Optimal. Leaf size=121 \[ \frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \tanh ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)} \]
[Out]
________________________________________________________________________________________
Rubi [A]
time = 0.05, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2188, 30}
\begin {gather*} -\frac {6 \tanh ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 30
Rule 2188
Rule 2199
Rubi steps
\begin {align*} \int x^3 \tanh ^{-1}(\tanh (a+b x))^n \, dx &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{1+n} \, dx}{b (1+n)}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 \int x \tanh ^{-1}(\tanh (a+b x))^{2+n} \, dx}{b^2 (1+n) (2+n)}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \int \tanh ^{-1}(\tanh (a+b x))^{3+n} \, dx}{b^3 (1+n) (2+n) (3+n)}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \text {Subst}\left (\int x^{3+n} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^4 (1+n) (2+n) (3+n)}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \tanh ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 0.06, size = 106, normalized size = 0.88 \begin {gather*} \frac {\tanh ^{-1}(\tanh (a+b x))^{1+n} \left (b^3 \left (24+26 n+9 n^2+n^3\right ) x^3-3 b^2 \left (12+7 n+n^2\right ) x^2 \tanh ^{-1}(\tanh (a+b x))+6 b (4+n) x \tanh ^{-1}(\tanh (a+b x))^2-6 \tanh ^{-1}(\tanh (a+b x))^3\right )}{b^4 (1+n) (2+n) (3+n) (4+n)} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(491\) vs.
\(2(121)=242\).
time = 2.54, size = 492, normalized size = 4.07
method | result | size |
default | \(\frac {x^{4} {\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}}{4+n}+\frac {n \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{3} {\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}}{b \left (n^{2}+7 n +12\right )}-\frac {6 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} a^{4}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {24 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {36 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {24 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {6 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {3 n \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) x^{2} {\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}}{b^{2} \left (n^{3}+9 n^{2}+26 n +24\right )}+\frac {6 n \left (a^{3}+3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) x \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}}{b^{3} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}\) | \(492\) |
risch | \(\text {Expression too large to display}\) | \(52442\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A]
time = 0.53, size = 101, normalized size = 0.83 \begin {gather*} \frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} + {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} x^{3} - 3 \, {\left (n^{2} + n\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b n x - 6 \, a^{4}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 255 vs.
\(2 (121) = 242\).
time = 0.36, size = 255, normalized size = 2.11 \begin {gather*} \frac {{\left (6 \, a^{3} b n x + {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} \cosh \left (n \log \left (b x + a\right )\right ) + {\left (6 \, a^{3} b n x + {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} \sinh \left (n \log \left (b x + a\right )\right )}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {x^{4} \operatorname {atanh}^{n}{\left (\tanh {\left (a \right )} \right )}}{4} & \text {for}\: b = 0 \\- \frac {x^{3}}{3 b \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x^{2}}{2 b^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x}{b^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{4}} & \text {for}\: n = -4 \\\int \frac {x^{3}}{\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -3 \\\int \frac {x^{3}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -2 \\\int \frac {x^{3}}{\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -1 \\\frac {b^{3} n^{3} x^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {9 b^{3} n^{2} x^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {26 b^{3} n x^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {24 b^{3} x^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {3 b^{2} n^{2} x^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {21 b^{2} n x^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {36 b^{2} x^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {6 b n x \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {24 b x \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {6 \operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A]
time = 0.39, size = 226, normalized size = 1.87 \begin {gather*} \frac {{\left (b x + a\right )}^{n} b^{4} n^{3} x^{4} + {\left (b x + a\right )}^{n} a b^{3} n^{3} x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} n^{2} x^{4} + 3 \, {\left (b x + a\right )}^{n} a b^{3} n^{2} x^{3} + 11 \, {\left (b x + a\right )}^{n} b^{4} n x^{4} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} n^{2} x^{2} + 2 \, {\left (b x + a\right )}^{n} a b^{3} n x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} x^{4} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} n x^{2} + 6 \, {\left (b x + a\right )}^{n} a^{3} b n x - 6 \, {\left (b x + a\right )}^{n} a^{4}}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Mupad [B]
time = 1.42, size = 418, normalized size = 3.45 \begin {gather*} -{\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}\right )}^n\,\left (\frac {3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}{8\,b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {x^4\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}+\frac {3\,n\,x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4\,b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {n\,x^3\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\,\left (n^2+3\,n+2\right )}{2\,b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {3\,n\,x^2\,\left (n+1\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________