3.528 \(\int e^{3 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^{7/2} \, dx\)

Optimal. Leaf size=217 \[ -\frac {a^{5/2} x^{7/2} \left (c-\frac {c}{a x}\right )^{7/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{7/2}}-\frac {a^3 x^4 \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{7/2}}{(1-a x)^{7/2}}+\frac {2 a^2 x^3 (a x+1)^{3/2} \left (c-\frac {c}{a x}\right )^{7/2}}{3 (1-a x)^{7/2}}+\frac {4 a x^2 (a x+1)^{5/2} \left (c-\frac {c}{a x}\right )^{7/2}}{3 (1-a x)^{7/2}}-\frac {2 x (a x+1)^{5/2} \left (c-\frac {c}{a x}\right )^{7/2}}{5 (1-a x)^{7/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6134, 6129, 89, 78, 47, 50, 54, 215} \[ -\frac {a^3 x^4 \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{7/2}}{(1-a x)^{7/2}}+\frac {2 a^2 x^3 (a x+1)^{3/2} \left (c-\frac {c}{a x}\right )^{7/2}}{3 (1-a x)^{7/2}}-\frac {a^{5/2} x^{7/2} \left (c-\frac {c}{a x}\right )^{7/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{7/2}}+\frac {4 a x^2 (a x+1)^{5/2} \left (c-\frac {c}{a x}\right )^{7/2}}{3 (1-a x)^{7/2}}-\frac {2 x (a x+1)^{5/2} \left (c-\frac {c}{a x}\right )^{7/2}}{5 (1-a x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*(c - c/(a*x))^(7/2)*x^4*Sqrt[1 + a*x])/(1 - a*x)^(7/2)) + (2*a^2*(c - c/(a*x))^(7/2)*x^3*(1 + a*x)^(3/2
))/(3*(1 - a*x)^(7/2)) - (2*(c - c/(a*x))^(7/2)*x*(1 + a*x)^(5/2))/(5*(1 - a*x)^(7/2)) + (4*a*(c - c/(a*x))^(7
/2)*x^2*(1 + a*x)^(5/2))/(3*(1 - a*x)^(7/2)) - (a^(5/2)*(c - c/(a*x))^(7/2)*x^(7/2)*ArcSinh[Sqrt[a]*Sqrt[x]])/
(1 - a*x)^(7/2)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, 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 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

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 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 6129

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

Rule 6134

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.06, size = 77, normalized size = 0.35 \[ -\frac {2 c^3 \sqrt {c-\frac {c}{a x}} \left (5 a^2 x^2 \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-a x\right )+(10 a x-3) (a x+1)^{5/2}\right )}{15 a^3 x^2 \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*c^3*Sqrt[c - c/(a*x)]*((1 + a*x)^(5/2)*(-3 + 10*a*x) + 5*a^2*x^2*Hypergeometric2F1[-3/2, -1/2, 1/2, -(a*x)
]))/(15*a^3*x^2*Sqrt[1 - a*x])

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 364, normalized size = 1.68 \[ \left [\frac {15 \, {\left (a^{3} c^{3} x^{3} - a^{2} c^{3} x^{2}\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (15 \, a^{3} c^{3} x^{3} + 44 \, a^{2} c^{3} x^{2} + 8 \, a c^{3} x - 6 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{60 \, {\left (a^{4} x^{3} - a^{3} x^{2}\right )}}, -\frac {15 \, {\left (a^{3} c^{3} x^{3} - a^{2} c^{3} x^{2}\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, {\left (15 \, a^{3} c^{3} x^{3} + 44 \, a^{2} c^{3} x^{2} + 8 \, a c^{3} x - 6 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{30 \, {\left (a^{4} x^{3} - a^{3} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(15*(a^3*c^3*x^3 - a^2*c^3*x^2)*sqrt(-c)*log(-(8*a^3*c*x^3 - 7*a*c*x + 4*(2*a^2*x^2 + a*x)*sqrt(-a^2*x^2
+ 1)*sqrt(-c)*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*(15*a^3*c^3*x^3 + 44*a^2*c^3*x^2 + 8*a*c^3*x - 6*c^
3)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^4*x^3 - a^3*x^2), -1/30*(15*(a^3*c^3*x^3 - a^2*c^3*x^2)*sqrt
(c)*arctan(2*sqrt(-a^2*x^2 + 1)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) - 2*(15*a^3*c^3
*x^3 + 44*a^2*c^3*x^2 + 8*a*c^3*x - 6*c^3)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^4*x^3 - a^3*x^2)]

________________________________________________________________________________________

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.06, size = 154, normalized size = 0.71 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{3} \sqrt {-a^{2} x^{2}+1}\, \left (30 a^{\frac {7}{2}} x^{3} \sqrt {-\left (a x +1\right ) x}+15 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) x^{3} a^{3}+88 a^{\frac {5}{2}} x^{2} \sqrt {-\left (a x +1\right ) x}+16 a^{\frac {3}{2}} x \sqrt {-\left (a x +1\right ) x}-12 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}\right )}{30 x^{2} a^{\frac {7}{2}} \left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(c*(a*x-1)/a/x)^(1/2)/x^2*c^3/a^(7/2)*(-a^2*x^2+1)^(1/2)*(30*a^(7/2)*x^3*(-(a*x+1)*x)^(1/2)+15*arctan(1/2
/a^(1/2)*(2*a*x+1)/(-(a*x+1)*x)^(1/2))*x^3*a^3+88*a^(5/2)*x^2*(-(a*x+1)*x)^(1/2)+16*a^(3/2)*x*(-(a*x+1)*x)^(1/
2)-12*a^(1/2)*(-(a*x+1)*x)^(1/2))/(a*x-1)/(-(a*x+1)*x)^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}^{\frac {7}{2}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c-\frac {c}{a\,x}\right )}^{7/2}\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (-1 + \frac {1}{a x}\right )\right )^{\frac {7}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________