Integrand size = 20, antiderivative size = 125 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{4 a c (c-a c x)^{5/2}}+\frac {3 \sqrt {1-a^2 x^2}}{16 a c^2 (c-a c x)^{3/2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{16 \sqrt {2} a c^{7/2}} \] Output:
1/4*(-a^2*x^2+1)^(1/2)/a/c/(-a*c*x+c)^(5/2)+3/16*(-a^2*x^2+1)^(1/2)/a/c^2/ (-a*c*x+c)^(3/2)+3/32*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)/a/c^(7/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.42 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {1}{2} (1+a x)\right )}{4 a c^3 \sqrt {c-a c x}} \] Input:
Integrate[1/(E^ArcTanh[a*x]*(c - a*c*x)^(7/2)),x]
Output:
(Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, 3, 3/2, (1 + a*x)/2])/(4*a*c^3*S qrt[c - a*c*x])
Time = 0.54 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6677, 470, 470, 471, 221}
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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle \frac {\int \frac {1}{(c-a c x)^{5/2} \sqrt {1-a^2 x^2}}dx}{c}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{(c-a c x)^{3/2} \sqrt {1-a^2 x^2}}dx}{8 c}+\frac {\sqrt {1-a^2 x^2}}{4 a (c-a c x)^{5/2}}}{c}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-a c x} \sqrt {1-a^2 x^2}}dx}{4 c}+\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}\right )}{8 c}+\frac {\sqrt {1-a^2 x^2}}{4 a (c-a c x)^{5/2}}}{c}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle \frac {\frac {3 \left (\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}-\frac {1}{2} a \int \frac {1}{\frac {a^2 c^2 \left (1-a^2 x^2\right )}{c-a c x}-2 a^2 c}d\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right )}{8 c}+\frac {\sqrt {1-a^2 x^2}}{4 a (c-a c x)^{5/2}}}{c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{2 \sqrt {2} a c^{3/2}}+\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}\right )}{8 c}+\frac {\sqrt {1-a^2 x^2}}{4 a (c-a c x)^{5/2}}}{c}\) |
Input:
Int[1/(E^ArcTanh[a*x]*(c - a*c*x)^(7/2)),x]
Output:
(Sqrt[1 - a^2*x^2]/(4*a*(c - a*c*x)^(5/2)) + (3*(Sqrt[1 - a^2*x^2]/(2*a*(c - a*c*x)^(3/2)) + ArcTanh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a *c*x])]/(2*Sqrt[2]*a*c^(3/2))))/(8*c))/c
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 2*p + 2)/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim p[2*d Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] ], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[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]
Time = 0.10 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.26
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a^{2} c \,x^{2}-6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a c x -6 a x \sqrt {c \left (a x +1\right )}\, \sqrt {c}+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +14 \sqrt {c \left (a x +1\right )}\, \sqrt {c}\right )}{32 c^{\frac {9}{2}} \left (a x -1\right )^{3} \sqrt {c \left (a x +1\right )}\, a}\) | \(158\) |
Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/32*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(9/2)*(3*2^(1/2)*arctanh(1/2 *(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*a^2*c*x^2-6*2^(1/2)*arctanh(1/2*(c*(a* x+1))^(1/2)*2^(1/2)/c^(1/2))*a*c*x-6*a*x*(c*(a*x+1))^(1/2)*c^(1/2)+3*2^(1/ 2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*c+14*(c*(a*x+1))^(1/2)*c ^(1/2))/(a*x-1)^3/(c*(a*x+1))^(1/2)/a
Time = 0.09 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.47 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (3 \, a x - 7\right )}}{64 \, {\left (a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x - a c^{4}\right )}}, \frac {3 \, \sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{2 \, {\left (a c x - c\right )}}\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (3 \, a x - 7\right )}}{32 \, {\left (a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x - a c^{4}\right )}}\right ] \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(7/2),x, algorithm="fric as")
Output:
[1/64*(3*sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*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)*sqrt(c) - 3*c)/ (a^2*x^2 - 2*a*x + 1)) + 4*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(3*a*x - 7) )/(a^4*c^4*x^3 - 3*a^3*c^4*x^2 + 3*a^2*c^4*x - a*c^4), 1/32*(3*sqrt(2)*(a^ 3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) + 2*sqrt(-a^2*x^2 + 1)*sqrt(-a* c*x + c)*(3*a*x - 7))/(a^4*c^4*x^3 - 3*a^3*c^4*x^2 + 3*a^2*c^4*x - a*c^4)]
\[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (a x - 1\right )\right )^{\frac {7}{2}} \left (a x + 1\right )}\, dx \] Input:
integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**(7/2),x)
Output:
Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(a*x - 1))**(7/2)*(a*x + 1)), x)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (-a c x + c\right )}^{\frac {7}{2}} {\left (a x + 1\right )}} \,d x } \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(7/2),x, algorithm="maxi ma")
Output:
integrate(sqrt(-a^2*x^2 + 1)/((-a*c*x + c)^(7/2)*(a*x + 1)), x)
Time = 0.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=-\frac {{\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{2}} + \frac {2 \, {\left (3 \, {\left (a c x + c\right )}^{\frac {3}{2}} - 10 \, \sqrt {a c x + c} c\right )}}{{\left (a c x - c\right )}^{2} c^{2}}\right )} {\left | c \right |}}{32 \, a c^{2}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(7/2),x, algorithm="giac ")
Output:
-1/32*(3*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c^ 2) + 2*(3*(a*c*x + c)^(3/2) - 10*sqrt(a*c*x + c)*c)/((a*c*x - c)^2*c^2))*a bs(c)/(a*c^2)
Timed out. \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-a\,c\,x\right )}^{7/2}\,\left (a\,x+1\right )} \,d x \] Input:
int((1 - a^2*x^2)^(1/2)/((c - a*c*x)^(7/2)*(a*x + 1)),x)
Output:
int((1 - a^2*x^2)^(1/2)/((c - a*c*x)^(7/2)*(a*x + 1)), x)
Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\frac {\sqrt {c}\, \left (-6 \sqrt {a x +1}\, a x +14 \sqrt {a x +1}-3 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a^{2} x^{2}+6 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a x -3 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )\right )}{32 a \,c^{4} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(7/2),x)
Output:
(sqrt(c)*( - 6*sqrt(a*x + 1)*a*x + 14*sqrt(a*x + 1) - 3*sqrt(2)*log(tan(as in(sqrt( - a*x + 1)/sqrt(2))/2))*a**2*x**2 + 6*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2))*a*x - 3*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt (2))/2))))/(32*a*c**4*(a**2*x**2 - 2*a*x + 1))