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]

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

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Rubi [A]  time = 0.09, antiderivative size = 76, 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, 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 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 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 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}{\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*}

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Mathematica [A]  time = 0.34, 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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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(-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="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):
Check [abs(t_nostep)]Unable to divide, perhaps due to rounding error%%%{1,[0,1,1,0]%%%} / %%%{1,[0,0,0,1]%%%}
Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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