\(\int \frac {(e x)^m (a c-b c x)}{a+b x} \, dx\) [76]

Optimal result

Integrand size = 22, antiderivative size = 52 \[ \int \frac {(e x)^m (a c-b c x)}{a+b x} \, dx=-\frac {c (e x)^{1+m}}{e (1+m)}+\frac {2 c (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )}{e (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {81, 66} \[ \int \frac {(e x)^m (a c-b c x)}{a+b x} \, dx=\frac {2 c (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {b x}{a}\right )}{e (m+1)}-\frac {c (e x)^{m+1}}{e (m+1)} \]

[In]

Int[((e*x)^m*(a*c - b*c*x))/(a + b*x),x]

[Out]

-((c*(e*x)^(1 + m))/(e*(1 + m))) + (2*c*(e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(e*(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 {c (e x)^{1+m}}{e (1+m)}+(2 a c) \int \frac {(e x)^m}{a+b x} \, dx \\ & = -\frac {c (e x)^{1+m}}{e (1+m)}+\frac {2 c (e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{e (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.62 \[ \int \frac {(e x)^m (a c-b c x)}{a+b x} \, dx=\frac {c x (e x)^m \left (-1+2 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )\right )}{1+m} \]

[In]

Integrate[((e*x)^m*(a*c - b*c*x))/(a + b*x),x]

[Out]

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

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (-b c x +a c \right )}{b x +a}d x\]

[In]

int((e*x)^m*(-b*c*x+a*c)/(b*x+a),x)

[Out]

int((e*x)^m*(-b*c*x+a*c)/(b*x+a),x)

Fricas [F]

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

[In]

integrate((e*x)^m*(-b*c*x+a*c)/(b*x+a),x, algorithm="fricas")

[Out]

integral(-(b*c*x - a*c)*(e*x)^m/(b*x + a), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.81 \[ \int \frac {(e x)^m (a c-b c x)}{a+b x} \, dx=\frac {c e^{m} m x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} + \frac {c e^{m} x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} - \frac {b c e^{m} m x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} - \frac {2 b c e^{m} x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} \]

[In]

integrate((e*x)**m*(-b*c*x+a*c)/(b*x+a),x)

[Out]

c*e**m*m*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/gamma(m + 2) + c*e**m*x**(m + 1)*le
rchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/gamma(m + 2) - b*c*e**m*m*x**(m + 2)*lerchphi(b*x*exp_pol
ar(I*pi)/a, 1, m + 2)*gamma(m + 2)/(a*gamma(m + 3)) - 2*b*c*e**m*x**(m + 2)*lerchphi(b*x*exp_polar(I*pi)/a, 1,
m + 2)*gamma(m + 2)/(a*gamma(m + 3))

Maxima [F]

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

[In]

integrate((e*x)^m*(-b*c*x+a*c)/(b*x+a),x, algorithm="maxima")

[Out]

-integrate((b*c*x - a*c)*(e*x)^m/(b*x + a), x)

Giac [F]

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

[In]

integrate((e*x)^m*(-b*c*x+a*c)/(b*x+a),x, algorithm="giac")

[Out]

integrate(-(b*c*x - a*c)*(e*x)^m/(b*x + a), x)

Mupad [F(-1)]

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

[In]

int(((a*c - b*c*x)*(e*x)^m)/(a + b*x),x)

[Out]

int(((a*c - b*c*x)*(e*x)^m)/(a + b*x), x)