3.957 \(\int (e x)^m (a-b x)^{2+n} (a+b x)^n \, dx\)

Optimal. Leaf size=211 \[ -\frac{2 a b (e x)^{m+2} (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \, _2F_1\left (\frac{m+2}{2},-n;\frac{m+4}{2};\frac{b^2 x^2}{a^2}\right )}{e^2 (m+2)}+\frac{2 a^2 (m+n+2) (e x)^{m+1} (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \, _2F_1\left (\frac{m+1}{2},-n;\frac{m+3}{2};\frac{b^2 x^2}{a^2}\right )}{e (m+1) (m+2 n+3)}-\frac{(e x)^{m+1} (a-b x)^{n+1} (a+b x)^{n+1}}{e (m+2 n+3)} \]

[Out]

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

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Rubi [A]  time = 0.173635, antiderivative size = 238, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {127, 126, 365, 364} \[ -\frac{2 a b (e x)^{m+2} (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \, _2F_1\left (\frac{m+2}{2},-n;\frac{m+4}{2};\frac{b^2 x^2}{a^2}\right )}{e^2 (m+2)}+\frac{b^2 (e x)^{m+3} (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \, _2F_1\left (\frac{m+3}{2},-n;\frac{m+5}{2};\frac{b^2 x^2}{a^2}\right )}{e^3 (m+3)}+\frac{a^2 (e x)^{m+1} (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \, _2F_1\left (\frac{m+1}{2},-n;\frac{m+3}{2};\frac{b^2 x^2}{a^2}\right )}{e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(e*x)^(1 + m)*(a - b*x)^n*(a + b*x)^n*Hypergeometric2F1[(1 + m)/2, -n, (3 + m)/2, (b^2*x^2)/a^2])/(e*(1 +
m)*(1 - (b^2*x^2)/a^2)^n) - (2*a*b*(e*x)^(2 + m)*(a - b*x)^n*(a + b*x)^n*Hypergeometric2F1[(2 + m)/2, -n, (4
+ m)/2, (b^2*x^2)/a^2])/(e^2*(2 + m)*(1 - (b^2*x^2)/a^2)^n) + (b^2*(e*x)^(3 + m)*(a - b*x)^n*(a + b*x)^n*Hyper
geometric2F1[(3 + m)/2, -n, (5 + m)/2, (b^2*x^2)/a^2])/(e^3*(3 + m)*(1 - (b^2*x^2)/a^2)^n)

Rule 127

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

Rule 126

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

Rule 365

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

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 (e x)^m (a-b x)^{2+n} (a+b x)^n \, dx &=\int \left (a^2 (e x)^m (a-b x)^n (a+b x)^n-\frac{2 a b (e x)^{1+m} (a-b x)^n (a+b x)^n}{e}+\frac{b^2 (e x)^{2+m} (a-b x)^n (a+b x)^n}{e^2}\right ) \, dx\\ &=a^2 \int (e x)^m (a-b x)^n (a+b x)^n \, dx+\frac{b^2 \int (e x)^{2+m} (a-b x)^n (a+b x)^n \, dx}{e^2}-\frac{(2 a b) \int (e x)^{1+m} (a-b x)^n (a+b x)^n \, dx}{e}\\ &=\left (a^2 (a-b x)^n (a+b x)^n \left (a^2-b^2 x^2\right )^{-n}\right ) \int (e x)^m \left (a^2-b^2 x^2\right )^n \, dx+\frac{\left (b^2 (a-b x)^n (a+b x)^n \left (a^2-b^2 x^2\right )^{-n}\right ) \int (e x)^{2+m} \left (a^2-b^2 x^2\right )^n \, dx}{e^2}-\frac{\left (2 a b (a-b x)^n (a+b x)^n \left (a^2-b^2 x^2\right )^{-n}\right ) \int (e x)^{1+m} \left (a^2-b^2 x^2\right )^n \, dx}{e}\\ &=\left (a^2 (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n}\right ) \int (e x)^m \left (1-\frac{b^2 x^2}{a^2}\right )^n \, dx+\frac{\left (b^2 (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n}\right ) \int (e x)^{2+m} \left (1-\frac{b^2 x^2}{a^2}\right )^n \, dx}{e^2}-\frac{\left (2 a b (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n}\right ) \int (e x)^{1+m} \left (1-\frac{b^2 x^2}{a^2}\right )^n \, dx}{e}\\ &=\frac{a^2 (e x)^{1+m} (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \, _2F_1\left (\frac{1+m}{2},-n;\frac{3+m}{2};\frac{b^2 x^2}{a^2}\right )}{e (1+m)}-\frac{2 a b (e x)^{2+m} (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \, _2F_1\left (\frac{2+m}{2},-n;\frac{4+m}{2};\frac{b^2 x^2}{a^2}\right )}{e^2 (2+m)}+\frac{b^2 (e x)^{3+m} (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \, _2F_1\left (\frac{3+m}{2},-n;\frac{5+m}{2};\frac{b^2 x^2}{a^2}\right )}{e^3 (3+m)}\\ \end{align*}

Mathematica [A]  time = 0.11497, size = 172, normalized size = 0.82 \[ \frac{x (e x)^m (a-b x)^n (a+b x)^n \left (1-\frac{b^2 x^2}{a^2}\right )^{-n} \left (a^2 \left (m^2+5 m+6\right ) \, _2F_1\left (\frac{m+1}{2},-n;\frac{m+3}{2};\frac{b^2 x^2}{a^2}\right )-b (m+1) x \left (2 a (m+3) \, _2F_1\left (\frac{m+2}{2},-n;\frac{m+4}{2};\frac{b^2 x^2}{a^2}\right )-b (m+2) x \, _2F_1\left (\frac{m+3}{2},-n;\frac{m+5}{2};\frac{b^2 x^2}{a^2}\right )\right )\right )}{(m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*(a - b*x)^n*(a + b*x)^n*(a^2*(6 + 5*m + m^2)*Hypergeometric2F1[(1 + m)/2, -n, (3 + m)/2, (b^2*x^2)/
a^2] - b*(1 + m)*x*(2*a*(3 + m)*Hypergeometric2F1[(2 + m)/2, -n, (4 + m)/2, (b^2*x^2)/a^2] - b*(2 + m)*x*Hyper
geometric2F1[(3 + m)/2, -n, (5 + m)/2, (b^2*x^2)/a^2])))/((1 + m)*(2 + m)*(3 + m)*(1 - (b^2*x^2)/a^2)^n)

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Maple [F]  time = 0.128, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( -bx+a \right ) ^{2+n} \left ( bx+a \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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