Integrand size = 16, antiderivative size = 47 \[ \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx=-\frac {b c}{3 x^{3/2}}+\frac {1}{3} b c^2 \text {arctanh}\left (c x^{3/2}\right )-\frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{3 x^3} \] Output:
-1/3*b*c/x^(3/2)+1/3*b*c^2*arctanh(c*x^(3/2))-1/3*(a+b*arctanh(c*x^(3/2))) /x^3
Leaf count is larger than twice the leaf count of optimal. \(140\) vs. \(2(47)=94\).
Time = 0.03 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.98 \[ \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {b c}{3 x^{3/2}}-\frac {b \text {arctanh}\left (c x^{3/2}\right )}{3 x^3}-\frac {1}{6} b c^2 \log \left (1-\sqrt [3]{c} \sqrt {x}\right )+\frac {1}{6} b c^2 \log \left (1+\sqrt [3]{c} \sqrt {x}\right )+\frac {1}{6} b c^2 \log \left (1-\sqrt [3]{c} \sqrt {x}+c^{2/3} x\right )-\frac {1}{6} b c^2 \log \left (1+\sqrt [3]{c} \sqrt {x}+c^{2/3} x\right ) \] Input:
Integrate[(a + b*ArcTanh[c*x^(3/2)])/x^4,x]
Output:
-1/3*a/x^3 - (b*c)/(3*x^(3/2)) - (b*ArcTanh[c*x^(3/2)])/(3*x^3) - (b*c^2*L og[1 - c^(1/3)*Sqrt[x]])/6 + (b*c^2*Log[1 + c^(1/3)*Sqrt[x]])/6 + (b*c^2*L og[1 - c^(1/3)*Sqrt[x] + c^(2/3)*x])/6 - (b*c^2*Log[1 + c^(1/3)*Sqrt[x] + c^(2/3)*x])/6
Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6452, 847, 851, 807, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{2} b c \int \frac {1}{x^{5/2} \left (1-c^2 x^3\right )}dx-\frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{3 x^3}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {1}{2} b c \left (c^2 \int \frac {\sqrt {x}}{1-c^2 x^3}dx-\frac {2}{3 x^{3/2}}\right )-\frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{3 x^3}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {1}{2} b c \left (2 c^2 \int \frac {x}{1-c^2 x^3}d\sqrt {x}-\frac {2}{3 x^{3/2}}\right )-\frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{3 x^3}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} b c \left (\frac {2}{3} c^2 \int \frac {1}{1-c^2 x}dx^{3/2}-\frac {2}{3 x^{3/2}}\right )-\frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{3 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} b c \left (\frac {2}{3} c \text {arctanh}\left (c x^{3/2}\right )-\frac {2}{3 x^{3/2}}\right )-\frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{3 x^3}\) |
Input:
Int[(a + b*ArcTanh[c*x^(3/2)])/x^4,x]
Output:
-1/3*(a + b*ArcTanh[c*x^(3/2)])/x^3 + (b*c*(-2/(3*x^(3/2)) + (2*c*ArcTanh[ c*x^(3/2)])/3))/2
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
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 = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(-\frac {a}{3 x^{3}}-\frac {b \,\operatorname {arctanh}\left (c \,x^{\frac {3}{2}}\right )}{3 x^{3}}-\frac {b c}{3 x^{\frac {3}{2}}}+\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}+1\right )}{6}-\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}-1\right )}{6}\) | \(55\) |
default | \(-\frac {a}{3 x^{3}}-\frac {b \,\operatorname {arctanh}\left (c \,x^{\frac {3}{2}}\right )}{3 x^{3}}-\frac {b c}{3 x^{\frac {3}{2}}}+\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}+1\right )}{6}-\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}-1\right )}{6}\) | \(55\) |
parts | \(-\frac {a}{3 x^{3}}-\frac {b \,\operatorname {arctanh}\left (c \,x^{\frac {3}{2}}\right )}{3 x^{3}}-\frac {b c}{3 x^{\frac {3}{2}}}+\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}+1\right )}{6}-\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}-1\right )}{6}\) | \(55\) |
Input:
int((a+b*arctanh(c*x^(3/2)))/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*a/x^3-1/3*b/x^3*arctanh(c*x^(3/2))-1/3*b*c/x^(3/2)+1/6*b*c^2*ln(c*x^( 3/2)+1)-1/6*b*c^2*ln(c*x^(3/2)-1)
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx=-\frac {2 \, b c x^{\frac {3}{2}} - {\left (b c^{2} x^{3} - b\right )} \log \left (-\frac {c^{2} x^{3} + 2 \, c x^{\frac {3}{2}} + 1}{c^{2} x^{3} - 1}\right ) + 2 \, a}{6 \, x^{3}} \] Input:
integrate((a+b*arctanh(c*x^(3/2)))/x^4,x, algorithm="fricas")
Output:
-1/6*(2*b*c*x^(3/2) - (b*c^2*x^3 - b)*log(-(c^2*x^3 + 2*c*x^(3/2) + 1)/(c^ 2*x^3 - 1)) + 2*a)/x^3
Timed out. \[ \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx=\text {Timed out} \] Input:
integrate((a+b*atanh(c*x**(3/2)))/x**4,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx=\frac {1}{6} \, {\left ({\left (c \log \left (c x^{\frac {3}{2}} + 1\right ) - c \log \left (c x^{\frac {3}{2}} - 1\right ) - \frac {2}{x^{\frac {3}{2}}}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x^{\frac {3}{2}}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \] Input:
integrate((a+b*arctanh(c*x^(3/2)))/x^4,x, algorithm="maxima")
Output:
1/6*((c*log(c*x^(3/2) + 1) - c*log(c*x^(3/2) - 1) - 2/x^(3/2))*c - 2*arcta nh(c*x^(3/2))/x^3)*b - 1/3*a/x^3
Time = 0.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.43 \[ \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx=\frac {1}{6} \, b c^{2} \log \left (c x^{\frac {3}{2}} + 1\right ) - \frac {1}{6} \, b c^{2} \log \left (c x^{\frac {3}{2}} - 1\right ) - \frac {b \log \left (-\frac {c x^{\frac {3}{2}} + 1}{c x^{\frac {3}{2}} - 1}\right )}{6 \, x^{3}} - \frac {b c x^{\frac {3}{2}} + a}{3 \, x^{3}} \] Input:
integrate((a+b*arctanh(c*x^(3/2)))/x^4,x, algorithm="giac")
Output:
1/6*b*c^2*log(c*x^(3/2) + 1) - 1/6*b*c^2*log(c*x^(3/2) - 1) - 1/6*b*log(-( c*x^(3/2) + 1)/(c*x^(3/2) - 1))/x^3 - 1/3*(b*c*x^(3/2) + a)/x^3
Time = 3.92 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.43 \[ \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx=\frac {b\,c^2\,\ln \left (\frac {c\,x^{3/2}+1}{c\,x^{3/2}-1}\right )}{6}-\frac {a}{3\,x^3}-\frac {b\,c}{3\,x^{3/2}}-\frac {b\,\ln \left (c\,x^{3/2}+1\right )}{6\,x^3}+\frac {b\,x\,\ln \left (1-c\,x^{3/2}\right )}{3\,\left (2\,x^4-2\,c^2\,x^7\right )}-\frac {b\,c^2\,x^4\,\ln \left (1-c\,x^{3/2}\right )}{3\,\left (2\,x^4-2\,c^2\,x^7\right )} \] Input:
int((a + b*atanh(c*x^(3/2)))/x^4,x)
Output:
(b*c^2*log((c*x^(3/2) + 1)/(c*x^(3/2) - 1)))/6 - a/(3*x^3) - (b*c)/(3*x^(3 /2)) - (b*log(c*x^(3/2) + 1))/(6*x^3) + (b*x*log(1 - c*x^(3/2)))/(3*(2*x^4 - 2*c^2*x^7)) - (b*c^2*x^4*log(1 - c*x^(3/2)))/(3*(2*x^4 - 2*c^2*x^7))
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \text {arctanh}\left (c x^{3/2}\right )}{x^4} \, dx=\frac {\mathit {atanh} \left (\sqrt {x}\, c x \right ) b \,c^{2} x^{3}-\mathit {atanh} \left (\sqrt {x}\, c x \right ) b -\sqrt {x}\, b c x -a}{3 x^{3}} \] Input:
int((a+b*atanh(c*x^(3/2)))/x^4,x)
Output:
(atanh(sqrt(x)*c*x)*b*c**2*x**3 - atanh(sqrt(x)*c*x)*b - sqrt(x)*b*c*x - a )/(3*x**3)