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

Optimal. Leaf size=124 \[ \frac{(a+b x)^{n+1} (a d-b c n) \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a^2 c^2 (n+1)}+\frac{d^2 (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;-\frac{d (a+b x)}{b c-a d}\right )}{c^2 (n+1) (b c-a d)}-\frac{(a+b x)^{n+1}}{a c x} \]

[Out]

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

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Rubi [A]  time = 0.0677802, antiderivative size = 124, 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} \[ \frac{(a+b x)^{n+1} (a d-b c n) \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a^2 c^2 (n+1)}+\frac{d^2 (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;-\frac{d (a+b x)}{b c-a d}\right )}{c^2 (n+1) (b c-a d)}-\frac{(a+b x)^{n+1}}{a c x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^n/(x^2*(c + d*x)),x]

[Out]

-((a + b*x)^(1 + n)/(a*c*x)) + (d^2*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, -((d*(a + b*x))/(b*c
- a*d))])/(c^2*(b*c - a*d)*(1 + n)) + ((a*d - b*c*n)*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 +
(b*x)/a])/(a^2*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^2 (c+d x)} \, dx &=-\frac{(a+b x)^{1+n}}{a c x}-\frac{\int \frac{(a+b x)^n (a d-b c n-b d n x)}{x (c+d x)} \, dx}{a c}\\ &=-\frac{(a+b x)^{1+n}}{a c x}+\frac{d^2 \int \frac{(a+b x)^n}{c+d x} \, dx}{c^2}-\frac{(a d-b c n) \int \frac{(a+b x)^n}{x} \, dx}{a c^2}\\ &=-\frac{(a+b x)^{1+n}}{a c x}+\frac{d^2 (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) (1+n)}+\frac{(a d-b c n) (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{b x}{a}\right )}{a^2 c^2 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.0518596, size = 113, normalized size = 0.91 \[ -\frac{(a+b x)^{n+1} \left (a^2 d^2 x \, _2F_1\left (1,n+1;n+2;\frac{d (a+b x)}{a d-b c}\right )+(a d-b c) \left (\, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right ) (b c n x-a d x)+a c (n+1)\right )\right )}{a^2 c^2 (n+1) x (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{n}}{{x}^{2} \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*x+a)^n/x^2/(d*x+c),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 )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^n/((d*x + c)*x^2), 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 x^{3} + c x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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