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

Optimal. Leaf size=51 \[ \frac{2 a^2 x^m \sqrt{a+b x} \left (-\frac{b x}{a}\right )^{-m} \, _2F_1\left (\frac{1}{2},-m-2;\frac{3}{2};\frac{b x}{a}+1\right )}{b^3} \]

[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)

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Rubi [A]  time = 0.0118605, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {67, 65} \[ \frac{2 a^2 x^m \sqrt{a+b x} \left (-\frac{b x}{a}\right )^{-m} \, _2F_1\left (\frac{1}{2},-m-2;\frac{3}{2};\frac{b x}{a}+1\right )}{b^3} \]

Antiderivative was successfully verified.

[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] :> 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 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])

Rubi steps

\begin{align*} \int \frac{x^{2+m}}{\sqrt{a+b x}} \, dx &=\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]  time = 0.0358965, size = 51, normalized size = 1. \[ \frac{2 a^2 x^m \sqrt{a+b x} \left (-\frac{b x}{a}\right )^{-m} \, _2F_1\left (\frac{1}{2},-m-2;\frac{3}{2};\frac{b x}{a}+1\right )}{b^3} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.022, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2+m}{\frac{1}{\sqrt{bx+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m + 2}}{\sqrt{b x + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m + 2)/sqrt(b*x + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m + 2}}{\sqrt{b x + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^(m + 2)/sqrt(b*x + a), x)

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Sympy [C]  time = 4.70403, size = 37, normalized size = 0.73 \begin{align*} \frac{x^{3} x^{m} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m + 2}}{\sqrt{b x + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m + 2)/sqrt(b*x + a), x)