\(\int \frac {(a+b x)^n (c+d x)^2}{x^2} \, dx\) [927]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 18, antiderivative size = 87 \[ \int \frac {(a+b x)^n (c+d x)^2}{x^2} \, dx=\frac {d^2 (a+b x)^{1+n}}{b (1+n)}-\frac {c^2 (a+b x)^{1+n}}{a x}-\frac {c (2 a d+b c n) (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b x}{a}\right )}{a^2 (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 87, 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 = {91, 81, 67} \[ \int \frac {(a+b x)^n (c+d x)^2}{x^2} \, dx=-\frac {c (a+b x)^{n+1} (2 a d+b c n) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b x}{a}+1\right )}{a^2 (n+1)}-\frac {c^2 (a+b x)^{n+1}}{a x}+\frac {d^2 (a+b x)^{n+1}}{b (n+1)} \]

[In]

Int[((a + b*x)^n*(c + d*x)^2)/x^2,x]

[Out]

(d^2*(a + b*x)^(1 + n))/(b*(1 + n)) - (c^2*(a + b*x)^(1 + n))/(a*x) - (c*(2*a*d + b*c*n)*(a + b*x)^(1 + n)*Hyp
ergeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a^2*(1 + n))

Rule 67

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

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 91

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x)^n (c+d x)^2}{x^2} \, dx=\frac {(a+b x)^{1+n} \left (a \left (-b c^2 (1+n)+a d^2 x\right )-b c (2 a d+b c n) x \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b x}{a}\right )\right )}{a^2 b (1+n) x} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.53 (sec) , antiderivative size = 265, normalized size of antiderivative = 3.05 \[ \int \frac {(a+b x)^n (c+d x)^2}{x^2} \, dx=d^{2} \left (\begin {cases} a^{n} x & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x\right )^{n + 1}}{n + 1} & \text {for}\: n \neq -1 \\\log {\left (a + b x \right )} & \text {otherwise} \end {cases}}{b} & \text {otherwise} \end {cases}\right ) - \frac {2 b^{n + 1} c d n \left (\frac {a}{b} + x\right )^{n + 1} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} - \frac {2 b^{n + 1} c d \left (\frac {a}{b} + x\right )^{n + 1} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} - \frac {b^{n + 2} c^{2} n \left (\frac {a}{b} + x\right )^{n + 1} \Gamma \left (n + 1\right )}{a b x \Gamma \left (n + 2\right )} - \frac {b^{n + 2} c^{2} \left (\frac {a}{b} + x\right )^{n + 1} \Gamma \left (n + 1\right )}{a b x \Gamma \left (n + 2\right )} - \frac {b^{n + 2} c^{2} n^{2} \left (\frac {a}{b} + x\right )^{n + 1} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a^{2} \Gamma \left (n + 2\right )} - \frac {b^{n + 2} c^{2} n \left (\frac {a}{b} + x\right )^{n + 1} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a^{2} \Gamma \left (n + 2\right )} \]

[In]

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

[Out]

d**2*Piecewise((a**n*x, Eq(b, 0)), (Piecewise(((a + b*x)**(n + 1)/(n + 1), Ne(n, -1)), (log(a + b*x), True))/b
, True)) - 2*b**(n + 1)*c*d*n*(a/b + x)**(n + 1)*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2)) -
 2*b**(n + 1)*c*d*(a/b + x)**(n + 1)*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2)) - b**(n + 2)*
c**2*n*(a/b + x)**(n + 1)*gamma(n + 1)/(a*b*x*gamma(n + 2)) - b**(n + 2)*c**2*(a/b + x)**(n + 1)*gamma(n + 1)/
(a*b*x*gamma(n + 2)) - b**(n + 2)*c**2*n**2*(a/b + x)**(n + 1)*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a**
2*gamma(n + 2)) - b**(n + 2)*c**2*n*(a/b + x)**(n + 1)*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a**2*gamma(
n + 2))

Maxima [F]

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

[In]

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

[Out]

(b*x + a)^(n + 1)*d^2/(b*(n + 1)) + integrate((2*c*d*x + c^2)*(b*x + a)^n/x^2, x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(((a + b*x)^n*(c + d*x)^2)/x^2,x)

[Out]

int(((a + b*x)^n*(c + d*x)^2)/x^2, x)