3.10.62 \(\int \frac {(a+b x)^n (c+d x)^p}{x^2} \, dx\) [962]

Optimal. Leaf size=85 \[ \frac {b (a+b x)^{1+n} (c+d x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-p} F_1\left (1+n;-p,2;2+n;-\frac {d (a+b x)}{b c-a d},\frac {a+b x}{a}\right )}{a^2 (1+n)} \]

[Out]

b*(b*x+a)^(1+n)*(d*x+c)^p*AppellF1(1+n,2,-p,2+n,(b*x+a)/a,-d*(b*x+a)/(-a*d+b*c))/a^2/(1+n)/((b*(d*x+c)/(-a*d+b
*c))^p)

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Rubi [A]
time = 0.03, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {142, 141} \begin {gather*} \frac {b (a+b x)^{n+1} (c+d x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac {d (a+b x)}{b c-a d},\frac {a+b x}{a}\right )}{a^2 (n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^n*(c + d*x)^p)/x^2,x]

[Out]

(b*(a + b*x)^(1 + n)*(c + d*x)^p*AppellF1[1 + n, -p, 2, 2 + n, -((d*(a + b*x))/(b*c - a*d)), (a + b*x)/a])/(a^
2*(1 + n)*((b*(c + d*x))/(b*c - a*d))^p)

Rule 141

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*e - a*f
)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 142

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*(b*(c/(b*c -
a*d)) + b*d*(x/(b*c - a*d)))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rubi steps

\begin {align*} \int \frac {(a+b x)^n (c+d x)^p}{x^2} \, dx &=\left ((c+d x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-p}\right ) \int \frac {(a+b x)^n \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^p}{x^2} \, dx\\ &=\frac {b (a+b x)^{1+n} (c+d x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-p} F_1\left (1+n;-p,2;2+n;-\frac {d (a+b x)}{b c-a d},\frac {a+b x}{a}\right )}{a^2 (1+n)}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 93, normalized size = 1.09 \begin {gather*} \frac {\left (1+\frac {a}{b x}\right )^{-n} \left (1+\frac {c}{d x}\right )^{-p} (a+b x)^n (c+d x)^p F_1\left (1-n-p;-n,-p;2-n-p;-\frac {a}{b x},-\frac {c}{d x}\right )}{(-1+n+p) x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^n*(c + d*x)^p*AppellF1[1 - n - p, -n, -p, 2 - n - p, -(a/(b*x)), -(c/(d*x))])/((-1 + n + p)*(1 + a/
(b*x))^n*(1 + c/(d*x))^p*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((b*x + a)^n*(d*x + c)^p/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} \left (c + d x\right )^{p}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^n*(c + d*x)^p)/x^2,x)

[Out]

int(((a + b*x)^n*(c + d*x)^p)/x^2, x)

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