3.959 \(\int x (a+b x)^n (c+d x)^p \, dx\)

Optimal. Leaf size=117 \[ \frac{(a+b x)^{n+1} (c+d x)^{p+1} (a d (p+1)+b c (n+1)) \, _2F_1\left (1,n+p+2;p+2;\frac{b (c+d x)}{b c-a d}\right )}{b d (p+1) (n+p+2) (b c-a d)}+\frac{(a+b x)^{n+1} (c+d x)^{p+1}}{b d (n+p+2)} \]

[Out]

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

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Rubi [A]  time = 0.0557113, antiderivative size = 129, normalized size of antiderivative = 1.1, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {80, 70, 69} \[ \frac{(a+b x)^{n+1} (c+d x)^{p+1}}{b d (n+p+2)}-\frac{(a+b x)^{n+1} (c+d x)^p (a d (p+1)+b c (n+1)) \left (\frac{b (c+d x)}{b c-a d}\right )^{-p} \, _2F_1\left (n+1,-p;n+2;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 d (n+1) (n+p+2)} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)^n*(c + d*x)^p,x]

[Out]

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

Rule 80

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.0689723, size = 105, normalized size = 0.9 \[ \frac{(a+b x)^{n+1} (c+d x)^p \left (b (c+d x)-\frac{(a d (p+1)+b c (n+1)) \left (\frac{b (c+d x)}{b c-a d}\right )^{-p} \, _2F_1\left (n+1,-p;n+2;\frac{d (a+b x)}{a d-b c}\right )}{n+1}\right )}{b^2 d (n+p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int x \left ( bx+a \right ) ^{n} \left ( dx+c \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x + a)^n*(d*x + c)^p*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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