3.244 \(\int e^{3 \coth ^{-1}(a x)} (c-a c x)^{7/2} \, dx\)

Optimal. Leaf size=137 \[ \frac {214 \left (\frac {1}{a x}+1\right )^{5/2} (c-a c x)^{7/2}}{315 a^2 x \left (1-\frac {1}{a x}\right )^{7/2}}-\frac {44 \left (\frac {1}{a x}+1\right )^{5/2} (c-a c x)^{7/2}}{63 a \left (1-\frac {1}{a x}\right )^{7/2}}+\frac {2 x \left (\frac {1}{a x}+1\right )^{5/2} (c-a c x)^{7/2}}{9 \left (1-\frac {1}{a x}\right )^{7/2}} \]

[Out]

-44/63*(1+1/a/x)^(5/2)*(-a*c*x+c)^(7/2)/a/(1-1/a/x)^(7/2)+214/315*(1+1/a/x)^(5/2)*(-a*c*x+c)^(7/2)/a^2/(1-1/a/
x)^(7/2)/x+2/9*(1+1/a/x)^(5/2)*x*(-a*c*x+c)^(7/2)/(1-1/a/x)^(7/2)

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Rubi [A]  time = 0.17, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6176, 6181, 89, 78, 37} \[ \frac {214 \left (\frac {1}{a x}+1\right )^{5/2} (c-a c x)^{7/2}}{315 a^2 x \left (1-\frac {1}{a x}\right )^{7/2}}-\frac {44 \left (\frac {1}{a x}+1\right )^{5/2} (c-a c x)^{7/2}}{63 a \left (1-\frac {1}{a x}\right )^{7/2}}+\frac {2 x \left (\frac {1}{a x}+1\right )^{5/2} (c-a c x)^{7/2}}{9 \left (1-\frac {1}{a x}\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcCoth[a*x])*(c - a*c*x)^(7/2),x]

[Out]

(-44*(1 + 1/(a*x))^(5/2)*(c - a*c*x)^(7/2))/(63*a*(1 - 1/(a*x))^(7/2)) + (214*(1 + 1/(a*x))^(5/2)*(c - a*c*x)^
(7/2))/(315*a^2*(1 - 1/(a*x))^(7/2)*x) + (2*(1 + 1/(a*x))^(5/2)*x*(c - a*c*x)^(7/2))/(9*(1 - 1/(a*x))^(7/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 6176

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^p/(x^p*(1 + c/(d
*x))^p), Int[u*x^p*(1 + c/(d*x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^
2, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6181

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Sub
st[Int[((1 + (d*x)/c)^p*(1 + x/a)^(n/2))/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x], x] /; FreeQ[{a, c, d, m, n,
p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{7/2} \, dx &=\frac {(c-a c x)^{7/2} \int e^{3 \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^{7/2} x^{7/2} \, dx}{\left (1-\frac {1}{a x}\right )^{7/2} x^{7/2}}\\ &=-\frac {\left (\left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^2 \left (1+\frac {x}{a}\right )^{3/2}}{x^{11/2}} \, dx,x,\frac {1}{x}\right )}{\left (1-\frac {1}{a x}\right )^{7/2}}\\ &=\frac {2 \left (1+\frac {1}{a x}\right )^{5/2} x (c-a c x)^{7/2}}{9 \left (1-\frac {1}{a x}\right )^{7/2}}-\frac {\left (2 \left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {11}{a}+\frac {9 x}{2 a^2}\right ) \left (1+\frac {x}{a}\right )^{3/2}}{x^{9/2}} \, dx,x,\frac {1}{x}\right )}{9 \left (1-\frac {1}{a x}\right )^{7/2}}\\ &=-\frac {44 \left (1+\frac {1}{a x}\right )^{5/2} (c-a c x)^{7/2}}{63 a \left (1-\frac {1}{a x}\right )^{7/2}}+\frac {2 \left (1+\frac {1}{a x}\right )^{5/2} x (c-a c x)^{7/2}}{9 \left (1-\frac {1}{a x}\right )^{7/2}}-\frac {\left (107 \left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}\right ) \operatorname {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^{3/2}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{63 a^2 \left (1-\frac {1}{a x}\right )^{7/2}}\\ &=-\frac {44 \left (1+\frac {1}{a x}\right )^{5/2} (c-a c x)^{7/2}}{63 a \left (1-\frac {1}{a x}\right )^{7/2}}+\frac {214 \left (1+\frac {1}{a x}\right )^{5/2} (c-a c x)^{7/2}}{315 a^2 \left (1-\frac {1}{a x}\right )^{7/2} x}+\frac {2 \left (1+\frac {1}{a x}\right )^{5/2} x (c-a c x)^{7/2}}{9 \left (1-\frac {1}{a x}\right )^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 69, normalized size = 0.50 \[ -\frac {2 c^3 \sqrt {\frac {1}{a x}+1} (a x+1)^2 \left (35 a^2 x^2-110 a x+107\right ) \sqrt {c-a c x}}{315 a \sqrt {1-\frac {1}{a x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - a*c*x)^(7/2),x]

[Out]

(-2*c^3*Sqrt[1 + 1/(a*x)]*(1 + a*x)^2*Sqrt[c - a*c*x]*(107 - 110*a*x + 35*a^2*x^2))/(315*a*Sqrt[1 - 1/(a*x)])

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fricas [A]  time = 0.60, size = 94, normalized size = 0.69 \[ -\frac {2 \, {\left (35 \, a^{5} c^{3} x^{5} - 5 \, a^{4} c^{3} x^{4} - 118 \, a^{3} c^{3} x^{3} + 26 \, a^{2} c^{3} x^{2} + 211 \, a c^{3} x + 107 \, c^{3}\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{315 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(7/2),x, algorithm="fricas")

[Out]

-2/315*(35*a^5*c^3*x^5 - 5*a^4*c^3*x^4 - 118*a^3*c^3*x^3 + 26*a^2*c^3*x^2 + 211*a*c^3*x + 107*c^3)*sqrt(-a*c*x
+ c)*sqrt((a*x - 1)/(a*x + 1))/(a^2*x - a)

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giac [A]  time = 0.21, size = 107, normalized size = 0.78 \[ -\frac {2 \, {\left (\frac {128 \, \sqrt {2} \sqrt {-c} c^{3}}{\mathrm {sgn}\relax (c)} + \frac {35 \, {\left (a c x + c\right )}^{4} \sqrt {-a c x - c} - 180 \, {\left (a c x + c\right )}^{3} \sqrt {-a c x - c} c + 252 \, {\left (a c x + c\right )}^{2} \sqrt {-a c x - c} c^{2}}{c \mathrm {sgn}\left (-a c x - c\right )}\right )}}{315 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(7/2),x, algorithm="giac")

[Out]

-2/315*(128*sqrt(2)*sqrt(-c)*c^3/sgn(c) + (35*(a*c*x + c)^4*sqrt(-a*c*x - c) - 180*(a*c*x + c)^3*sqrt(-a*c*x -
c)*c + 252*(a*c*x + c)^2*sqrt(-a*c*x - c)*c^2)/(c*sgn(-a*c*x - c)))/a

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maple [A]  time = 0.04, size = 56, normalized size = 0.41 \[ \frac {2 \left (a x +1\right ) \left (35 a^{2} x^{2}-110 a x +107\right ) \left (-a c x +c \right )^{\frac {7}{2}}}{315 a \left (a x -1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(7/2),x)

[Out]

2/315*(a*x+1)*(35*a^2*x^2-110*a*x+107)*(-a*c*x+c)^(7/2)/a/(a*x-1)^2/((a*x-1)/(a*x+1))^(3/2)

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maxima [A]  time = 0.34, size = 90, normalized size = 0.66 \[ -\frac {2 \, {\left (35 \, a^{4} \sqrt {-c} c^{3} x^{4} - 110 \, a^{3} \sqrt {-c} c^{3} x^{3} + 72 \, a^{2} \sqrt {-c} c^{3} x^{2} + 110 \, a \sqrt {-c} c^{3} x - 107 \, \sqrt {-c} c^{3}\right )} {\left (a x + 1\right )}^{\frac {3}{2}}}{315 \, {\left (a x - 1\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(7/2),x, algorithm="maxima")

[Out]

-2/315*(35*a^4*sqrt(-c)*c^3*x^4 - 110*a^3*sqrt(-c)*c^3*x^3 + 72*a^2*sqrt(-c)*c^3*x^2 + 110*a*sqrt(-c)*c^3*x -
107*sqrt(-c)*c^3)*(a*x + 1)^(3/2)/((a*x - 1)*a)

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mupad [B]  time = 1.44, size = 102, normalized size = 0.74 \[ -\frac {2\,c^3\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (35\,a^4\,x^4+30\,a^3\,x^3-88\,a^2\,x^2-62\,a\,x+149\right )}{315\,a}-\frac {512\,c^3\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{315\,a\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a*c*x)^(7/2)/((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

- (2*c^3*(c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1))^(1/2)*(30*a^3*x^3 - 88*a^2*x^2 - 62*a*x + 35*a^4*x^4 + 149))/
(315*a) - (512*c^3*(c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1))^(1/2))/(315*a*(a*x - 1))

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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(1/((a*x-1)/(a*x+1))**(3/2)*(-a*c*x+c)**(7/2),x)

[Out]

Timed out

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