Optimal. Leaf size=294 \[ \frac {2 a^3 (10439 A+9230 B+8368 C) \tan (c+d x)}{6435 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (10439 A+9230 B+8368 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {2 a (10439 A+9230 B+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d} \]
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Rubi [A]
time = 0.63, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4173, 4103,
4101, 3885, 4086, 3877} \begin {gather*} \frac {2 a^3 (2717 A+2522 B+2224 C) \tan (c+d x) \sec ^3(c+d x)}{9009 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (10439 A+9230 B+8368 C) \tan (c+d x)}{6435 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (143 A+182 B+136 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{1287 d}-\frac {4 a^2 (10439 A+9230 B+8368 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{45045 d}+\frac {2 a (10439 A+9230 B+8368 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{15015 d}+\frac {2 a (13 B+5 C) \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 3877
Rule 3885
Rule 4086
Rule 4101
Rule 4103
Rule 4173
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac {2 \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (13 A+6 C)+\frac {1}{2} a (13 B+5 C) \sec (c+d x)\right ) \, dx}{13 a}\\ &=\frac {2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac {4 \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (143 A+78 B+96 C)+\frac {1}{4} a^2 (143 A+182 B+136 C) \sec (c+d x)\right ) \, dx}{143 a}\\ &=\frac {2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac {8 \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{8} a^3 (715 A+598 B+560 C)+\frac {1}{8} a^3 (2717 A+2522 B+2224 C) \sec (c+d x)\right ) \, dx}{1287 a}\\ &=\frac {2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac {\left (a^2 (10439 A+9230 B+8368 C)\right ) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{3003}\\ &=\frac {2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {2 a (10439 A+9230 B+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac {(2 a (10439 A+9230 B+8368 C)) \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx}{15015}\\ &=\frac {2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (10439 A+9230 B+8368 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {2 a (10439 A+9230 B+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}+\frac {\left (a^2 (10439 A+9230 B+8368 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{6435}\\ &=\frac {2 a^3 (10439 A+9230 B+8368 C) \tan (c+d x)}{6435 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2717 A+2522 B+2224 C) \sec ^3(c+d x) \tan (c+d x)}{9009 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (10439 A+9230 B+8368 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{45045 d}+\frac {2 a^2 (143 A+182 B+136 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1287 d}+\frac {2 a (10439 A+9230 B+8368 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{15015 d}+\frac {2 a (13 B+5 C) \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{143 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{13 d}\\ \end {align*}
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Mathematica [A]
time = 2.24, size = 222, normalized size = 0.76 \begin {gather*} \frac {a^2 (322751 A+325910 B+343612 C+70 (4576 A+5083 B+5552 C) \cos (c+d x)+14 (32747 A+31850 B+30334 C) \cos (2 (c+d x))+141570 A \cos (3 (c+d x))+138450 B \cos (3 (c+d x))+125520 C \cos (3 (c+d x))+156585 A \cos (4 (c+d x))+138450 B \cos (4 (c+d x))+125520 C \cos (4 (c+d x))+20878 A \cos (5 (c+d x))+18460 B \cos (5 (c+d x))+16736 C \cos (5 (c+d x))+20878 A \cos (6 (c+d x))+18460 B \cos (6 (c+d x))+16736 C \cos (6 (c+d x))) \sec ^6(c+d x) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{180180 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 15.07, size = 240, normalized size = 0.82
method | result | size |
default | \(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (83512 A \left (\cos ^{6}\left (d x +c \right )\right )+73840 B \left (\cos ^{6}\left (d x +c \right )\right )+66944 C \left (\cos ^{6}\left (d x +c \right )\right )+41756 A \left (\cos ^{5}\left (d x +c \right )\right )+36920 B \left (\cos ^{5}\left (d x +c \right )\right )+33472 C \left (\cos ^{5}\left (d x +c \right )\right )+31317 A \left (\cos ^{4}\left (d x +c \right )\right )+27690 B \left (\cos ^{4}\left (d x +c \right )\right )+25104 C \left (\cos ^{4}\left (d x +c \right )\right )+18590 A \left (\cos ^{3}\left (d x +c \right )\right )+23075 B \left (\cos ^{3}\left (d x +c \right )\right )+20920 C \left (\cos ^{3}\left (d x +c \right )\right )+5005 A \left (\cos ^{2}\left (d x +c \right )\right )+14560 B \left (\cos ^{2}\left (d x +c \right )\right )+18305 C \left (\cos ^{2}\left (d x +c \right )\right )+4095 B \cos \left (d x +c \right )+11970 C \cos \left (d x +c \right )+3465 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{45045 d \cos \left (d x +c \right )^{6} \sin \left (d x +c \right )}\) | \(240\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.73, size = 192, normalized size = 0.65 \begin {gather*} \frac {2 \, {\left (8 \, {\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 4 \, {\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 3 \, {\left (10439 \, A + 9230 \, B + 8368 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (3718 \, A + 4615 \, B + 4184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \, {\left (143 \, A + 416 \, B + 523 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \, {\left (13 \, B + 38 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.67, size = 472, normalized size = 1.61 \begin {gather*} \frac {8 \, {\left (45045 \, \sqrt {2} A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 45045 \, \sqrt {2} B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 45045 \, \sqrt {2} C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (180180 \, \sqrt {2} A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 150150 \, \sqrt {2} B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 120120 \, \sqrt {2} C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (342342 \, \sqrt {2} A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 300300 \, \sqrt {2} B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 294294 \, \sqrt {2} C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (391248 \, \sqrt {2} A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 356070 \, \sqrt {2} B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 310596 \, \sqrt {2} C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (265837 \, \sqrt {2} A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 232375 \, \sqrt {2} B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 212069 \, \sqrt {2} C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 4 \, {\left (24167 \, \sqrt {2} A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 21125 \, \sqrt {2} B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 19279 \, \sqrt {2} C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 2 \, {\left (1859 \, \sqrt {2} A a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1625 \, \sqrt {2} B a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1483 \, \sqrt {2} C a^{9} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{45045 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{6} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 17.37, size = 1201, normalized size = 4.09 \begin {gather*} -\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {a^2\,\left (A-16\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {C\,a^2\,128{}\mathrm {i}}{143\,d}-\frac {a^2\,\left (3\,A+4\,B+4\,C\right )\,40{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (11\,A+10\,B+20\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,8{}\mathrm {i}}{11\,d}\right )-\frac {A\,a^2\,8{}\mathrm {i}}{11\,d}-\frac {C\,a^2\,128{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (5\,A+2\,B-16\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (11\,A+10\,B+4\,C\right )\,8{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (15\,A+20\,B+36\,C\right )\,8{}\mathrm {i}}{11\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (-{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {a^2\,\left (A-8\,B\right )\,8{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (5\,A+9\,B+10\,C\right )\,16{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,8{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (13\,B-6\,C\right )\,64{}\mathrm {i}}{1287\,d}\right )+\frac {A\,a^2\,8{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (5\,A+5\,B+2\,C\right )\,16{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (5\,A+10\,B+32\,C\right )\,8{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (B+6\,C\right )\,64{}\mathrm {i}}{9\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {a^2\,\left (403\,B-572\,A+1046\,C\right )\,16{}\mathrm {i}}{15015\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,8{}\mathrm {i}}{5\,d}\right )+\frac {A\,a^2\,8{}\mathrm {i}}{5\,d}-\frac {a^2\,\left (4\,A+5\,B+2\,C\right )\,16{}\mathrm {i}}{5\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,a^2\,8{}\mathrm {i}}{13\,d}-\frac {a^2\,\left (6\,A+5\,B+2\,C\right )\,16{}\mathrm {i}}{13\,d}-\frac {a^2\,\left (10\,A+11\,B+10\,C\right )\,16{}\mathrm {i}}{13\,d}+\frac {a^2\,\left (13\,A+15\,B+20\,C\right )\,16{}\mathrm {i}}{13\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,8{}\mathrm {i}}{13\,d}\right )-\frac {A\,a^2\,8{}\mathrm {i}}{13\,d}+\frac {a^2\,\left (6\,A+5\,B+2\,C\right )\,16{}\mathrm {i}}{13\,d}+\frac {a^2\,\left (10\,A+11\,B+10\,C\right )\,16{}\mathrm {i}}{13\,d}-\frac {a^2\,\left (13\,A+15\,B+20\,C\right )\,16{}\mathrm {i}}{13\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,8{}\mathrm {i}}{13\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {a^2\,\left (5\,A+16\,B+20\,C\right )\,8{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (143\,A+650\,B+811\,C\right )\,32{}\mathrm {i}}{9009\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,8{}\mathrm {i}}{7\,d}\right )+\frac {A\,a^2\,8{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (A-7\,C\right )\,32{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (9\,A+10\,B+4\,C\right )\,8{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\left (\frac {A\,a^2\,8{}\mathrm {i}}{3\,d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (10439\,A+9230\,B+8368\,C\right )\,8{}\mathrm {i}}{45045\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (10439\,A+9230\,B+8368\,C\right )\,16{}\mathrm {i}}{45045\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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