3.878 \(\int e^{n \tanh ^{-1}(a+b x)} x \, dx\)

Optimal. Leaf size=114 \[ \frac {2^{n/2} (2 a-n) (-a-b x+1)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (-a-b x+1)\right )}{b^2 (2-n)}-\frac {(-a-b x+1)^{1-\frac {n}{2}} (a+b x+1)^{\frac {n+2}{2}}}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6163, 80, 69} \[ \frac {2^{n/2} (2 a-n) (-a-b x+1)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (-a-b x+1)\right )}{b^2 (2-n)}-\frac {(-a-b x+1)^{1-\frac {n}{2}} (a+b x+1)^{\frac {n+2}{2}}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a + b*x])*x,x]

[Out]

-((1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^((2 + n)/2))/(2*b^2) + (2^(n/2)*(2*a - n)*(1 - a - b*x)^(1 - n/2)*Hype
rgeometric2F1[1 - n/2, -n/2, 2 - n/2, (1 - a - b*x)/2])/(b^2*(2 - n))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 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]

Rule 6163

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1
+ a*c + b*c*x)^(n/2))/(1 - a*c - b*c*x)^(n/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 96, normalized size = 0.84 \[ \frac {(-a-b x+1)^{1-\frac {n}{2}} \left (\frac {b 2^{\frac {n}{2}+1} (n-2 a) \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (-a-b x+1)\right )}{n-2}-b (a+b x+1)^{\frac {n}{2}+1}\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a + b*x])*x,x]

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \left (\frac {b x + a + 1}{b x + a - 1}\right )^{\frac {1}{2} \, n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a))*x,x, algorithm="fricas")

[Out]

integral(x*((b*x + a + 1)/(b*x + a - 1))^(1/2*n), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (\frac {b x + a + 1}{b x + a - 1}\right )^{\frac {1}{2} \, n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a))*x,x, algorithm="giac")

[Out]

integrate(x*((b*x + a + 1)/(b*x + a - 1))^(1/2*n), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \arctanh \left (b x +a \right )} x\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(b*x+a))*x,x)

[Out]

int(exp(n*arctanh(b*x+a))*x,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (\frac {b x + a + 1}{b x + a - 1}\right )^{\frac {1}{2} \, n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a))*x,x, algorithm="maxima")

[Out]

integrate(x*((b*x + a + 1)/(b*x + a - 1))^(1/2*n), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*exp(n*atanh(a + b*x)),x)

[Out]

int(x*exp(n*atanh(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x e^{n \operatorname {atanh}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(b*x+a))*x,x)

[Out]

Integral(x*exp(n*atanh(a + b*x)), x)

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