3.5.56 \(\int e^{2 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^p \, dx\) [456]

Optimal. Leaf size=59 \[ -\left (c-\frac {c}{a x}\right )^p x-\frac {(2-p) \left (c-\frac {c}{a x}\right )^p \, _2F_1\left (1,p;1+p;1-\frac {1}{a x}\right )}{a p} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, 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 = {6268, 25, 528, 382, 79, 67} \begin {gather*} -\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 \end {gather*}

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 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 67

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

Rule 79

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] || L
tQ[p, n]))))

Rule 382

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 528

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 6268

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]

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 \text {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)) \text {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 = 0.02, size = 46, normalized size = 0.78 \begin {gather*} \frac {\left (c-\frac {c}{a x}\right )^p \left (-a p x+(-2+p) \, _2F_1\left (1,p;1+p;1-\frac {1}{a x}\right )\right )}{a p} \end {gather*}

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 = 0.75, size = 0, normalized size = 0.00 \[\int \frac {\left (a x +1\right )^{2} \left (c -\frac {c}{a x}\right )^{p}}{-a^{2} x^{2}+1}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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] Result contains complex when optimal does not.
time = 5.23, size = 274, normalized size = 4.64 \begin {gather*} - 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 {gather*}

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int -\frac {{\left (c-\frac {c}{a\,x}\right )}^p\,{\left (a\,x+1\right )}^2}{a^2\,x^2-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________