3.10.46 \(\int \frac {(a+b x)^n}{x (c+d x)^2} \, dx\) [946]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {105, 162, 67, 70} \begin {gather*} \frac {d (a+b x)^{n+1} (a d-b c (1-n)) \, _2F_1\left (1,n+1;n+2;-\frac {d (a+b x)}{b c-a d}\right )}{c^2 (n+1) (b c-a d)^2}-\frac {(a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a c^2 (n+1)}-\frac {d (a+b x)^{n+1}}{c (c+d x) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^n/(x*(c + d*x)^2),x]

[Out]

-((d*(a + b*x)^(1 + n))/(c*(b*c - a*d)*(c + d*x))) + (d*(a*d - b*c*(1 - n))*(a + b*x)^(1 + n)*Hypergeometric2F
1[1, 1 + n, 2 + n, -((d*(a + b*x))/(b*c - a*d))])/(c^2*(b*c - a*d)^2*(1 + n)) - ((a + b*x)^(1 + n)*Hypergeomet
ric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*c^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]

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 111, normalized size = 0.80 \begin {gather*} \frac {(a+b x)^{1+n} \left (-\frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b x}{a}\right )}{a+a n}+\frac {d \left (\frac {c (-b c+a d)}{c+d x}+\frac {(a d+b c (-1+n)) \, _2F_1\left (1,1+n;2+n;\frac {d (a+b x)}{-b c+a d}\right )}{1+n}\right )}{(b c-a d)^2}\right )}{c^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)*(-(Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*x)/a]/(a + a*n)) + (d*((c*(-(b*c) + a*d))/(c +
d*x) + ((a*d + b*c*(-1 + n))*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)])/(1 + n)))/(b*c
- a*d)^2))/c^2

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{n}}{x \left (d x +c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^n/(x*(c + d*x)^2),x)

[Out]

int((a + b*x)^n/(x*(c + d*x)^2), x)

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