3.4.8 \(\int \frac {1}{(c x)^{11/2} (\frac {a}{x^3}+b x^n)^{3/2}} \, dx\)

Optimal. Leaf size=90 \[ \frac {2}{a c^4 (n+3) (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}-\frac {2 \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {a}}{x^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}\right )}{a^{3/2} c^5 (n+3) \sqrt {c x}} \]

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Rubi [A]  time = 0.22, antiderivative size = 90, 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 = {2030, 2031, 2029, 206} \begin {gather*} \frac {2}{a c^4 (n+3) (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}-\frac {2 \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {a}}{x^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}\right )}{a^{3/2} c^5 (n+3) \sqrt {c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(11/2)*(a/x^3 + b*x^n)^(3/2)),x]

[Out]

2/(a*c^4*(3 + n)*(c*x)^(3/2)*Sqrt[a/x^3 + b*x^n]) - (2*Sqrt[x]*ArcTanh[Sqrt[a]/(x^(3/2)*Sqrt[a/x^3 + b*x^n])])
/(a^(3/2)*c^5*(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 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 2030

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

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 \frac {1}{(c x)^{11/2} \left (\frac {a}{x^3}+b x^n\right )^{3/2}} \, dx &=\frac {2}{a c^4 (3+n) (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}+\frac {\int \frac {1}{(c x)^{5/2} \sqrt {\frac {a}{x^3}+b x^n}} \, dx}{a c^3}\\ &=\frac {2}{a c^4 (3+n) (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}+\frac {\sqrt {x} \int \frac {1}{x^{5/2} \sqrt {\frac {a}{x^3}+b x^n}} \, dx}{a c^5 \sqrt {c x}}\\ &=\frac {2}{a c^4 (3+n) (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}-\frac {\left (2 \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 )}{a c^5 (3+n) \sqrt {c x}}\\ &=\frac {2}{a c^4 (3+n) (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}-\frac {2 \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {a}}{x^{3/2} \sqrt {\frac {a}{x^3}+b x^n}}\right )}{a^{3/2} c^5 (3+n) \sqrt {c x}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 55, normalized size = 0.61 \begin {gather*} \frac {2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b x^{n+3}}{a}+1\right )}{a c^4 (n+3) (c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(11/2)*(a/x^3 + b*x^n)^(3/2)),x]

[Out]

(2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*x^(3 + n))/a])/(a*c^4*(3 + n)*(c*x)^(3/2)*Sqrt[a/x^3 + b*x^n])

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IntegrateAlgebraic [A]  time = 1.64, size = 107, normalized size = 1.19 \begin {gather*} \frac {(c x)^{3/2} \sqrt {\frac {a}{x^3}+b x^n} \left (\frac {2}{a c^{11/2} (n+3) \sqrt {a+b x^{n+3}}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^{n+3}}}{\sqrt {a}}\right )}{a^{3/2} c^{11/2} (n+3)}\right )}{c^{3/2} \sqrt {a+b x^{n+3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((c*x)^(11/2)*(a/x^3 + b*x^n)^(3/2)),x]

[Out]

((c*x)^(3/2)*Sqrt[a/x^3 + b*x^n]*(2/(a*c^(11/2)*(3 + n)*Sqrt[a + b*x^(3 + n)]) - (2*ArcTanh[Sqrt[a + b*x^(3 +
n)]/Sqrt[a]])/(a^(3/2)*c^(11/2)*(3 + n))))/(c^(3/2)*Sqrt[a + b*x^(3 + n)])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(a/x^3+b*x^n)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{n} + \frac {a}{x^{3}}\right )}^{\frac {3}{2}} \left (c x\right )^{\frac {11}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(a/x^3+b*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^n + a/x^3)^(3/2)*(c*x)^(11/2)), x)

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maple [F]  time = 0.69, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (c x \right )^{\frac {11}{2}} \left (b \,x^{n}+\frac {a}{x^{3}}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(11/2)/(b*x^n+a/x^3)^(3/2),x)

[Out]

int(1/(c*x)^(11/2)/(b*x^n+a/x^3)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{n} + \frac {a}{x^{3}}\right )}^{\frac {3}{2}} \left (c x\right )^{\frac {11}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(a/x^3+b*x^n)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^n + a/x^3)^(3/2)*(c*x)^(11/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (c\,x\right )}^{11/2}\,{\left (b\,x^n+\frac {a}{x^3}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(11/2)*(b*x^n + a/x^3)^(3/2)),x)

[Out]

int(1/((c*x)^(11/2)*(b*x^n + a/x^3)^(3/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(11/2)/(a/x**3+b*x**n)**(3/2),x)

[Out]

Timed out

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