3.2922 \(\int (d x)^m \sqrt {a+\frac {b}{(c x)^{3/2}}} \, dx\)

Optimal. Leaf size=78 \[ \frac {x (d x)^m \sqrt {a+\frac {b}{(c x)^{3/2}}} \, _2F_1\left (-\frac {1}{2},-\frac {2}{3} (m+1);\frac {1}{3} (1-2 m);-\frac {b}{a (c x)^{3/2}}\right )}{(m+1) \sqrt {\frac {b}{a (c x)^{3/2}}+1}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {367, 343, 341, 339, 365, 364} \[ \frac {x (d x)^m \sqrt {a+\frac {b}{(c x)^{3/2}}} \, _2F_1\left (-\frac {1}{2},-\frac {2}{3} (m+1);\frac {1}{3} (1-2 m);-\frac {b}{a (c x)^{3/2}}\right )}{(m+1) \sqrt {\frac {b}{a (c x)^{3/2}}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, Subst
[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]

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 367

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_)*(x_))^(n_))^(p_.), x_Symbol] :> Dist[1/c, Subst[Int[((d*x)/c)^m*(a
+ b*x^n)^p, x], x, c*x], x] /; FreeQ[{a, b, c, d, m, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 84, normalized size = 1.08 \[ \frac {4 x (d x)^m \sqrt {a+\frac {b}{(c x)^{3/2}}} \, _2F_1\left (-\frac {1}{2},\frac {1}{6} (4 m+1);\frac {1}{6} (4 m+7);-\frac {a (c x)^{3/2}}{b}\right )}{(4 m+1) \sqrt {\frac {a (c x)^{3/2}+b}{b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x*(d*x)^m*Sqrt[a + b/(c*x)^(3/2)]*Hypergeometric2F1[-1/2, (1 + 4*m)/6, (7 + 4*m)/6, -((a*(c*x)^(3/2))/b)])/
((1 + 4*m)*Sqrt[(b + a*(c*x)^(3/2))/b])

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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 \[ \int \left (d x\right )^{m} \sqrt {a + \frac {b}{\left (c x\right )^{\frac {3}{2}}}}\,{d x} \]

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 = 0.24, size = 0, normalized size = 0.00 \[ \int \sqrt {a +\frac {b}{\left (c x \right )^{\frac {3}{2}}}}\, \left (d x \right )^{m}\, dx \]

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

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 \[ \int \sqrt {a+\frac {b}{{\left (c\,x\right )}^{3/2}}}\,{\left (d\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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