Integrand size = 25, antiderivative size = 129 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {c x \sqrt {c-a^2 c x^2}}{4 a}+\frac {1}{2} a c x^3 \sqrt {c-a^2 c x^2}-\frac {7 \left (c-a^2 c x^2\right )^{3/2}}{15 a^2}-\frac {1}{5} x^2 \left (c-a^2 c x^2\right )^{3/2}+\frac {c^{3/2} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{4 a^2} \] Output:
-1/4*c*x*(-a^2*c*x^2+c)^(1/2)/a+1/2*a*c*x^3*(-a^2*c*x^2+c)^(1/2)-7/15*(-a^ 2*c*x^2+c)^(3/2)/a^2-1/5*x^2*(-a^2*c*x^2+c)^(3/2)+1/4*c^(3/2)*arctan(a*c^( 1/2)*x/(-a^2*c*x^2+c)^(1/2))/a^2
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.75 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {c \sqrt {c-a^2 c x^2} \left (-28-15 a x+16 a^2 x^2+30 a^3 x^3+12 a^4 x^4\right )-15 c^{3/2} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )}{60 a^2} \] Input:
Integrate[E^(2*ArcTanh[a*x])*x*(c - a^2*c*x^2)^(3/2),x]
Output:
(c*Sqrt[c - a^2*c*x^2]*(-28 - 15*a*x + 16*a^2*x^2 + 30*a^3*x^3 + 12*a^4*x^ 4) - 15*c^(3/2)*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))] )/(60*a^2)
Time = 0.64 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6701, 541, 25, 27, 533, 27, 455, 211, 224, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 6701 |
\(\displaystyle c \int x (a x+1)^2 \sqrt {c-a^2 c x^2}dx\) |
\(\Big \downarrow \) 541 |
\(\displaystyle c \left (-\frac {\int -a^2 c x (10 a x+7) \sqrt {c-a^2 c x^2}dx}{5 a^2 c}-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle c \left (\frac {\int a^2 c x (10 a x+7) \sqrt {c-a^2 c x^2}dx}{5 a^2 c}-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (\frac {1}{5} \int x (10 a x+7) \sqrt {c-a^2 c x^2}dx-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
\(\Big \downarrow \) 533 |
\(\displaystyle c \left (\frac {1}{5} \left (\frac {\int 2 a c (14 a x+5) \sqrt {c-a^2 c x^2}dx}{4 a^2 c}-\frac {5 x \left (c-a^2 c x^2\right )^{3/2}}{2 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (\frac {1}{5} \left (\frac {\int (14 a x+5) \sqrt {c-a^2 c x^2}dx}{2 a}-\frac {5 x \left (c-a^2 c x^2\right )^{3/2}}{2 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c \left (\frac {1}{5} \left (\frac {5 \int \sqrt {c-a^2 c x^2}dx-\frac {14 \left (c-a^2 c x^2\right )^{3/2}}{3 a c}}{2 a}-\frac {5 x \left (c-a^2 c x^2\right )^{3/2}}{2 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle c \left (\frac {1}{5} \left (\frac {5 \left (\frac {1}{2} c \int \frac {1}{\sqrt {c-a^2 c x^2}}dx+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )-\frac {14 \left (c-a^2 c x^2\right )^{3/2}}{3 a c}}{2 a}-\frac {5 x \left (c-a^2 c x^2\right )^{3/2}}{2 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle c \left (\frac {1}{5} \left (\frac {5 \left (\frac {1}{2} c \int \frac {1}{\frac {a^2 c x^2}{c-a^2 c x^2}+1}d\frac {x}{\sqrt {c-a^2 c x^2}}+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )-\frac {14 \left (c-a^2 c x^2\right )^{3/2}}{3 a c}}{2 a}-\frac {5 x \left (c-a^2 c x^2\right )^{3/2}}{2 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle c \left (\frac {1}{5} \left (\frac {5 \left (\frac {\sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a}+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )-\frac {14 \left (c-a^2 c x^2\right )^{3/2}}{3 a c}}{2 a}-\frac {5 x \left (c-a^2 c x^2\right )^{3/2}}{2 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 c}\right )\) |
Input:
Int[E^(2*ArcTanh[a*x])*x*(c - a^2*c*x^2)^(3/2),x]
Output:
c*(-1/5*(x^2*(c - a^2*c*x^2)^(3/2))/c + ((-5*x*(c - a^2*c*x^2)^(3/2))/(2*a *c) + ((-14*(c - a^2*c*x^2)^(3/2))/(3*a*c) + 5*((x*Sqrt[c - a^2*c*x^2])/2 + (Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(2*a)))/(2*a))/5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x ] + Simp[1/(b*(m + n + 2*p + 1)) Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) *x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^(n/2) Int[x^m*(c + d*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ [c, 0]) && IGtQ[n/2, 0]
Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {\left (12 a^{4} x^{4}+30 a^{3} x^{3}+16 a^{2} x^{2}-15 a x -28\right ) \left (a^{2} x^{2}-1\right ) c^{2}}{60 a^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}}+\frac {\arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c^{2}}{4 a \sqrt {a^{2} c}}\) | \(101\) |
default | \(\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{5 a^{2} c}-\frac {2 \left (\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4}\right )}{a}-\frac {2 \left (\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}{3}-a c \left (-\frac {\left (-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c \right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{4 a^{2} c}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{2 \sqrt {a^{2} c}}\right )\right )}{a^{2}}\) | \(244\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*x*(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/60*(12*a^4*x^4+30*a^3*x^3+16*a^2*x^2-15*a*x-28)*(a^2*x^2-1)/a^2/(-c*(a^ 2*x^2-1))^(1/2)*c^2+1/4/a/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2 +c)^(1/2))*c^2
Time = 0.15 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.53 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx=\left [\frac {15 \, \sqrt {-c} c \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) + 2 \, {\left (12 \, a^{4} c x^{4} + 30 \, a^{3} c x^{3} + 16 \, a^{2} c x^{2} - 15 \, a c x - 28 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{120 \, a^{2}}, -\frac {15 \, c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - {\left (12 \, a^{4} c x^{4} + 30 \, a^{3} c x^{3} + 16 \, a^{2} c x^{2} - 15 \, a c x - 28 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{60 \, a^{2}}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x*(-a^2*c*x^2+c)^(3/2),x, algorithm="fric as")
Output:
[1/120*(15*sqrt(-c)*c*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)* x - c) + 2*(12*a^4*c*x^4 + 30*a^3*c*x^3 + 16*a^2*c*x^2 - 15*a*c*x - 28*c)* sqrt(-a^2*c*x^2 + c))/a^2, -1/60*(15*c^(3/2)*arctan(sqrt(-a^2*c*x^2 + c)*a *sqrt(c)*x/(a^2*c*x^2 - c)) - (12*a^4*c*x^4 + 30*a^3*c*x^3 + 16*a^2*c*x^2 - 15*a*c*x - 28*c)*sqrt(-a^2*c*x^2 + c))/a^2]
Time = 3.69 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.60 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx=a^{2} c \left (\begin {cases} \sqrt {- a^{2} c x^{2} + c} \left (\frac {x^{4}}{5} - \frac {x^{2}}{15 a^{2}} - \frac {2}{15 a^{4}}\right ) & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + 2 a c \left (\begin {cases} \left (\frac {x^{3}}{4} - \frac {x}{8 a^{2}}\right ) \sqrt {- a^{2} c x^{2} + c} + \frac {c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} c x + 2 \sqrt {- a^{2} c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- a^{2} c}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{8 a^{2}} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \left (\frac {x^{2}}{3} - \frac {1}{3 a^{2}}\right ) \sqrt {- a^{2} c x^{2} + c} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*x*(-a**2*c*x**2+c)**(3/2),x)
Output:
a**2*c*Piecewise((sqrt(-a**2*c*x**2 + c)*(x**4/5 - x**2/(15*a**2) - 2/(15* a**4)), Ne(a**2*c, 0)), (sqrt(c)*x**4/4, True)) + 2*a*c*Piecewise(((x**3/4 - x/(8*a**2))*sqrt(-a**2*c*x**2 + c) + c*Piecewise((log(-2*a**2*c*x + 2*s qrt(-a**2*c)*sqrt(-a**2*c*x**2 + c))/sqrt(-a**2*c), Ne(c, 0)), (x*log(x)/s qrt(-a**2*c*x**2), True))/(8*a**2), Ne(a**2*c, 0)), (sqrt(c)*x**3/3, True) ) + c*Piecewise(((x**2/3 - 1/(3*a**2))*sqrt(-a**2*c*x**2 + c), Ne(a**2*c, 0)), (sqrt(c)*x**2/2, True))
Time = 0.20 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.29 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {1}{60} \, a {\left (\frac {30 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {60 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c x}{a^{2}} + \frac {45 \, \sqrt {-a^{2} c x^{2} + c} c x}{a^{2}} + \frac {45 \, c^{\frac {3}{2}} \arcsin \left (a x\right )}{a^{3}} + \frac {40 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{a^{3}} - \frac {12 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}{a^{3} c} + \frac {120 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c}{a^{3}} - \frac {60 \, c^{3} \arcsin \left (a x - 2\right )}{a^{6} \left (-\frac {c}{a^{2}}\right )^{\frac {3}{2}}}\right )} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x*(-a^2*c*x^2+c)^(3/2),x, algorithm="maxi ma")
Output:
-1/60*a*(30*(-a^2*c*x^2 + c)^(3/2)*x/a^2 - 60*sqrt(a^2*c*x^2 - 4*a*c*x + 3 *c)*c*x/a^2 + 45*sqrt(-a^2*c*x^2 + c)*c*x/a^2 + 45*c^(3/2)*arcsin(a*x)/a^3 + 40*(-a^2*c*x^2 + c)^(3/2)/a^3 - 12*(-a^2*c*x^2 + c)^(5/2)/(a^3*c) + 120 *sqrt(a^2*c*x^2 - 4*a*c*x + 3*c)*c/a^3 - 60*c^3*arcsin(a*x - 2)/(a^6*(-c/a ^2)^(3/2)))
Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.76 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {1}{60} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (2 \, {\left (3 \, {\left (2 \, a^{2} c x + 5 \, a c\right )} x + 8 \, c\right )} x - \frac {15 \, c}{a}\right )} x - \frac {28 \, c}{a^{2}}\right )} - \frac {c^{2} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{4 \, a \sqrt {-c} {\left | a \right |}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x*(-a^2*c*x^2+c)^(3/2),x, algorithm="giac ")
Output:
1/60*sqrt(-a^2*c*x^2 + c)*((2*(3*(2*a^2*c*x + 5*a*c)*x + 8*c)*x - 15*c/a)* x - 28*c/a^2) - 1/4*c^2*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/( a*sqrt(-c)*abs(a))
Timed out. \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx=\int -\frac {x\,{\left (c-a^2\,c\,x^2\right )}^{3/2}\,{\left (a\,x+1\right )}^2}{a^2\,x^2-1} \,d x \] Input:
int(-(x*(c - a^2*c*x^2)^(3/2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)
Output:
int(-(x*(c - a^2*c*x^2)^(3/2)*(a*x + 1)^2)/(a^2*x^2 - 1), x)
Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.78 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {\sqrt {c}\, c \left (15 \mathit {asin} \left (a x \right )+12 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+30 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+16 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-15 \sqrt {-a^{2} x^{2}+1}\, a x -28 \sqrt {-a^{2} x^{2}+1}+28\right )}{60 a^{2}} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*x*(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*c*(15*asin(a*x) + 12*sqrt( - a**2*x**2 + 1)*a**4*x**4 + 30*sqrt( - a**2*x**2 + 1)*a**3*x**3 + 16*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 15*sqrt ( - a**2*x**2 + 1)*a*x - 28*sqrt( - a**2*x**2 + 1) + 28))/(60*a**2)