3.2957 \(\int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x^2}}} \, dx\)

Optimal. Leaf size=87 \[ -\frac {2 b (d x)^{m+1} \left (a+\frac {b}{\sqrt {c x^2}}\right )^{3/2} \left (-\frac {b}{a \sqrt {c x^2}}\right )^m \, _2F_1\left (\frac {3}{2},m+2;\frac {5}{2};\frac {b}{a \sqrt {c x^2}}+1\right )}{3 a^2 d \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {368, 339, 67, 65} \[ -\frac {2 b (d x)^{m+1} \left (a+\frac {b}{\sqrt {c x^2}}\right )^{3/2} \left (-\frac {b}{a \sqrt {c x^2}}\right )^m \, _2F_1\left (\frac {3}{2},m+2;\frac {5}{2};\frac {b}{a \sqrt {c x^2}}+1\right )}{3 a^2 d \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 368

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 81, normalized size = 0.93 \[ \frac {2 x (d x)^m \sqrt {a+\frac {b}{\sqrt {c x^2}}} \, _2F_1\left (-\frac {1}{2},m+\frac {1}{2};m+\frac {3}{2};-\frac {a \sqrt {c x^2}}{b}\right )}{(2 m+1) \sqrt {\frac {a \sqrt {c x^2}}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.21, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d x\right )^{m} \sqrt {\frac {a c x^{2} + \sqrt {c x^{2}} b}{c x^{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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