3.177 \(\int e^{3 \tanh ^{-1}(a x)} (c-a c x)^p \, dx\)

Optimal. Leaf size=65 \[ \frac {4 \sqrt {2} (c-a c x)^{p+1} \, _2F_1\left (-\frac {3}{2},p-\frac {1}{2};p+\frac {1}{2};\frac {1}{2} (1-a x)\right )}{a c (1-2 p) (1-a x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, 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 = {6130, 23, 69} \[ \frac {4 \sqrt {2} (c-a c x)^{p+1} \, _2F_1\left (-\frac {3}{2},p-\frac {1}{2};p+\frac {1}{2};\frac {1}{2} (1-a x)\right )}{a c (1-2 p) (1-a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 6130

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 {(c-a c x)^{3/2} \int (1+a x)^{3/2} (c-a c x)^{-\frac {3}{2}+p} \, dx}{(1-a x)^{3/2}}\\ &=\frac {4 \sqrt {2} (c-a c x)^{1+p} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2}+p;\frac {1}{2}+p;\frac {1}{2} (1-a x)\right )}{a c (1-2 p) (1-a x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 58, normalized size = 0.89 \[ \frac {4 \sqrt {2} (c-a c x)^p \, _2F_1\left (-\frac {3}{2},p-\frac {1}{2};p+\frac {1}{2};\frac {1}{2}-\frac {a x}{2}\right )}{(a-2 a p) \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x + 1\right )} {\left (-a c x + c\right )}^{p}}{a^{2} x^{2} - 2 \, a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right )^{3} \left (-a c x +c \right )^{p}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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