\(\int c^5 x^5 (\frac {a}{x^4}+b x^n)^{3/2} \, dx\) [381]

Optimal result

Integrand size = 22, antiderivative size = 100 \[ \int c^5 x^5 \left (\frac {a}{x^4}+b x^n\right )^{3/2} \, dx=\frac {2 a c^5 x^2 \sqrt {\frac {a}{x^4}+b x^n}}{4+n}+\frac {2 c^5 x^6 \left (\frac {a}{x^4}+b x^n\right )^{3/2}}{3 (4+n)}-\frac {2 a^{3/2} c^5 \text {arctanh}\left (\frac {\sqrt {a}}{x^2 \sqrt {\frac {a}{x^4}+b x^n}}\right )}{4+n} \]

[Out]

2/3*c^5*x^6*(a/x^4+b*x^n)^(3/2)/(4+n)-2*a^(3/2)*c^5*arctanh(a^(1/2)/x^2/(a/x^4+b*x^n)^(1/2))/(4+n)+2*a*c^5*x^2
*(a/x^4+b*x^n)^(1/2)/(4+n)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 2053, 2054, 212} \[ \int c^5 x^5 \left (\frac {a}{x^4}+b x^n\right )^{3/2} \, dx=-\frac {2 a^{3/2} c^5 \text {arctanh}\left (\frac {\sqrt {a}}{x^2 \sqrt {\frac {a}{x^4}+b x^n}}\right )}{n+4}+\frac {2 c^5 x^6 \left (\frac {a}{x^4}+b x^n\right )^{3/2}}{3 (n+4)}+\frac {2 a c^5 x^2 \sqrt {\frac {a}{x^4}+b x^n}}{n+4} \]

[In]

Int[c^5*x^5*(a/x^4 + b*x^n)^(3/2),x]

[Out]

(2*a*c^5*x^2*Sqrt[a/x^4 + b*x^n])/(4 + n) + (2*c^5*x^6*(a/x^4 + b*x^n)^(3/2))/(3*(4 + n)) - (2*a^(3/2)*c^5*Arc
Tanh[Sqrt[a]/(x^2*Sqrt[a/x^4 + b*x^n])])/(4 + n)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2053

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*p*(n - j))), x] + Dist[a/c^j, Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c,
j, m, n}, x] && IGtQ[p + 1/2, 0] && NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = c^5 \int x^5 \left (\frac {a}{x^4}+b x^n\right )^{3/2} \, dx \\ & = \frac {2 c^5 x^6 \left (\frac {a}{x^4}+b x^n\right )^{3/2}}{3 (4+n)}+\left (a c^5\right ) \int x \sqrt {\frac {a}{x^4}+b x^n} \, dx \\ & = \frac {2 a c^5 x^2 \sqrt {\frac {a}{x^4}+b x^n}}{4+n}+\frac {2 c^5 x^6 \left (\frac {a}{x^4}+b x^n\right )^{3/2}}{3 (4+n)}+\left (a^2 c^5\right ) \int \frac {1}{x^3 \sqrt {\frac {a}{x^4}+b x^n}} \, dx \\ & = \frac {2 a c^5 x^2 \sqrt {\frac {a}{x^4}+b x^n}}{4+n}+\frac {2 c^5 x^6 \left (\frac {a}{x^4}+b x^n\right )^{3/2}}{3 (4+n)}-\frac {\left (2 a^2 c^5\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {1}{x^2 \sqrt {\frac {a}{x^4}+b x^n}}\right )}{4+n} \\ & = \frac {2 a c^5 x^2 \sqrt {\frac {a}{x^4}+b x^n}}{4+n}+\frac {2 c^5 x^6 \left (\frac {a}{x^4}+b x^n\right )^{3/2}}{3 (4+n)}-\frac {2 a^{3/2} c^5 \tanh ^{-1}\left (\frac {\sqrt {a}}{x^2 \sqrt {\frac {a}{x^4}+b x^n}}\right )}{4+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96 \[ \int c^5 x^5 \left (\frac {a}{x^4}+b x^n\right )^{3/2} \, dx=\frac {2 c^5 x^2 \sqrt {\frac {a}{x^4}+b x^n} \left (\sqrt {a+b x^{4+n}} \left (4 a+b x^{4+n}\right )-3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x^{4+n}}}{\sqrt {a}}\right )\right )}{3 (4+n) \sqrt {a+b x^{4+n}}} \]

[In]

Integrate[c^5*x^5*(a/x^4 + b*x^n)^(3/2),x]

[Out]

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

Maple [F]

\[\int c^{5} x^{5} \left (\frac {a}{x^{4}}+b \,x^{n}\right )^{\frac {3}{2}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int c^5 x^5 \left (\frac {a}{x^4}+b x^n\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(c^5*x^5*(a/x^4+b*x^n)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

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

[In]

integrate(c**5*x**5*(a/x**4+b*x**n)**(3/2),x)

[Out]

c**5*(Integral(a*x*sqrt(a/x**4 + b*x**n), x) + Integral(b*x**5*x**n*sqrt(a/x**4 + b*x**n), x))

Maxima [F]

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

[In]

integrate(c^5*x^5*(a/x^4+b*x^n)^(3/2),x, algorithm="maxima")

[Out]

c^5*integrate((b*x^n + a/x^4)^(3/2)*x^5, x)

Giac [F]

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

[In]

integrate(c^5*x^5*(a/x^4+b*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^n + a/x^4)^(3/2)*c^5*x^5, x)

Mupad [F(-1)]

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

[In]

int(c^5*x^5*(b*x^n + a/x^4)^(3/2),x)

[Out]

int(c^5*x^5*(b*x^n + a/x^4)^(3/2), x)