3.3001 \(\int \frac {(d x)^m}{\sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}}} \, dx\)

Optimal. Leaf size=76 \[ -\frac {4 a c (d x)^m \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} \, _2F_1\left (\frac {1}{2},-2 m-1;\frac {3}{2};\frac {b}{a \sqrt {\frac {c}{x}}}+1\right )}{b^2} \]

[Out]

-4*a*c*(d*x)^m*hypergeom([1/2, -1-2*m],[3/2],1+b/a/(c/x)^(1/2))*(a+b/(c/x)^(1/2))^(1/2)/b^2/((-b/a/(c/x)^(1/2)
)^(2*m))

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Rubi [A]  time = 0.07, antiderivative size = 76, 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, 67, 65} \[ -\frac {4 a c (d x)^m \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} \, _2F_1\left (\frac {1}{2},-2 m-1;\frac {3}{2};\frac {b}{a \sqrt {\frac {c}{x}}}+1\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/Sqrt[a + b/Sqrt[c/x]],x]

[Out]

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

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-((b*c)/d))^IntPart[m]*(b*x)^FracPart[m])/
(-((d*x)/c))^FracPart[m], Int[(-((d*x)/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-(d/(b*c)), 0]

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 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}{\sqrt {\frac {c}{x}}}}} \, dx &=\operatorname {Subst}\left (\int \frac {(d x)^m}{\sqrt {a+\frac {b \sqrt {x}}{\sqrt {c}}}} \, 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 \sqrt {x}}{\sqrt {c}}}} \, 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}{\sqrt {c}}}} \, dx,x,\sqrt {x}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=-\operatorname {Subst}\left (\frac {\left (2 a \sqrt {c} \left (-\frac {b \sqrt {x}}{a \sqrt {c}}\right )^{-2 m} (d x)^m\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {b x}{a \sqrt {c}}\right )^{-1+2 (1+m)}}{\sqrt {a+\frac {b x}{\sqrt {c}}}} \, dx,x,\sqrt {x}\right )}{b},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=-\frac {4 a c \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} (d x)^m \, _2F_1\left (\frac {1}{2},-1-2 m;\frac {3}{2};1+\frac {b}{a \sqrt {\frac {c}{x}}}\right )}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 116, normalized size = 1.53 \[ \frac {a^2 c (d x)^m \left (\frac {a \sqrt {\frac {c}{x}}}{a \sqrt {\frac {c}{x}}+b}\right )^{2 m-\frac {1}{2}} \, _2F_1\left (2 m+2,2 m+\frac {5}{2};2 m+3;\frac {b}{\sqrt {\frac {c}{x}} a+b}\right )}{(m+1) \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} \left (a \sqrt {\frac {c}{x}}+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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)^(1/2))^(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 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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