Optimal. Leaf size=426 \[ \frac {a x^4}{4}-\frac {10080 i b \text {Li}_8\left (-i e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 i b \text {Li}_8\left (i e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 i b \sqrt {x} \text {Li}_7\left (-i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \sqrt {x} \text {Li}_7\left (i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {5040 i b x \text {Li}_6\left (-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 i b x \text {Li}_6\left (i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {1680 i b x^{3/2} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {84 i b x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \]
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Rubi [A] time = 0.40, antiderivative size = 426, 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, 5436, 4180, 2531, 6609, 2282, 6589} \[ -\frac {14 i b x^3 \text {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \text {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 i b x^2 \text {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \text {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 i b x^{3/2} \text {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \text {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 i b x \text {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 i b x \text {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 i b \sqrt {x} \text {PolyLog}\left (7,-i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \sqrt {x} \text {PolyLog}\left (7,i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \text {PolyLog}\left (8,-i e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 i b \text {PolyLog}\left (8,i e^{c+d \sqrt {x}}\right )}{d^8}+\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 2282
Rule 2531
Rule 4180
Rule 5436
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^3+b x^3 \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {a x^4}{4}+b \int x^3 \text {sech}\left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {a x^4}{4}+(2 b) \operatorname {Subst}\left (\int x^7 \text {sech}(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {(14 i b) \operatorname {Subst}\left (\int x^6 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(14 i b) \operatorname {Subst}\left (\int x^6 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(84 i b) \operatorname {Subst}\left (\int x^5 \text {Li}_2\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(84 i b) \operatorname {Subst}\left (\int x^5 \text {Li}_2\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(420 i b) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(420 i b) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(1680 i b) \operatorname {Subst}\left (\int x^3 \text {Li}_4\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(1680 i b) \operatorname {Subst}\left (\int x^3 \text {Li}_4\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 i b x^{3/2} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(5040 i b) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(5040 i b) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 i b x^{3/2} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 i b x \text {Li}_6\left (-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 i b x \text {Li}_6\left (i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {(10080 i b) \operatorname {Subst}\left (\int x \text {Li}_6\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^6}-\frac {(10080 i b) \operatorname {Subst}\left (\int x \text {Li}_6\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^6}\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 i b x^{3/2} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 i b x \text {Li}_6\left (-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 i b x \text {Li}_6\left (i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 i b \sqrt {x} \text {Li}_7\left (-i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \sqrt {x} \text {Li}_7\left (i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {(10080 i b) \operatorname {Subst}\left (\int \text {Li}_7\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^7}+\frac {(10080 i b) \operatorname {Subst}\left (\int \text {Li}_7\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^7}\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 i b x^{3/2} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 i b x \text {Li}_6\left (-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 i b x \text {Li}_6\left (i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 i b \sqrt {x} \text {Li}_7\left (-i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \sqrt {x} \text {Li}_7\left (i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {(10080 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_7(-i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^8}+\frac {(10080 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_7(i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^8}\\ &=\frac {a x^4}{4}+\frac {4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 i b x^{3/2} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 i b x \text {Li}_6\left (-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 i b x \text {Li}_6\left (i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 i b \sqrt {x} \text {Li}_7\left (-i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \sqrt {x} \text {Li}_7\left (i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \text {Li}_8\left (-i e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 i b \text {Li}_8\left (i e^{c+d \sqrt {x}}\right )}{d^8}\\ \end {align*}
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Mathematica [A] time = 1.95, size = 415, normalized size = 0.97 \[ \frac {a x^4}{4}+\frac {2 i b \left (d^7 x^{7/2} \log \left (1-i e^{c+d \sqrt {x}}\right )-d^7 x^{7/2} \log \left (1+i e^{c+d \sqrt {x}}\right )-7 d^6 x^3 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )+7 d^6 x^3 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )+42 d^5 x^{5/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )-42 d^5 x^{5/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )-210 d^4 x^2 \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )+210 d^4 x^2 \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )+840 d^3 x^{3/2} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )-840 d^3 x^{3/2} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )-2520 d^2 x \text {Li}_6\left (-i e^{c+d \sqrt {x}}\right )+2520 d^2 x \text {Li}_6\left (i e^{c+d \sqrt {x}}\right )+5040 d \sqrt {x} \text {Li}_7\left (-i e^{c+d \sqrt {x}}\right )-5040 d \sqrt {x} \text {Li}_7\left (i e^{c+d \sqrt {x}}\right )-5040 \text {Li}_8\left (-i e^{c+d \sqrt {x}}\right )+5040 \text {Li}_8\left (i e^{c+d \sqrt {x}}\right )\right )}{d^8} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{3} \operatorname {sech}\left (d \sqrt {x} + c\right ) + a x^{3}, x\right ) \]
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 \[ \int {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )} x^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.57, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \,\mathrm {sech}\left (c +d \sqrt {x}\right )\right )\, dx \]
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 \[ \frac {1}{4} \, a x^{4} + 2 \, b \int \frac {x^{3} e^{\left (d \sqrt {x} + c\right )}}{e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} + 1}\,{d x} \]
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 \[ \int x^3\,\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \]
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 \[ \int x^{3} \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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