}\right )} c \]

[In]

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

[Out]

1/16*(2*sqrt(c*x^4 + b*x^2 + a)*(2*x^2 + b/c) + (b^2 - 4*a*c)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a)
)*sqrt(c) + b))/c^(3/2) - (b^2*log(abs(b - 2*sqrt(a)*sqrt(c))) - 4*a*c*log(abs(b - 2*sqrt(a)*sqrt(c))) + 2*sqr
t(a)*b*sqrt(c))/c^(3/2))*a + 1/96*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*x^2 + b/c)*x^2 - (3*b^2 - 8*a*c)/c^2) - 3*(
b^3 - 4*a*b*c)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(5/2) + (3*b^3*log(abs(b - 2*
sqrt(a)*sqrt(c))) - 12*a*b*c*log(abs(b - 2*sqrt(a)*sqrt(c))) + 6*sqrt(a)*b^2*sqrt(c) - 16*a^(3/2)*c^(3/2))/c^(
5/2))*b + 1/768*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*x^2 + b/c)*x^2 - (5*b^2*c - 12*a*c^2)/c^3)*x^2 + (15*b^3 -
52*a*b*c)/c^3) + 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c
) + b))/c^(7/2) - (15*b^4*log(abs(b - 2*sqrt(a)*sqrt(c))) - 72*a*b^2*c*log(abs(b - 2*sqrt(a)*sqrt(c))) + 48*a^
2*c^2*log(abs(b - 2*sqrt(a)*sqrt(c))) + 30*sqrt(a)*b^3*sqrt(c) - 104*a^(3/2)*b*c^(3/2))/c^(7/2))*c

Mupad [F(-1)]

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

[In]

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

[Out]

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