Integrand size = 20, antiderivative size = 157 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{4 a (c-a c x)^{5/2}}+\frac {\sqrt {1-a^2 x^2}}{16 a c (c-a c x)^{3/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a (c-a c x)^{9/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{16 \sqrt {2} a c^{5/2}} \] Output:
-1/4*(-a^2*x^2+1)^(1/2)/a/(-a*c*x+c)^(5/2)+1/16*(-a^2*x^2+1)^(1/2)/a/c/(-a *c*x+c)^(3/2)+1/3*c^2*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(9/2)+1/32*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^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {(1+a x)^{5/2} (c-a c x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},4,\frac {7}{2},\frac {1}{2} (1+a x)\right )}{40 a c^4 (1-a x)^{3/2}} \] Input:
Integrate[E^(3*ArcTanh[a*x])/(c - a*c*x)^(5/2),x]
Output:
((1 + a*x)^(5/2)*(c - a*c*x)^(3/2)*Hypergeometric2F1[5/2, 4, 7/2, (1 + a*x )/2])/(40*a*c^4*(1 - a*x)^(3/2))
Time = 0.57 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6677, 465, 465, 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.
| \(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6677 |
| \(\displaystyle c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(c-a c x)^{11/2}}dx\) |
| \(\Big \downarrow \) 465 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c (c-a c x)^{9/2}}-\frac {\int \frac {\sqrt {1-a^2 x^2}}{(c-a c x)^{7/2}}dx}{2 c^2}\right )\) |
| \(\Big \downarrow \) 465 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c (c-a c x)^{9/2}}-\frac {\frac {\sqrt {1-a^2 x^2}}{2 a c (c-a c x)^{5/2}}-\frac {\int \frac {1}{(c-a c x)^{3/2} \sqrt {1-a^2 x^2}}dx}{4 c^2}}{2 c^2}\right )\) |
| \(\Big \downarrow \) 470 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c (c-a c x)^{9/2}}-\frac {\frac {\sqrt {1-a^2 x^2}}{2 a c (c-a c x)^{5/2}}-\frac {\frac {\int \frac {1}{\sqrt {c-a c x} \sqrt {1-a^2 x^2}}dx}{4 c}+\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}}{4 c^2}}{2 c^2}\right )\) |
| \(\Big \downarrow \) 471 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c (c-a c x)^{9/2}}-\frac {\frac {\sqrt {1-a^2 x^2}}{2 a c (c-a c x)^{5/2}}-\frac {\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}-\frac {1}{2} a \int \frac {1}{\frac {a^2 c^2 \left (1-a^2 x^2\right )}{c-a c x}-2 a^2 c}d\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}}{4 c^2}}{2 c^2}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c (c-a c x)^{9/2}}-\frac {\frac {\sqrt {1-a^2 x^2}}{2 a c (c-a c x)^{5/2}}-\frac {\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{2 \sqrt {2} a c^{3/2}}+\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}}{4 c^2}}{2 c^2}\right )\) |
Input:
Int[E^(3*ArcTanh[a*x])/(c - a*c*x)^(5/2),x]
Output:
c^3*((1 - a^2*x^2)^(3/2)/(3*a*c*(c - a*c*x)^(9/2)) - (Sqrt[1 - a^2*x^2]/(2 *a*c*(c - a*c*x)^(5/2)) - (Sqrt[1 - a^2*x^2]/(2*a*(c - a*c*x)^(3/2)) + Arc Tanh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])]/(2*Sqrt[2]*a*c ^(3/2)))/(4*c^2))/(2*c^2))
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[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + p + 1))), x] - Simp[b*(p/(d^2*(n + p + 1))) Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n + 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
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 = 0.12 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.32
| method | result | size |
| default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a^{3} c \,x^{3}-9 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a^{2} c \,x^{2}-6 a^{2} x^{2} \sqrt {c}\, \sqrt {c \left (a x +1\right )}+9 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a c x -44 a x \sqrt {c \left (a x +1\right )}\, \sqrt {c}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -14 \sqrt {c \left (a x +1\right )}\, \sqrt {c}\right )}{96 c^{\frac {7}{2}} \left (a x -1\right )^{4} \sqrt {c \left (a x +1\right )}\, a}\) | \(208\) |
Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/96*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(7/2)*(3*2^(1/2)*arctanh(1/2 *(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*a^3*c*x^3-9*2^(1/2)*arctanh(1/2*(c*(a* x+1))^(1/2)*2^(1/2)/c^(1/2))*a^2*c*x^2-6*a^2*x^2*c^(1/2)*(c*(a*x+1))^(1/2) +9*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*a*c*x-44*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)^4/(c*(a*x+1))^(1/2)/a
Time = 0.09 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.30 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, {\left (3 \, a^{2} x^{2} + 22 \, a x + 7\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{192 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}}, \frac {3 \, \sqrt {2} {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{2 \, {\left (a c x - c\right )}}\right ) + 2 \, {\left (3 \, a^{2} x^{2} + 22 \, a x + 7\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{96 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}}\right ] \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(5/2),x, algorithm="fric as")
Output:
[1/192*(3*sqrt(2)*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1)*sqrt(c)*lo g(-(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)) + 4*(3*a^2*x^2 + 22*a*x + 7)*sqrt(-a^2 *x^2 + 1)*sqrt(-a*c*x + c))/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3), 1/96*(3*sqrt(2)*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*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*(3*a^2*x^2 + 22*a*x + 7)*sqrt(-a^2*x^2 + 1)*sqr t(-a*c*x + c))/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)]
\[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int \frac {\left (a x + 1\right )^{3}}{\left (- c \left (a x - 1\right )\right )^{\frac {5}{2}} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)/(-a*c*x+c)**(5/2),x)
Output:
Integral((a*x + 1)**3/((-c*(a*x - 1))**(5/2)*(-(a*x - 1)*(a*x + 1))**(3/2) ), x)
\[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (-a c x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(5/2),x, algorithm="maxi ma")
Output:
integrate((a*x + 1)^3/((-a^2*x^2 + 1)^(3/2)*(-a*c*x + c)^(5/2)), x)
Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.59 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {2 \, {\left (3 \, {\left (a c x + c\right )}^{\frac {5}{2}} + 16 \, {\left (a c x + c\right )}^{\frac {3}{2}} c - 12 \, \sqrt {a c x + c} c^{2}\right )}}{{\left (a c x - c\right )}^{3} c}}{96 \, a {\left | c \right |}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(5/2),x, algorithm="giac ")
Output:
-1/96*(3*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c) + 2*(3*(a*c*x + c)^(5/2) + 16*(a*c*x + c)^(3/2)*c - 12*sqrt(a*c*x + c)*c^ 2)/((a*c*x - c)^3*c))/(a*abs(c))
Timed out. \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int \frac {{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}\,{\left (c-a\,c\,x\right )}^{5/2}} \,d x \] Input:
int((a*x + 1)^3/((1 - a^2*x^2)^(3/2)*(c - a*c*x)^(5/2)),x)
Output:
int((a*x + 1)^3/((1 - a^2*x^2)^(3/2)*(c - a*c*x)^(5/2)), x)
Time = 0.17 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.04 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-6 \sqrt {a x +1}\, a^{2} x^{2}-44 \sqrt {a x +1}\, a x -14 \sqrt {a x +1}-3 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a^{3} x^{3}+9 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a^{2} x^{2}-9 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a x +3 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )\right )}{96 a \,c^{3} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(5/2),x)
Output:
(sqrt(c)*( - 6*sqrt(a*x + 1)*a**2*x**2 - 44*sqrt(a*x + 1)*a*x - 14*sqrt(a* x + 1) - 3*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2))*a**3*x**3 + 9*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2))*a**2*x**2 - 9*sqrt(2) *log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2))*a*x + 3*sqrt(2)*log(tan(asin(s qrt( - a*x + 1)/sqrt(2))/2))))/(96*a*c**3*(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))