3.971 \(\int x^2 (a+b x)^n (c+d x)^{-n} \, dx\)

Optimal. Leaf size=199 \[ \frac {(a+b x)^{n+1} (c+d x)^{-n} \left (a^2 d^2 \left (n^2-3 n+2\right )+2 a b c d \left (1-n^2\right )+b^2 c^2 \left (n^2+3 n+2\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n+1;n+2;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (n+1)}-\frac {(a+b x)^{n+1} (c+d x)^{1-n} (a d (2-n)+b c (n+2))}{6 b^2 d^2}+\frac {x (a+b x)^{n+1} (c+d x)^{1-n}}{3 b d} \]

[Out]

-1/6*(a*d*(2-n)+b*c*(2+n))*(b*x+a)^(1+n)*(d*x+c)^(1-n)/b^2/d^2+1/3*x*(b*x+a)^(1+n)*(d*x+c)^(1-n)/b/d+1/6*(2*a*
b*c*d*(-n^2+1)+a^2*d^2*(n^2-3*n+2)+b^2*c^2*(n^2+3*n+2))*(b*x+a)^(1+n)*(b*(d*x+c)/(-a*d+b*c))^n*hypergeom([n, 1
+n],[2+n],-d*(b*x+a)/(-a*d+b*c))/b^3/d^2/(1+n)/((d*x+c)^n)

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Rubi [A]  time = 0.14, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {90, 80, 70, 69} \[ \frac {(a+b x)^{n+1} (c+d x)^{-n} \left (a^2 d^2 \left (n^2-3 n+2\right )+2 a b c d \left (1-n^2\right )+b^2 c^2 \left (n^2+3 n+2\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n+1;n+2;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (n+1)}-\frac {(a+b x)^{n+1} (c+d x)^{1-n} (a d (2-n)+b c (n+2))}{6 b^2 d^2}+\frac {x (a+b x)^{n+1} (c+d x)^{1-n}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^n)/(c + d*x)^n,x]

[Out]

-((a*d*(2 - n) + b*c*(2 + n))*(a + b*x)^(1 + n)*(c + d*x)^(1 - n))/(6*b^2*d^2) + (x*(a + b*x)^(1 + n)*(c + d*x
)^(1 - n))/(3*b*d) + ((2*a*b*c*d*(1 - n^2) + a^2*d^2*(2 - 3*n + n^2) + b^2*c^2*(2 + 3*n + n^2))*(a + b*x)^(1 +
n)*((b*(c + d*x))/(b*c - a*d))^n*Hypergeometric2F1[n, 1 + n, 2 + n, -((d*(a + b*x))/(b*c - a*d))])/(6*b^3*d^2
*(1 + n)*(c + d*x)^n)

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]))

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 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 90

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)^{-n} \, dx &=\frac {x (a+b x)^{1+n} (c+d x)^{1-n}}{3 b d}+\frac {\int (a+b x)^n (c+d x)^{-n} (-a c-(a d (2-n)+b c (2+n)) x) \, dx}{3 b d}\\ &=-\frac {(a d (2-n)+b c (2+n)) (a+b x)^{1+n} (c+d x)^{1-n}}{6 b^2 d^2}+\frac {x (a+b x)^{1+n} (c+d x)^{1-n}}{3 b d}+\frac {\left (2 a b c d \left (1-n^2\right )+a^2 d^2 \left (2-3 n+n^2\right )+b^2 c^2 \left (2+3 n+n^2\right )\right ) \int (a+b x)^n (c+d x)^{-n} \, dx}{6 b^2 d^2}\\ &=-\frac {(a d (2-n)+b c (2+n)) (a+b x)^{1+n} (c+d x)^{1-n}}{6 b^2 d^2}+\frac {x (a+b x)^{1+n} (c+d x)^{1-n}}{3 b d}+\frac {\left (\left (2 a b c d \left (1-n^2\right )+a^2 d^2 \left (2-3 n+n^2\right )+b^2 c^2 \left (2+3 n+n^2\right )\right ) (c+d x)^{-n} \left (\frac {b (c+d x)}{b c-a d}\right )^n\right ) \int (a+b x)^n \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-n} \, dx}{6 b^2 d^2}\\ &=-\frac {(a d (2-n)+b c (2+n)) (a+b x)^{1+n} (c+d x)^{1-n}}{6 b^2 d^2}+\frac {x (a+b x)^{1+n} (c+d x)^{1-n}}{3 b d}+\frac {\left (2 a b c d \left (1-n^2\right )+a^2 d^2 \left (2-3 n+n^2\right )+b^2 c^2 \left (2+3 n+n^2\right )\right ) (a+b x)^{1+n} (c+d x)^{-n} \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,1+n;2+n;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 154, normalized size = 0.77 \[ \frac {(a+b x)^{n+1} (c+d x)^{-n} \left (\frac {\left (a^2 d^2 \left (n^2-3 n+2\right )-2 a b c d \left (n^2-1\right )+b^2 c^2 \left (n^2+3 n+2\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n+1;n+2;\frac {d (a+b x)}{a d-b c}\right )}{n+1}+b (c+d x) (a d (n-2)-b c (n+2))+2 b^2 d x (c+d x)\right )}{6 b^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + n)*(b*(a*d*(-2 + n) - b*c*(2 + n))*(c + d*x) + 2*b^2*d*x*(c + d*x) + ((-2*a*b*c*d*(-1 + n^2) +
a^2*d^2*(2 - 3*n + n^2) + b^2*c^2*(2 + 3*n + n^2))*((b*(c + d*x))/(b*c - a*d))^n*Hypergeometric2F1[n, 1 + n,
2 + n, (d*(a + b*x))/(-(b*c) + a*d)])/(1 + n)))/(6*b^3*d^2*(c + d*x)^n)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{n} x^{2}}{{\left (d x + c\right )}^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int x^{2} \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*x)^n)/(c + d*x)^n,x)

[Out]

int((x^2*(a + b*x)^n)/(c + d*x)^n, x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed

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