\(\int \frac {x^m}{\sqrt {2+3 x}} \, dx\) [719]

Optimal result

Integrand size = 13, antiderivative size = 31 \[ \int \frac {x^m}{\sqrt {2+3 x}} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,-\frac {3 x}{2}\right )}{\sqrt {2} (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {66} \[ \int \frac {x^m}{\sqrt {2+3 x}} \, dx=\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},m+1,m+2,-\frac {3 x}{2}\right )}{\sqrt {2} (m+1)} \]

[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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {x^m}{\sqrt {2+3 x}} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,-\frac {3 x}{2}\right )}{\sqrt {2} (1+m)} \]

[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))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94

method	result	size
meijerg	\(\frac {x^{1+m} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},1+m ;2+m ;-\frac {3 x}{2}\right ) \sqrt {2}}{2+2 m}\)	\(29\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {x^m}{\sqrt {2+3 x}} \, dx=\int { \frac {x^{m}}{\sqrt {3 \, x + 2}} \,d x } \]

[In]

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

[Out]

integral(x^m/sqrt(3*x + 2), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {x^m}{\sqrt {2+3 x}} \, dx=\frac {2 \cdot 2^{m} \sqrt {3} \cdot 3^{- m} \sqrt {x + \frac {2}{3}} e^{i \pi m} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - m \\ \frac {3}{2} \end {matrix}\middle | {\frac {3 x}{2} + 1} \right )}}{3} \]

[In]

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

[Out]

2*2**m*sqrt(3)*sqrt(x + 2/3)*exp(I*pi*m)*hyper((1/2, -m), (3/2,), 3*x/2 + 1)/(3*3**m)

Maxima [F]

\[ \int \frac {x^m}{\sqrt {2+3 x}} \, dx=\int { \frac {x^{m}}{\sqrt {3 \, x + 2}} \,d x } \]

[In]

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

[Out]

integrate(x^m/sqrt(3*x + 2), x)

Giac [F]

\[ \int \frac {x^m}{\sqrt {2+3 x}} \, dx=\int { \frac {x^{m}}{\sqrt {3 \, x + 2}} \,d x } \]

[In]

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

[Out]

integrate(x^m/sqrt(3*x + 2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m}{\sqrt {2+3 x}} \, dx=\int \frac {x^m}{\sqrt {3\,x+2}} \,d x \]

[In]

int(x^m/(3*x + 2)^(1/2),x)

[Out]

int(x^m/(3*x + 2)^(1/2), x)