3.261 \(\int e^{-2 \tanh ^{-1}(a x)} (c-a c x)^{7/2} \, dx\)

Optimal. Leaf size=137 \[ -\frac {32 \sqrt {2} c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}+\frac {32 c^3 \sqrt {c-a c x}}{a}+\frac {16 c^2 (c-a c x)^{3/2}}{3 a}+\frac {2 (c-a c x)^{9/2}}{9 a c}+\frac {4 (c-a c x)^{7/2}}{7 a}+\frac {8 c (c-a c x)^{5/2}}{5 a} \]

[Out]

16/3*c^2*(-a*c*x+c)^(3/2)/a+8/5*c*(-a*c*x+c)^(5/2)/a+4/7*(-a*c*x+c)^(7/2)/a+2/9*(-a*c*x+c)^(9/2)/a/c-32*c^(7/2
)*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a+32*c^3*(-a*c*x+c)^(1/2)/a

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Rubi [A]  time = 0.11, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6130, 21, 50, 63, 206} \[ \frac {16 c^2 (c-a c x)^{3/2}}{3 a}+\frac {32 c^3 \sqrt {c-a c x}}{a}-\frac {32 \sqrt {2} c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}+\frac {2 (c-a c x)^{9/2}}{9 a c}+\frac {4 (c-a c x)^{7/2}}{7 a}+\frac {8 c (c-a c x)^{5/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^(7/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(32*c^3*Sqrt[c - a*c*x])/a + (16*c^2*(c - a*c*x)^(3/2))/(3*a) + (8*c*(c - a*c*x)^(5/2))/(5*a) + (4*(c - a*c*x)
^(7/2))/(7*a) + (2*(c - a*c*x)^(9/2))/(9*a*c) - (32*Sqrt[2]*c^(7/2)*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]
)

Rubi steps

\begin {align*} \int e^{-2 \tanh ^{-1}(a x)} (c-a c x)^{7/2} \, dx &=\int \frac {(1-a x) (c-a c x)^{7/2}}{1+a x} \, dx\\ &=\frac {\int \frac {(c-a c x)^{9/2}}{1+a x} \, dx}{c}\\ &=\frac {2 (c-a c x)^{9/2}}{9 a c}+2 \int \frac {(c-a c x)^{7/2}}{1+a x} \, dx\\ &=\frac {4 (c-a c x)^{7/2}}{7 a}+\frac {2 (c-a c x)^{9/2}}{9 a c}+(4 c) \int \frac {(c-a c x)^{5/2}}{1+a x} \, dx\\ &=\frac {8 c (c-a c x)^{5/2}}{5 a}+\frac {4 (c-a c x)^{7/2}}{7 a}+\frac {2 (c-a c x)^{9/2}}{9 a c}+\left (8 c^2\right ) \int \frac {(c-a c x)^{3/2}}{1+a x} \, dx\\ &=\frac {16 c^2 (c-a c x)^{3/2}}{3 a}+\frac {8 c (c-a c x)^{5/2}}{5 a}+\frac {4 (c-a c x)^{7/2}}{7 a}+\frac {2 (c-a c x)^{9/2}}{9 a c}+\left (16 c^3\right ) \int \frac {\sqrt {c-a c x}}{1+a x} \, dx\\ &=\frac {32 c^3 \sqrt {c-a c x}}{a}+\frac {16 c^2 (c-a c x)^{3/2}}{3 a}+\frac {8 c (c-a c x)^{5/2}}{5 a}+\frac {4 (c-a c x)^{7/2}}{7 a}+\frac {2 (c-a c x)^{9/2}}{9 a c}+\left (32 c^4\right ) \int \frac {1}{(1+a x) \sqrt {c-a c x}} \, dx\\ &=\frac {32 c^3 \sqrt {c-a c x}}{a}+\frac {16 c^2 (c-a c x)^{3/2}}{3 a}+\frac {8 c (c-a c x)^{5/2}}{5 a}+\frac {4 (c-a c x)^{7/2}}{7 a}+\frac {2 (c-a c x)^{9/2}}{9 a c}-\frac {\left (64 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )}{a}\\ &=\frac {32 c^3 \sqrt {c-a c x}}{a}+\frac {16 c^2 (c-a c x)^{3/2}}{3 a}+\frac {8 c (c-a c x)^{5/2}}{5 a}+\frac {4 (c-a c x)^{7/2}}{7 a}+\frac {2 (c-a c x)^{9/2}}{9 a c}-\frac {32 \sqrt {2} c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 88, normalized size = 0.64 \[ \frac {2 c^3 \left (\left (35 a^4 x^4-230 a^3 x^3+732 a^2 x^2-1754 a x+6257\right ) \sqrt {c-a c x}-5040 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )\right )}{315 a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a*c*x)^(7/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(2*c^3*(Sqrt[c - a*c*x]*(6257 - 1754*a*x + 732*a^2*x^2 - 230*a^3*x^3 + 35*a^4*x^4) - 5040*Sqrt[2]*Sqrt[c]*ArcT
anh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]))/(315*a)

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fricas [A]  time = 0.51, size = 203, normalized size = 1.48 \[ \left [\frac {2 \, {\left (2520 \, \sqrt {2} c^{\frac {7}{2}} \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + {\left (35 \, a^{4} c^{3} x^{4} - 230 \, a^{3} c^{3} x^{3} + 732 \, a^{2} c^{3} x^{2} - 1754 \, a c^{3} x + 6257 \, c^{3}\right )} \sqrt {-a c x + c}\right )}}{315 \, a}, \frac {2 \, {\left (5040 \, \sqrt {2} \sqrt {-c} c^{3} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) + {\left (35 \, a^{4} c^{3} x^{4} - 230 \, a^{3} c^{3} x^{3} + 732 \, a^{2} c^{3} x^{2} - 1754 \, a c^{3} x + 6257 \, c^{3}\right )} \sqrt {-a c x + c}\right )}}{315 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="fricas")

[Out]

[2/315*(2520*sqrt(2)*c^(7/2)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (35*a^4*c^3*x
^4 - 230*a^3*c^3*x^3 + 732*a^2*c^3*x^2 - 1754*a*c^3*x + 6257*c^3)*sqrt(-a*c*x + c))/a, 2/315*(5040*sqrt(2)*sqr
t(-c)*c^3*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) + (35*a^4*c^3*x^4 - 230*a^3*c^3*x^3 + 732*a^2*c^3*x^
2 - 1754*a*c^3*x + 6257*c^3)*sqrt(-a*c*x + c))/a]

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giac [A]  time = 0.20, size = 161, normalized size = 1.18 \[ \frac {32 \, \sqrt {2} c^{4} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c}} + \frac {2 \, {\left (35 \, {\left (a c x - c\right )}^{4} \sqrt {-a c x + c} a^{8} c^{8} - 90 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} a^{8} c^{9} + 252 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{8} c^{10} + 840 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{8} c^{11} + 5040 \, \sqrt {-a c x + c} a^{8} c^{12}\right )}}{315 \, a^{9} c^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="giac")

[Out]

32*sqrt(2)*c^4*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)) + 2/315*(35*(a*c*x - c)^4*sqrt(-a*c*
x + c)*a^8*c^8 - 90*(a*c*x - c)^3*sqrt(-a*c*x + c)*a^8*c^9 + 252*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^8*c^10 + 840
*(-a*c*x + c)^(3/2)*a^8*c^11 + 5040*sqrt(-a*c*x + c)*a^8*c^12)/(a^9*c^9)

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maple [A]  time = 0.03, size = 101, normalized size = 0.74 \[ \frac {\frac {2 \left (-a c x +c \right )^{\frac {9}{2}}}{9}+\frac {4 \left (-a c x +c \right )^{\frac {7}{2}} c}{7}+\frac {8 \left (-a c x +c \right )^{\frac {5}{2}} c^{2}}{5}+\frac {16 c^{3} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+32 \sqrt {-a c x +c}\, c^{4}-32 c^{\frac {9}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a*c*x+c)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x)

[Out]

2/c/a*(1/9*(-a*c*x+c)^(9/2)+2/7*(-a*c*x+c)^(7/2)*c+4/5*(-a*c*x+c)^(5/2)*c^2+8/3*c^3*(-a*c*x+c)^(3/2)+16*(-a*c*
x+c)^(1/2)*c^4-16*c^(9/2)*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 = 0.46, size = 123, normalized size = 0.90 \[ \frac {2 \, {\left (2520 \, \sqrt {2} c^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 35 \, {\left (-a c x + c\right )}^{\frac {9}{2}} + 90 \, {\left (-a c x + c\right )}^{\frac {7}{2}} c + 252 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c^{2} + 840 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{3} + 5040 \, \sqrt {-a c x + c} c^{4}\right )}}{315 \, a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="maxima")

[Out]

2/315*(2520*sqrt(2)*c^(9/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c))) +
35*(-a*c*x + c)^(9/2) + 90*(-a*c*x + c)^(7/2)*c + 252*(-a*c*x + c)^(5/2)*c^2 + 840*(-a*c*x + c)^(3/2)*c^3 + 50
40*sqrt(-a*c*x + c)*c^4)/(a*c)

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mupad [B]  time = 0.84, size = 112, normalized size = 0.82 \[ \frac {4\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a}+\frac {8\,c\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a}+\frac {32\,c^3\,\sqrt {c-a\,c\,x}}{a}+\frac {16\,c^2\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a}+\frac {2\,{\left (c-a\,c\,x\right )}^{9/2}}{9\,a\,c}+\frac {\sqrt {2}\,c^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,32{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((a^2*x^2 - 1)*(c - a*c*x)^(7/2))/(a*x + 1)^2,x)

[Out]

(4*(c - a*c*x)^(7/2))/(7*a) + (8*c*(c - a*c*x)^(5/2))/(5*a) + (32*c^3*(c - a*c*x)^(1/2))/a + (16*c^2*(c - a*c*
x)^(3/2))/(3*a) + (2*(c - a*c*x)^(9/2))/(9*a*c) + (2^(1/2)*c^(7/2)*atan((2^(1/2)*(c - a*c*x)^(1/2)*1i)/(2*c^(1
/2)))*32i)/a

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sympy [A]  time = 128.96, size = 128, normalized size = 0.93 \[ \frac {32 \sqrt {2} c^{4} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{a \sqrt {- c}} + \frac {32 c^{3} \sqrt {- a c x + c}}{a} + \frac {16 c^{2} \left (- a c x + c\right )^{\frac {3}{2}}}{3 a} + \frac {8 c \left (- a c x + c\right )^{\frac {5}{2}}}{5 a} + \frac {4 \left (- a c x + c\right )^{\frac {7}{2}}}{7 a} + \frac {2 \left (- a c x + c\right )^{\frac {9}{2}}}{9 a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)**(7/2)/(a*x+1)**2*(-a**2*x**2+1),x)

[Out]

32*sqrt(2)*c**4*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/(a*sqrt(-c)) + 32*c**3*sqrt(-a*c*x + c)/a + 16*c**
2*(-a*c*x + c)**(3/2)/(3*a) + 8*c*(-a*c*x + c)**(5/2)/(5*a) + 4*(-a*c*x + c)**(7/2)/(7*a) + 2*(-a*c*x + c)**(9
/2)/(9*a*c)

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