3.2921 \(\int (d x)^m \sqrt{a+\frac{b}{\sqrt{c x}}} \, dx\)

Optimal. Leaf size=76 \[ \frac{4 b^2 (d x)^m \left (a+\frac{b}{\sqrt{c x}}\right )^{3/2} \left (-\frac{b}{a \sqrt{c x}}\right )^{2 m} \, _2F_1\left (\frac{3}{2},2 m+3;\frac{5}{2};\frac{b}{a \sqrt{c x}}+1\right )}{3 a^3 c} \]

[Out]

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

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Rubi [A]  time = 0.0928944, antiderivative size = 76, normalized size of antiderivative = 1., 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, 67, 65} \[ \frac{4 b^2 (d x)^m \left (a+\frac{b}{\sqrt{c x}}\right )^{3/2} \left (-\frac{b}{a \sqrt{c x}}\right )^{2 m} \, _2F_1\left (\frac{3}{2},2 m+3;\frac{5}{2};\frac{b}{a \sqrt{c x}}+1\right )}{3 a^3 c} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*Sqrt[a + b/Sqrt[c*x]],x]

[Out]

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

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]

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 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 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 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 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])

Rubi steps

\begin{align*} \int (d x)^m \sqrt{a+\frac{b}{\sqrt{c x}}} \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{a+\frac{b}{\sqrt{x}}} \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}{\sqrt{x}}} 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}} 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} \, dx,x,\frac{1}{\sqrt{c x}}\right )}{c}\\ &=\frac{\left (2 b^3 (d x)^m \left (-\frac{b}{a \sqrt{c x}}\right )^{2 m}\right ) \operatorname{Subst}\left (\int \left (-\frac{b x}{a}\right )^{-1-2 (1+m)} \sqrt{a+b x} \, dx,x,\frac{1}{\sqrt{c x}}\right )}{a^3 c}\\ &=\frac{4 b^2 (d x)^m \left (-\frac{b}{a \sqrt{c x}}\right )^{2 m} \left (a+\frac{b}{\sqrt{c x}}\right )^{3/2} \, _2F_1\left (\frac{3}{2},3+2 m;\frac{5}{2};1+\frac{b}{a \sqrt{c x}}\right )}{3 a^3 c}\\ \end{align*}

Mathematica [A]  time = 0.282285, size = 135, normalized size = 1.78 \[ \frac{4 (d x)^m \sqrt{a+\frac{b}{\sqrt{c x}}} \left (a \sqrt{c x}+b\right ) \left (-\frac{a \sqrt{c x}}{b}\right )^{\frac{1}{2}-2 m} \left (3 \left (a \sqrt{c x}+b\right ) \, _2F_1\left (\frac{5}{2},\frac{1}{2}-2 m;\frac{7}{2};\frac{\sqrt{c x} a}{b}+1\right )-5 b \, _2F_1\left (\frac{3}{2},\frac{1}{2}-2 m;\frac{5}{2};\frac{\sqrt{c x} a}{b}+1\right )\right )}{15 a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(d*x)^m*(-((a*Sqrt[c*x])/b))^(1/2 - 2*m)*Sqrt[a + b/Sqrt[c*x]]*(b + a*Sqrt[c*x])*(-5*b*Hypergeometric2F1[3/
2, 1/2 - 2*m, 5/2, 1 + (a*Sqrt[c*x])/b] + 3*(b + a*Sqrt[c*x])*Hypergeometric2F1[5/2, 1/2 - 2*m, 7/2, 1 + (a*Sq
rt[c*x])/b]))/(15*a^2*c)

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m}\sqrt{a+{b{\frac{1}{\sqrt{cx}}}}}\, dx \end{align*}

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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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \sqrt{a + \frac{b}{\sqrt{c x}}}\,{d x} \end{align*}

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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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \sqrt{a + \frac{b}{\sqrt{c x}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \sqrt{a + \frac{b}{\sqrt{c x}}}\,{d x} \end{align*}

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]

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