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)*(-b*x+a)^(1+n)*(b*x+a)^(1+n)/e/(3+m+2*n)+2*a^2*(2+m+n)*(e*x)^(1+m)*(-b*x+a)^n*(b*x+a)^n*hypergeom
([-n, 1/2+1/2*m],[3/2+1/2*m],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)*(-b*x+a)^n*(
b*x+a)^n*hypergeom([-n, 1+1/2*m],[2+1/2*m],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 = 0.17, 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 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 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 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])

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])

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*}

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Mathematica [A]  time = 0.12, 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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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x + a\right )}^{n} {\left (-b x + a\right )}^{n + 2} \left (e x\right )^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{n} {\left (-b x + a\right )}^{n + 2} \left (e x\right )^{m}\,{d x} \]

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)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (-b x +a \right )^{n +2} \left (b x +a \right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{n} {\left (-b x + a\right )}^{n + 2} \left (e x\right )^{m}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,x\right )}^m\,{\left (a+b\,x\right )}^n\,{\left (a-b\,x\right )}^{n+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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