3.10.47 \(\int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx\) [947]

Optimal. Leaf size=190 \[ -\frac {d (b c-2 a d) (a+b x)^{1+n}}{a c^2 (b c-a d) (c+d x)}-\frac {(a+b x)^{1+n}}{a c x (c+d x)}-\frac {d^2 (2 a d-b c (2-n)) (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;-\frac {d (a+b x)}{b c-a d}\right )}{c^3 (b c-a d)^2 (1+n)}+\frac {(2 a d-b c n) (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )}{a^2 c^3 (1+n)} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {105, 156, 162, 67, 70} \begin {gather*} \frac {(a+b x)^{n+1} (2 a d-b c n) \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a^2 c^3 (n+1)}-\frac {d^2 (a+b x)^{n+1} (2 a d-b c (2-n)) \, _2F_1\left (1,n+1;n+2;-\frac {d (a+b x)}{b c-a d}\right )}{c^3 (n+1) (b c-a d)^2}-\frac {d (b c-2 a d) (a+b x)^{n+1}}{a c^2 (c+d x) (b c-a d)}-\frac {(a+b x)^{n+1}}{a c x (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^n/(x^2*(c + d*x)^2),x]

[Out]

-((d*(b*c - 2*a*d)*(a + b*x)^(1 + n))/(a*c^2*(b*c - a*d)*(c + d*x))) - (a + b*x)^(1 + n)/(a*c*x*(c + d*x)) - (
d^2*(2*a*d - b*c*(2 - n))*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, -((d*(a + b*x))/(b*c - a*d))])/
(c^3*(b*c - a*d)^2*(1 + n)) + ((2*a*d - b*c*n)*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/
a])/(a^2*c^3*(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 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

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

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Mathematica [A]
time = 0.16, size = 176, normalized size = 0.93 \begin {gather*} -\frac {(a+b x)^{1+n} \left (a c^2 (b c-a d)^2 (1+n)+a c d (-b c+a d) (-b c+2 a d) (1+n) x-x (c+d x) \left (-a^2 d^2 (2 a d+b c (-2+n)) \, _2F_1\left (1,1+n;2+n;\frac {d (a+b x)}{-b c+a d}\right )+(b c-a d)^2 (2 a d-b c n) \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )\right )\right )}{a^2 c^3 (b c-a d)^2 (1+n) x (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b*x)^(1 + n)*(a*c^2*(b*c - a*d)^2*(1 + n) + a*c*d*(-(b*c) + a*d)*(-(b*c) + 2*a*d)*(1 + n)*x - x*(c + d
*x)*(-(a^2*d^2*(2*a*d + b*c*(-2 + n))*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)]) + (b*c
- a*d)^2*(2*a*d - b*c*n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])))/(a^2*c^3*(b*c - a*d)^2*(1 + n)*x*
(c + d*x)))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{n}}{x^{2} \left (d x +c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((b*x + a)^n/(d^2*x^4 + 2*c*d*x^3 + c^2*x^2), 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^{2} \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**2/(d*x+c)**2,x)

[Out]

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

[Out]

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

[Out]

int((a + b*x)^n/(x^2*(c + d*x)^2), x)

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