Optimal. Leaf size=157 \[ -\frac {3 \text {CosIntegral}(1-\tanh (a+b x)) \sin (1)}{8 b}-\frac {3 \text {CosIntegral}(1+\tanh (a+b x)) \sin (1)}{8 b}+\frac {\text {CosIntegral}(3-3 \tanh (a+b x)) \sin (3)}{8 b}+\frac {\text {CosIntegral}(3+3 \tanh (a+b x)) \sin (3)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1+\tanh (a+b x))}{8 b}-\frac {\cos (3) \text {Si}(3+3 \tanh (a+b x))}{8 b} \]
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Rubi [A]
time = 0.28, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps
used = 19, number of rules used = 5, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6857, 3393,
3384, 3380, 3383} \begin {gather*} \frac {\sin (3) \text {CosIntegral}(3-3 \tanh (a+b x))}{8 b}+\frac {\sin (3) \text {CosIntegral}(3 \tanh (a+b x)+3)}{8 b}-\frac {3 \sin (1) \text {CosIntegral}(1-\tanh (a+b x))}{8 b}-\frac {3 \sin (1) \text {CosIntegral}(\tanh (a+b x)+1)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(\tanh (a+b x)+1)}{8 b}-\frac {\cos (3) \text {Si}(3 \tanh (a+b x)+3)}{8 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 3380
Rule 3383
Rule 3384
Rule 3393
Rule 6857
Rubi steps
\begin {align*} \int \sin ^3(\tanh (a+b x)) \, dx &=\frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{1-x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {\sin ^3(x)}{2 (-1+x)}+\frac {\sin ^3(x)}{2 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{b}\\ &=-\frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=-\frac {\text {Subst}\left (\int \left (\frac {3 \sin (x)}{4 (-1+x)}-\frac {\sin (3 x)}{4 (-1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \left (\frac {3 \sin (x)}{4 (1+x)}-\frac {\sin (3 x)}{4 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=\frac {\text {Subst}\left (\int \frac {\sin (3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {\text {Subst}\left (\int \frac {\sin (3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {3 \text {Subst}\left (\int \frac {\sin (x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac {3 \text {Subst}\left (\int \frac {\sin (x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}\\ &=\frac {(3 \cos (1)) \text {Subst}\left (\int \frac {\sin (1-x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac {(3 \cos (1)) \text {Subst}\left (\int \frac {\sin (1+x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {\cos (3) \text {Subst}\left (\int \frac {\sin (3-3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {\cos (3) \text {Subst}\left (\int \frac {\sin (3+3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {(3 \sin (1)) \text {Subst}\left (\int \frac {\cos (1-x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac {(3 \sin (1)) \text {Subst}\left (\int \frac {\cos (1+x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac {\sin (3) \text {Subst}\left (\int \frac {\cos (3-3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac {\sin (3) \text {Subst}\left (\int \frac {\cos (3+3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}\\ &=-\frac {3 \text {Ci}(1-\tanh (a+b x)) \sin (1)}{8 b}-\frac {3 \text {Ci}(1+\tanh (a+b x)) \sin (1)}{8 b}+\frac {\text {Ci}(3-3 \tanh (a+b x)) \sin (3)}{8 b}+\frac {\text {Ci}(3+3 \tanh (a+b x)) \sin (3)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\tanh (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1+\tanh (a+b x))}{8 b}-\frac {\cos (3) \text {Si}(3+3 \tanh (a+b x))}{8 b}\\ \end {align*}
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Mathematica [A]
time = 0.17, size = 124, normalized size = 0.79 \begin {gather*} \frac {-6 \text {CosIntegral}(1-\tanh (a+b x)) \sin (1)-6 \text {CosIntegral}(1+\tanh (a+b x)) \sin (1)+2 \text {CosIntegral}(3-3 \tanh (a+b x)) \sin (3)+2 \text {CosIntegral}(3+3 \tanh (a+b x)) \sin (3)-2 \cos (3) \text {Si}(3-3 \tanh (a+b x))+6 \cos (1) \text {Si}(1-\tanh (a+b x))+6 \cos (1) \text {Si}(1+\tanh (a+b x))-2 \cos (3) \text {Si}(3+3 \tanh (a+b x))}{16 b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.29, size = 118, normalized size = 0.75
method | result | size |
derivativedivides | \(\frac {\frac {\sinIntegral \left (-3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\cosineIntegral \left (-3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}-\frac {\sinIntegral \left (3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\cosineIntegral \left (3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}-\frac {3 \sinIntegral \left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}-\frac {3 \cosineIntegral \left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}+\frac {3 \sinIntegral \left (\tanh \left (b x +a \right )+1\right ) \cos \left (1\right )}{8}-\frac {3 \cosineIntegral \left (\tanh \left (b x +a \right )+1\right ) \sin \left (1\right )}{8}}{b}\) | \(118\) |
default | \(\frac {\frac {\sinIntegral \left (-3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\cosineIntegral \left (-3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}-\frac {\sinIntegral \left (3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\cosineIntegral \left (3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}-\frac {3 \sinIntegral \left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}-\frac {3 \cosineIntegral \left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}+\frac {3 \sinIntegral \left (\tanh \left (b x +a \right )+1\right ) \cos \left (1\right )}{8}-\frac {3 \cosineIntegral \left (\tanh \left (b x +a \right )+1\right ) \sin \left (1\right )}{8}}{b}\) | \(118\) |
risch | \(\frac {i {\mathrm e}^{3 i} \expIntegral \left (1, \frac {6 i}{{\mathrm e}^{2 b x +2 a}+1}\right )}{16 b}-\frac {i {\mathrm e}^{-3 i} \expIntegral \left (1, \frac {6 i}{{\mathrm e}^{2 b x +2 a}+1}-6 i\right )}{16 b}-\frac {i {\mathrm e}^{-3 i} \expIntegral \left (1, -\frac {6 i}{{\mathrm e}^{2 b x +2 a}+1}\right )}{16 b}+\frac {i {\mathrm e}^{3 i} \expIntegral \left (1, -\frac {6 i}{{\mathrm e}^{2 b x +2 a}+1}+6 i\right )}{16 b}+\frac {3 i {\mathrm e}^{-i} \expIntegral \left (1, -\frac {2 i}{{\mathrm e}^{2 b x +2 a}+1}\right )}{16 b}-\frac {3 i {\mathrm e}^{i} \expIntegral \left (1, -\frac {2 i}{{\mathrm e}^{2 b x +2 a}+1}+2 i\right )}{16 b}-\frac {3 i {\mathrm e}^{i} \expIntegral \left (1, \frac {2 i}{{\mathrm e}^{2 b x +2 a}+1}\right )}{16 b}+\frac {3 i {\mathrm e}^{-i} \expIntegral \left (1, \frac {2 i}{{\mathrm e}^{2 b x +2 a}+1}-2 i\right )}{16 b}\) | \(230\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 296 vs.
\(2 (139) = 278\).
time = 0.38, size = 296, normalized size = 1.89 \begin {gather*} \frac {\operatorname {Ci}\left (\frac {6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (3\right ) + \operatorname {Ci}\left (-\frac {6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (3\right ) + \operatorname {Ci}\left (\frac {6}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (3\right ) + \operatorname {Ci}\left (-\frac {6}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (3\right ) - 3 \, \operatorname {Ci}\left (\frac {2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (1\right ) - 3 \, \operatorname {Ci}\left (-\frac {2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (1\right ) - 3 \, \operatorname {Ci}\left (\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (1\right ) - 3 \, \operatorname {Ci}\left (-\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (1\right ) - 2 \, \cos \left (3\right ) \operatorname {Si}\left (\frac {6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + 6 \, \cos \left (1\right ) \operatorname {Si}\left (\frac {2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - 2 \, \cos \left (3\right ) \operatorname {Si}\left (\frac {6}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + 6 \, \cos \left (1\right ) \operatorname {Si}\left (\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right )}{16 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{3}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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