3.284 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\)

Optimal. Leaf size=78 \[ \frac {2^{\frac {n}{2}+1} (1-a x)^{-n/2} \, _2F_1\left (\frac {1}{2} (-n-3),-\frac {n}{2};\frac {1}{2} (-n-1);\frac {1}{2} (1-a x)\right )}{a c (n+3) (c-a c x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6130, 23, 69} \[ \frac {2^{\frac {n}{2}+1} (1-a x)^{-n/2} \, _2F_1\left (\frac {1}{2} (-n-3),-\frac {n}{2};\frac {1}{2} (-n-1);\frac {1}{2} (1-a x)\right )}{a c (n+3) (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a*x])/(c - a*c*x)^(5/2),x]

[Out]

(2^(1 + n/2)*Hypergeometric2F1[(-3 - n)/2, -n/2, (-1 - n)/2, (1 - a*x)/2])/(a*c*(3 + n)*(1 - a*x)^(n/2)*(c - a
*c*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 \frac {e^{n \tanh ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx &=\int \frac {(1-a x)^{-n/2} (1+a x)^{n/2}}{(c-a c x)^{5/2}} \, dx\\ &=\left ((1-a x)^{-n/2} (c-a c x)^{n/2}\right ) \int (1+a x)^{n/2} (c-a c x)^{-\frac {5}{2}-\frac {n}{2}} \, dx\\ &=\frac {2^{1+\frac {n}{2}} (1-a x)^{-n/2} \, _2F_1\left (\frac {1}{2} (-3-n),-\frac {n}{2};\frac {1}{2} (-1-n);\frac {1}{2} (1-a x)\right )}{a c (3+n) (c-a c x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 80, normalized size = 1.03 \[ \frac {2^{\frac {n}{2}+1} (1-a x)^{-\frac {n}{2}-1} \, _2F_1\left (-\frac {n}{2}-\frac {3}{2},-\frac {n}{2};-\frac {n}{2}-\frac {1}{2};\frac {1}{2}-\frac {a x}{2}\right )}{a c^2 (n+3) \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a*x])/(c - a*c*x)^(5/2),x]

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a c x + c} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{3} c^{3} x^{3} - 3 \, a^{2} c^{3} x^{2} + 3 \, a c^{3} x - c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a*c*x + c)*((a*x + 1)/(a*x - 1))^(1/2*n)/(a^3*c^3*x^3 - 3*a^2*c^3*x^2 + 3*a*c^3*x - c^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a c x + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate(((a*x + 1)/(a*x - 1))^(1/2*n)/(-a*c*x + c)^(5/2), x)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )}}{\left (-a c x +c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x)

[Out]

int(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a c x + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(((a*x + 1)/(a*x - 1))^(1/2*n)/(-a*c*x + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{{\left (c-a\,c\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a*x))/(c - a*c*x)^(5/2),x)

[Out]

int(exp(n*atanh(a*x))/(c - a*c*x)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))/(-a*c*x+c)**(5/2),x)

[Out]

Timed out

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