3.10.41 \(\int \frac {(a+b x)^n}{x^2 (c+d x)} \, dx\) [941]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 124, 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 {(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} \end {gather*}

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 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^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*}

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Mathematica [A]
time = 0.09, size = 113, normalized size = 0.91 \begin {gather*} -\frac {(a+b x)^{1+n} \left (a^2 d^2 x \, _2F_1\left (1,1+n;2+n;\frac {d (a+b x)}{-b c+a d}\right )+(-b c+a d) \left (a c (1+n)+(-a d x+b c n x) \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )\right )\right )}{a^2 c^2 (-b c+a d) (1+n) x} \end {gather*}

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

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 = 0.00, 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),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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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),x, algorithm="giac")

[Out]

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

[Out]

int((a + b*x)^n/(x^2*(c + d*x)), x)

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