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

Optimal. Leaf size=139 \[ -\frac{\left (a+b \sqrt{c+d x}\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{a+b \sqrt{c+d x}}{a-b \sqrt{c}}\right )}{(p+1) \left (a-b \sqrt{c}\right )}-\frac{\left (a+b \sqrt{c+d x}\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{a+b \sqrt{c+d x}}{a+b \sqrt{c}}\right )}{(p+1) \left (a+b \sqrt{c}\right )} \]

[Out]

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

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Rubi [A]  time = 0.13054, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {371, 1398, 831, 68} \[ -\frac{\left (a+b \sqrt{c+d x}\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{a+b \sqrt{c+d x}}{a-b \sqrt{c}}\right )}{(p+1) \left (a-b \sqrt{c}\right )}-\frac{\left (a+b \sqrt{c+d x}\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{a+b \sqrt{c+d x}}{a+b \sqrt{c}}\right )}{(p+1) \left (a+b \sqrt{c}\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[c + d*x])^p/x,x]

[Out]

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

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rule 831

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.0995351, size = 136, normalized size = 0.98 \[ -\frac{\left (a+b \sqrt{c+d x}\right )^{p+1} \left (\left (a+b \sqrt{c}\right ) \, _2F_1\left (1,p+1;p+2;\frac{a+b \sqrt{c+d x}}{a-b \sqrt{c}}\right )+\left (a-b \sqrt{c}\right ) \, _2F_1\left (1,p+1;p+2;\frac{a+b \sqrt{c+d x}}{a+b \sqrt{c}}\right )\right )}{(p+1) \left (a-b \sqrt{c}\right ) \left (a+b \sqrt{c}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[c + d*x])^p/x,x]

[Out]

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

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Maple [F]  time = 0.005, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x} \left ( a+b\sqrt{dx+c} \right ) ^{p}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (\sqrt{d x + c} b + a\right )}^{p}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((sqrt(d*x + c)*b + a)^p/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((sqrt(d*x + c)*b + a)^p/x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sqrt{c + d x}\right )^{p}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sqrt(c + d*x))**p/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (\sqrt{d x + c} b + a\right )}^{p}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((sqrt(d*x + c)*b + a)^p/x, x)