\(\int \frac {\sqrt {\frac {a}{x}+b x^n}}{\sqrt {c x}} \, dx\) [370]

Optimal result

Integrand size = 23, antiderivative size = 84 \[ \int \frac {\sqrt {\frac {a}{x}+b x^n}}{\sqrt {c x}} \, dx=\frac {2 \sqrt {c x} \sqrt {\frac {a}{x}+b x^n}}{c (1+n)}-\frac {2 \sqrt {a} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {a}}{\sqrt {x} \sqrt {\frac {a}{x}+b x^n}}\right )}{(1+n) \sqrt {c x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2053, 2056, 2054, 212} \[ \int \frac {\sqrt {\frac {a}{x}+b x^n}}{\sqrt {c x}} \, dx=\frac {2 \sqrt {c x} \sqrt {\frac {a}{x}+b x^n}}{c (n+1)}-\frac {2 \sqrt {a} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {a}}{\sqrt {x} \sqrt {\frac {a}{x}+b x^n}}\right )}{(n+1) \sqrt {c x}} \]

[In]

Int[Sqrt[a/x + b*x^n]/Sqrt[c*x],x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {\frac {a}{x}+b x^n}}{\sqrt {c x}} \, dx=\frac {2 x \sqrt {\frac {a}{x}+b x^n} \left (\sqrt {a+b x^{1+n}}-\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^{1+n}}}{\sqrt {a}}\right )\right )}{(1+n) \sqrt {c x} \sqrt {a+b x^{1+n}}} \]

[In]

Integrate[Sqrt[a/x + b*x^n]/Sqrt[c*x],x]

[Out]

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

Maple [F]

\[\int \frac {\sqrt {\frac {a}{x}+b \,x^{n}}}{\sqrt {c x}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {\frac {a}{x}+b x^n}}{\sqrt {c x}} \, dx=\int \frac {\sqrt {\frac {a}{x} + b x^{n}}}{\sqrt {c x}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {\frac {a}{x}+b x^n}}{\sqrt {c x}} \, dx=\int { \frac {\sqrt {b x^{n} + \frac {a}{x}}}{\sqrt {c x}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\sqrt {\frac {a}{x}+b x^n}}{\sqrt {c x}} \, dx=\int { \frac {\sqrt {b x^{n} + \frac {a}{x}}}{\sqrt {c x}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\frac {a}{x}+b x^n}}{\sqrt {c x}} \, dx=\int \frac {\sqrt {b\,x^n+\frac {a}{x}}}{\sqrt {c\,x}} \,d x \]

[In]

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

[Out]

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