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

Optimal. Leaf size=181 \[ \frac {5 a^{5/2} x^{7/2} \left (c-\frac {c}{a x}\right )^{7/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{7/2}}-\frac {a^2 x^3 \sqrt {a x+1} (31 a x+18) \left (c-\frac {c}{a x}\right )^{7/2}}{15 (1-a x)^{7/2}}+\frac {2 a x^2 \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{7/2}}{3 (1-a x)^{3/2}}-\frac {2 x \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{7/2}}{5 \sqrt {1-a x}} \]

[Out]

5*a^(5/2)*(c-c/a/x)^(7/2)*x^(7/2)*arcsinh(a^(1/2)*x^(1/2))/(-a*x+1)^(7/2)+2/3*a*(c-c/a/x)^(7/2)*x^2*(a*x+1)^(1
/2)/(-a*x+1)^(3/2)-1/15*a^2*(c-c/a/x)^(7/2)*x^3*(31*a*x+18)*(a*x+1)^(1/2)/(-a*x+1)^(7/2)-2/5*(c-c/a/x)^(7/2)*x
*(a*x+1)^(1/2)/(-a*x+1)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6134, 6129, 97, 150, 143, 54, 215} \[ -\frac {a^2 x^3 \sqrt {a x+1} (31 a x+18) \left (c-\frac {c}{a x}\right )^{7/2}}{15 (1-a x)^{7/2}}+\frac {5 a^{5/2} x^{7/2} \left (c-\frac {c}{a x}\right )^{7/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{7/2}}+\frac {2 a x^2 \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{7/2}}{3 (1-a x)^{3/2}}-\frac {2 x \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{7/2}}{5 \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(c - c/(a*x))^(7/2)*x^2*Sqrt[1 + a*x])/(3*(1 - a*x)^(3/2)) - (2*(c - c/(a*x))^(7/2)*x*Sqrt[1 + a*x])/(5*S
qrt[1 - a*x]) - (a^2*(c - c/(a*x))^(7/2)*x^3*Sqrt[1 + a*x]*(18 + 31*a*x))/(15*(1 - a*x)^(7/2)) + (5*a^(5/2)*(c
- c/(a*x))^(7/2)*x^(7/2)*ArcSinh[Sqrt[a]*Sqrt[x]])/(1 - a*x)^(7/2)

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 97

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

Rule 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

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^{\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^{\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)^3 \sqrt {1+a x}}{x^{7/2}} \, dx}{(1-a x)^{7/2}}\\ &=-\frac {2 \left (c-\frac {c}{a x}\right )^{7/2} x \sqrt {1+a x}}{5 \sqrt {1-a x}}+\frac {\left (2 \left (c-\frac {c}{a x}\right )^{7/2} x^{7/2}\right ) \int \frac {(1-a x)^2 \left (-\frac {5 a}{2}-\frac {7 a^2 x}{2}\right )}{x^{5/2} \sqrt {1+a x}} \, dx}{5 (1-a x)^{7/2}}\\ &=\frac {2 a \left (c-\frac {c}{a x}\right )^{7/2} x^2 \sqrt {1+a x}}{3 (1-a x)^{3/2}}-\frac {2 \left (c-\frac {c}{a x}\right )^{7/2} x \sqrt {1+a x}}{5 \sqrt {1-a x}}+\frac {\left (4 \left (c-\frac {c}{a x}\right )^{7/2} x^{7/2}\right ) \int \frac {(1-a x) \left (\frac {9 a^2}{4}+\frac {31 a^3 x}{4}\right )}{x^{3/2} \sqrt {1+a x}} \, dx}{15 (1-a x)^{7/2}}\\ &=\frac {2 a \left (c-\frac {c}{a x}\right )^{7/2} x^2 \sqrt {1+a x}}{3 (1-a x)^{3/2}}-\frac {2 \left (c-\frac {c}{a x}\right )^{7/2} x \sqrt {1+a x}}{5 \sqrt {1-a x}}-\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2} x^3 \sqrt {1+a x} (18+31 a x)}{15 (1-a x)^{7/2}}+\frac {\left (5 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 {2 a \left (c-\frac {c}{a x}\right )^{7/2} x^2 \sqrt {1+a x}}{3 (1-a x)^{3/2}}-\frac {2 \left (c-\frac {c}{a x}\right )^{7/2} x \sqrt {1+a x}}{5 \sqrt {1-a x}}-\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2} x^3 \sqrt {1+a x} (18+31 a x)}{15 (1-a x)^{7/2}}+\frac {\left (5 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 {2 a \left (c-\frac {c}{a x}\right )^{7/2} x^2 \sqrt {1+a x}}{3 (1-a x)^{3/2}}-\frac {2 \left (c-\frac {c}{a x}\right )^{7/2} x \sqrt {1+a x}}{5 \sqrt {1-a x}}-\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2} x^3 \sqrt {1+a x} (18+31 a x)}{15 (1-a x)^{7/2}}+\frac {5 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*}

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Mathematica [A]  time = 0.07, size = 95, normalized size = 0.52 \[ \frac {c^3 \sqrt {c-\frac {c}{a x}} \left (\sqrt {a x+1} \left (15 a^3 x^3+56 a^2 x^2-28 a x+6\right )-75 a^{5/2} x^{5/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )\right )}{15 a^3 x^2 \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*Sqrt[c - c/(a*x)]*(Sqrt[1 + a*x]*(6 - 28*a*x + 56*a^2*x^2 + 15*a^3*x^3) - 75*a^(5/2)*x^(5/2)*ArcSinh[Sqrt
[a]*Sqrt[x]]))/(15*a^3*x^2*Sqrt[1 - a*x])

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fricas [A]  time = 0.70, size = 364, normalized size = 2.01 \[ \left [\frac {75 \, {\left (a^{3} c^{3} x^{3} - a^{2} c^{3} x^{2}\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) - 4 \, {\left (15 \, a^{3} c^{3} x^{3} + 56 \, a^{2} c^{3} x^{2} - 28 \, a c^{3} x + 6 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{60 \, {\left (a^{4} x^{3} - a^{3} x^{2}\right )}}, \frac {75 \, {\left (a^{3} c^{3} x^{3} - a^{2} c^{3} x^{2}\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, {\left (15 \, a^{3} c^{3} x^{3} + 56 \, a^{2} c^{3} x^{2} - 28 \, a c^{3} x + 6 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{30 \, {\left (a^{4} x^{3} - a^{3} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(75*(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 + 56*a^2*c^3*x^2 - 28*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*(75*(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 + 56*a^2*c^3*x^2 - 28*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)]

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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)/(-a^2*x^2+1)^(1/2)*(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

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maple [A]  time = 0.06, size = 154, normalized size = 0.85 \[ -\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{3} \sqrt {-a^{2} x^{2}+1}\, \left (30 a^{\frac {7}{2}} x^{3} \sqrt {-\left (a x +1\right ) x}+75 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) x^{3} a^{3}+112 a^{\frac {5}{2}} x^{2} \sqrt {-\left (a x +1\right ) x}-56 a^{\frac {3}{2}} x \sqrt {-\left (a x +1\right ) x}+12 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}\right )}{30 x^{2} a^{\frac {7}{2}} \left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(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)+75*arctan(1/
2/a^(1/2)*(2*a*x+1)/(-(a*x+1)*x)^(1/2))*x^3*a^3+112*a^(5/2)*x^2*(-(a*x+1)*x)^(1/2)-56*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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-\frac {c}{a\,x}\right )}^{7/2}\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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