Integrand size = 22, antiderivative size = 104 \[ \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx=-\frac {2 a \sqrt {a x^2+b x^n}}{c^4 (2-n) x}-\frac {2 \left (a x^2+b x^n\right )^{3/2}}{3 c^4 (2-n) x^3}+\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^n}}\right )}{c^4 (2-n)} \] Output:
-2*a*(a*x^2+b*x^n)^(1/2)/c^4/(2-n)/x-2/3*(a*x^2+b*x^n)^(3/2)/c^4/(2-n)/x^3 +2*a^(3/2)*arctanh(a^(1/2)*x/(a*x^2+b*x^n)^(1/2))/c^4/(2-n)
Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx=\frac {2 \left (4 a^2 x^4+b^2 x^{2 n}+5 a b x^{2+n}-3 a^{3/2} \sqrt {b} x^{3+\frac {n}{2}} \sqrt {1+\frac {a x^{2-n}}{b}} \text {arcsinh}\left (\frac {\sqrt {a} x^{1-\frac {n}{2}}}{\sqrt {b}}\right )\right )}{3 c^4 (-2+n) x^3 \sqrt {a x^2+b x^n}} \] Input:
Integrate[(a*x^2 + b*x^n)^(3/2)/(c^4*x^4),x]
Output:
(2*(4*a^2*x^4 + b^2*x^(2*n) + 5*a*b*x^(2 + n) - 3*a^(3/2)*Sqrt[b]*x^(3 + n /2)*Sqrt[1 + (a*x^(2 - n))/b]*ArcSinh[(Sqrt[a]*x^(1 - n/2))/Sqrt[b]]))/(3* c^4*(-2 + n)*x^3*Sqrt[a*x^2 + b*x^n])
Time = 0.42 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {27, 1934, 1934, 1914, 219}
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.
\(\displaystyle \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\left (b x^n+a x^2\right )^{3/2}}{x^4}dx}{c^4}\) |
\(\Big \downarrow \) 1934 |
\(\displaystyle \frac {a \int \frac {\sqrt {b x^n+a x^2}}{x^2}dx-\frac {2 \left (a x^2+b x^n\right )^{3/2}}{3 (2-n) x^3}}{c^4}\) |
\(\Big \downarrow \) 1934 |
\(\displaystyle \frac {a \left (a \int \frac {1}{\sqrt {b x^n+a x^2}}dx-\frac {2 \sqrt {a x^2+b x^n}}{(2-n) x}\right )-\frac {2 \left (a x^2+b x^n\right )^{3/2}}{3 (2-n) x^3}}{c^4}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {a \left (\frac {2 a \int \frac {1}{1-\frac {a x^2}{b x^n+a x^2}}d\frac {x}{\sqrt {b x^n+a x^2}}}{2-n}-\frac {2 \sqrt {a x^2+b x^n}}{(2-n) x}\right )-\frac {2 \left (a x^2+b x^n\right )^{3/2}}{3 (2-n) x^3}}{c^4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a \left (\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^n}}\right )}{2-n}-\frac {2 \sqrt {a x^2+b x^n}}{(2-n) x}\right )-\frac {2 \left (a x^2+b x^n\right )^{3/2}}{3 (2-n) x^3}}{c^4}\) |
Input:
Int[(a*x^2 + b*x^n)^(3/2)/(c^4*x^4),x]
Output:
((-2*(a*x^2 + b*x^n)^(3/2))/(3*(2 - n)*x^3) + a*((-2*Sqrt[a*x^2 + b*x^n])/ ((2 - n)*x) + (2*Sqrt[a]*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^n]])/(2 - n) ))/c^4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[2/(2 - n) Subst[Int[1/(1 - a*x^2), x], x, x/Sqrt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]
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] + Simp[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])
\[\int \frac {\left (a \,x^{2}+b \,x^{n}\right )^{\frac {3}{2}}}{c^{4} x^{4}}d x\]
Input:
int((a*x^2+b*x^n)^(3/2)/c^4/x^4,x)
Output:
int((a*x^2+b*x^n)^(3/2)/c^4/x^4,x)
Exception generated. \[ \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a*x^2+b*x^n)^(3/2)/c^4/x^4,x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx=\frac {\int \frac {a \sqrt {a x^{2} + b x^{n}}}{x^{2}}\, dx + \int \frac {b x^{n} \sqrt {a x^{2} + b x^{n}}}{x^{4}}\, dx}{c^{4}} \] Input:
integrate((a*x**2+b*x**n)**(3/2)/c**4/x**4,x)
Output:
(Integral(a*sqrt(a*x**2 + b*x**n)/x**2, x) + Integral(b*x**n*sqrt(a*x**2 + b*x**n)/x**4, x))/c**4
\[ \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx=\int { \frac {{\left (a x^{2} + b x^{n}\right )}^{\frac {3}{2}}}{c^{4} x^{4}} \,d x } \] Input:
integrate((a*x^2+b*x^n)^(3/2)/c^4/x^4,x, algorithm="maxima")
Output:
integrate((a*x^2 + b*x^n)^(3/2)/x^4, x)/c^4
\[ \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx=\int { \frac {{\left (a x^{2} + b x^{n}\right )}^{\frac {3}{2}}}{c^{4} x^{4}} \,d x } \] Input:
integrate((a*x^2+b*x^n)^(3/2)/c^4/x^4,x, algorithm="giac")
Output:
integrate((a*x^2 + b*x^n)^(3/2)/(c^4*x^4), x)
Timed out. \[ \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx=\int \frac {{\left (b\,x^n+a\,x^2\right )}^{3/2}}{c^4\,x^4} \,d x \] Input:
int((b*x^n + a*x^2)^(3/2)/(c^4*x^4),x)
Output:
int((b*x^n + a*x^2)^(3/2)/(c^4*x^4), x)
\[ \int \frac {\left (a x^2+b x^n\right )^{3/2}}{c^4 x^4} \, dx=\frac {2 x^{n} \sqrt {x^{n} b +a \,x^{2}}\, b +2 \sqrt {x^{n} b +a \,x^{2}}\, a \,x^{2}+3 \left (\int \frac {\sqrt {x^{n} b +a \,x^{2}}}{x^{2}}d x \right ) a n \,x^{3}-6 \left (\int \frac {\sqrt {x^{n} b +a \,x^{2}}}{x^{2}}d x \right ) a \,x^{3}}{3 c^{4} x^{3} \left (n -2\right )} \] Input:
int((a*x^2+b*x^n)^(3/2)/c^4/x^4,x)
Output:
(2*x**n*sqrt(x**n*b + a*x**2)*b + 2*sqrt(x**n*b + a*x**2)*a*x**2 + 3*int(s qrt(x**n*b + a*x**2)/x**2,x)*a*n*x**3 - 6*int(sqrt(x**n*b + a*x**2)/x**2,x )*a*x**3)/(3*c**4*x**3*(n - 2))