Optimal. Leaf size=426 \[ -\frac{14 i b x^3 \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{14 i b x^3 \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{84 i b x^{5/2} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{84 i b x^{5/2} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{420 i b x^2 \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{420 i b x^2 \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{1680 i b x^{3/2} \text{PolyLog}\left (5,-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{1680 i b x^{3/2} \text{PolyLog}\left (5,i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{5040 i b x \text{PolyLog}\left (6,-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{5040 i b x \text{PolyLog}\left (6,i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{10080 i b \sqrt{x} \text{PolyLog}\left (7,-i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{10080 i b \sqrt{x} \text{PolyLog}\left (7,i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{10080 i b \text{PolyLog}\left (8,-i e^{c+d \sqrt{x}}\right )}{d^8}+\frac{10080 i b \text{PolyLog}\left (8,i e^{c+d \sqrt{x}}\right )}{d^8}+\frac{a x^4}{4}+\frac{4 b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d} \]
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Rubi [A] time = 0.402782, antiderivative size = 426, normalized size of antiderivative = 1., 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 5436
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
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*}
Mathematica [A] time = 1.86076, size = 415, normalized size = 0.97 \[ \frac{a x^4}{4}+\frac{2 i b \left (-7 d^6 x^3 \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )+7 d^6 x^3 \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )+42 d^5 x^{5/2} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )-42 d^5 x^{5/2} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )-210 d^4 x^2 \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )+210 d^4 x^2 \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )+840 d^3 x^{3/2} \text{PolyLog}\left (5,-i e^{c+d \sqrt{x}}\right )-840 d^3 x^{3/2} \text{PolyLog}\left (5,i e^{c+d \sqrt{x}}\right )-2520 d^2 x \text{PolyLog}\left (6,-i e^{c+d \sqrt{x}}\right )+2520 d^2 x \text{PolyLog}\left (6,i e^{c+d \sqrt{x}}\right )+5040 d \sqrt{x} \text{PolyLog}\left (7,-i e^{c+d \sqrt{x}}\right )-5040 d \sqrt{x} \text{PolyLog}\left (7,i e^{c+d \sqrt{x}}\right )-5040 \text{PolyLog}\left (8,-i e^{c+d \sqrt{x}}\right )+5040 \text{PolyLog}\left (8,i e^{c+d \sqrt{x}}\right )+d^7 x^{7/2} \log \left (1-i e^{c+d \sqrt{x}}\right )-d^7 x^{7/2} \log \left (1+i e^{c+d \sqrt{x}}\right )\right )}{d^8} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.097, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b{\rm sech} \left (c+d\sqrt{x}\right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b x^{3} \operatorname{sech}\left (d \sqrt{x} + c\right ) + a x^{3}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (a + b \operatorname{sech}{\left (c + d \sqrt{x} \right )}\right )\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{sech}\left (d \sqrt{x} + c\right ) + a\right )} x^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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