3.719 \(\int \frac {x^m}{\sqrt {2+3 x}} \, dx\)

Optimal. Leaf size=31 \[ \frac {x^{m+1} \, _2F_1\left (\frac {1}{2},m+1;m+2;-\frac {3 x}{2}\right )}{\sqrt {2} (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {64} \[ \frac {x^{m+1} \, _2F_1\left (\frac {1}{2},m+1;m+2;-\frac {3 x}{2}\right )}{\sqrt {2} (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/Sqrt[2 + 3*x],x]

[Out]

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

Rule 64

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.00 \[ \frac {x^{m+1} \, _2F_1\left (\frac {1}{2},m+1;m+2;-\frac {3 x}{2}\right )}{\sqrt {2} (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/Sqrt[2 + 3*x],x]

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{m}}{\sqrt {3 \, x + 2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(2+3*x)^(1/2),x, algorithm="fricas")

[Out]

integral(x^m/sqrt(3*x + 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\sqrt {3 \, x + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(2+3*x)^(1/2),x, algorithm="giac")

[Out]

integrate(x^m/sqrt(3*x + 2), x)

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maple [A]  time = 0.05, size = 29, normalized size = 0.94 \[ \frac {\sqrt {2}\, x^{m +1} \hypergeom \left (\left [\frac {1}{2}, m +1\right ], \left [m +2\right ], -\frac {3 x}{2}\right )}{2 m +2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(3*x+2)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\sqrt {3 \, x + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(2+3*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^m/sqrt(3*x + 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^m}{\sqrt {3\,x+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(3*x + 2)^(1/2),x)

[Out]

int(x^m/(3*x + 2)^(1/2), x)

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sympy [C]  time = 1.20, size = 37, normalized size = 1.19 \[ \frac {\sqrt {2} x x^{m} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {3 x e^{i \pi }}{2}} \right )}}{2 \Gamma \left (m + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(2+3*x)**(1/2),x)

[Out]

sqrt(2)*x*x**m*gamma(m + 1)*hyper((1/2, m + 1), (m + 2,), 3*x*exp_polar(I*pi)/2)/(2*gamma(m + 2))

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