3.245 \(\int e^{3 \tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx\)

Optimal. Leaf size=71 \[ \frac {8 c^5 \left (1-a^2 x^2\right )^{5/2}}{35 a (c-a c x)^{5/2}}+\frac {2 c^4 \left (1-a^2 x^2\right )^{5/2}}{7 a (c-a c x)^{3/2}} \]

[Out]

8/35*c^5*(-a^2*x^2+1)^(5/2)/a/(-a*c*x+c)^(5/2)+2/7*c^4*(-a^2*x^2+1)^(5/2)/a/(-a*c*x+c)^(3/2)

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Rubi [A]  time = 0.07, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6127, 657, 649} \[ \frac {8 c^5 \left (1-a^2 x^2\right )^{5/2}}{35 a (c-a c x)^{5/2}}+\frac {2 c^4 \left (1-a^2 x^2\right )^{5/2}}{7 a (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])*(c - a*c*x)^(5/2),x]

[Out]

(8*c^5*(1 - a^2*x^2)^(5/2))/(35*a*(c - a*c*x)^(5/2)) + (2*c^4*(1 - a^2*x^2)^(5/2))/(7*a*(c - a*c*x)^(3/2))

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*Simplify[m + p])/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rule 6127

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

Rubi steps

\begin {align*} \int e^{3 \tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx &=c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{\sqrt {c-a c x}} \, dx\\ &=\frac {2 c^4 \left (1-a^2 x^2\right )^{5/2}}{7 a (c-a c x)^{3/2}}+\frac {1}{7} \left (4 c^4\right ) \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(c-a c x)^{3/2}} \, dx\\ &=\frac {8 c^5 \left (1-a^2 x^2\right )^{5/2}}{35 a (c-a c x)^{5/2}}+\frac {2 c^4 \left (1-a^2 x^2\right )^{5/2}}{7 a (c-a c x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 46, normalized size = 0.65 \[ -\frac {2 c^2 (a x+1)^{5/2} (5 a x-9) \sqrt {c-a c x}}{35 a \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a*x])*(c - a*c*x)^(5/2),x]

[Out]

(-2*c^2*(1 + a*x)^(5/2)*(-9 + 5*a*x)*Sqrt[c - a*c*x])/(35*a*Sqrt[1 - a*x])

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fricas [A]  time = 0.42, size = 68, normalized size = 0.96 \[ \frac {2 \, {\left (5 \, a^{3} c^{2} x^{3} + a^{2} c^{2} x^{2} - 13 \, a c^{2} x - 9 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{35 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/35*(5*a^3*c^2*x^3 + a^2*c^2*x^2 - 13*a*c^2*x - 9*c^2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^2*x - a)

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giac [A]  time = 0.16, size = 47, normalized size = 0.66 \[ -\frac {2 \, {\left (16 \, \sqrt {2} c^{\frac {3}{2}} + \frac {5 \, {\left (a c x + c\right )}^{\frac {7}{2}} - 14 \, {\left (a c x + c\right )}^{\frac {5}{2}} c}{c^{2}}\right )} c^{2}}{35 \, a {\left | c \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="giac")

[Out]

-2/35*(16*sqrt(2)*c^(3/2) + (5*(a*c*x + c)^(7/2) - 14*(a*c*x + c)^(5/2)*c)/c^2)*c^2/(a*abs(c))

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maple [A]  time = 0.03, size = 47, normalized size = 0.66 \[ \frac {2 \left (a x +1\right )^{4} \left (5 a x -9\right ) \left (-a c x +c \right )^{\frac {5}{2}}}{35 a \left (a x -1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(5/2),x)

[Out]

2/35*(a*x+1)^4*(5*a*x-9)*(-a*c*x+c)^(5/2)/a/(a*x-1)/(-a^2*x^2+1)^(3/2)

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maxima [B]  time = 0.40, size = 164, normalized size = 2.31 \[ -\frac {2 \, {\left (a^{4} c^{\frac {5}{2}} x^{4} - 3 \, a^{3} c^{\frac {5}{2}} x^{3} + 6 \, a^{2} c^{\frac {5}{2}} x^{2} - 24 \, a c^{\frac {5}{2}} x - 48 \, c^{\frac {5}{2}}\right )}}{7 \, \sqrt {a x + 1} a} - \frac {2 \, {\left (3 \, a^{3} c^{\frac {5}{2}} x^{3} - 11 \, a^{2} c^{\frac {5}{2}} x^{2} + 44 \, a c^{\frac {5}{2}} x + 88 \, c^{\frac {5}{2}}\right )}}{5 \, \sqrt {a x + 1} a} - \frac {2 \, {\left (a^{2} c^{\frac {5}{2}} x^{2} - 7 \, a c^{\frac {5}{2}} x - 14 \, c^{\frac {5}{2}}\right )}}{\sqrt {a x + 1} a} - \frac {2 \, {\left (a c^{\frac {5}{2}} x + 3 \, c^{\frac {5}{2}}\right )}}{\sqrt {a x + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="maxima")

[Out]

-2/7*(a^4*c^(5/2)*x^4 - 3*a^3*c^(5/2)*x^3 + 6*a^2*c^(5/2)*x^2 - 24*a*c^(5/2)*x - 48*c^(5/2))/(sqrt(a*x + 1)*a)
- 2/5*(3*a^3*c^(5/2)*x^3 - 11*a^2*c^(5/2)*x^2 + 44*a*c^(5/2)*x + 88*c^(5/2))/(sqrt(a*x + 1)*a) - 2*(a^2*c^(5/
2)*x^2 - 7*a*c^(5/2)*x - 14*c^(5/2))/(sqrt(a*x + 1)*a) - 2*(a*c^(5/2)*x + 3*c^(5/2))/(sqrt(a*x + 1)*a)

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mupad [B]  time = 0.97, size = 68, normalized size = 0.96 \[ \frac {\sqrt {c-a\,c\,x}\,\left (\frac {44\,c^2\,x}{35}+\frac {18\,c^2}{35\,a}+\frac {24\,a\,c^2\,x^2}{35}-\frac {12\,a^2\,c^2\,x^3}{35}-\frac {2\,a^3\,c^2\,x^4}{7}\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - a*c*x)^(5/2)*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)

[Out]

((c - a*c*x)^(1/2)*((44*c^2*x)/35 + (18*c^2)/(35*a) + (24*a*c^2*x^2)/35 - (12*a^2*c^2*x^3)/35 - (2*a^3*c^2*x^4
)/7))/(1 - a^2*x^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (a x - 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(-a*c*x+c)**(5/2),x)

[Out]

Integral((-c*(a*x - 1))**(5/2)*(a*x + 1)**3/(-(a*x - 1)*(a*x + 1))**(3/2), x)

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