Optimal. Leaf size=175 \[ \frac {64 a^3 (15 B+13 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (15 B+13 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {2 a (15 B+13 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]
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Rubi [A] time = 0.41, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4072, 4010, 4001, 3793, 3792} \[ \frac {16 a^2 (15 B+13 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {64 a^3 (15 B+13 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {2 a (15 B+13 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]
Antiderivative was successfully verified.
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Rule 3792
Rule 3793
Rule 4001
Rule 4010
Rule 4072
Rubi steps
\begin {align*} \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (B+C \sec (c+d x)) \, dx\\ &=\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {2 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {7 a C}{2}+\frac {1}{2} a (9 B-2 C) \sec (c+d x)\right ) \, dx}{9 a}\\ &=\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{21} (15 B+13 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 a (15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{105} (8 a (15 B+13 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 (15 B+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{315} \left (32 a^2 (15 B+13 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {64 a^3 (15 B+13 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (15 B+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}\\ \end {align*}
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Mathematica [A] time = 0.65, size = 96, normalized size = 0.55 \[ \frac {2 a^3 \tan (c+d x) \left (5 (9 B+26 C) \sec ^3(c+d x)+3 (60 B+73 C) \sec ^2(c+d x)+(345 B+292 C) \sec (c+d x)+690 B+35 C \sec ^4(c+d x)+584 C\right )}{315 d \sqrt {a (\sec (c+d x)+1)}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 136, normalized size = 0.78 \[ \frac {2 \, {\left (2 \, {\left (345 \, B + 292 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (345 \, B + 292 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (60 \, B + 73 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, B + 26 \, C\right )} a^{2} \cos \left (d x + c\right ) + 35 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.12, size = 261, normalized size = 1.49 \[ \frac {8 \, {\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (15 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \sqrt {2} {\left (15 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 63 \, \sqrt {2} {\left (15 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 210 \, \sqrt {2} {\left (4 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, \sqrt {2} {\left (B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.59, size = 141, normalized size = 0.81 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (690 B \left (\cos ^{4}\left (d x +c \right )\right )+584 C \left (\cos ^{4}\left (d x +c \right )\right )+345 B \left (\cos ^{3}\left (d x +c \right )\right )+292 C \left (\cos ^{3}\left (d x +c \right )\right )+180 B \left (\cos ^{2}\left (d x +c \right )\right )+219 C \left (\cos ^{2}\left (d x +c \right )\right )+45 B \cos \left (d x +c \right )+130 C \cos \left (d x +c \right )+35 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{315 d \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.94, size = 723, normalized size = 4.13 \[ \frac {\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {B\,a^2\,4{}\mathrm {i}}{3\,d}-\frac {a^2\,\left (60\,B+73\,C\right )\,8{}\mathrm {i}}{315\,d}\right )+\frac {a^2\,\left (5\,B+2\,C\right )\,4{}\mathrm {i}}{3\,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 {\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 {B\,a^2\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (3\,B+4\,C\right )\,16{}\mathrm {i}}{105\,d}+\frac {a^2\,\left (9\,B+10\,C\right )\,4{}\mathrm {i}}{5\,d}\right )-\frac {a^2\,\left (5\,B+2\,C\right )\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (5\,B+16\,C\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 {B\,a^2\,4{}\mathrm {i}}{7\,d}+\frac {C\,a^2\,32{}\mathrm {i}}{63\,d}+\frac {a^2\,\left (B+2\,C\right )\,20{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (B+C\right )\,40{}\mathrm {i}}{7\,d}\right )+\frac {a^2\,\left (B-8\,C\right )\,4{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (5\,B+2\,C\right )\,4{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,B+9\,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 {\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 {B\,a^2\,4{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (3\,B+4\,C\right )\,20{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (5\,B+2\,C\right )\,4{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (11\,B+10\,C\right )\,4{}\mathrm {i}}{9\,d}\right )-\frac {B\,a^2\,4{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (3\,B+4\,C\right )\,20{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (5\,B+2\,C\right )\,4{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (11\,B+10\,C\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 {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 (345\,B+292\,C\right )\,4{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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