3.946 \(\int \frac{(a+b x)^n}{x (c+d x)^2} \, dx\)

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

[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))

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Rubi [A]  time = 0.0814086, antiderivative size = 139, normalized size of antiderivative = 1., 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 = {103, 156, 65, 68} \[ \frac{d (a+b x)^{n+1} (a d-b c (1-n)) \, _2F_1\left (1,n+1;n+2;-\frac{d (a+b x)}{b c-a d}\right )}{c^2 (n+1) (b c-a d)^2}-\frac{(a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a c^2 (n+1)}-\frac{d (a+b x)^{n+1}}{c (c+d x) (b c-a d)} \]

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 103

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] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

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 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]

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*}

Mathematica [A]  time = 0.129849, size = 111, normalized size = 0.8 \[ \frac{(a+b x)^{n+1} \left (\frac{d \left (\frac{(a d+b c (n-1)) \, _2F_1\left (1,n+1;n+2;\frac{d (a+b x)}{a d-b c}\right )}{n+1}+\frac{c (a d-b c)}{c+d x}\right )}{(b c-a d)^2}-\frac{\, _2F_1\left (1,n+1;n+2;\frac{a+b x}{a}\right )}{a n+a}\right )}{c^2} \]

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 = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{n}}{x \left ( dx+c \right ) ^{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{2} x}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{n}}{d^{2} x^{3} + 2 \, c d x^{2} + c^{2} x}, x\right ) \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

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)