3.372 \(\int \sqrt {c x} \sqrt {\frac {a}{x^3}+b x^n} \, dx\)

Optimal. Leaf size=85 \[ \frac {2 (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}{c (n+3)}-\frac {2 \sqrt {a} c \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {a}}{x^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}\right )}{(n+3) \sqrt {c x}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2028, 2031, 2029, 206} \[ \frac {2 (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}{c (n+3)}-\frac {2 \sqrt {a} c \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {a}}{x^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}\right )}{(n+3) \sqrt {c x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]*Sqrt[a/x^3 + b*x^n],x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2028

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 2029

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 2031

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

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Mathematica [A]  time = 0.07, size = 85, normalized size = 1.00 \[ \frac {x \sqrt {c x} \sqrt {\frac {a}{x^3}+b x^n} \left (2 \sqrt {a+b x^{n+3}}-2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^{n+3}}}{\sqrt {a}}\right )\right )}{(n+3) \sqrt {a+b x^{n+3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*x]*Sqrt[a/x^3 + b*x^n],x]

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)*(a/x^3+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{n} + \frac {a}{x^{3}}} \sqrt {c x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)*(a/x^3+b*x^n)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \sqrt {c x}\, \sqrt {b \,x^{n}+\frac {a}{x^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)*(a/x^3+b*x^n)^(1/2),x)

[Out]

int((c*x)^(1/2)*(a/x^3+b*x^n)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{n} + \frac {a}{x^{3}}} \sqrt {c x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)*(a/x^3+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {c\,x}\,\sqrt {b\,x^n+\frac {a}{x^3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)*(b*x^n + a/x^3)^(1/2),x)

[Out]

int((c*x)^(1/2)*(b*x^n + a/x^3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x} \sqrt {\frac {a}{x^{3}} + b x^{n}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(1/2)*(a/x**3+b*x**n)**(1/2),x)

[Out]

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

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