3.2268 \(\int \frac {(a+b \sqrt {x})^p}{x} \, dx\)

Optimal. Leaf size=43 \[ -\frac {2 \left (a+b \sqrt {x}\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {\sqrt {x} b}{a}+1\right )}{a (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 43, 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 = {266, 65} \[ -\frac {2 \left (a+b \sqrt {x}\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {\sqrt {x} b}{a}+1\right )}{a (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^p/x,x]

[Out]

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

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])

Rule 266

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.00 \[ -\frac {2 \left (a+b \sqrt {x}\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {\sqrt {x} b}{a}+1\right )}{a (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^p/x,x]

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \sqrt {x} + a\right )}^{p}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sqrt(x) + a)^p/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sqrt {x} + a\right )}^{p}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sqrt(x) + a)^p/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sqrt {x} + a\right )}^{p}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sqrt(x) + a)^p/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+b\,\sqrt {x}\right )}^p}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/2))^p/x,x)

[Out]

int((a + b*x^(1/2))^p/x, x)

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sympy [C]  time = 2.04, size = 41, normalized size = 0.95 \[ - \frac {2 b^{p} x^{\frac {p}{2}} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b \sqrt {x}}} \right )}}{\Gamma \left (1 - p\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b**p*x**(p/2)*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*sqrt(x)))/gamma(1 - p)

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