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

Optimal. Leaf size=71 \[ \frac {8 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 a (c-a c x)^{3/2}}+\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt {c-a c x}} \]

[Out]

8/15*c^3*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(3/2)+2/5*c^2*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6127, 657, 649} \[ \frac {8 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 a (c-a c x)^{3/2}}+\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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^{\tanh ^{-1}(a x)} (c-a c x)^{3/2} \, dx &=c \int \sqrt {c-a c x} \sqrt {1-a^2 x^2} \, dx\\ &=\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt {c-a c x}}+\frac {1}{5} \left (4 c^2\right ) \int \frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}} \, dx\\ &=\frac {8 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 a (c-a c x)^{3/2}}+\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt {c-a c x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.62 \[ -\frac {2 c (a x+1)^{3/2} (3 a x-7) \sqrt {c-a c x}}{15 a \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*c*(1 + a*x)^(3/2)*(-7 + 3*a*x)*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - a*x])

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fricas [A]  time = 0.45, size = 52, normalized size = 0.73 \[ \frac {2 \, {\left (3 \, a^{2} c x^{2} - 4 \, a c x - 7 \, c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{15 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^2*c*x^2 - 4*a*c*x - 7*c)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^2*x - a)

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.03, size = 47, normalized size = 0.66 \[ \frac {2 \left (a x +1\right )^{2} \left (3 a x -7\right ) \left (-a c x +c \right )^{\frac {3}{2}}}{15 a \left (a x -1\right ) \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(a*x+1)^2*(3*a*x-7)*(-a*c*x+c)^(3/2)/a/(a*x-1)/(-a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.35, size = 82, normalized size = 1.15 \[ -\frac {2 \, {\left (a^{3} c^{\frac {3}{2}} x^{3} - 2 \, a^{2} c^{\frac {3}{2}} x^{2} + 3 \, a c^{\frac {3}{2}} x + 6 \, c^{\frac {3}{2}}\right )}}{5 \, \sqrt {a x + 1} a} - \frac {2 \, {\left (a^{2} c^{\frac {3}{2}} x^{2} - 4 \, a c^{\frac {3}{2}} x - 5 \, c^{\frac {3}{2}}\right )}}{3 \, \sqrt {a x + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(a^3*c^(3/2)*x^3 - 2*a^2*c^(3/2)*x^2 + 3*a*c^(3/2)*x + 6*c^(3/2))/(sqrt(a*x + 1)*a) - 2/3*(a^2*c^(3/2)*x^
2 - 4*a*c^(3/2)*x - 5*c^(3/2))/(sqrt(a*x + 1)*a)

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mupad [B]  time = 0.93, size = 49, normalized size = 0.69 \[ \frac {\sqrt {c-a\,c\,x}\,\left (\frac {22\,c\,x}{15}+\frac {14\,c}{15\,a}-\frac {2\,a^2\,c\,x^3}{5}+\frac {2\,a\,c\,x^2}{15}\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c - a*c*x)^(1/2)*((22*c*x)/15 + (14*c)/(15*a) - (2*a^2*c*x^3)/5 + (2*a*c*x^2)/15))/(1 - a^2*x^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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