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]

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

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Rubi [A]  time = 0.13, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (\sqrt {d x + c} b + a\right )}^{p}}{x}, x\right ) \]

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

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\sqrt {c+d\,x}\right )}^p}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c + d*x)^(1/2))^p/x,x)

[Out]

int((a + b*(c + d*x)^(1/2))^p/x, x)

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

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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