3.10.58 \(\int x^2 (a+b x)^n (c+d x)^p \, dx\) [958]

Optimal. Leaf size=206 \[ -\frac {(b c (2+n)+a d (2+p)) (a+b x)^{1+n} (c+d x)^{1+p}}{b^2 d^2 (2+n+p) (3+n+p)}+\frac {x (a+b x)^{1+n} (c+d x)^{1+p}}{b d (3+n+p)}-\frac {\left (b^2 c^2 \left (2+3 n+n^2\right )+2 a b c d (1+n) (1+p)+a^2 d^2 \left (2+3 p+p^2\right )\right ) (a+b x)^{1+n} (c+d x)^{1+p} \, _2F_1\left (1,2+n+p;2+p;\frac {b (c+d x)}{b c-a d}\right )}{b^2 d^2 (b c-a d) (1+p) (2+n+p) (3+n+p)} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 216, normalized size of antiderivative = 1.05, 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 = {92, 81, 72, 71} \begin {gather*} \frac {(a+b x)^{n+1} (c+d x)^p \left (a^2 d^2 \left (p^2+3 p+2\right )+2 a b c d (n+1) (p+1)+b^2 c^2 \left (n^2+3 n+2\right )\right ) \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^3 d^2 (n+1) (n+p+2) (n+p+3)}-\frac {(a+b x)^{n+1} (c+d x)^{p+1} (a d (p+2)+b c (n+2))}{b^2 d^2 (n+p+2) (n+p+3)}+\frac {x (a+b x)^{n+1} (c+d x)^{p+1}}{b d (n+p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)^n*(c + d*x)^p,x]

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
 - a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 72

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 81

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 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
 f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 178, normalized size = 0.86 \begin {gather*} \frac {(a+b x)^{1+n} (c+d x)^p \left (-\frac {(b c (2+n)+a d (2+p)) (c+d x)}{b d (2+n+p)}+x (c+d x)+\frac {\left (b^2 c^2 \left (2+3 n+n^2\right )+2 a b c d (1+n) (1+p)+a^2 d^2 \left (2+3 p+p^2\right )\right ) \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)}\right )}{b d (3+n+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + n)*(c + d*x)^p*(-(((b*c*(2 + n) + a*d*(2 + p))*(c + d*x))/(b*d*(2 + n + p))) + x*(c + d*x) + (
(b^2*c^2*(2 + 3*n + n^2) + 2*a*b*c*d*(1 + n)*(1 + p) + a^2*d^2*(2 + 3*p + p^2))*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)))/(b*d*(3 + n +
 p))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integral((b*x + a)^n*(d*x + c)^p*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(x**2*(b*x+a)**n*(d*x+c)**p,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(x^2*(b*x+a)^n*(d*x+c)^p,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 = -0.00 \begin {gather*} \int x^2\,{\left (a+b\,x\right )}^n\,{\left (c+d\,x\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x)^n*(c + d*x)^p,x)

[Out]

int(x^2*(a + b*x)^n*(c + d*x)^p, x)

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