3.247 \(\int e^{3 \tanh ^{-1}(a x)} \sqrt {c-a c x} \, dx\)

Optimal. Leaf size=119 \[ -\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a (c-a c x)^{3/2}}-\frac {4 c \sqrt {1-a^2 x^2}}{a \sqrt {c-a c x}}+\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{a} \]

[Out]

-2/3*c^2*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(3/2)+4*arctanh(1/2*c^(1/2)*(-a^2*x^2+1)^(1/2)*2^(1/2)/(-a*c*x+c)^(1/
2))*2^(1/2)*c^(1/2)/a-4*c*(-a^2*x^2+1)^(1/2)/a/(-a*c*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6127, 665, 661, 208} \[ -\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a (c-a c x)^{3/2}}-\frac {4 c \sqrt {1-a^2 x^2}}{a \sqrt {c-a c x}}+\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 661

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

Rule 665

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 67, normalized size = 0.56 \[ -\frac {2 \sqrt {c-a c x} \left (\sqrt {a x+1} (a x+7)-6 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a x+1}}{\sqrt {2}}\right )\right )}{3 a \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[c - a*c*x]*(Sqrt[1 + a*x]*(7 + a*x) - 6*Sqrt[2]*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]]))/(3*a*Sqrt[1 - a*x])

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 220, normalized size = 1.85 \[ \left [\frac {2 \, {\left (3 \, \sqrt {2} {\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 7\right )}\right )}}{3 \, {\left (a^{2} x - a\right )}}, \frac {2 \, {\left (6 \, \sqrt {2} {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 7\right )}\right )}}{3 \, {\left (a^{2} x - a\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/3*(3*sqrt(2)*(a*x - 1)*sqrt(c)*log(-(a^2*c*x^2 + 2*a*c*x - 2*sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sq
rt(c) - 3*c)/(a^2*x^2 - 2*a*x + 1)) + sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 7))/(a^2*x - a), 2/3*(6*sqrt(
2)*(a*x - 1)*sqrt(-c)*arctan(sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + sqrt(-a^2
*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 7))/(a^2*x - a)]

________________________________________________________________________________________

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 95, normalized size = 0.80 \[ -\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (6 \sqrt {c}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-x a \sqrt {c \left (a x +1\right )}-7 \sqrt {c \left (a x +1\right )}\right )}{3 \left (a x -1\right ) \sqrt {c \left (a x +1\right )}\, a} \]

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)^(1/2),x)

[Out]

-2/3*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(6*c^(1/2)*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))-x
*a*(c*(a*x+1))^(1/2)-7*(c*(a*x+1))^(1/2))/(a*x-1)/(c*(a*x+1))^(1/2)/a

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c-a\,c\,x}\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)**(1/2),x)

[Out]

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

________________________________________________________________________________________