3.3002 \(\int \frac {(d x)^m}{\sqrt {a+\frac {b}{(\frac {c}{x})^{3/2}}}} \, dx\)

Optimal. Leaf size=102 \[ \frac {x (d x)^m \sqrt {\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{a c^3}+1} \, _2F_1\left (\frac {1}{2},\frac {2 (m+1)}{3};\frac {1}{3} (2 m+5);-\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a c^3}\right )}{(m+1) \sqrt {a+\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{c^3}}} \]

[Out]

x*(d*x)^m*hypergeom([1/2, 2/3+2/3*m],[5/3+2/3*m],-b*(c/x)^(3/2)*x^3/a/c^3)*(1+b*(c/x)^(3/2)*x^3/a/c^3)^(1/2)/(
1+m)/(a+b*(c/x)^(3/2)*x^3/c^3)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {369, 343, 341, 365, 364} \[ \frac {x (d x)^m \sqrt {\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{a c^3}+1} \, _2F_1\left (\frac {1}{2},\frac {2 (m+1)}{3};\frac {1}{3} (2 m+5);-\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a c^3}\right )}{(m+1) \sqrt {a+\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{c^3}}} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/Sqrt[a + b/(c/x)^(3/2)],x]

[Out]

(x*(d*x)^m*Sqrt[1 + (b*(c/x)^(3/2)*x^3)/(a*c^3)]*Hypergeometric2F1[1/2, (2*(1 + m))/3, (5 + 2*m)/3, -((b*(c/x)
^(3/2)*x^3)/(a*c^3))])/((1 + m)*Sqrt[a + (b*(c/x)^(3/2)*x^3)/c^3])

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

Rule 343

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracPart[m])/x^FracP
art[m], Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && FractionQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b
, c, d, m, p, q}, x] && FractionQ[n]

Rubi steps

\begin {align*} \int \frac {(d x)^m}{\sqrt {a+\frac {b}{\left (\frac {c}{x}\right )^{3/2}}}} \, dx &=\operatorname {Subst}\left (\int \frac {(d x)^m}{\sqrt {a+\frac {b x^{3/2}}{c^{3/2}}}} \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\operatorname {Subst}\left (\left (x^{-m} (d x)^m\right ) \int \frac {x^m}{\sqrt {a+\frac {b x^{3/2}}{c^{3/2}}}} \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\operatorname {Subst}\left (\left (2 x^{-m} (d x)^m\right ) \operatorname {Subst}\left (\int \frac {x^{-1+2 (1+m)}}{\sqrt {a+\frac {b x^3}{c^{3/2}}}} \, dx,x,\sqrt {x}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\operatorname {Subst}\left (\frac {\left (2 x^{-m} (d x)^m \sqrt {1+\frac {b x^{3/2}}{a c^{3/2}}}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+2 (1+m)}}{\sqrt {1+\frac {b x^3}{a c^{3/2}}}} \, dx,x,\sqrt {x}\right )}{\sqrt {a+\frac {b x^{3/2}}{c^{3/2}}}},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\frac {x (d x)^m \sqrt {1+\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a c^3}} \, _2F_1\left (\frac {1}{2},\frac {2 (1+m)}{3};\frac {1}{3} (5+2 m);-\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a c^3}\right )}{(1+m) \sqrt {a+\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{c^3}}}\\ \end {align*}

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Mathematica [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {(d x)^m}{\sqrt {a+\frac {b}{\left (\frac {c}{x}\right )^{3/2}}}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(d*x)^m/Sqrt[a + b/(c/x)^(3/2)],x]

[Out]

Integrate[(d*x)^m/Sqrt[a + b/(c/x)^(3/2)], x]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   alglogextint: unimplemented

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{\sqrt {a + \frac {b}{\left (\frac {c}{x}\right )^{\frac {3}{2}}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^m/sqrt(a + b/(c/x)^(3/2)), x)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x \right )^{m}}{\sqrt {a +\frac {b}{\left (\frac {c}{x}\right )^{\frac {3}{2}}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{\sqrt {a + \frac {b}{\left (\frac {c}{x}\right )^{\frac {3}{2}}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^m/sqrt(a + b/(c/x)^(3/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{\sqrt {a + \frac {b}{\left (\frac {c}{x}\right )^{\frac {3}{2}}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*x)**m/sqrt(a + b/(c/x)**(3/2)), x)

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