\(\int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx\) [942]

Optimal result

Integrand size = 18, antiderivative size = 203 \[ \int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx=\frac {x^2 (a+b x)^{1+n}}{b d (2+n) (c+d x)}-\frac {(a+b x)^{1+n} (c (b c (2+n) (a d+b c (3+n))-a d (a d+b c (5+3 n)))+d (b c-a d) (a d+b c (3+n)) x)}{b^2 d^3 (b c-a d) (1+n) (2+n) (c+d x)}-\frac {c^2 (3 a d-b c (3+n)) (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {d (a+b x)}{b c-a d}\right )}{d^3 (b c-a d)^2 (1+n)} \]

[Out]

x^2*(b*x+a)^(1+n)/b/d/(2+n)/(d*x+c)-(b*x+a)^(1+n)*(c*(b*c*(2+n)*(a*d+b*c*(3+n))-a*d*(a*d+b*c*(5+3*n)))+d*(-a*d
+b*c)*(a*d+b*c*(3+n))*x)/b^2/d^3/(-a*d+b*c)/(1+n)/(2+n)/(d*x+c)-c^2*(3*a*d-b*c*(3+n))*(b*x+a)^(1+n)*hypergeom(
[1, 1+n],[2+n],-d*(b*x+a)/(-a*d+b*c))/d^3/(-a*d+b*c)^2/(1+n)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {102, 151, 70} \[ \int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx=-\frac {(a+b x)^{n+1} (d x (b c-a d) (a d+b c (n+3))+c (b c (n+2) (a d+b c (n+3))-a d (a d+b c (3 n+5))))}{b^2 d^3 (n+1) (n+2) (c+d x) (b c-a d)}-\frac {c^2 (a+b x)^{n+1} (3 a d-b c (n+3)) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {d (a+b x)}{b c-a d}\right )}{d^3 (n+1) (b c-a d)^2}+\frac {x^2 (a+b x)^{n+1}}{b d (n+2) (c+d x)} \]

[In]

Int[(x^3*(a + b*x)^n)/(c + d*x)^2,x]

[Out]

(x^2*(a + b*x)^(1 + n))/(b*d*(2 + n)*(c + d*x)) - ((a + b*x)^(1 + n)*(c*(b*c*(2 + n)*(a*d + b*c*(3 + n)) - a*d
*(a*d + b*c*(5 + 3*n))) + d*(b*c - a*d)*(a*d + b*c*(3 + n))*x))/(b^2*d^3*(b*c - a*d)*(1 + n)*(2 + n)*(c + d*x)
) - (c^2*(3*a*d - b*c*(3 + n))*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, -((d*(a + b*x))/(b*c - a*d
))])/(d^3*(b*c - a*d)^2*(1 + n))

Rule 70

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

Rule 102

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 151

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

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 (a+b x)^{1+n}}{b d (2+n) (c+d x)}+\frac {\int \frac {x (a+b x)^n (-2 a c+(-a d-b c (3+n)) x)}{(c+d x)^2} \, dx}{b d (2+n)} \\ & = \frac {x^2 (a+b x)^{1+n}}{b d (2+n) (c+d x)}-\frac {(a+b x)^{1+n} (c (b c (2+n) (a d+b c (3+n))-a d (a d+b c (5+3 n)))+d (b c-a d) (a d+b c (3+n)) x)}{b^2 d^3 (b c-a d) (1+n) (2+n) (c+d x)}-\frac {\left (c^2 (3 a d-b c (3+n))\right ) \int \frac {(a+b x)^n}{c+d x} \, dx}{d^3 (b c-a d)} \\ & = \frac {x^2 (a+b x)^{1+n}}{b d (2+n) (c+d x)}-\frac {(a+b x)^{1+n} (c (b c (2+n) (a d+b c (3+n))-a d (a d+b c (5+3 n)))+d (b c-a d) (a d+b c (3+n)) x)}{b^2 d^3 (b c-a d) (1+n) (2+n) (c+d x)}-\frac {c^2 (3 a d-b c (3+n)) (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;-\frac {d (a+b x)}{b c-a d}\right )}{d^3 (b c-a d)^2 (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx=\frac {(a+b x)^{1+n} \left (\frac {x^2}{c+d x}+\frac {a^2 d^2 (c+d x)-b^2 c^2 (3+n) (c (2+n)+d x)+a b c d (c (3+2 n)+d (2+n) x)}{b d^2 (b c-a d) (1+n) (c+d x)}+\frac {b c^2 (2+n) (-3 a d+b c (3+n)) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (a+b x)}{-b c+a d}\right )}{d^2 (b c-a d)^2 (1+n)}\right )}{b d (2+n)} \]

[In]

Integrate[(x^3*(a + b*x)^n)/(c + d*x)^2,x]

[Out]

((a + b*x)^(1 + n)*(x^2/(c + d*x) + (a^2*d^2*(c + d*x) - b^2*c^2*(3 + n)*(c*(2 + n) + d*x) + a*b*c*d*(c*(3 + 2
*n) + d*(2 + n)*x))/(b*d^2*(b*c - a*d)*(1 + n)*(c + d*x)) + (b*c^2*(2 + n)*(-3*a*d + b*c*(3 + n))*Hypergeometr
ic2F1[1, 1 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)])/(d^2*(b*c - a*d)^2*(1 + n))))/(b*d*(2 + n))

Maple [F]

\[\int \frac {x^{3} \left (b x +a \right )^{n}}{\left (d x +c \right )^{2}}d x\]

[In]

int(x^3*(b*x+a)^n/(d*x+c)^2,x)

[Out]

int(x^3*(b*x+a)^n/(d*x+c)^2,x)

Fricas [F]

\[ \int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

integrate(x^3*(b*x+a)^n/(d*x+c)^2,x, algorithm="fricas")

[Out]

integral((b*x + a)^n*x^3/(d^2*x^2 + 2*c*d*x + c^2), x)

Sympy [F(-2)]

Exception generated. \[ \int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate(x**3*(b*x+a)**n/(d*x+c)**2,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

integrate(x^3*(b*x+a)^n/(d*x+c)^2,x, algorithm="maxima")

[Out]

integrate((b*x + a)^n*x^3/(d*x + c)^2, x)

Giac [F]

\[ \int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

integrate(x^3*(b*x+a)^n/(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((b*x + a)^n*x^3/(d*x + c)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b x)^n}{(c+d x)^2} \, dx=\int \frac {x^3\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^2} \,d x \]

[In]

int((x^3*(a + b*x)^n)/(c + d*x)^2,x)

[Out]

int((x^3*(a + b*x)^n)/(c + d*x)^2, x)