Integrand size = 27, antiderivative size = 130 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx=-\frac {75 \sqrt {c-\frac {c}{a x}} x}{64 a^3}-\frac {25 \sqrt {c-\frac {c}{a x}} x^2}{32 a^2}-\frac {5 \sqrt {c-\frac {c}{a x}} x^3}{8 a}-\frac {1}{4} \sqrt {c-\frac {c}{a x}} x^4-\frac {75 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{64 a^4} \] Output:
-75/64*(c-c/a/x)^(1/2)*x/a^3-25/32*(c-c/a/x)^(1/2)*x^2/a^2-5/8*(c-c/a/x)^( 1/2)*x^3/a-1/4*(c-c/a/x)^(1/2)*x^4-75/64*c^(1/2)*arctanh((c-c/a/x)^(1/2)/c ^(1/2))/a^4
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.38 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx=-\frac {\sqrt {c-\frac {c}{a x}} \left (a^4 x^4+15 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\frac {1}{a x}\right )\right )}{4 a^4} \] Input:
Integrate[E^(2*ArcTanh[a*x])*Sqrt[c - c/(a*x)]*x^3,x]
Output:
-1/4*(Sqrt[c - c/(a*x)]*(a^4*x^4 + 15*Hypergeometric2F1[1/2, 4, 3/2, 1 - 1 /(a*x)]))/a^4
Time = 0.91 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.20, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {6683, 1070, 281, 948, 87, 52, 52, 52, 73, 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.
| \(\displaystyle \int x^3 e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} \, dx\) |
| \(\Big \downarrow \) 6683 |
| \(\displaystyle \int \frac {x^3 (a x+1) \sqrt {c-\frac {c}{a x}}}{1-a x}dx\) |
| \(\Big \downarrow \) 1070 |
| \(\displaystyle \int \frac {x^3 \left (a+\frac {1}{x}\right ) \sqrt {c-\frac {c}{a x}}}{\frac {1}{x}-a}dx\) |
| \(\Big \downarrow \) 281 |
| \(\displaystyle -\frac {c \int \frac {\left (a+\frac {1}{x}\right ) x^3}{\sqrt {c-\frac {c}{a x}}}dx}{a}\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {c \int \frac {\left (a+\frac {1}{x}\right ) x^5}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}}{a}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {c \left (\frac {15}{8} \int \frac {x^4}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}-\frac {a x^4 \sqrt {c-\frac {c}{a x}}}{4 c}\right )}{a}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {c \left (\frac {15}{8} \left (\frac {5 \int \frac {x^3}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}}{6 a}-\frac {x^3 \sqrt {c-\frac {c}{a x}}}{3 c}\right )-\frac {a x^4 \sqrt {c-\frac {c}{a x}}}{4 c}\right )}{a}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {c \left (\frac {15}{8} \left (\frac {5 \left (\frac {3 \int \frac {x^2}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}}{4 a}-\frac {x^2 \sqrt {c-\frac {c}{a x}}}{2 c}\right )}{6 a}-\frac {x^3 \sqrt {c-\frac {c}{a x}}}{3 c}\right )-\frac {a x^4 \sqrt {c-\frac {c}{a x}}}{4 c}\right )}{a}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {c \left (\frac {15}{8} \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {x}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}}{2 a}-\frac {x \sqrt {c-\frac {c}{a x}}}{c}\right )}{4 a}-\frac {x^2 \sqrt {c-\frac {c}{a x}}}{2 c}\right )}{6 a}-\frac {x^3 \sqrt {c-\frac {c}{a x}}}{3 c}\right )-\frac {a x^4 \sqrt {c-\frac {c}{a x}}}{4 c}\right )}{a}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {c \left (\frac {15}{8} \left (\frac {5 \left (\frac {3 \left (-\frac {\int \frac {1}{a-\frac {a}{c x^2}}d\sqrt {c-\frac {c}{a x}}}{c}-\frac {x \sqrt {c-\frac {c}{a x}}}{c}\right )}{4 a}-\frac {x^2 \sqrt {c-\frac {c}{a x}}}{2 c}\right )}{6 a}-\frac {x^3 \sqrt {c-\frac {c}{a x}}}{3 c}\right )-\frac {a x^4 \sqrt {c-\frac {c}{a x}}}{4 c}\right )}{a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {c \left (\frac {15}{8} \left (\frac {5 \left (\frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a \sqrt {c}}-\frac {x \sqrt {c-\frac {c}{a x}}}{c}\right )}{4 a}-\frac {x^2 \sqrt {c-\frac {c}{a x}}}{2 c}\right )}{6 a}-\frac {x^3 \sqrt {c-\frac {c}{a x}}}{3 c}\right )-\frac {a x^4 \sqrt {c-\frac {c}{a x}}}{4 c}\right )}{a}\) |
Input:
Int[E^(2*ArcTanh[a*x])*Sqrt[c - c/(a*x)]*x^3,x]
Output:
(c*(-1/4*(a*Sqrt[c - c/(a*x)]*x^4)/c + (15*(-1/3*(Sqrt[c - c/(a*x)]*x^3)/c + (5*(-1/2*(Sqrt[c - c/(a*x)]*x^2)/c + (3*(-((Sqrt[c - c/(a*x)]*x)/c) - A rcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]]/(a*Sqrt[c])))/(4*a)))/(6*a)))/8))/a
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ Symbol] :> Simp[(b/d)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] && !(IntegerQ[q] & & SimplerQ[a + b*x^n, c + d*x^n])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_.) + (b_.)*(x_)^(n_.))^ (p_.)*((e_) + (f_.)*(x_)^(n_.))^(r_.), x_Symbol] :> Int[x^(m + n*(p + r))*( b + a/x^n)^p*(c + d/x^n)^q*(f + e/x^n)^r, x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && EqQ[mn, -n] && IntegerQ[p] && IntegerQ[r]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Int[u*(c + d/x)^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c, d, p}, x] && EqQ[c^2 - a^2*d^2, 0] && !IntegerQ[p] && IntegerQ[n/2] && !G tQ[c, 0]
Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98
| method | result | size |
| risch | \(-\frac {\left (16 a^{3} x^{3}+40 a^{2} x^{2}+50 a x +75\right ) x \sqrt {\frac {c \left (a x -1\right )}{a x}}}{64 a^{3}}-\frac {75 \ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-a c x}\right ) \sqrt {c \left (a x -1\right ) a x}\, \sqrt {\frac {c \left (a x -1\right )}{a x}}}{128 a^{3} \sqrt {a^{2} c}\, \left (a x -1\right )}\) | \(127\) |
| default | \(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (32 x \left (a \,x^{2}-x \right )^{\frac {3}{2}} a^{\frac {7}{2}}+112 \left (a \,x^{2}-x \right )^{\frac {3}{2}} a^{\frac {5}{2}}+212 \sqrt {a \,x^{2}-x}\, a^{\frac {5}{2}} x +256 \sqrt {x \left (a x -1\right )}\, a^{\frac {3}{2}}-106 \sqrt {a \,x^{2}-x}\, a^{\frac {3}{2}}+128 a \ln \left (\frac {2 \sqrt {x \left (a x -1\right )}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right )-53 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \right )}{128 \sqrt {x \left (a x -1\right )}\, a^{\frac {9}{2}}}\) | \(172\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x^3,x,method=_RETURNVERBOSE)
Output:
-1/64*(16*a^3*x^3+40*a^2*x^2+50*a*x+75)/a^3*x*(c*(a*x-1)/a/x)^(1/2)-75/128 /a^3*ln((-1/2*a*c+a^2*c*x)/(a^2*c)^(1/2)+(a^2*c*x^2-a*c*x)^(1/2))/(a^2*c)^ (1/2)*(c*(a*x-1)*a*x)^(1/2)*(c*(a*x-1)/a/x)^(1/2)/(a*x-1)
Time = 0.08 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.45 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx=\left [-\frac {2 \, {\left (16 \, a^{4} x^{4} + 40 \, a^{3} x^{3} + 50 \, a^{2} x^{2} + 75 \, a x\right )} \sqrt {\frac {a c x - c}{a x}} - 75 \, \sqrt {c} \log \left (-2 \, a c x + 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right )}{128 \, a^{4}}, -\frac {{\left (16 \, a^{4} x^{4} + 40 \, a^{3} x^{3} + 50 \, a^{2} x^{2} + 75 \, a x\right )} \sqrt {\frac {a c x - c}{a x}} - 75 \, \sqrt {-c} \arctan \left (\frac {a \sqrt {-c} x \sqrt {\frac {a c x - c}{a x}}}{a c x - c}\right )}{64 \, a^{4}}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x^3,x, algorithm="fricas" )
Output:
[-1/128*(2*(16*a^4*x^4 + 40*a^3*x^3 + 50*a^2*x^2 + 75*a*x)*sqrt((a*c*x - c )/(a*x)) - 75*sqrt(c)*log(-2*a*c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c))/a^4, -1/64*((16*a^4*x^4 + 40*a^3*x^3 + 50*a^2*x^2 + 75*a*x)*sqrt((a *c*x - c)/(a*x)) - 75*sqrt(-c)*arctan(a*sqrt(-c)*x*sqrt((a*c*x - c)/(a*x)) /(a*c*x - c)))/a^4]
\[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx=- \int \frac {x^{3} \sqrt {c - \frac {c}{a x}}}{a x - 1}\, dx - \int \frac {a x^{4} \sqrt {c - \frac {c}{a x}}}{a x - 1}\, dx \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(c-c/a/x)**(1/2)*x**3,x)
Output:
-Integral(x**3*sqrt(c - c/(a*x))/(a*x - 1), x) - Integral(a*x**4*sqrt(c - c/(a*x))/(a*x - 1), x)
\[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx=\int { -\frac {{\left (a x + 1\right )}^{2} \sqrt {c - \frac {c}{a x}} x^{3}}{a^{2} x^{2} - 1} \,d x } \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x^3,x, algorithm="maxima" )
Output:
-integrate((a*x + 1)^2*sqrt(c - c/(a*x))*x^3/(a^2*x^2 - 1), x)
Time = 0.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.09 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx=-\frac {1}{64} \, \sqrt {a^{2} c x^{2} - a c x} {\left (2 \, {\left (4 \, x {\left (\frac {2 \, x {\left | a \right |}}{a^{2} \mathrm {sgn}\left (x\right )} + \frac {5 \, {\left | a \right |}}{a^{3} \mathrm {sgn}\left (x\right )}\right )} + \frac {25 \, {\left | a \right |}}{a^{4} \mathrm {sgn}\left (x\right )}\right )} x + \frac {75 \, {\left | a \right |}}{a^{5} \mathrm {sgn}\left (x\right )}\right )} - \frac {75 \, \sqrt {c} \log \left ({\left | a \right |} {\left | c \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, a^{4}} + \frac {75 \, \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} \sqrt {c} {\left | a \right |} + a c \right |}\right )}{128 \, a^{4} \mathrm {sgn}\left (x\right )} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x^3,x, algorithm="giac")
Output:
-1/64*sqrt(a^2*c*x^2 - a*c*x)*(2*(4*x*(2*x*abs(a)/(a^2*sgn(x)) + 5*abs(a)/ (a^3*sgn(x))) + 25*abs(a)/(a^4*sgn(x)))*x + 75*abs(a)/(a^5*sgn(x))) - 75/1 28*sqrt(c)*log(abs(a)*abs(c))*sgn(x)/a^4 + 75/128*sqrt(c)*log(abs(-2*(sqrt (a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*sqrt(c)*abs(a) + a*c))/(a^4*sgn(x))
Time = 23.93 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.85 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx=\frac {365\,x^4\,{\left (c-\frac {c}{a\,x}\right )}^{3/2}}{64\,c}-\frac {181\,x^4\,\sqrt {c-\frac {c}{a\,x}}}{64}-\frac {275\,x^4\,{\left (c-\frac {c}{a\,x}\right )}^{5/2}}{64\,c^2}+\frac {75\,x^4\,{\left (c-\frac {c}{a\,x}\right )}^{7/2}}{64\,c^3}+\frac {\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c-\frac {c}{a\,x}}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,75{}\mathrm {i}}{64\,a^4} \] Input:
int(-(x^3*(c - c/(a*x))^(1/2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)
Output:
(365*x^4*(c - c/(a*x))^(3/2))/(64*c) - (181*x^4*(c - c/(a*x))^(1/2))/64 - (275*x^4*(c - c/(a*x))^(5/2))/(64*c^2) + (75*x^4*(c - c/(a*x))^(7/2))/(64* c^3) + (c^(1/2)*atan(((c - c/(a*x))^(1/2)*1i)/c^(1/2))*75i)/(64*a^4)
Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.65 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^3 \, dx=\frac {\sqrt {c}\, \left (-16 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}\, a^{3} x^{3}-40 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}\, a^{2} x^{2}-50 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}\, a x -75 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}-75 \,\mathrm {log}\left (\sqrt {a x -1}+\sqrt {x}\, \sqrt {a}\right )\right )}{64 a^{4}} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x^3,x)
Output:
(sqrt(c)*( - 16*sqrt(x)*sqrt(a)*sqrt(a*x - 1)*a**3*x**3 - 40*sqrt(x)*sqrt( a)*sqrt(a*x - 1)*a**2*x**2 - 50*sqrt(x)*sqrt(a)*sqrt(a*x - 1)*a*x - 75*sqr t(x)*sqrt(a)*sqrt(a*x - 1) - 75*log(sqrt(a*x - 1) + sqrt(x)*sqrt(a))))/(64 *a**4)