3.3.94 \(\int \frac {(a+\frac {b}{x})^n x}{(c+d x)^2} \, dx\) [294]

Optimal. Leaf size=150 \[ -\frac {c \left (a+\frac {b}{x}\right )^{1+n}}{d (a c-b d) \left (d+\frac {c}{x}\right )}-\frac {c (a c-b d (1-n)) \left (a+\frac {b}{x}\right )^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{d^2 (a c-b d)^2 (1+n)}+\frac {\left (a+\frac {b}{x}\right )^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {b}{a x}\right )}{a d^2 (1+n)} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {528, 457, 105, 162, 67, 70} \begin {gather*} -\frac {c \left (a+\frac {b}{x}\right )^{n+1} (a c-b d (1-n)) \, _2F_1\left (1,n+1;n+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{d^2 (n+1) (a c-b d)^2}-\frac {c \left (a+\frac {b}{x}\right )^{n+1}}{d \left (\frac {c}{x}+d\right ) (a c-b d)}+\frac {\left (a+\frac {b}{x}\right )^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b}{a x}+1\right )}{a d^2 (n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b/x)^n*x)/(c + d*x)^2,x]

[Out]

-((c*(a + b/x)^(1 + n))/(d*(a*c - b*d)*(d + c/x))) - (c*(a*c - b*d*(1 - n))*(a + b/x)^(1 + n)*Hypergeometric2F
1[1, 1 + n, 2 + n, (c*(a + b/x))/(a*c - b*d)])/(d^2*(a*c - b*d)^2*(1 + n)) + ((a + b/x)^(1 + n)*Hypergeometric
2F1[1, 1 + n, 2 + n, 1 + b/(a*x)])/(a*d^2*(1 + n))

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 162

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

Rule 457

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]]

Rule 528

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

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 120, normalized size = 0.80 \begin {gather*} \frac {\left (a+\frac {b}{x}\right )^{1+n} \left (-\frac {c d x}{(a c-b d) (c+d x)}-\frac {c (a c+b d (-1+n)) \, _2F_1\left (1,1+n;2+n;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{(a c-b d)^2 (1+n)}+\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b}{a x}\right )}{a (1+n)}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b/x)^n*x)/(c + d*x)^2,x]

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (a +\frac {b}{x}\right )^{n} x}{\left (d x +c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^n*x/(d*x + c)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (a + \frac {b}{x}\right )^{n}}{\left (c + d x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b/x)**n/(c + d*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^n*x/(d*x + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (a+\frac {b}{x}\right )}^n}{{\left (c+d\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b/x)^n)/(c + d*x)^2,x)

[Out]

int((x*(a + b/x)^n)/(c + d*x)^2, x)

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