3.262 \(\int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx\)

Optimal. Leaf size=116 \[ \frac {16 \sqrt {2} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}-\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {8 c (c-a c x)^{3/2}}{3 a} \]

[Out]

-8/3*c*(-a*c*x+c)^(3/2)/a-4/5*(-a*c*x+c)^(5/2)/a-2/7*(-a*c*x+c)^(7/2)/a/c+16*c^(5/2)*arctanh(1/2*(-a*c*x+c)^(1
/2)*2^(1/2)/c^(1/2))*2^(1/2)/a-16*c^2*(-a*c*x+c)^(1/2)/a

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Rubi [A]  time = 0.13, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6167, 6130, 21, 50, 63, 206} \[ -\frac {16 c^2 \sqrt {c-a c x}}{a}+\frac {16 \sqrt {2} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {8 c (c-a c x)^{3/2}}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*c^2*Sqrt[c - a*c*x])/a - (8*c*(c - a*c*x)^(3/2))/(3*a) - (4*(c - a*c*x)^(5/2))/(5*a) - (2*(c - a*c*x)^(7/
2))/(7*a*c) + (16*Sqrt[2]*c^(5/2)*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/a

Rule 21

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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]
)

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx\\ &=-\int \frac {(1-a x) (c-a c x)^{5/2}}{1+a x} \, dx\\ &=-\frac {\int \frac {(c-a c x)^{7/2}}{1+a x} \, dx}{c}\\ &=-\frac {2 (c-a c x)^{7/2}}{7 a c}-2 \int \frac {(c-a c x)^{5/2}}{1+a x} \, dx\\ &=-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-(4 c) \int \frac {(c-a c x)^{3/2}}{1+a x} \, dx\\ &=-\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-\left (8 c^2\right ) \int \frac {\sqrt {c-a c x}}{1+a x} \, dx\\ &=-\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-\left (16 c^3\right ) \int \frac {1}{(1+a x) \sqrt {c-a c x}} \, dx\\ &=-\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}+\frac {\left (32 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )}{a}\\ &=-\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}+\frac {16 \sqrt {2} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 80, normalized size = 0.69 \[ \frac {2 c^2 \left (\left (15 a^3 x^3-87 a^2 x^2+269 a x-1037\right ) \sqrt {c-a c x}+840 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )\right )}{105 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*(Sqrt[c - a*c*x]*(-1037 + 269*a*x - 87*a^2*x^2 + 15*a^3*x^3) + 840*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c
*x]/(Sqrt[2]*Sqrt[c])]))/(105*a)

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fricas [A]  time = 0.52, size = 182, normalized size = 1.57 \[ \left [\frac {2 \, {\left (420 \, \sqrt {2} c^{\frac {5}{2}} \log \left (\frac {a c x - 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + {\left (15 \, a^{3} c^{2} x^{3} - 87 \, a^{2} c^{2} x^{2} + 269 \, a c^{2} x - 1037 \, c^{2}\right )} \sqrt {-a c x + c}\right )}}{105 \, a}, -\frac {2 \, {\left (840 \, \sqrt {2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) - {\left (15 \, a^{3} c^{2} x^{3} - 87 \, a^{2} c^{2} x^{2} + 269 \, a c^{2} x - 1037 \, c^{2}\right )} \sqrt {-a c x + c}\right )}}{105 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(5/2)*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

[2/105*(420*sqrt(2)*c^(5/2)*log((a*c*x - 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (15*a^3*c^2*x^
3 - 87*a^2*c^2*x^2 + 269*a*c^2*x - 1037*c^2)*sqrt(-a*c*x + c))/a, -2/105*(840*sqrt(2)*sqrt(-c)*c^2*arctan(1/2*
sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) - (15*a^3*c^2*x^3 - 87*a^2*c^2*x^2 + 269*a*c^2*x - 1037*c^2)*sqrt(-a*c*x
+ c))/a]

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giac [A]  time = 0.16, size = 134, normalized size = 1.16 \[ -\frac {16 \, \sqrt {2} c^{3} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c}} + \frac {2 \, {\left (15 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} a^{6} c^{6} - 42 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{6} c^{7} - 140 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{6} c^{8} - 840 \, \sqrt {-a c x + c} a^{6} c^{9}\right )}}{105 \, a^{7} c^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(5/2)*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

-16*sqrt(2)*c^3*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)) + 2/105*(15*(a*c*x - c)^3*sqrt(-a*c
*x + c)*a^6*c^6 - 42*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^6*c^7 - 140*(-a*c*x + c)^(3/2)*a^6*c^8 - 840*sqrt(-a*c*x
+ c)*a^6*c^9)/(a^7*c^7)

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maple [A]  time = 0.04, size = 87, normalized size = 0.75 \[ -\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {2 c \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {4 \left (-a c x +c \right )^{\frac {3}{2}} c^{2}}{3}+8 \sqrt {-a c x +c}\, c^{3}-8 c^{\frac {7}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{c a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/c/a*(1/7*(-a*c*x+c)^(7/2)+2/5*c*(-a*c*x+c)^(5/2)+4/3*(-a*c*x+c)^(3/2)*c^2+8*(-a*c*x+c)^(1/2)*c^3-8*c^(7/2)*
2^(1/2)*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2)))

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maxima [A]  time = 0.41, size = 109, normalized size = 0.94 \[ -\frac {2 \, {\left (420 \, \sqrt {2} c^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 15 \, {\left (-a c x + c\right )}^{\frac {7}{2}} + 42 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c + 140 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} + 840 \, \sqrt {-a c x + c} c^{3}\right )}}{105 \, a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(5/2)*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

-2/105*(420*sqrt(2)*c^(7/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c))) +
15*(-a*c*x + c)^(7/2) + 42*(-a*c*x + c)^(5/2)*c + 140*(-a*c*x + c)^(3/2)*c^2 + 840*sqrt(-a*c*x + c)*c^3)/(a*c)

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mupad [B]  time = 0.07, size = 95, normalized size = 0.82 \[ -\frac {4\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a}-\frac {8\,c\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a}-\frac {16\,c^2\,\sqrt {c-a\,c\,x}}{a}-\frac {2\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a\,c}-\frac {\sqrt {2}\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,16{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (4*(c - a*c*x)^(5/2))/(5*a) - (8*c*(c - a*c*x)^(3/2))/(3*a) - (16*c^2*(c - a*c*x)^(1/2))/a - (2*(c - a*c*x)^
(7/2))/(7*a*c) - (2^(1/2)*c^(5/2)*atan((2^(1/2)*(c - a*c*x)^(1/2)*1i)/(2*c^(1/2)))*16i)/a

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sympy [A]  time = 51.78, size = 110, normalized size = 0.95 \[ - \frac {16 \sqrt {2} c^{3} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{a \sqrt {- c}} - \frac {16 c^{2} \sqrt {- a c x + c}}{a} - \frac {8 c \left (- a c x + c\right )^{\frac {3}{2}}}{3 a} - \frac {4 \left (- a c x + c\right )^{\frac {5}{2}}}{5 a} - \frac {2 \left (- a c x + c\right )^{\frac {7}{2}}}{7 a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*sqrt(2)*c**3*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/(a*sqrt(-c)) - 16*c**2*sqrt(-a*c*x + c)/a - 8*c*(
-a*c*x + c)**(3/2)/(3*a) - 4*(-a*c*x + c)**(5/2)/(5*a) - 2*(-a*c*x + c)**(7/2)/(7*a*c)

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