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

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

[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)

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Rubi [A]  time = 0.0373016, antiderivative size = 85, normalized size of antiderivative = 1., 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 = {137, 136} \[ \frac{b (a+b x)^{n+1} (c+d x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac{d (a+b x)}{b c-a d},\frac{a+b x}{a}\right )}{a^2 (n+1)} \]

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 137

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]

Rule 136

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

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*}

Mathematica [A]  time = 0.0549415, size = 93, normalized size = 1.09 \[ \frac{\left (\frac{a}{b x}+1\right )^{-n} (a+b x)^n \left (\frac{c}{d x}+1\right )^{-p} (c+d x)^p F_1\left (-n-p+1;-n,-p;-n-p+2;-\frac{a}{b x},-\frac{c}{d x}\right )}{x (n+p-1)} \]

Warning: Unable to verify antiderivative.

[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 = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{n} \left ( dx+c \right ) ^{p}}{{x}^{2}}}\, dx \end{align*}

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

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

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

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

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)