3.593 \(\int \frac {(a+\frac {b}{x})^m}{(c+d x)^3} \, dx\)

Optimal. Leaf size=112 \[ -\frac {b \left (a+\frac {b}{x}\right )^{m+1} (2 a c-b d (m+1)) \, _2F_1\left (2,m+1;m+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{2 c (m+1) (a c-b d)^3}-\frac {d \left (a+\frac {b}{x}\right )^{m+1}}{2 c \left (\frac {c}{x}+d\right )^2 (a c-b d)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {434, 446, 78, 68} \[ -\frac {b \left (a+\frac {b}{x}\right )^{m+1} (2 a c-b d (m+1)) \, _2F_1\left (2,m+1;m+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{2 c (m+1) (a c-b d)^3}-\frac {d \left (a+\frac {b}{x}\right )^{m+1}}{2 c \left (\frac {c}{x}+d\right )^2 (a c-b d)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^m/(c + d*x)^3,x]

[Out]

-(d*(a + b/x)^(1 + m))/(2*c*(a*c - b*d)*(d + c/x)^2) - (b*(2*a*c - b*d*(1 + m))*(a + b/x)^(1 + m)*Hypergeometr
ic2F1[2, 1 + m, 2 + m, (c*(a + b/x))/(a*c - b*d)])/(2*c*(a*c - b*d)^3*(1 + 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]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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)^3} \, dx &=\int \frac {\left (a+\frac {b}{x}\right )^m}{\left (d+\frac {c}{x}\right )^3 x^3} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {x (a+b x)^m}{(d+c x)^3} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {d \left (a+\frac {b}{x}\right )^{1+m}}{2 c (a c-b d) \left (d+\frac {c}{x}\right )^2}-\frac {(2 a c-b d (1+m)) \operatorname {Subst}\left (\int \frac {(a+b x)^m}{(d+c x)^2} \, dx,x,\frac {1}{x}\right )}{2 c (a c-b d)}\\ &=-\frac {d \left (a+\frac {b}{x}\right )^{1+m}}{2 c (a c-b d) \left (d+\frac {c}{x}\right )^2}-\frac {b (2 a c-b d (1+m)) \left (a+\frac {b}{x}\right )^{1+m} \, _2F_1\left (2,1+m;2+m;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{2 c (a c-b d)^3 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 99, normalized size = 0.88 \[ \frac {\left (a+\frac {b}{x}\right )^{m+1} \left (\frac {b (b d (m+1)-2 a c) \, _2F_1\left (2,m+1;m+2;\frac {b c+a x c}{a c x-b d x}\right )}{(m+1) (a c-b d)^2}-\frac {d x^2}{(c+d x)^2}\right )}{2 c (a c-b d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^m/(c + d*x)^3,x]

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x + b}{x}\right )^{m}}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m/(d*x+c)^3,x, algorithm="fricas")

[Out]

integral(((a*x + b)/x)^m/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m/(d*x+c)^3,x, algorithm="giac")

[Out]

integrate((a + b/x)^m/(d*x + c)^3, x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +\frac {b}{x}\right )^{m}}{\left (d x +c \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^m/(d*x+c)^3,x)

[Out]

int((a+b/x)^m/(d*x+c)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m/(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate((a + b/x)^m/(d*x + c)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{x}\right )}^m}{{\left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^m/(c + d*x)^3,x)

[Out]

int((a + b/x)^m/(c + d*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + \frac {b}{x}\right )^{m}}{\left (c + d x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**m/(d*x+c)**3,x)

[Out]

Integral((a + b/x)**m/(c + d*x)**3, x)

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