3.591 \(\int \frac {(a+\frac {b}{x})^m}{c+d x} \, dx\)

Optimal. Leaf size=101 \[ \frac {\left (a+\frac {b}{x}\right )^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {b}{a x}+1\right )}{a d (m+1)}-\frac {c \left (a+\frac {b}{x}\right )^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{d (m+1) (a c-b d)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {434, 446, 86, 65, 68} \[ \frac {\left (a+\frac {b}{x}\right )^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {b}{a x}+1\right )}{a d (m+1)}-\frac {c \left (a+\frac {b}{x}\right )^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{d (m+1) (a c-b d)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^m/(c + d*x),x]

[Out]

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

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 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 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 434

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[((a + b*x^n)^p*(d + c*x
^n)^q)/x^(n*q), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 97, normalized size = 0.96 \[ \frac {(a x+b) \left (a+\frac {b}{x}\right )^m \left (a c \, _2F_1\left (1,m+1;m+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )+(b d-a c) \, _2F_1\left (1,m+1;m+2;\frac {b}{a x}+1\right )\right )}{a d (m+1) x (b d-a c)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^m/(c + d*x),x]

[Out]

((a + b/x)^m*(b + a*x)*(a*c*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(a + b/x))/(a*c - b*d)] + (-(a*c) + b*d)*Hyp
ergeometric2F1[1, 1 + m, 2 + m, 1 + b/(a*x)]))/(a*d*(-(a*c) + b*d)*(1 + m)*x)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x + b}{x}\right )^{m}}{d x + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^m/(d*x + c), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +\frac {b}{x}\right )^{m}}{d x +c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^m/(d*x+c),x)

[Out]

int((a+b/x)^m/(d*x+c),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^m/(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^m/(c + d*x),x)

[Out]

int((a + b/x)^m/(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**m/(d*x+c),x)

[Out]

Integral((a + b/x)**m/(c + d*x), x)

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