Optimal. Leaf size=677 \[ -\frac{28 i a b x^3 \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{168 i a b x^{5/2} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{840 i a b x^2 \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{3360 i a b x^{3/2} \text{PolyLog}\left (5,-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{3360 i a b x^{3/2} \text{PolyLog}\left (5,i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{10080 i a b x \text{PolyLog}\left (6,-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{10080 i a b x \text{PolyLog}\left (6,i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{20160 i a b \sqrt{x} \text{PolyLog}\left (7,-i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \sqrt{x} \text{PolyLog}\left (7,i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \text{PolyLog}\left (8,-i e^{c+d \sqrt{x}}\right )}{d^8}+\frac{20160 i a b \text{PolyLog}\left (8,i e^{c+d \sqrt{x}}\right )}{d^8}-\frac{42 b^2 x^{5/2} \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{105 b^2 x^2 \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{210 b^2 x^{3/2} \text{PolyLog}\left (4,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{315 b^2 x \text{PolyLog}\left (5,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{315 b^2 \sqrt{x} \text{PolyLog}\left (6,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{315 b^2 \text{PolyLog}\left (7,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{2 d^8}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )}{d^2}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{2 b^2 x^{7/2}}{d} \]
[Out]
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Rubi [A] time = 0.816235, antiderivative size = 677, normalized size of antiderivative = 1., number of steps used = 30, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5436, 4190, 4180, 2531, 6609, 2282, 6589, 4184, 3718, 2190} \[ -\frac{28 i a b x^3 \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{168 i a b x^{5/2} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{840 i a b x^2 \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{3360 i a b x^{3/2} \text{PolyLog}\left (5,-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{3360 i a b x^{3/2} \text{PolyLog}\left (5,i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{10080 i a b x \text{PolyLog}\left (6,-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{10080 i a b x \text{PolyLog}\left (6,i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{20160 i a b \sqrt{x} \text{PolyLog}\left (7,-i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \sqrt{x} \text{PolyLog}\left (7,i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \text{PolyLog}\left (8,-i e^{c+d \sqrt{x}}\right )}{d^8}+\frac{20160 i a b \text{PolyLog}\left (8,i e^{c+d \sqrt{x}}\right )}{d^8}-\frac{42 b^2 x^{5/2} \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{105 b^2 x^2 \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{210 b^2 x^{3/2} \text{PolyLog}\left (4,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{315 b^2 x \text{PolyLog}\left (5,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{315 b^2 \sqrt{x} \text{PolyLog}\left (6,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{315 b^2 \text{PolyLog}\left (7,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{2 d^8}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )}{d^2}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{2 b^2 x^{7/2}}{d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 5436
Rule 4190
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 4184
Rule 3718
Rule 2190
Rubi steps
\begin{align*} \int x^3 \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^7 (a+b \text{sech}(c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^7+2 a b x^7 \text{sech}(c+d x)+b^2 x^7 \text{sech}^2(c+d x)\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^4}{4}+(4 a b) \operatorname{Subst}\left (\int x^7 \text{sech}(c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^7 \text{sech}^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(28 i a b) \operatorname{Subst}\left (\int x^6 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(28 i a b) \operatorname{Subst}\left (\int x^6 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}-\frac{\left (14 b^2\right ) \operatorname{Subst}\left (\int x^6 \tanh (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{(168 i a 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{(168 i a 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{\left (28 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 (c+d x)} x^6}{1+e^{2 (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{168 i a b x^{5/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(840 i a 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{(840 i a 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{\left (84 b^2\right ) \operatorname{Subst}\left (\int x^5 \log \left (1+e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{42 b^2 x^{5/2} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{168 i a b x^{5/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{840 i a b x^2 \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{(3360 i a 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{(3360 i a 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{\left (210 b^2\right ) \operatorname{Subst}\left (\int x^4 \text{Li}_2\left (-e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{42 b^2 x^{5/2} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{168 i a b x^{5/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{105 b^2 x^2 \text{Li}_3\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{840 i a b x^2 \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{3360 i a b x^{3/2} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{3360 i a b x^{3/2} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(10080 i a 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{(10080 i a 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{\left (420 b^2\right ) \operatorname{Subst}\left (\int x^3 \text{Li}_3\left (-e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{42 b^2 x^{5/2} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{168 i a b x^{5/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{105 b^2 x^2 \text{Li}_3\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{840 i a b x^2 \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}-\frac{210 b^2 x^{3/2} \text{Li}_4\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{3360 i a b x^{3/2} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{3360 i a b x^{3/2} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{10080 i a b x \text{Li}_6\left (-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{10080 i a b x \text{Li}_6\left (i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{(20160 i a b) \operatorname{Subst}\left (\int x \text{Li}_6\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^6}-\frac{(20160 i a b) \operatorname{Subst}\left (\int x \text{Li}_6\left (i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^6}+\frac{\left (630 b^2\right ) \operatorname{Subst}\left (\int x^2 \text{Li}_4\left (-e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{42 b^2 x^{5/2} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{168 i a b x^{5/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{105 b^2 x^2 \text{Li}_3\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{840 i a b x^2 \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}-\frac{210 b^2 x^{3/2} \text{Li}_4\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{3360 i a b x^{3/2} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{3360 i a b x^{3/2} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}+\frac{315 b^2 x \text{Li}_5\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 i a b x \text{Li}_6\left (-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{10080 i a b x \text{Li}_6\left (i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{20160 i a b \sqrt{x} \text{Li}_7\left (-i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \sqrt{x} \text{Li}_7\left (i e^{c+d \sqrt{x}}\right )}{d^7}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(20160 i a b) \operatorname{Subst}\left (\int \text{Li}_7\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^7}+\frac{(20160 i a b) \operatorname{Subst}\left (\int \text{Li}_7\left (i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^7}-\frac{\left (630 b^2\right ) \operatorname{Subst}\left (\int x \text{Li}_5\left (-e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^6}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{42 b^2 x^{5/2} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{168 i a b x^{5/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{105 b^2 x^2 \text{Li}_3\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{840 i a b x^2 \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}-\frac{210 b^2 x^{3/2} \text{Li}_4\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{3360 i a b x^{3/2} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{3360 i a b x^{3/2} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}+\frac{315 b^2 x \text{Li}_5\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 i a b x \text{Li}_6\left (-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{10080 i a b x \text{Li}_6\left (i e^{c+d \sqrt{x}}\right )}{d^6}-\frac{315 b^2 \sqrt{x} \text{Li}_6\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{20160 i a b \sqrt{x} \text{Li}_7\left (-i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \sqrt{x} \text{Li}_7\left (i e^{c+d \sqrt{x}}\right )}{d^7}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(20160 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_7(-i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^8}+\frac{(20160 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_7(i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^8}+\frac{\left (315 b^2\right ) \operatorname{Subst}\left (\int \text{Li}_6\left (-e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^7}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{42 b^2 x^{5/2} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{168 i a b x^{5/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{105 b^2 x^2 \text{Li}_3\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{840 i a b x^2 \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}-\frac{210 b^2 x^{3/2} \text{Li}_4\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{3360 i a b x^{3/2} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{3360 i a b x^{3/2} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}+\frac{315 b^2 x \text{Li}_5\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 i a b x \text{Li}_6\left (-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{10080 i a b x \text{Li}_6\left (i e^{c+d \sqrt{x}}\right )}{d^6}-\frac{315 b^2 \sqrt{x} \text{Li}_6\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{20160 i a b \sqrt{x} \text{Li}_7\left (-i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \sqrt{x} \text{Li}_7\left (i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \text{Li}_8\left (-i e^{c+d \sqrt{x}}\right )}{d^8}+\frac{20160 i a b \text{Li}_8\left (i e^{c+d \sqrt{x}}\right )}{d^8}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{\left (315 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_6(-x)}{x} \, dx,x,e^{2 \left (c+d \sqrt{x}\right )}\right )}{2 d^8}\\ &=\frac{2 b^2 x^{7/2}}{d}+\frac{a^2 x^4}{4}+\frac{8 a b x^{7/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{14 b^2 x^3 \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{28 i a b x^3 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{28 i a b x^3 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{42 b^2 x^{5/2} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{168 i a b x^{5/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{168 i a b x^{5/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{105 b^2 x^2 \text{Li}_3\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{840 i a b x^2 \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{840 i a b x^2 \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}-\frac{210 b^2 x^{3/2} \text{Li}_4\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{3360 i a b x^{3/2} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{3360 i a b x^{3/2} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}+\frac{315 b^2 x \text{Li}_5\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 i a b x \text{Li}_6\left (-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{10080 i a b x \text{Li}_6\left (i e^{c+d \sqrt{x}}\right )}{d^6}-\frac{315 b^2 \sqrt{x} \text{Li}_6\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{20160 i a b \sqrt{x} \text{Li}_7\left (-i e^{c+d \sqrt{x}}\right )}{d^7}-\frac{20160 i a b \sqrt{x} \text{Li}_7\left (i e^{c+d \sqrt{x}}\right )}{d^7}+\frac{315 b^2 \text{Li}_7\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{2 d^8}-\frac{20160 i a b \text{Li}_8\left (-i e^{c+d \sqrt{x}}\right )}{d^8}+\frac{20160 i a b \text{Li}_8\left (i e^{c+d \sqrt{x}}\right )}{d^8}+\frac{2 b^2 x^{7/2} \tanh \left (c+d \sqrt{x}\right )}{d}\\ \end{align*}
Mathematica [A] time = 8.67537, size = 739, normalized size = 1.09 \[ \frac{\cosh \left (c+d \sqrt{x}\right ) \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )^2 \left (\frac{2 b \cosh \left (c+d \sqrt{x}\right ) \left (\frac{8 b e^{2 c} d^7 x^{7/2}}{e^{2 c}+1}+i \left (-56 a d^6 x^3 \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )+56 a d^6 x^3 \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )+336 a d^5 x^{5/2} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )-336 a d^5 x^{5/2} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )-1680 a d^4 x^2 \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )+1680 a d^4 x^2 \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )+6720 a d^3 x^{3/2} \text{PolyLog}\left (5,-i e^{c+d \sqrt{x}}\right )-6720 a d^3 x^{3/2} \text{PolyLog}\left (5,i e^{c+d \sqrt{x}}\right )-20160 a d^2 x \text{PolyLog}\left (6,-i e^{c+d \sqrt{x}}\right )+20160 a d^2 x \text{PolyLog}\left (6,i e^{c+d \sqrt{x}}\right )+40320 a d \sqrt{x} \text{PolyLog}\left (7,-i e^{c+d \sqrt{x}}\right )-40320 a d \sqrt{x} \text{PolyLog}\left (7,i e^{c+d \sqrt{x}}\right )-40320 a \text{PolyLog}\left (8,-i e^{c+d \sqrt{x}}\right )+40320 a \text{PolyLog}\left (8,i e^{c+d \sqrt{x}}\right )+84 i b d^5 x^{5/2} \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )-210 i b d^4 x^2 \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )+420 i b d^3 x^{3/2} \text{PolyLog}\left (4,-e^{2 \left (c+d \sqrt{x}\right )}\right )-630 i b d^2 x \text{PolyLog}\left (5,-e^{2 \left (c+d \sqrt{x}\right )}\right )+630 i b d \sqrt{x} \text{PolyLog}\left (6,-e^{2 \left (c+d \sqrt{x}\right )}\right )-315 i b \text{PolyLog}\left (7,-e^{2 \left (c+d \sqrt{x}\right )}\right )+8 a d^7 x^{7/2} \log \left (1-i e^{c+d \sqrt{x}}\right )-8 a d^7 x^{7/2} \log \left (1+i e^{c+d \sqrt{x}}\right )+28 i b d^6 x^3 \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )\right )\right )}{d^8}+a^2 x^4 \cosh \left (c+d \sqrt{x}\right )+\frac{8 b^2 x^{7/2} \text{sech}(c) \sinh \left (d \sqrt{x}\right )}{d}\right )}{4 \left (a \cosh \left (c+d \sqrt{x}\right )+b\right )^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b{\rm sech} \left (c+d\sqrt{x}\right ) \right ) ^{2}\, 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{a^{2} d x^{4} e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + a^{2} d x^{4} - 16 \, b^{2} x^{\frac{7}{2}}}{4 \,{\left (d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + d\right )}} + \int \frac{2 \,{\left (2 \, a b d x^{3} e^{\left (d \sqrt{x} + c\right )} + 7 \, b^{2} x^{\frac{5}{2}}\right )}}{d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + d}\,{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^{2} x^{3} \operatorname{sech}\left (d \sqrt{x} + c\right )^{2} + 2 \, a b x^{3} \operatorname{sech}\left (d \sqrt{x} + c\right ) + a^{2} 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 )^{2}\, 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 )}^{2} x^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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