Integrand size = 16, antiderivative size = 60 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx=-\frac {b c}{6 x^{3/2}}-\frac {b c^3}{2 \sqrt {x}}+\frac {1}{2} b c^4 \text {arctanh}\left (c \sqrt {x}\right )-\frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{2 x^2} \] Output:
-1/6*b*c/x^(3/2)-1/2*b*c^3/x^(1/2)+1/2*b*c^4*arctanh(c*x^(1/2))-1/2*(a+b*a rctanh(c*x^(1/2)))/x^2
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.43 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {b c}{6 x^{3/2}}-\frac {b c^3}{2 \sqrt {x}}-\frac {b \text {arctanh}\left (c \sqrt {x}\right )}{2 x^2}-\frac {1}{4} b c^4 \log \left (1-c \sqrt {x}\right )+\frac {1}{4} b c^4 \log \left (1+c \sqrt {x}\right ) \] Input:
Integrate[(a + b*ArcTanh[c*Sqrt[x]])/x^3,x]
Output:
-1/2*a/x^2 - (b*c)/(6*x^(3/2)) - (b*c^3)/(2*Sqrt[x]) - (b*ArcTanh[c*Sqrt[x ]])/(2*x^2) - (b*c^4*Log[1 - c*Sqrt[x]])/4 + (b*c^4*Log[1 + c*Sqrt[x]])/4
Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6452, 61, 61, 73, 219}
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.
\(\displaystyle \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{4} b c \int \frac {1}{x^{5/2} \left (1-c^2 x\right )}dx-\frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{4} b c \left (c^2 \int \frac {1}{x^{3/2} \left (1-c^2 x\right )}dx-\frac {2}{3 x^{3/2}}\right )-\frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{4} b c \left (c^2 \left (c^2 \int \frac {1}{\sqrt {x} \left (1-c^2 x\right )}dx-\frac {2}{\sqrt {x}}\right )-\frac {2}{3 x^{3/2}}\right )-\frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} b c \left (c^2 \left (2 c^2 \int \frac {1}{1-c^2 x}d\sqrt {x}-\frac {2}{\sqrt {x}}\right )-\frac {2}{3 x^{3/2}}\right )-\frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} b c \left (c^2 \left (2 c \text {arctanh}\left (c \sqrt {x}\right )-\frac {2}{\sqrt {x}}\right )-\frac {2}{3 x^{3/2}}\right )-\frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{2 x^2}\) |
Input:
Int[(a + b*ArcTanh[c*Sqrt[x]])/x^3,x]
Output:
-1/2*(a + b*ArcTanh[c*Sqrt[x]])/x^2 + (b*c*(-2/(3*x^(3/2)) + c^2*(-2/Sqrt[ x] + 2*c*ArcTanh[c*Sqrt[x]])))/4
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] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
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_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08
method | result | size |
parts | \(-\frac {a}{2 x^{2}}+2 b \,c^{4} \left (-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right )}{4 c^{4} x^{2}}-\frac {\ln \left (c \sqrt {x}-1\right )}{8}+\frac {\ln \left (1+c \sqrt {x}\right )}{8}-\frac {1}{12 c^{3} x^{\frac {3}{2}}}-\frac {1}{4 c \sqrt {x}}\right )\) | \(65\) |
derivativedivides | \(2 c^{4} \left (-\frac {a}{4 c^{4} x^{2}}+b \left (-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right )}{4 c^{4} x^{2}}-\frac {\ln \left (c \sqrt {x}-1\right )}{8}+\frac {\ln \left (1+c \sqrt {x}\right )}{8}-\frac {1}{12 c^{3} x^{\frac {3}{2}}}-\frac {1}{4 c \sqrt {x}}\right )\right )\) | \(69\) |
default | \(2 c^{4} \left (-\frac {a}{4 c^{4} x^{2}}+b \left (-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right )}{4 c^{4} x^{2}}-\frac {\ln \left (c \sqrt {x}-1\right )}{8}+\frac {\ln \left (1+c \sqrt {x}\right )}{8}-\frac {1}{12 c^{3} x^{\frac {3}{2}}}-\frac {1}{4 c \sqrt {x}}\right )\right )\) | \(69\) |
Input:
int((a+b*arctanh(c*x^(1/2)))/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2*a/x^2+2*b*c^4*(-1/4/c^4/x^2*arctanh(c*x^(1/2))-1/8*ln(c*x^(1/2)-1)+1/ 8*ln(1+c*x^(1/2))-1/12/c^3/x^(3/2)-1/4/c/x^(1/2))
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx=\frac {3 \, {\left (b c^{4} x^{2} - b\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) - 2 \, {\left (3 \, b c^{3} x + b c\right )} \sqrt {x} - 6 \, a}{12 \, x^{2}} \] Input:
integrate((a+b*arctanh(c*x^(1/2)))/x^3,x, algorithm="fricas")
Output:
1/12*(3*(b*c^4*x^2 - b)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) - 2*(3 *b*c^3*x + b*c)*sqrt(x) - 6*a)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (54) = 108\).
Time = 7.34 (sec) , antiderivative size = 342, normalized size of antiderivative = 5.70 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx=\begin {cases} - \frac {a}{2 x^{2}} + \frac {b \operatorname {atanh}{\left (\sqrt {x} \sqrt {\frac {1}{x}} \right )}}{2 x^{2}} & \text {for}\: c = - \sqrt {\frac {1}{x}} \\- \frac {a}{2 x^{2}} - \frac {b \operatorname {atanh}{\left (\sqrt {x} \sqrt {\frac {1}{x}} \right )}}{2 x^{2}} & \text {for}\: c = \sqrt {\frac {1}{x}} \\- \frac {3 a c^{2} x^{\frac {3}{2}}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {3 a \sqrt {x}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {3 b c^{6} x^{\frac {7}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 b c^{5} x^{3}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 b c^{4} x^{\frac {5}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {2 b c^{3} x^{2}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 b c^{2} x^{\frac {3}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {b c x}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {3 b \sqrt {x} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atanh(c*x**(1/2)))/x**3,x)
Output:
Piecewise((-a/(2*x**2) + b*atanh(sqrt(x)*sqrt(1/x))/(2*x**2), Eq(c, -sqrt( 1/x))), (-a/(2*x**2) - b*atanh(sqrt(x)*sqrt(1/x))/(2*x**2), Eq(c, sqrt(1/x ))), (-3*a*c**2*x**(3/2)/(6*c**2*x**(7/2) - 6*x**(5/2)) + 3*a*sqrt(x)/(6*c **2*x**(7/2) - 6*x**(5/2)) + 3*b*c**6*x**(7/2)*atanh(c*sqrt(x))/(6*c**2*x* *(7/2) - 6*x**(5/2)) - 3*b*c**5*x**3/(6*c**2*x**(7/2) - 6*x**(5/2)) - 3*b* c**4*x**(5/2)*atanh(c*sqrt(x))/(6*c**2*x**(7/2) - 6*x**(5/2)) + 2*b*c**3*x **2/(6*c**2*x**(7/2) - 6*x**(5/2)) - 3*b*c**2*x**(3/2)*atanh(c*sqrt(x))/(6 *c**2*x**(7/2) - 6*x**(5/2)) + b*c*x/(6*c**2*x**(7/2) - 6*x**(5/2)) + 3*b* sqrt(x)*atanh(c*sqrt(x))/(6*c**2*x**(7/2) - 6*x**(5/2)), True))
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx=\frac {1}{12} \, {\left ({\left (3 \, c^{3} \log \left (c \sqrt {x} + 1\right ) - 3 \, c^{3} \log \left (c \sqrt {x} - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x + 1\right )}}{x^{\frac {3}{2}}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c \sqrt {x}\right )}{x^{2}}\right )} b - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*arctanh(c*x^(1/2)))/x^3,x, algorithm="maxima")
Output:
1/12*((3*c^3*log(c*sqrt(x) + 1) - 3*c^3*log(c*sqrt(x) - 1) - 2*(3*c^2*x + 1)/x^(3/2))*c - 6*arctanh(c*sqrt(x))/x^2)*b - 1/2*a/x^2
Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (44) = 88\).
Time = 0.14 (sec) , antiderivative size = 356, normalized size of antiderivative = 5.93 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx=\frac {2}{3} \, c {\left (\frac {3 \, {\left (\frac {{\left (c \sqrt {x} + 1\right )}^{3} b c^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {{\left (c \sqrt {x} + 1\right )} b c^{3}}{c \sqrt {x} - 1}\right )} \log \left (-\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1}\right )}{\frac {{\left (c \sqrt {x} + 1\right )}^{4}}{{\left (c \sqrt {x} - 1\right )}^{4}} + \frac {4 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {4 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 1} + \frac {\frac {6 \, {\left (c \sqrt {x} + 1\right )}^{3} a c^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )} a c^{3}}{c \sqrt {x} - 1} + \frac {3 \, {\left (c \sqrt {x} + 1\right )}^{3} b c^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}^{2} b c^{3}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {5 \, {\left (c \sqrt {x} + 1\right )} b c^{3}}{c \sqrt {x} - 1} + 2 \, b c^{3}}{\frac {{\left (c \sqrt {x} + 1\right )}^{4}}{{\left (c \sqrt {x} - 1\right )}^{4}} + \frac {4 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {4 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 1}\right )} \] Input:
integrate((a+b*arctanh(c*x^(1/2)))/x^3,x, algorithm="giac")
Output:
2/3*c*(3*((c*sqrt(x) + 1)^3*b*c^3/(c*sqrt(x) - 1)^3 + (c*sqrt(x) + 1)*b*c^ 3/(c*sqrt(x) - 1))*log(-(c*sqrt(x) + 1)/(c*sqrt(x) - 1))/((c*sqrt(x) + 1)^ 4/(c*sqrt(x) - 1)^4 + 4*(c*sqrt(x) + 1)^3/(c*sqrt(x) - 1)^3 + 6*(c*sqrt(x) + 1)^2/(c*sqrt(x) - 1)^2 + 4*(c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 1) + (6*(c *sqrt(x) + 1)^3*a*c^3/(c*sqrt(x) - 1)^3 + 6*(c*sqrt(x) + 1)*a*c^3/(c*sqrt( x) - 1) + 3*(c*sqrt(x) + 1)^3*b*c^3/(c*sqrt(x) - 1)^3 + 6*(c*sqrt(x) + 1)^ 2*b*c^3/(c*sqrt(x) - 1)^2 + 5*(c*sqrt(x) + 1)*b*c^3/(c*sqrt(x) - 1) + 2*b* c^3)/((c*sqrt(x) + 1)^4/(c*sqrt(x) - 1)^4 + 4*(c*sqrt(x) + 1)^3/(c*sqrt(x) - 1)^3 + 6*(c*sqrt(x) + 1)^2/(c*sqrt(x) - 1)^2 + 4*(c*sqrt(x) + 1)/(c*sqr t(x) - 1) + 1))
Time = 4.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx=\frac {b\,c^4\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{2}-\frac {b\,\left (3\,\ln \left (c\,\sqrt {x}+1\right )-3\,\ln \left (1-c\,\sqrt {x}\right )+2\,c\,\sqrt {x}+6\,c^3\,x^{3/2}\right )}{12\,x^2}-\frac {a}{2\,x^2} \] Input:
int((a + b*atanh(c*x^(1/2)))/x^3,x)
Output:
(b*c^4*atanh(c*x^(1/2)))/2 - (b*(3*log(c*x^(1/2) + 1) - 3*log(1 - c*x^(1/2 )) + 2*c*x^(1/2) + 6*c^3*x^(3/2)))/(12*x^2) - a/(2*x^2)
Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{x^3} \, dx=\frac {3 \mathit {atanh} \left (\sqrt {x}\, c \right ) b \,c^{4} x^{2}-3 \mathit {atanh} \left (\sqrt {x}\, c \right ) b -3 \sqrt {x}\, b \,c^{3} x -\sqrt {x}\, b c -3 a}{6 x^{2}} \] Input:
int((a+b*atanh(c*x^(1/2)))/x^3,x)
Output:
(3*atanh(sqrt(x)*c)*b*c**4*x**2 - 3*atanh(sqrt(x)*c)*b - 3*sqrt(x)*b*c**3* x - sqrt(x)*b*c - 3*a)/(6*x**2)