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

Optimal. Leaf size=55 \[ \frac {2 (e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {b x}{a}\right )}{c e (m+1)}-\frac {(e x)^{m+1}}{c e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {80, 64} \[ \frac {2 (e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {b x}{a}\right )}{c e (m+1)}-\frac {(e x)^{m+1}}{c e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), 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 80

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.60 \[ \frac {x (e x)^m \left (2 \, _2F_1\left (1,m+1;m+2;\frac {b x}{a}\right )-1\right )}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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(-(b*x + a)*(e*x)^m/(b*c*x - a*c), x)

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

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(-(b*x + a)*(e*x)^m/(b*c*x - a*c), x)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right ) \left (e x \right )^{m}}{-b c x +a c}\, dx \]

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)

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

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((b*x + a)*(e*x)^m/(b*c*x - a*c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 5.69, size = 129, normalized size = 2.35 \[ \frac {e^{m} m x x^{m} \Phi \left (\frac {b x}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{c \Gamma \left (m + 2\right )} + \frac {e^{m} x x^{m} \Phi \left (\frac {b x}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{c \Gamma \left (m + 2\right )} + \frac {b e^{m} m x^{2} x^{m} \Phi \left (\frac {b x}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a c \Gamma \left (m + 3\right )} + \frac {2 b e^{m} x^{2} x^{m} \Phi \left (\frac {b x}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a c \Gamma \left (m + 3\right )} \]

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*x**m*lerchphi(b*x/a, 1, m + 1)*gamma(m + 1)/(c*gamma(m + 2)) + e**m*x*x**m*lerchphi(b*x/a, 1, m + 1)*
gamma(m + 1)/(c*gamma(m + 2)) + b*e**m*m*x**2*x**m*lerchphi(b*x/a, 1, m + 2)*gamma(m + 2)/(a*c*gamma(m + 3)) +
2*b*e**m*x**2*x**m*lerchphi(b*x/a, 1, m + 2)*gamma(m + 2)/(a*c*gamma(m + 3))

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