Optimal. Leaf size=229 \[ \frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d} \]
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Rubi [A]
time = 0.39, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {4173, 4095,
4086, 3878, 3877} \begin {gather*} \frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (165 A+143 B+125 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac {2 a (165 A+143 B+125 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 3877
Rule 3878
Rule 4086
Rule 4095
Rule 4173
Rubi steps
\begin {align*} \int \sec ^2(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 ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (11 A+4 C)+\frac {1}{2} a (11 B+5 C) \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {4 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {7}{4} a^2 (11 B+5 C)+\frac {1}{4} a^2 (99 A-22 B+26 C) \sec (c+d x)\right ) \, dx}{99 a^2}\\ &=\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {1}{231} (165 A+143 B+125 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {(8 a (165 A+143 B+125 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx}{1155}\\ &=\frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {\left (32 a^2 (165 A+143 B+125 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{3465}\\ &=\frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}\\ \end {align*}
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Mathematica [A]
time = 1.73, size = 188, normalized size = 0.82 \begin {gather*} \frac {a^2 (13365 A+15356 B+18140 C+(49830 A+49654 B+50140 C) \cos (c+d x)+4 (4290 A+4642 B+4615 C) \cos (2 (c+d x))+22935 A \cos (3 (c+d x))+20878 B \cos (3 (c+d x))+18460 C \cos (3 (c+d x))+3795 A \cos (4 (c+d x))+3212 B \cos (4 (c+d x))+2840 C \cos (4 (c+d x))+3795 A \cos (5 (c+d x))+3212 B \cos (5 (c+d x))+2840 C \cos (5 (c+d x))) \sec ^5(c+d x) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{13860 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 14.90, size = 207, normalized size = 0.90
method | result | size |
default | \(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (7590 A \left (\cos ^{5}\left (d x +c \right )\right )+6424 B \left (\cos ^{5}\left (d x +c \right )\right )+5680 C \left (\cos ^{5}\left (d x +c \right )\right )+3795 A \left (\cos ^{4}\left (d x +c \right )\right )+3212 B \left (\cos ^{4}\left (d x +c \right )\right )+2840 C \left (\cos ^{4}\left (d x +c \right )\right )+1980 A \left (\cos ^{3}\left (d x +c \right )\right )+2409 B \left (\cos ^{3}\left (d x +c \right )\right )+2130 C \left (\cos ^{3}\left (d x +c \right )\right )+495 A \left (\cos ^{2}\left (d x +c \right )\right )+1430 B \left (\cos ^{2}\left (d x +c \right )\right )+1775 C \left (\cos ^{2}\left (d x +c \right )\right )+385 B \cos \left (d x +c \right )+1120 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )}\) | \(207\) |
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 = 2.82, size = 168, normalized size = 0.73 \begin {gather*} \frac {2 \, {\left (2 \, {\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (660 \, A + 803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (99 \, A + 286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right ) + 315 \, 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 )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\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 = 1.64, size = 383, normalized size = 1.67 \begin {gather*} \frac {8 \, {\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (165 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} {\left (165 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 99 \, \sqrt {2} {\left (165 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 231 \, \sqrt {2} {\left (85 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 69 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 65 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1155 \, \sqrt {2} {\left (11 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 9 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3465 \, \sqrt {2} {\left (A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \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 = 17.05, size = 1034, normalized size = 4.52 \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\,a^2\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (9\,A+10\,B+4\,C\right )\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (33\,A+44\,B-31\,C\right )\,16{}\mathrm {i}}{1155\,d}\right )+\frac {a^2\,\left (5\,A+16\,B+20\,C\right )\,4{}\mathrm {i}}{5\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\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\,4{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (6\,A+5\,B+2\,C\right )\,8{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (10\,A+11\,B+10\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (13\,A+15\,B+20\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{11\,d}\right )+\frac {A\,a^2\,4{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (6\,A+5\,B+2\,C\right )\,8{}\mathrm {i}}{11\,d}-\frac {a^2\,\left (10\,A+11\,B+10\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (13\,A+15\,B+20\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\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\,a^2\,4{}\mathrm {i}}{9\,d}+\frac {C\,a^2\,64{}\mathrm {i}}{99\,d}-\frac {a^2\,\left (5\,A+2\,B-16\,C\right )\,4{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (11\,A+10\,B+4\,C\right )\,4{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (15\,A+20\,B+36\,C\right )\,4{}\mathrm {i}}{9\,d}\right )-\frac {a^2\,\left (A-16\,C\right )\,4{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (3\,A+4\,B+4\,C\right )\,20{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (11\,A+10\,B+20\,C\right )\,4{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\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\,a^2\,4{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,A+5\,B+2\,C\right )\,8{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (5\,A+10\,B+32\,C\right )\,4{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (11\,B+50\,C\right )\,32{}\mathrm {i}}{693\,d}\right )+\frac {a^2\,\left (A-8\,B\right )\,4{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,A+9\,B+10\,C\right )\,8{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\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 {\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\,4{}\mathrm {i}}{3\,d}-\frac {a^2\,\left (660\,A+803\,B+710\,C\right )\,8{}\mathrm {i}}{3465\,d}\right )+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{3\,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 )}-\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 (3795\,A+3212\,B+2840\,C\right )\,4{}\mathrm {i}}{3465\,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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