3.6.79 \(\int \frac {(a+b x^n+c x^{2 n})^{3/2}}{x} \, dx\) [579]

Optimal. Leaf size=173 \[ \frac {\left (b^2+8 a c+2 b c x^n\right ) \sqrt {a+b x^n+c x^{2 n}}}{8 c n}+\frac {\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}-\frac {a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^n}{2 \sqrt {a} \sqrt {a+b x^n+c x^{2 n}}}\right )}{n}-\frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^n}{2 \sqrt {c} \sqrt {a+b x^n+c x^{2 n}}}\right )}{16 c^{3/2} n} \]

[Out]

1/3*(a+b*x^n+c*x^(2*n))^(3/2)/n-a^(3/2)*arctanh(1/2*(2*a+b*x^n)/a^(1/2)/(a+b*x^n+c*x^(2*n))^(1/2))/n-1/16*b*(-
12*a*c+b^2)*arctanh(1/2*(b+2*c*x^n)/c^(1/2)/(a+b*x^n+c*x^(2*n))^(1/2))/c^(3/2)/n+1/8*(b^2+8*a*c+2*b*c*x^n)*(a+
b*x^n+c*x^(2*n))^(1/2)/c/n

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Rubi [A]
time = 0.11, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1371, 748, 828, 857, 635, 212, 738} \begin {gather*} -\frac {a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^n}{2 \sqrt {a} \sqrt {a+b x^n+c x^{2 n}}}\right )}{n}-\frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^n}{2 \sqrt {c} \sqrt {a+b x^n+c x^{2 n}}}\right )}{16 c^{3/2} n}+\frac {\left (8 a c+b^2+2 b c x^n\right ) \sqrt {a+b x^n+c x^{2 n}}}{8 c n}+\frac {\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 + 8*a*c + 2*b*c*x^n)*Sqrt[a + b*x^n + c*x^(2*n)])/(8*c*n) + (a + b*x^n + c*x^(2*n))^(3/2)/(3*n) - (a^(3/
2)*ArcTanh[(2*a + b*x^n)/(2*Sqrt[a]*Sqrt[a + b*x^n + c*x^(2*n)])])/n - (b*(b^2 - 12*a*c)*ArcTanh[(b + 2*c*x^n)
/(2*Sqrt[c]*Sqrt[a + b*x^n + c*x^(2*n)])])/(16*c^(3/2)*n)

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 738

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

Rule 748

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 828

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^
2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^n+c x^{2 n}\right )^{3/2}}{x} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x} \, dx,x,x^n\right )}{n}\\ &=\frac {\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}-\frac {\text {Subst}\left (\int \frac {(-2 a-b x) \sqrt {a+b x+c x^2}}{x} \, dx,x,x^n\right )}{2 n}\\ &=\frac {\left (b^2+8 a c+2 b c x^n\right ) \sqrt {a+b x^n+c x^{2 n}}}{8 c n}+\frac {\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}+\frac {\text {Subst}\left (\int \frac {8 a^2 c-\frac {1}{2} b \left (b^2-12 a c\right ) x}{x \sqrt {a+b x+c x^2}} \, dx,x,x^n\right )}{8 c n}\\ &=\frac {\left (b^2+8 a c+2 b c x^n\right ) \sqrt {a+b x^n+c x^{2 n}}}{8 c n}+\frac {\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}+\frac {a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^n\right )}{n}-\frac {\left (b \left (b^2-12 a c\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^n\right )}{16 c n}\\ &=\frac {\left (b^2+8 a c+2 b c x^n\right ) \sqrt {a+b x^n+c x^{2 n}}}{8 c n}+\frac {\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^n}{\sqrt {a+b x^n+c x^{2 n}}}\right )}{n}-\frac {\left (b \left (b^2-12 a c\right )\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^n}{\sqrt {a+b x^n+c x^{2 n}}}\right )}{8 c n}\\ &=\frac {\left (b^2+8 a c+2 b c x^n\right ) \sqrt {a+b x^n+c x^{2 n}}}{8 c n}+\frac {\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}-\frac {a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^n}{2 \sqrt {a} \sqrt {a+b x^n+c x^{2 n}}}\right )}{n}-\frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^n}{2 \sqrt {c} \sqrt {a+b x^n+c x^{2 n}}}\right )}{16 c^{3/2} n}\\ \end {align*}

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Mathematica [A]
time = 0.63, size = 159, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {c} \sqrt {a+x^n \left (b+c x^n\right )} \left (3 b^2+14 b c x^n+8 c \left (4 a+c x^{2 n}\right )\right )+96 a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x^n-\sqrt {a+x^n \left (b+c x^n\right )}}{\sqrt {a}}\right )+3 \left (b^3-12 a b c\right ) \log \left (c n \left (b+2 c x^n-2 \sqrt {c} \sqrt {a+x^n \left (b+c x^n\right )}\right )\right )}{48 c^{3/2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x^n*(b + c*x^n)]*(3*b^2 + 14*b*c*x^n + 8*c*(4*a + c*x^(2*n))) + 96*a^(3/2)*c^(3/2)*ArcTanh
[(Sqrt[c]*x^n - Sqrt[a + x^n*(b + c*x^n)])/Sqrt[a]] + 3*(b^3 - 12*a*b*c)*Log[c*n*(b + 2*c*x^n - 2*Sqrt[c]*Sqrt
[a + x^n*(b + c*x^n)])])/(48*c^(3/2)*n)

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Maple [A]
time = 0.10, size = 209, normalized size = 1.21

method	result	size
risch	\(\frac {\left (8 c^{2} {\mathrm e}^{2 n \ln \left (x \right )}+14 b c \,{\mathrm e}^{n \ln \left (x \right )}+32 a c +3 b^{2}\right ) \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}+c \,{\mathrm e}^{2 n \ln \left (x \right )}}}{24 c n}+\frac {3 a b \ln \left (\frac {\frac {b}{2}+c \,{\mathrm e}^{n \ln \left (x \right )}}{\sqrt {c}}+\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}+c \,{\mathrm e}^{2 n \ln \left (x \right )}}\right )}{4 \sqrt {c}\, n}-\frac {b^{3} \ln \left (\frac {\frac {b}{2}+c \,{\mathrm e}^{n \ln \left (x \right )}}{\sqrt {c}}+\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}+c \,{\mathrm e}^{2 n \ln \left (x \right )}}\right )}{16 c^{\frac {3}{2}} n}-\frac {a^{\frac {3}{2}} \ln \left (\left (2 a +b \,{\mathrm e}^{n \ln \left (x \right )}+2 \sqrt {a}\, \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}+c \,{\mathrm e}^{2 n \ln \left (x \right )}}\right ) {\mathrm e}^{-n \ln \left (x \right )}\right )}{n}\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*c^2*exp(n*ln(x))^2+14*b*c*exp(n*ln(x))+32*a*c+3*b^2)*(a+b*exp(n*ln(x))+c*exp(n*ln(x))^2)^(1/2)/c/n+3/4
/c^(1/2)/n*a*b*ln((1/2*b+c*exp(n*ln(x)))/c^(1/2)+(a+b*exp(n*ln(x))+c*exp(n*ln(x))^2)^(1/2))-1/16/c^(3/2)/n*b^3
*ln((1/2*b+c*exp(n*ln(x)))/c^(1/2)+(a+b*exp(n*ln(x))+c*exp(n*ln(x))^2)^(1/2))-1/n*a^(3/2)*ln((2*a+b*exp(n*ln(x
))+2*a^(1/2)*(a+b*exp(n*ln(x))+c*exp(n*ln(x))^2)^(1/2))/exp(n*ln(x)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.45, size = 827, normalized size = 4.78 \begin {gather*} \left [\frac {48 \, a^{\frac {3}{2}} c^{2} \log \left (-\frac {8 \, a b x^{n} + 8 \, a^{2} + {\left (b^{2} + 4 \, a c\right )} x^{2 \, n} - 4 \, {\left (\sqrt {a} b x^{n} + 2 \, a^{\frac {3}{2}}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{x^{2 \, n}}\right ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2 \, n} - 8 \, b c x^{n} - b^{2} - 4 \, a c - 4 \, {\left (2 \, c^{\frac {3}{2}} x^{n} + b \sqrt {c}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}\right ) + 4 \, {\left (8 \, c^{3} x^{2 \, n} + 14 \, b c^{2} x^{n} + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{96 \, c^{2} n}, \frac {24 \, a^{\frac {3}{2}} c^{2} \log \left (-\frac {8 \, a b x^{n} + 8 \, a^{2} + {\left (b^{2} + 4 \, a c\right )} x^{2 \, n} - 4 \, {\left (\sqrt {a} b x^{n} + 2 \, a^{\frac {3}{2}}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{x^{2 \, n}}\right ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-c} \arctan \left (\frac {{\left (2 \, \sqrt {-c} c x^{n} + b \sqrt {-c}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{2 \, {\left (c^{2} x^{2 \, n} + b c x^{n} + a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{2 \, n} + 14 \, b c^{2} x^{n} + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{48 \, c^{2} n}, \frac {96 \, \sqrt {-a} a c^{2} \arctan \left (\frac {{\left (\sqrt {-a} b x^{n} + 2 \, \sqrt {-a} a\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{2 \, {\left (a c x^{2 \, n} + a b x^{n} + a^{2}\right )}}\right ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2 \, n} - 8 \, b c x^{n} - b^{2} - 4 \, a c - 4 \, {\left (2 \, c^{\frac {3}{2}} x^{n} + b \sqrt {c}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}\right ) + 4 \, {\left (8 \, c^{3} x^{2 \, n} + 14 \, b c^{2} x^{n} + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{96 \, c^{2} n}, \frac {48 \, \sqrt {-a} a c^{2} \arctan \left (\frac {{\left (\sqrt {-a} b x^{n} + 2 \, \sqrt {-a} a\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{2 \, {\left (a c x^{2 \, n} + a b x^{n} + a^{2}\right )}}\right ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-c} \arctan \left (\frac {{\left (2 \, \sqrt {-c} c x^{n} + b \sqrt {-c}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{2 \, {\left (c^{2} x^{2 \, n} + b c x^{n} + a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{2 \, n} + 14 \, b c^{2} x^{n} + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{48 \, c^{2} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(48*a^(3/2)*c^2*log(-(8*a*b*x^n + 8*a^2 + (b^2 + 4*a*c)*x^(2*n) - 4*(sqrt(a)*b*x^n + 2*a^(3/2))*sqrt(c*x
^(2*n) + b*x^n + a))/x^(2*n)) - 3*(b^3 - 12*a*b*c)*sqrt(c)*log(-8*c^2*x^(2*n) - 8*b*c*x^n - b^2 - 4*a*c - 4*(2
*c^(3/2)*x^n + b*sqrt(c))*sqrt(c*x^(2*n) + b*x^n + a)) + 4*(8*c^3*x^(2*n) + 14*b*c^2*x^n + 3*b^2*c + 32*a*c^2)
*sqrt(c*x^(2*n) + b*x^n + a))/(c^2*n), 1/48*(24*a^(3/2)*c^2*log(-(8*a*b*x^n + 8*a^2 + (b^2 + 4*a*c)*x^(2*n) -
4*(sqrt(a)*b*x^n + 2*a^(3/2))*sqrt(c*x^(2*n) + b*x^n + a))/x^(2*n)) + 3*(b^3 - 12*a*b*c)*sqrt(-c)*arctan(1/2*(
2*sqrt(-c)*c*x^n + b*sqrt(-c))*sqrt(c*x^(2*n) + b*x^n + a)/(c^2*x^(2*n) + b*c*x^n + a*c)) + 2*(8*c^3*x^(2*n) +
14*b*c^2*x^n + 3*b^2*c + 32*a*c^2)*sqrt(c*x^(2*n) + b*x^n + a))/(c^2*n), 1/96*(96*sqrt(-a)*a*c^2*arctan(1/2*(
sqrt(-a)*b*x^n + 2*sqrt(-a)*a)*sqrt(c*x^(2*n) + b*x^n + a)/(a*c*x^(2*n) + a*b*x^n + a^2)) - 3*(b^3 - 12*a*b*c)
*sqrt(c)*log(-8*c^2*x^(2*n) - 8*b*c*x^n - b^2 - 4*a*c - 4*(2*c^(3/2)*x^n + b*sqrt(c))*sqrt(c*x^(2*n) + b*x^n +
a)) + 4*(8*c^3*x^(2*n) + 14*b*c^2*x^n + 3*b^2*c + 32*a*c^2)*sqrt(c*x^(2*n) + b*x^n + a))/(c^2*n), 1/48*(48*sq
rt(-a)*a*c^2*arctan(1/2*(sqrt(-a)*b*x^n + 2*sqrt(-a)*a)*sqrt(c*x^(2*n) + b*x^n + a)/(a*c*x^(2*n) + a*b*x^n + a
^2)) + 3*(b^3 - 12*a*b*c)*sqrt(-c)*arctan(1/2*(2*sqrt(-c)*c*x^n + b*sqrt(-c))*sqrt(c*x^(2*n) + b*x^n + a)/(c^2
*x^(2*n) + b*c*x^n + a*c)) + 2*(8*c^3*x^(2*n) + 14*b*c^2*x^n + 3*b^2*c + 32*a*c^2)*sqrt(c*x^(2*n) + b*x^n + a)
)/(c^2*n)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{n} + c x^{2 n}\right )^{\frac {3}{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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