Optimal. Leaf size=551 \[ \frac {a^2 x^3}{3}+\frac {480 i a b \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {240 i a b x \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {80 a b x^{3/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {20 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {15 b^2 \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{5/2}}{d} \]
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Rubi [A] time = 0.63, antiderivative size = 551, normalized size of antiderivative = 1.00, number of steps used = 24, 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} \[ \frac {20 i a b x^2 \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {80 a b x^{3/2} \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {240 i a b x \text {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {480 a b \sqrt {x} \text {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 i a b \text {PolyLog}\left (6,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \text {PolyLog}\left (6,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {20 i b^2 x^{3/2} \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 i b^2 \sqrt {x} \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {15 b^2 \text {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{5/2}}{d} \]
Antiderivative was successfully verified.
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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^2 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^5 (a+b \sec (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (a^2 x^5+2 a b x^5 \sec (c+d x)+b^2 x^5 \sec ^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}+(4 a b) \operatorname {Subst}\left (\int x^5 \sec (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {Subst}\left (\int x^5 \sec ^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(20 a b) \operatorname {Subst}\left (\int x^4 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(20 a b) \operatorname {Subst}\left (\int x^4 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}-\frac {\left (10 b^2\right ) \operatorname {Subst}\left (\int x^4 \tan (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {20 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(80 i a b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(80 i a b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (20 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^4}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {80 a b x^{3/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(240 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(240 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (40 b^2\right ) \operatorname {Subst}\left (\int x^3 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {240 i a b x \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(480 i a b) \operatorname {Subst}\left (\int x \text {Li}_4\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(480 i a b) \operatorname {Subst}\left (\int x \text {Li}_4\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}+\frac {\left (60 i b^2\right ) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(480 a b) \operatorname {Subst}\left (\int \text {Li}_5\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(480 a b) \operatorname {Subst}\left (\int \text {Li}_5\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}-\frac {\left (60 b^2\right ) \operatorname {Subst}\left (\int x \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(480 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {(480 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {\left (30 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 i a b \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {15 b^2 \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {480 i a b \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}\\ \end {align*}
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Mathematica [A] time = 1.26, size = 543, normalized size = 0.99 \[ \frac {a^2 d^6 x^3-24 i a b d^5 x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )+60 i a b d^4 x^2 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-60 i a b d^4 x^2 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-240 a b d^3 x^{3/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+240 a b d^3 x^{3/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-720 i a b d^2 x \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+720 i a b d^2 x \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+1440 a b d \sqrt {x} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-1440 a b d \sqrt {x} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+1440 i a b \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-1440 i a b \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+6 b^2 d^5 x^{5/2} \tan \left (c+d \sqrt {x}\right )+30 b^2 d^4 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )-60 i b^2 d^3 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )+90 b^2 d^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )+90 i b^2 d \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )-45 b^2 \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )-6 i b^2 d^5 x^{5/2}}{3 d^6} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{2} \sec \left (d \sqrt {x} + c\right )^{2} + 2 \, a b x^{2} \sec \left (d \sqrt {x} + c\right ) + a^{2} x^{2}, x\right ) \]
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 \[ \int {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{2}\,{d x} \]
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 \[ \int x^{2} \left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.43, size = 3898, normalized size = 7.07 \[ \text {result too large to display} \]
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 \[ \int x^2\,{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]
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 \[ \int x^{2} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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