3.456 \(\int e^{2 \tanh ^{-1}(a x)} (c-\frac{c}{a x})^p \, dx\)

Optimal. Leaf size=59 \[ -\frac{(2-p) \left (c-\frac{c}{a x}\right )^p \text{Hypergeometric2F1}\left (1,p,p+1,1-\frac{1}{a x}\right )}{a p}-x \left (c-\frac{c}{a x}\right )^p \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0803001, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6133, 25, 514, 375, 78, 65} \[ -\frac{(2-p) \left (c-\frac{c}{a x}\right )^p \, _2F_1\left (1,p;p+1;1-\frac{1}{a x}\right )}{a p}-x \left (c-\frac{c}{a x}\right )^p \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])*(c - c/(a*x))^p,x]

[Out]

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

Rule 6133

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Int[(u*(c + d/x)^p*(1 + a*x)^(n/
2))/(1 - a*x)^(n/2), x] /; FreeQ[{a, c, d, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &
&  !GtQ[c, 0]

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 65

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

Rubi steps

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

Mathematica [A]  time = 0.0275809, size = 46, normalized size = 0.78 \[ \frac{\left (c-\frac{c}{a x}\right )^p \left ((p-2) \text{Hypergeometric2F1}\left (1,p,p+1,1-\frac{1}{a x}\right )-a p x\right )}{a p} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])*(c - c/(a*x))^p,x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.335, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1} \left ( c-{\frac{c}{ax}} \right ) ^{p}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^p,x)

[Out]

int((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^p,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (a x + 1\right )}^{2}{\left (c - \frac{c}{a x}\right )}^{p}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^p,x, algorithm="maxima")

[Out]

-integrate((a*x + 1)^2*(c - c/(a*x))^p/(a^2*x^2 - 1), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a x + 1\right )} \left (\frac{a c x - c}{a x}\right )^{p}}{a x - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^p,x, algorithm="fricas")

[Out]

integral(-(a*x + 1)*((a*c*x - c)/(a*x))^p/(a*x - 1), x)

________________________________________________________________________________________

Sympy [C]  time = 7.34323, size = 274, normalized size = 4.64 \begin{align*} - a \left (\begin{cases} \frac{0^{p} x}{a} + \frac{0^{p} \log{\left (a x - 1 \right )}}{a^{2}} - \frac{a^{- p} c^{p} p x^{2} x^{- p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (2 - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, 2 - p \\ 3 - p \end{matrix}\middle |{a x} \right )}}{\Gamma \left (3 - p\right ) \Gamma \left (p + 1\right )} & \text{for}\: \left |{a x}\right | > 1 \\\frac{0^{p} x}{a} + \frac{0^{p} \log{\left (- a x + 1 \right )}}{a^{2}} - \frac{a^{- p} c^{p} p x^{2} x^{- p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (2 - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, 2 - p \\ 3 - p \end{matrix}\middle |{a x} \right )}}{\Gamma \left (3 - p\right ) \Gamma \left (p + 1\right )} & \text{otherwise} \end{cases}\right ) - \begin{cases} \frac{0^{p} \log{\left (a x - 1 \right )}}{a} - \frac{a^{- p} c^{p} p x x^{- p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (1 - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, 1 - p \\ 2 - p \end{matrix}\middle |{a x} \right )}}{\Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} & \text{for}\: \left |{a x}\right | > 1 \\\frac{0^{p} \log{\left (- a x + 1 \right )}}{a} - \frac{a^{- p} c^{p} p x x^{- p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (1 - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, 1 - p \\ 2 - p \end{matrix}\middle |{a x} \right )}}{\Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*(c-c/a/x)**p,x)

[Out]

-a*Piecewise((0**p*x/a + 0**p*log(a*x - 1)/a**2 - a**(-p)*c**p*p*x**2*x**(-p)*exp(I*pi*p)*gamma(p)*gamma(2 - p
)*hyper((1 - p, 2 - p), (3 - p,), a*x)/(gamma(3 - p)*gamma(p + 1)), Abs(a*x) > 1), (0**p*x/a + 0**p*log(-a*x +
1)/a**2 - a**(-p)*c**p*p*x**2*x**(-p)*exp(I*pi*p)*gamma(p)*gamma(2 - p)*hyper((1 - p, 2 - p), (3 - p,), a*x)/
(gamma(3 - p)*gamma(p + 1)), True)) - Piecewise((0**p*log(a*x - 1)/a - a**(-p)*c**p*p*x*x**(-p)*exp(I*pi*p)*ga
mma(p)*gamma(1 - p)*hyper((1 - p, 1 - p), (2 - p,), a*x)/(gamma(2 - p)*gamma(p + 1)), Abs(a*x) > 1), (0**p*log
(-a*x + 1)/a - a**(-p)*c**p*p*x*x**(-p)*exp(I*pi*p)*gamma(p)*gamma(1 - p)*hyper((1 - p, 1 - p), (2 - p,), a*x)
/(gamma(2 - p)*gamma(p + 1)), True))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (a x + 1\right )}^{2}{\left (c - \frac{c}{a x}\right )}^{p}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^p,x, algorithm="giac")

[Out]

integrate(-(a*x + 1)^2*(c - c/(a*x))^p/(a^2*x^2 - 1), x)