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

Optimal. Leaf size=48 \[ \frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};\frac {b^2 x^2}{a^2}\right )}{a^2 c e (m+1)} \]

[Out]

(e*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],b^2*x^2/a^2)/a^2/c/e/(1+m)

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {73, 364} \[ \frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};\frac {b^2 x^2}{a^2}\right )}{a^2 c e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (b^2*x^2)/a^2])/(a^2*c*e*(1 + m))

Rule 73

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {(e x)^m}{(a+b x) (a c-b c x)} \, dx &=\int \frac {(e x)^m}{a^2 c-b^2 c x^2} \, dx\\ &=\frac {(e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};\frac {b^2 x^2}{a^2}\right )}{a^2 c e (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 46, normalized size = 0.96 \[ \frac {x (e x)^m \, _2F_1\left (1,\frac {m+1}{2};\frac {m+1}{2}+1;\frac {b^2 x^2}{a^2}\right )}{a^2 c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (e x\right )^{m}}{b^{2} c x^{2} - a^{2} c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(e*x)^m/(b^2*c*x^2 - a^2*c), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{m}}{\left (b x +a \right ) \left (-b c x +a c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (e\,x\right )}^m}{\left (a\,c-b\,c\,x\right )\,\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 2.22, size = 73, normalized size = 1.52 \[ - \frac {e^{m} m x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{2 a b c \Gamma \left (1 - m\right )} + \frac {e^{m} m x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m\right ) \Gamma \left (- m\right )}{2 a b c \Gamma \left (1 - m\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________