3.2261 \(\int \frac{x^m}{a+b \sqrt{x}} \, dx\)

Optimal. Leaf size=37 \[ \frac{x^{m+1} \, _2F_1\left (1,2 (m+1);2 m+3;-\frac{b \sqrt{x}}{a}\right )}{a (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.0148449, antiderivative size = 37, 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 = {341, 64} \[ \frac{x^{m+1} \, _2F_1\left (1,2 (m+1);2 m+3;-\frac{b \sqrt{x}}{a}\right )}{a (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*Sqrt[x]),x]

[Out]

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

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

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

Mathematica [A]  time = 0.0114614, size = 36, normalized size = 0.97 \[ \frac{x^{m+1} \, _2F_1\left (1,2 m+2;2 m+3;-\frac{b \sqrt{x}}{a}\right )}{a m+a} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/(a + b*Sqrt[x]),x]

[Out]

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

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m} \left ( a+b\sqrt{x} \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(b*sqrt(x) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sqrt(x)*x^m - a*x^m)/(b^2*x - a^2), x)

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Sympy [C]  time = 0.852701, size = 82, normalized size = 2.22 \begin{align*} \frac{4 m x x^{m} \Phi \left (\frac{b \sqrt{x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a \Gamma \left (2 m + 3\right )} + \frac{4 x x^{m} \Phi \left (\frac{b \sqrt{x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a \Gamma \left (2 m + 3\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(b*sqrt(x) + a), x)