Integrand size = 14, antiderivative size = 54 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {b x^3}{12 c^3}+\frac {b x^9}{36 c}-\frac {b \text {arctanh}\left (c x^3\right )}{12 c^4}+\frac {1}{12} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \] Output:
1/12*b*x^3/c^3+1/36*b*x^9/c-1/12*b*arctanh(c*x^3)/c^4+1/12*x^12*(a+b*arcta nh(c*x^3))
Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {b x^3}{12 c^3}+\frac {b x^9}{36 c}+\frac {a x^{12}}{12}+\frac {1}{12} b x^{12} \text {arctanh}\left (c x^3\right )+\frac {b \log \left (1-c x^3\right )}{24 c^4}-\frac {b \log \left (1+c x^3\right )}{24 c^4} \] Input:
Integrate[x^11*(a + b*ArcTanh[c*x^3]),x]
Output:
(b*x^3)/(12*c^3) + (b*x^9)/(36*c) + (a*x^12)/12 + (b*x^12*ArcTanh[c*x^3])/ 12 + (b*Log[1 - c*x^3])/(24*c^4) - (b*Log[1 + c*x^3])/(24*c^4)
Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6452, 807, 254, 2009}
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.
\(\displaystyle \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{12} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{4} b c \int \frac {x^{14}}{1-c^2 x^6}dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{12} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{12} b c \int \frac {x^{12}}{1-c^2 x^6}dx^3\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {1}{12} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{12} b c \int \left (-\frac {x^6}{c^2}+\frac {1}{c^4 \left (1-c^2 x^6\right )}-\frac {1}{c^4}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{12} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{12} b c \left (\frac {\text {arctanh}\left (c x^3\right )}{c^5}-\frac {x^3}{c^4}-\frac {x^9}{3 c^2}\right )\) |
Input:
Int[x^11*(a + b*ArcTanh[c*x^3]),x]
Output:
(x^12*(a + b*ArcTanh[c*x^3]))/12 - (b*c*(-(x^3/c^4) - x^9/(3*c^2) + ArcTan h[c*x^3]/c^5))/12
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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[((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 = 0.61 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(-\frac {-3 b \,\operatorname {arctanh}\left (c \,x^{3}\right ) x^{12} c^{4}-3 a \,c^{4} x^{12}-b \,c^{3} x^{9}-3 b c \,x^{3}+3 b \,\operatorname {arctanh}\left (c \,x^{3}\right )}{36 c^{4}}\) | \(56\) |
default | \(\frac {a \,x^{12}}{12}+\frac {b \,x^{12} \operatorname {arctanh}\left (c \,x^{3}\right )}{12}+\frac {b \,x^{9}}{36 c}+\frac {b \,x^{3}}{12 c^{3}}-\frac {b \ln \left (c \,x^{3}+1\right )}{24 c^{4}}+\frac {b \ln \left (c \,x^{3}-1\right )}{24 c^{4}}\) | \(66\) |
parts | \(\frac {a \,x^{12}}{12}+\frac {b \,x^{12} \operatorname {arctanh}\left (c \,x^{3}\right )}{12}+\frac {b \,x^{9}}{36 c}+\frac {b \,x^{3}}{12 c^{3}}-\frac {b \ln \left (c \,x^{3}+1\right )}{24 c^{4}}+\frac {b \ln \left (c \,x^{3}-1\right )}{24 c^{4}}\) | \(66\) |
risch | \(\frac {x^{12} b \ln \left (c \,x^{3}+1\right )}{24}-\frac {x^{12} b \ln \left (-c \,x^{3}+1\right )}{24}+\frac {a \,x^{12}}{12}+\frac {b \,x^{9}}{36 c}+\frac {b \,x^{3}}{12 c^{3}}-\frac {b \ln \left (c \,x^{3}+1\right )}{24 c^{4}}+\frac {b \ln \left (c \,x^{3}-1\right )}{24 c^{4}}\) | \(83\) |
orering | \(\frac {\left (10 c^{4} x^{12}+11 c^{2} x^{6}-21\right ) \left (a +b \,\operatorname {arctanh}\left (c \,x^{3}\right )\right )}{54 c^{4}}-\frac {\left (c^{2} x^{6}+3\right ) \left (c \,x^{3}-1\right ) \left (c \,x^{3}+1\right ) \left (11 x^{10} \left (a +b \,\operatorname {arctanh}\left (c \,x^{3}\right )\right )+\frac {3 x^{13} b c}{-c^{2} x^{6}+1}\right )}{108 x^{10} c^{4}}\) | \(101\) |
Input:
int(x^11*(a+b*arctanh(c*x^3)),x,method=_RETURNVERBOSE)
Output:
-1/36*(-3*b*arctanh(c*x^3)*x^12*c^4-3*a*c^4*x^12-b*c^3*x^9-3*b*c*x^3+3*b*a rctanh(c*x^3))/c^4
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.19 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {6 \, a c^{4} x^{12} + 2 \, b c^{3} x^{9} + 6 \, b c x^{3} + 3 \, {\left (b c^{4} x^{12} - b\right )} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )}{72 \, c^{4}} \] Input:
integrate(x^11*(a+b*arctanh(c*x^3)),x, algorithm="fricas")
Output:
1/72*(6*a*c^4*x^12 + 2*b*c^3*x^9 + 6*b*c*x^3 + 3*(b*c^4*x^12 - b)*log(-(c* x^3 + 1)/(c*x^3 - 1)))/c^4
Timed out. \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x**11*(a+b*atanh(c*x**3)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.28 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {1}{12} \, a x^{12} + \frac {1}{72} \, {\left (6 \, x^{12} \operatorname {artanh}\left (c x^{3}\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{9} + 3 \, x^{3}\right )}}{c^{4}} - \frac {3 \, \log \left (c x^{3} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x^{3} - 1\right )}{c^{5}}\right )}\right )} b \] Input:
integrate(x^11*(a+b*arctanh(c*x^3)),x, algorithm="maxima")
Output:
1/12*a*x^12 + 1/72*(6*x^12*arctanh(c*x^3) + c*(2*(c^2*x^9 + 3*x^3)/c^4 - 3 *log(c*x^3 + 1)/c^5 + 3*log(c*x^3 - 1)/c^5))*b
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {1}{24} \, b x^{12} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + \frac {1}{12} \, a x^{12} + \frac {b x^{9}}{36 \, c} + \frac {b x^{3}}{12 \, c^{3}} - \frac {b \log \left (c x^{3} + 1\right )}{24 \, c^{4}} + \frac {b \log \left (c x^{3} - 1\right )}{24 \, c^{4}} \] Input:
integrate(x^11*(a+b*arctanh(c*x^3)),x, algorithm="giac")
Output:
1/24*b*x^12*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 1/12*a*x^12 + 1/36*b*x^9/c + 1 /12*b*x^3/c^3 - 1/24*b*log(c*x^3 + 1)/c^4 + 1/24*b*log(c*x^3 - 1)/c^4
Time = 3.82 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.28 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {a\,x^{12}}{12}+\frac {b\,x^3}{12\,c^3}+\frac {b\,x^9}{36\,c}+\frac {b\,x^{12}\,\ln \left (c\,x^3+1\right )}{24}-\frac {b\,x^{12}\,\ln \left (1-c\,x^3\right )}{24}+\frac {b\,\mathrm {atan}\left (c\,x^3\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{12\,c^4} \] Input:
int(x^11*(a + b*atanh(c*x^3)),x)
Output:
(a*x^12)/12 + (b*x^3)/(12*c^3) + (b*x^9)/(36*c) + (b*atan(c*x^3*1i)*1i)/(1 2*c^4) + (b*x^12*log(c*x^3 + 1))/24 - (b*x^12*log(1 - c*x^3))/24
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {3 \mathit {atanh} \left (c \,x^{3}\right ) b \,c^{4} x^{12}-3 \mathit {atanh} \left (c \,x^{3}\right ) b +3 a \,c^{4} x^{12}+b \,c^{3} x^{9}+3 b c \,x^{3}}{36 c^{4}} \] Input:
int(x^11*(a+b*atanh(c*x^3)),x)
Output:
(3*atanh(c*x**3)*b*c**4*x**12 - 3*atanh(c*x**3)*b + 3*a*c**4*x**12 + b*c** 3*x**9 + 3*b*c*x**3)/(36*c**4)