\(\int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx\) [365]

Optimal result

Integrand size = 27, antiderivative size = 99 \[ \int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx=-\frac {2 (c x)^{-j/2} \sqrt {a x^j+b x^n}}{c (j-n)}+\frac {2 \sqrt {a} x^{j/2} (c x)^{-j/2} \text {arctanh}\left (\frac {\sqrt {a} x^{j/2}}{\sqrt {a x^j+b x^n}}\right )}{c (j-n)} \]

[Out]

2*x^(1/2*j)*arctanh(x^(1/2*j)*a^(1/2)/(a*x^j+b*x^n)^(1/2))*a^(1/2)/c/(j-n)/((c*x)^(1/2*j))-2*(a*x^j+b*x^n)^(1/
2)/c/(j-n)/((c*x)^(1/2*j))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2056, 2053, 2054, 212} \[ \int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx=\frac {2 \sqrt {a} x^{j/2} (c x)^{-j/2} \text {arctanh}\left (\frac {\sqrt {a} x^{j/2}}{\sqrt {a x^j+b x^n}}\right )}{c (j-n)}-\frac {2 (c x)^{-j/2} \sqrt {a x^j+b x^n}}{c (j-n)} \]

[In]

Int[(c*x)^(-1 - j/2)*Sqrt[a*x^j + b*x^n],x]

[Out]

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

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]

Rule 2056

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

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10 \[ \int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx=-\frac {2 (c x)^{-j/2} \left (a x^j+b x^n-\sqrt {a} \sqrt {b} x^{\frac {j+n}{2}} \sqrt {1+\frac {a x^{j-n}}{b}} \text {arcsinh}\left (\frac {\sqrt {a} x^{\frac {j-n}{2}}}{\sqrt {b}}\right )\right )}{c (j-n) \sqrt {a x^j+b x^n}} \]

[In]

Integrate[(c*x)^(-1 - j/2)*Sqrt[a*x^j + b*x^n],x]

[Out]

(-2*(a*x^j + b*x^n - Sqrt[a]*Sqrt[b]*x^((j + n)/2)*Sqrt[1 + (a*x^(j - n))/b]*ArcSinh[(Sqrt[a]*x^((j - n)/2))/S
qrt[b]]))/(c*(j - n)*(c*x)^(j/2)*Sqrt[a*x^j + b*x^n])

Maple [F]

\[\int \left (c x \right )^{-1-\frac {j}{2}} \sqrt {a \,x^{j}+b \,x^{n}}d x\]

[In]

int((c*x)^(-1-1/2*j)*(a*x^j+b*x^n)^(1/2),x)

[Out]

int((c*x)^(-1-1/2*j)*(a*x^j+b*x^n)^(1/2),x)

Fricas [F(-2)]

Exception generated. \[ \int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c*x)^(-1-1/2*j)*(a*x^j+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx=\int \left (c x\right )^{- \frac {j}{2} - 1} \sqrt {a x^{j} + b x^{n}}\, dx \]

[In]

integrate((c*x)**(-1-1/2*j)*(a*x**j+b*x**n)**(1/2),x)

[Out]

Integral((c*x)**(-j/2 - 1)*sqrt(a*x**j + b*x**n), x)

Maxima [F]

\[ \int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx=\int { \sqrt {a x^{j} + b x^{n}} \left (c x\right )^{-\frac {1}{2} \, j - 1} \,d x } \]

[In]

integrate((c*x)^(-1-1/2*j)*(a*x^j+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*x^j + b*x^n)*(c*x)^(-1/2*j - 1), x)

Giac [F]

\[ \int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx=\int { \sqrt {a x^{j} + b x^{n}} \left (c x\right )^{-\frac {1}{2} \, j - 1} \,d x } \]

[In]

integrate((c*x)^(-1-1/2*j)*(a*x^j+b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*x^j + b*x^n)*(c*x)^(-1/2*j - 1), x)

Mupad [F(-1)]

Timed out. \[ \int (c x)^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx=\int \frac {\sqrt {a\,x^j+b\,x^n}}{{\left (c\,x\right )}^{\frac {j}{2}+1}} \,d x \]

[In]

int((a*x^j + b*x^n)^(1/2)/(c*x)^(j/2 + 1),x)

[Out]

int((a*x^j + b*x^n)^(1/2)/(c*x)^(j/2 + 1), x)