3.970 \(\int x^3 (a+b x)^n (c+d x)^{-n} \, dx\)

Optimal. Leaf size=295 \[ \frac {(a+b x)^{n+1} (c+d x)^{1-n} \left (a^2 d^2 \left (n^2-5 n+6\right )+2 a b c d \left (3-n^2\right )-2 b d x (a d (3-n)+b c (n+3))+b^2 c^2 \left (n^2+5 n+6\right )\right )}{24 b^3 d^3}-\frac {(a+b x)^{n+1} (c+d x)^{-n} \left (a^3 d^3 \left (-n^3+6 n^2-11 n+6\right )+3 a^2 b c d^2 \left (n^3-2 n^2-n+2\right )+3 a b^2 c^2 d \left (-n^3-2 n^2+n+2\right )+b^3 c^3 \left (n^3+6 n^2+11 n+6\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n+1;n+2;-\frac {d (a+b x)}{b c-a d}\right )}{24 b^4 d^3 (n+1)}+\frac {x^2 (a+b x)^{n+1} (c+d x)^{1-n}}{4 b d} \]

[Out]

1/4*x^2*(b*x+a)^(1+n)*(d*x+c)^(1-n)/b/d+1/24*(b*x+a)^(1+n)*(d*x+c)^(1-n)*(2*a*b*c*d*(-n^2+3)+a^2*d^2*(n^2-5*n+
6)+b^2*c^2*(n^2+5*n+6)-2*b*d*(a*d*(3-n)+b*c*(3+n))*x)/b^3/d^3-1/24*(3*a*b^2*c^2*d*(-n^3-2*n^2+n+2)+a^3*d^3*(-n
^3+6*n^2-11*n+6)+3*a^2*b*c*d^2*(n^3-2*n^2-n+2)+b^3*c^3*(n^3+6*n^2+11*n+6))*(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^4/d^3/(1+n)/((d*x+c)^n)

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Rubi [A]  time = 0.25, antiderivative size = 295, 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 = {100, 147, 70, 69} \[ -\frac {(a+b x)^{n+1} (c+d x)^{-n} \left (3 a^2 b c d^2 \left (n^3-2 n^2-n+2\right )+a^3 d^3 \left (-n^3+6 n^2-11 n+6\right )+3 a b^2 c^2 d \left (-n^3-2 n^2+n+2\right )+b^3 c^3 \left (n^3+6 n^2+11 n+6\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n+1;n+2;-\frac {d (a+b x)}{b c-a d}\right )}{24 b^4 d^3 (n+1)}+\frac {(a+b x)^{n+1} (c+d x)^{1-n} \left (a^2 d^2 \left (n^2-5 n+6\right )+2 a b c d \left (3-n^2\right )-2 b d x (a d (3-n)+b c (n+3))+b^2 c^2 \left (n^2+5 n+6\right )\right )}{24 b^3 d^3}+\frac {x^2 (a+b x)^{n+1} (c+d x)^{1-n}}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*x)^n)/(c + d*x)^n,x]

[Out]

(x^2*(a + b*x)^(1 + n)*(c + d*x)^(1 - n))/(4*b*d) + ((a + b*x)^(1 + n)*(c + d*x)^(1 - n)*(2*a*b*c*d*(3 - n^2)
+ a^2*d^2*(6 - 5*n + n^2) + b^2*c^2*(6 + 5*n + n^2) - 2*b*d*(a*d*(3 - n) + b*c*(3 + n))*x))/(24*b^3*d^3) - ((3
*a*b^2*c^2*d*(2 + n - 2*n^2 - n^3) + a^3*d^3*(6 - 11*n + 6*n^2 - n^3) + 3*a^2*b*c*d^2*(2 - n - 2*n^2 + n^3) +
b^3*c^3*(6 + 11*n + 6*n^2 + n^3))*(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))])/(24*b^4*d^3*(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 100

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

Rule 147

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

Rubi steps

\begin {align*} \int x^3 (a+b x)^n (c+d x)^{-n} \, dx &=\frac {x^2 (a+b x)^{1+n} (c+d x)^{1-n}}{4 b d}+\frac {\int x (a+b x)^n (c+d x)^{-n} (-2 a c+(-a d (3-n)-b c (3+n)) x) \, dx}{4 b d}\\ &=\frac {x^2 (a+b x)^{1+n} (c+d x)^{1-n}}{4 b d}+\frac {(a+b x)^{1+n} (c+d x)^{1-n} \left (2 a b c d \left (3-n^2\right )+a^2 d^2 \left (6-5 n+n^2\right )+b^2 c^2 \left (6+5 n+n^2\right )-2 b d (a d (3-n)+b c (3+n)) x\right )}{24 b^3 d^3}-\frac {\left (3 a b^2 c^2 d \left (2+n-2 n^2-n^3\right )+a^3 d^3 \left (6-11 n+6 n^2-n^3\right )+3 a^2 b c d^2 \left (2-n-2 n^2+n^3\right )+b^3 c^3 \left (6+11 n+6 n^2+n^3\right )\right ) \int (a+b x)^n (c+d x)^{-n} \, dx}{24 b^3 d^3}\\ &=\frac {x^2 (a+b x)^{1+n} (c+d x)^{1-n}}{4 b d}+\frac {(a+b x)^{1+n} (c+d x)^{1-n} \left (2 a b c d \left (3-n^2\right )+a^2 d^2 \left (6-5 n+n^2\right )+b^2 c^2 \left (6+5 n+n^2\right )-2 b d (a d (3-n)+b c (3+n)) x\right )}{24 b^3 d^3}-\frac {\left (\left (3 a b^2 c^2 d \left (2+n-2 n^2-n^3\right )+a^3 d^3 \left (6-11 n+6 n^2-n^3\right )+3 a^2 b c d^2 \left (2-n-2 n^2+n^3\right )+b^3 c^3 \left (6+11 n+6 n^2+n^3\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}{24 b^3 d^3}\\ &=\frac {x^2 (a+b x)^{1+n} (c+d x)^{1-n}}{4 b d}+\frac {(a+b x)^{1+n} (c+d x)^{1-n} \left (2 a b c d \left (3-n^2\right )+a^2 d^2 \left (6-5 n+n^2\right )+b^2 c^2 \left (6+5 n+n^2\right )-2 b d (a d (3-n)+b c (3+n)) x\right )}{24 b^3 d^3}-\frac {\left (3 a b^2 c^2 d \left (2+n-2 n^2-n^3\right )+a^3 d^3 \left (6-11 n+6 n^2-n^3\right )+3 a^2 b c d^2 \left (2-n-2 n^2+n^3\right )+b^3 c^3 \left (6+11 n+6 n^2+n^3\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 )}{24 b^4 d^3 (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 262, normalized size = 0.89 \[ \frac {(a+b x)^{n+1} (c+d x)^{-n} \left (-b^2 c^2 (b c (n+3)-a d (n-1)) \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n+1;n+2;\frac {d (a+b x)}{a d-b c}\right )-(b c-a d)^2 (b c (n+3)-a d (n-3)) \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n-2,n+1;n+2;\frac {d (a+b x)}{a d-b c}\right )+2 b c (b c-a d) (b c (n+3)-a d (n-2)) \left (\frac {b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n-1,n+1;n+2;\frac {d (a+b x)}{a d-b c}\right )+b^3 d^2 (n+1) x^2 (c+d x)\right )}{4 b^4 d^3 (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*x)^n)/(c + d*x)^n,x]

[Out]

((a + b*x)^(1 + n)*(b^3*d^2*(1 + n)*x^2*(c + d*x) - (b*c - a*d)^2*(-(a*d*(-3 + n)) + b*c*(3 + n))*((b*(c + d*x
))/(b*c - a*d))^n*Hypergeometric2F1[-2 + n, 1 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)] + 2*b*c*(b*c - a*d)*(-
(a*d*(-2 + n)) + b*c*(3 + n))*((b*(c + d*x))/(b*c - a*d))^n*Hypergeometric2F1[-1 + n, 1 + n, 2 + n, (d*(a + b*
x))/(-(b*c) + a*d)] - b^2*c^2*(-(a*d*(-1 + n)) + b*c*(3 + n))*((b*(c + d*x))/(b*c - a*d))^n*Hypergeometric2F1[
n, 1 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)]))/(4*b^4*d^3*(1 + n)*(c + d*x)^n)

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^n/((d*x+c)^n),x, algorithm="fricas")

[Out]

integral((b*x + a)^n*x^3/(d*x + c)^n, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^n/((d*x+c)^n),x, algorithm="giac")

[Out]

integrate((b*x + a)^n*x^3/(d*x + c)^n, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int x^{3} \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x+a)^n/((d*x+c)^n),x)

[Out]

int(x^3*(b*x+a)^n/((d*x+c)^n),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^n/((d*x+c)^n),x, algorithm="maxima")

[Out]

integrate((b*x + a)^n*x^3/(d*x + c)^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*x)^n)/(c + d*x)^n,x)

[Out]

int((x^3*(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**3*(b*x+a)**n/((d*x+c)**n),x)

[Out]

Exception raised: HeuristicGCDFailed

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