3.75 \(\int \frac{(e x)^m (a c-b c x)^2}{a+b x} \, dx\)

Optimal. Leaf size=78 \[ \frac{4 a c^2 (e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{e (m+1)}-\frac{3 a c^2 (e x)^{m+1}}{e (m+1)}+\frac{b c^2 (e x)^{m+2}}{e^2 (m+2)} \]

[Out]

(-3*a*c^2*(e*x)^(1 + m))/(e*(1 + m)) + (b*c^2*(e*x)^(2 + m))/(e^2*(2 + m)) + (4*a*c^2*(e*x)^(1 + m)*Hypergeome
tric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(e*(1 + m))

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Rubi [A]  time = 0.0576958, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {88, 64, 43} \[ \frac{4 a c^2 (e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{e (m+1)}-\frac{3 a c^2 (e x)^{m+1}}{e (m+1)}+\frac{b c^2 (e x)^{m+2}}{e^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*c^2*(e*x)^(1 + m))/(e*(1 + m)) + (b*c^2*(e*x)^(2 + m))/(e^2*(2 + m)) + (4*a*c^2*(e*x)^(1 + m)*Hypergeome
tric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(e*(1 + m))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(e x)^m (a c-b c x)^2}{a+b x} \, dx &=\int \left (-2 a c^2 (e x)^m+\frac{4 a^2 c^2 (e x)^m}{a+b x}-c (e x)^m (a c-b c x)\right ) \, dx\\ &=-\frac{2 a c^2 (e x)^{1+m}}{e (1+m)}-c \int (e x)^m (a c-b c x) \, dx+\left (4 a^2 c^2\right ) \int \frac{(e x)^m}{a+b x} \, dx\\ &=-\frac{2 a c^2 (e x)^{1+m}}{e (1+m)}+\frac{4 a c^2 (e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{b x}{a}\right )}{e (1+m)}-c \int \left (a c (e x)^m-\frac{b c (e x)^{1+m}}{e}\right ) \, dx\\ &=-\frac{3 a c^2 (e x)^{1+m}}{e (1+m)}+\frac{b c^2 (e x)^{2+m}}{e^2 (2+m)}+\frac{4 a c^2 (e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{b x}{a}\right )}{e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0284785, size = 54, normalized size = 0.69 \[ \frac{c^2 x (e x)^m \left (4 a (m+2) \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )-3 a (m+2)+b (m+1) x\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x*(e*x)^m*(-3*a*(2 + m) + b*(1 + m)*x + 4*a*(2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)]))/((1
+ m)*(2 + m))

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( -bcx+ac \right ) ^{2}}{bx+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b c x - a c\right )}^{2} \left (e x\right )^{m}}{b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(-b*c*x+a*c)^2/(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x + a^{2} c^{2}\right )} \left (e x\right )^{m}}{b x + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(-b*c*x+a*c)^2/(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 3.80036, size = 246, normalized size = 3.15 \begin{align*} \frac{a c^{2} e^{m} m x x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} + \frac{a c^{2} e^{m} x x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} - \frac{2 b c^{2} e^{m} m x^{2} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} - \frac{4 b c^{2} e^{m} x^{2} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} + \frac{b^{2} c^{2} e^{m} m x^{3} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} + \frac{3 b^{2} c^{2} e^{m} x^{3} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(-b*c*x+a*c)**2/(b*x+a),x)

[Out]

a*c**2*e**m*m*x*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/gamma(m + 2) + a*c**2*e**m*x*x**m*
lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/gamma(m + 2) - 2*b*c**2*e**m*m*x**2*x**m*lerchphi(b*x*e
xp_polar(I*pi)/a, 1, m + 2)*gamma(m + 2)/gamma(m + 3) - 4*b*c**2*e**m*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/a
, 1, m + 2)*gamma(m + 2)/gamma(m + 3) + b**2*c**2*e**m*m*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 3)*g
amma(m + 3)/(a*gamma(m + 4)) + 3*b**2*c**2*e**m*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 3)*gamma(m +
3)/(a*gamma(m + 4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b c x - a c\right )}^{2} \left (e x\right )^{m}}{b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(-b*c*x+a*c)^2/(b*x+a),x, algorithm="giac")

[Out]

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