Optimal. Leaf size=356 \[ \frac {a x^4}{4}-\frac {10080 b \text {Li}_8\left (-e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 b \text {Li}_8\left (e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 b \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )}{d^7}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \]
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Rubi [A] time = 0.40, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {14, 5437, 4182, 2531, 6609, 2282, 6589} \[ -\frac {14 b x^3 \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 b \sqrt {x} \text {PolyLog}\left (7,-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {PolyLog}\left (7,e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \text {PolyLog}\left (8,-e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 b \text {PolyLog}\left (8,e^{c+d \sqrt {x}}\right )}{d^8}+\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 2282
Rule 2531
Rule 4182
Rule 5437
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^3+b x^3 \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {a x^4}{4}+b \int x^3 \text {csch}\left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {a x^4}{4}+(2 b) \operatorname {Subst}\left (\int x^7 \text {csch}(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {(14 b) \operatorname {Subst}\left (\int x^6 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(14 b) \operatorname {Subst}\left (\int x^6 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(84 b) \operatorname {Subst}\left (\int x^5 \text {Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(84 b) \operatorname {Subst}\left (\int x^5 \text {Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(420 b) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(420 b) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(1680 b) \operatorname {Subst}\left (\int x^3 \text {Li}_4\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(1680 b) \operatorname {Subst}\left (\int x^3 \text {Li}_4\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(5040 b) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(5040 b) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {(10080 b) \operatorname {Subst}\left (\int x \text {Li}_6\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^6}-\frac {(10080 b) \operatorname {Subst}\left (\int x \text {Li}_6\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^6}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 b \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )}{d^7}-\frac {(10080 b) \operatorname {Subst}\left (\int \text {Li}_7\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^7}+\frac {(10080 b) \operatorname {Subst}\left (\int \text {Li}_7\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^7}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 b \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )}{d^7}-\frac {(10080 b) \operatorname {Subst}\left (\int \frac {\text {Li}_7(-x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^8}+\frac {(10080 b) \operatorname {Subst}\left (\int \frac {\text {Li}_7(x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^8}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 b \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \text {Li}_8\left (-e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 b \text {Li}_8\left (e^{c+d \sqrt {x}}\right )}{d^8}\\ \end {align*}
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Mathematica [A] time = 2.90, size = 365, normalized size = 1.03 \[ \frac {a x^4}{4}+\frac {2 b \left (d^7 x^{7/2} \log \left (1-e^{c+d \sqrt {x}}\right )-d^7 x^{7/2} \log \left (e^{c+d \sqrt {x}}+1\right )-7 d^6 x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )+7 d^6 x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )+42 d^5 x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )-42 d^5 x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )-210 d^4 x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )+210 d^4 x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )+840 d^3 x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )-840 d^3 x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )-2520 d^2 x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )+2520 d^2 x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )+5040 d \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )-5040 d \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )-5040 \text {Li}_8\left (-e^{c+d \sqrt {x}}\right )+5040 \text {Li}_8\left (e^{c+d \sqrt {x}}\right )\right )}{d^8} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{3} \operatorname {csch}\left (d \sqrt {x} + c\right ) + a x^{3}, x\right ) \]
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 \[ \int {\left (b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a\right )} x^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.64, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.90, size = 349, normalized size = 0.98 \[ \frac {1}{4} \, a x^{4} - \frac {2 \, {\left (\log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{7} + 7 \, {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{6} - 42 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{5} {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 210 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{4} {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )}) - 840 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} {\rm Li}_{5}(-e^{\left (d \sqrt {x} + c\right )}) + 2520 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} {\rm Li}_{6}(-e^{\left (d \sqrt {x} + c\right )}) - 5040 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{7}(-e^{\left (d \sqrt {x} + c\right )}) + 5040 \, {\rm Li}_{8}(-e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{8}} + \frac {2 \, {\left (\log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{7} + 7 \, {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{6} - 42 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{5} {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 210 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{4} {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )}) - 840 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} {\rm Li}_{5}(e^{\left (d \sqrt {x} + c\right )}) + 2520 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} {\rm Li}_{6}(e^{\left (d \sqrt {x} + c\right )}) - 5040 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{7}(e^{\left (d \sqrt {x} + c\right )}) + 5040 \, {\rm Li}_{8}(e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{8}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,\left (a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \]
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 \[ \int x^{3} \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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