3.3.16 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a c x)^p \, dx\) [216]

Optimal. Leaf size=65 \[ -\frac {(1-a x)^{3/2} (c-a c x)^{1+p} \, _2F_1\left (\frac {3}{2},\frac {5}{2}+p;\frac {7}{2}+p;\frac {1}{2} (1-a x)\right )}{\sqrt {2} a c (5+2 p)} \]

[Out]

-1/2*(-a*x+1)^(3/2)*(-a*c*x+c)^(1+p)*hypergeom([3/2, 5/2+p],[7/2+p],-1/2*a*x+1/2)/a/c/(5+2*p)*2^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6265, 23, 71} \begin {gather*} -\frac {(1-a x)^{3/2} (c-a c x)^{p+1} \, _2F_1\left (\frac {3}{2},p+\frac {5}{2};p+\frac {7}{2};\frac {1}{2} (1-a x)\right )}{\sqrt {2} a c (2 p+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^p/E^(3*ArcTanh[a*x]),x]

[Out]

-(((1 - a*x)^(3/2)*(c - a*c*x)^(1 + p)*Hypergeometric2F1[3/2, 5/2 + p, 7/2 + p, (1 - a*x)/2])/(Sqrt[2]*a*c*(5
+ 2*p)))

Rule 23

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 6265

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, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !(IntegerQ[p] || GtQ[c, 0]
)

Rubi steps

\begin {align*} \int e^{-3 \tanh ^{-1}(a x)} (c-a c x)^p \, dx &=\int \frac {(1-a x)^{3/2} (c-a c x)^p}{(1+a x)^{3/2}} \, dx\\ &=\frac {(1-a x)^{3/2} \int \frac {(c-a c x)^{\frac {3}{2}+p}}{(1+a x)^{3/2}} \, dx}{(c-a c x)^{3/2}}\\ &=-\frac {(1-a x)^{3/2} (c-a c x)^{1+p} \, _2F_1\left (\frac {3}{2},\frac {5}{2}+p;\frac {7}{2}+p;\frac {1}{2} (1-a x)\right )}{\sqrt {2} a c (5+2 p)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 60, normalized size = 0.92 \begin {gather*} -\frac {(1-a x)^{5/2} (c-a c x)^p \, _2F_1\left (\frac {3}{2},\frac {5}{2}+p;\frac {7}{2}+p;\frac {1}{2}-\frac {a x}{2}\right )}{\sqrt {2} a (5+2 p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - a*c*x)^p/E^(3*ArcTanh[a*x]),x]

[Out]

-(((1 - a*x)^(5/2)*(c - a*c*x)^p*Hypergeometric2F1[3/2, 5/2 + p, 7/2 + p, 1/2 - (a*x)/2])/(Sqrt[2]*a*(5 + 2*p)
))

________________________________________________________________________________________

Maple [F]
time = 1.08, size = 0, normalized size = 0.00 \[\int \frac {\left (-c x a +c \right )^{p} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{\left (a x +1\right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a*c*x+c)^p/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

int((-a*c*x+c)^p/(a*x+1)^3*(-a^2*x^2+1)^(3/2),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*c*x+c)^p/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*(-a*c*x + c)^p/(a*x + 1)^3, 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*c*x+c)^p/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- c \left (a x - 1\right )\right )^{p} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (a x + 1\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)**p/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)

[Out]

Integral((-c*(a*x - 1))**p*(-(a*x - 1)*(a*x + 1))**(3/2)/(a*x + 1)**3, x)

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^p/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (1-a^2\,x^2\right )}^{3/2}\,{\left (c-a\,c\,x\right )}^p}{{\left (a\,x+1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - a^2*x^2)^(3/2)*(c - a*c*x)^p)/(a*x + 1)^3,x)

[Out]

int(((1 - a^2*x^2)^(3/2)*(c - a*c*x)^p)/(a*x + 1)^3, x)

________________________________________________________________________________________