\(\int \frac {x^{2+m}}{\sqrt {a+b x}} \, dx\) [713]

Optimal result

Integrand size = 15, antiderivative size = 51 \[ \int \frac {x^{2+m}}{\sqrt {a+b x}} \, dx=\frac {2 a^2 x^m \left (-\frac {b x}{a}\right )^{-m} \sqrt {a+b x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2-m,\frac {3}{2},1+\frac {b x}{a}\right )}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {69, 67} \[ \int \frac {x^{2+m}}{\sqrt {a+b x}} \, dx=\frac {2 a^2 x^m \sqrt {a+b x} \left (-\frac {b x}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m-2,\frac {3}{2},\frac {b x}{a}+1\right )}{b^3} \]

[In]

Int[x^(2 + m)/Sqrt[a + b*x],x]

[Out]

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

Rule 67

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

Rule 69

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]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {x^{2+m}}{\sqrt {a+b x}} \, dx=\frac {2 a^2 x^m \left (-\frac {b x}{a}\right )^{-m} \sqrt {a+b x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2-m,\frac {3}{2},1+\frac {b x}{a}\right )}{b^3} \]

[In]

Integrate[x^(2 + m)/Sqrt[a + b*x],x]

[Out]

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

Maple [F]

\[\int \frac {x^{2+m}}{\sqrt {b x +a}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(x^(m + 2)/sqrt(b*x + a), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71 \[ \int \frac {x^{2+m}}{\sqrt {a+b x}} \, dx=\frac {x^{m + 3} \Gamma \left (m + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m + 3 \\ m + 4 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{\sqrt {a} \Gamma \left (m + 4\right )} \]

[In]

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

[Out]

x**(m + 3)*gamma(m + 3)*hyper((1/2, m + 3), (m + 4,), b*x*exp_polar(I*pi)/a)/(sqrt(a)*gamma(m + 4))

Maxima [F]

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

[In]

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

[Out]

integrate(x^(m + 2)/sqrt(b*x + a), x)

Giac [F]

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

[In]

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

[Out]

integrate(x^(m + 2)/sqrt(b*x + a), x)

Mupad [F(-1)]

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

[In]

int(x^(m + 2)/(a + b*x)^(1/2),x)

[Out]

int(x^(m + 2)/(a + b*x)^(1/2), x)