\(\int \frac {x^m (A+B x)}{a+b x} \, dx\) [375]

Optimal result

Integrand size = 16, antiderivative size = 56 \[ \int \frac {x^m (A+B x)}{a+b x} \, dx=\frac {B x^{1+m}}{b (1+m)}+\frac {(A b-a B) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )}{a b (1+m)} \]

[Out]

B*x^(1+m)/b/(1+m)+(A*b-B*a)*x^(1+m)*hypergeom([1, 1+m],[2+m],-b*x/a)/a/b/(1+m)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {81, 66} \[ \int \frac {x^m (A+B x)}{a+b x} \, dx=\frac {x^{m+1} (A b-a B) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {b x}{a}\right )}{a b (m+1)}+\frac {B x^{m+1}}{b (m+1)} \]

[In]

Int[(x^m*(A + B*x))/(a + b*x),x]

[Out]

(B*x^(1 + m))/(b*(1 + m)) + ((A*b - a*B)*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(a*b*(1 + m
))

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && 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]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80 \[ \int \frac {x^m (A+B x)}{a+b x} \, dx=\frac {x^{1+m} \left (a B+(A b-a B) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )\right )}{a b (1+m)} \]

[In]

Integrate[(x^m*(A + B*x))/(a + b*x),x]

[Out]

(x^(1 + m)*(a*B + (A*b - a*B)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)]))/(a*b*(1 + m))

Maple [F]

\[\int \frac {x^{m} \left (B x +A \right )}{b x +a}d x\]

[In]

int(x^m*(B*x+A)/(b*x+a),x)

[Out]

int(x^m*(B*x+A)/(b*x+a),x)

Fricas [F]

\[ \int \frac {x^m (A+B x)}{a+b x} \, dx=\int { \frac {{\left (B x + A\right )} x^{m}}{b x + a} \,d x } \]

[In]

integrate(x^m*(B*x+A)/(b*x+a),x, algorithm="fricas")

[Out]

integral((B*x + A)*x^m/(b*x + a), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.38 \[ \int \frac {x^m (A+B x)}{a+b x} \, dx=\frac {A m x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac {A x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac {B m x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac {2 B x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} \]

[In]

integrate(x**m*(B*x+A)/(b*x+a),x)

[Out]

A*m*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a*gamma(m + 2)) + A*x**(m + 1)*lerchphi
(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a*gamma(m + 2)) + B*m*x**(m + 2)*lerchphi(b*x*exp_polar(I*pi)/
a, 1, m + 2)*gamma(m + 2)/(a*gamma(m + 3)) + 2*B*x**(m + 2)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m
+ 2)/(a*gamma(m + 3))

Maxima [F]

\[ \int \frac {x^m (A+B x)}{a+b x} \, dx=\int { \frac {{\left (B x + A\right )} x^{m}}{b x + a} \,d x } \]

[In]

integrate(x^m*(B*x+A)/(b*x+a),x, algorithm="maxima")

[Out]

integrate((B*x + A)*x^m/(b*x + a), x)

Giac [F]

\[ \int \frac {x^m (A+B x)}{a+b x} \, dx=\int { \frac {{\left (B x + A\right )} x^{m}}{b x + a} \,d x } \]

[In]

integrate(x^m*(B*x+A)/(b*x+a),x, algorithm="giac")

[Out]

integrate((B*x + A)*x^m/(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m (A+B x)}{a+b x} \, dx=\int \frac {x^m\,\left (A+B\,x\right )}{a+b\,x} \,d x \]

[In]

int((x^m*(A + B*x))/(a + b*x),x)

[Out]

int((x^m*(A + B*x))/(a + b*x), x)