Optimal. Leaf size=749 \[ \frac {a^2 x^4}{4}-\frac {20160 i a b \text {Li}_8\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {20160 i a b \text {Li}_8\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}-\frac {20160 a b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {20160 a b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 i a b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 i a b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {3360 a b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {840 i a b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {315 b^2 \text {Li}_7\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^8}-\frac {315 i b^2 \sqrt {x} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {315 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {210 i b^2 x^{3/2} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {105 b^2 x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {42 i b^2 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{7/2}}{d} \]
[Out]
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Rubi [A] time = 0.86, antiderivative size = 749, normalized size of antiderivative = 1.00, number of steps used = 30, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4204, 4190, 4181, 2531, 6609, 2282, 6589, 4184, 3719, 2190} \[ \frac {28 i a b x^3 \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {168 a b x^{5/2} \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {840 i a b x^2 \text {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {3360 a b x^{3/2} \text {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \text {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {10080 i a b x \text {PolyLog}\left (6,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 i a b x \text {PolyLog}\left (6,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {20160 a b \sqrt {x} \text {PolyLog}\left (7,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {20160 a b \sqrt {x} \text {PolyLog}\left (7,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {20160 i a b \text {PolyLog}\left (8,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {20160 i a b \text {PolyLog}\left (8,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}-\frac {42 i b^2 x^{5/2} \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {105 b^2 x^2 \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {210 i b^2 x^{3/2} \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {315 b^2 x \text {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {315 i b^2 \sqrt {x} \text {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {315 b^2 \text {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^8}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{7/2}}{d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2190
Rule 2282
Rule 2531
Rule 3719
Rule 4181
Rule 4184
Rule 4190
Rule 4204
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^7 (a+b \sec (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (a^2 x^7+2 a b x^7 \sec (c+d x)+b^2 x^7 \sec ^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 \sec (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {Subst}\left (\int x^7 \sec ^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(28 a b) \operatorname {Subst}\left (\int x^6 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(28 a b) \operatorname {Subst}\left (\int x^6 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}-\frac {\left (14 b^2\right ) \operatorname {Subst}\left (\int x^6 \tan (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(168 i a b) \operatorname {Subst}\left (\int x^5 \text {Li}_2\left (-i e^{i (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^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (28 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^6}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(840 a b) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(840 a b) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (i e^{i (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 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {42 i b^2 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {840 i a b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(3360 i a b) \operatorname {Subst}\left (\int x^3 \text {Li}_4\left (-i e^{i (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^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}+\frac {\left (210 i b^2\right ) \operatorname {Subst}\left (\int x^4 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {42 i b^2 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {105 b^2 x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {840 i a b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {3360 a b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(10080 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(10080 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (i e^{i (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 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {42 i b^2 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {105 b^2 x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {840 i a b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {210 i b^2 x^{3/2} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {3360 a b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {10080 i a b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 i a b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(20160 i a b) \operatorname {Subst}\left (\int x \text {Li}_6\left (-i e^{i (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^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^6}-\frac {\left (630 i b^2\right ) \operatorname {Subst}\left (\int x^2 \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {42 i b^2 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {105 b^2 x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {840 i a b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {210 i b^2 x^{3/2} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {3360 a b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {315 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {10080 i a b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 i a b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {20160 a b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {20160 a b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(20160 a b) \operatorname {Subst}\left (\int \text {Li}_7\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^7}-\frac {(20160 a b) \operatorname {Subst}\left (\int \text {Li}_7\left (i e^{i (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 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^6}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {42 i b^2 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {105 b^2 x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {840 i a b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {210 i b^2 x^{3/2} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {3360 a b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {315 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {10080 i a b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 i a b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {315 i b^2 \sqrt {x} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {20160 a b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {20160 a b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {2 b^2 x^{7/2} \tan \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^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {(20160 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_7(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {\left (315 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_6\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^7}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {42 i b^2 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {105 b^2 x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {840 i a b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {210 i b^2 x^{3/2} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {3360 a b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {315 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {10080 i a b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 i a b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {315 i b^2 \sqrt {x} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {20160 a b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {20160 a b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {20160 i a b \text {Li}_8\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {20160 i a b \text {Li}_8\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {2 b^2 x^{7/2} \tan \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 i \left (c+d \sqrt {x}\right )}\right )}{2 d^8}\\ &=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 i a b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {42 i b^2 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {168 a b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {105 b^2 x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {840 i a b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {210 i b^2 x^{3/2} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {3360 a b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {315 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {10080 i a b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 i a b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {315 i b^2 \sqrt {x} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {20160 a b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {20160 a b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {315 b^2 \text {Li}_7\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^8}-\frac {20160 i a b \text {Li}_8\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {20160 i a b \text {Li}_8\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {2 b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}\\ \end {align*}
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Mathematica [A] time = 2.20, size = 739, normalized size = 0.99 \[ \frac {a^2 d^8 x^4-32 i a b d^7 x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )+112 i a b d^6 x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-112 i a b d^6 x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-672 a b d^5 x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+672 a b d^5 x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-3360 i a b d^4 x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+3360 i a b d^4 x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+13440 a b d^3 x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-13440 a b d^3 x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+40320 i a b d^2 x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-40320 i a b d^2 x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-80640 a b d \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+80640 a b d \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-80640 i a b \text {Li}_8\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+80640 i a b \text {Li}_8\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+8 b^2 d^7 x^{7/2} \tan \left (c+d \sqrt {x}\right )+56 b^2 d^6 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )-168 i b^2 d^5 x^{5/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )+420 b^2 d^4 x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )+840 i b^2 d^3 x^{3/2} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )-1260 b^2 d^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )-1260 i b^2 d \sqrt {x} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )+630 b^2 \text {Li}_7\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )-8 i b^2 d^7 x^{7/2}}{4 d^8} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{3} \sec \left (d \sqrt {x} + c\right )^{2} + 2 \, a b x^{3} \sec \left (d \sqrt {x} + c\right ) + a^{2} x^{3}, x\right ) \]
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 \[ \int {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.38, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.62, size = 6373, normalized size = 8.51 \[ \text {result too large to display} \]
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 \[ \int x^3\,{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]
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 \[ \int x^{3} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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