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

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Rubi [A]  time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 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])))

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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \sqrt {x} x^{m} - a x^{m}}{b^{2} x - a^{2}}, x\right ) \]

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

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)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{b \sqrt {x}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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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sympy [C]  time = 1.11, size = 82, normalized size = 2.22 \[ \frac {4 m x x^{m} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a \Gamma \left (2 m + 3\right )} + \frac {4 x x^{m} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a \Gamma \left (2 m + 3\right )} \]

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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