Optimal. Leaf size=319 \[ \frac {2 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]
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Rubi [A]
time = 0.56, antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4179, 4181,
4161, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (-\left (a^2 (10 A-42 C)\right )+45 a b B+3 b^2 (5 A+3 C)\right )}{15 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right )}{5 d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{15 d}-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{15 d}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 3856
Rule 4131
Rule 4132
Rule 4161
Rule 4179
Rule 4181
Rubi steps
\begin {align*} \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2}{3} \int \frac {(a+b \sec (c+d x))^2 \left (\frac {3}{2} (2 A b+a B)+\frac {1}{2} (3 b B+a (A+3 C)) \sec (c+d x)-\frac {1}{2} b (5 A-3 C) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {4}{15} \int \frac {(a+b \sec (c+d x)) \left (\frac {1}{4} a (35 A b+15 a B-3 b C)+\frac {1}{4} \left (30 a b B+5 a^2 (A+3 C)+3 b^2 (5 A+3 C)\right ) \sec (c+d x)-\frac {3}{4} b (5 a A-5 b B-9 a C) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {8}{45} \int \frac {\frac {3}{8} a^2 (35 A b+15 a B-3 b C)+\frac {15}{8} \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sec (c+d x)+\frac {3}{8} b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {8}{45} \int \frac {\frac {3}{8} a^2 (35 A b+15 a B-3 b C)+\frac {3}{8} b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{5} \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{5} \left (\left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 3.68, size = 311, normalized size = 0.97 \begin {gather*} \frac {2 (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (6 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+30 A b^3 \sin (c+d x)+90 a b^2 B \sin (c+d x)+90 a^2 b C \sin (c+d x)+18 b^3 C \sin (c+d x)+5 a^3 A \sin (2 (c+d x))+10 b^3 B \tan (c+d x)+30 a b^2 C \tan (c+d x)+6 b^3 C \sec (c+d x) \tan (c+d x)\right )}{15 d (b+a \cos (c+d x))^3 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {9}{2}}(c+d x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1391\) vs.
\(2(343)=686\).
time = 0.26, size = 1392, normalized size = 4.36
method | result | size |
default | \(\text {Expression too large to display}\) | \(1392\) |
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 [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.75, size = 382, normalized size = 1.20 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (i \, {\left (A + 3 \, C\right )} a^{3} + 9 i \, B a^{2} b + 3 i \, {\left (3 \, A + C\right )} a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (A + 3 \, C\right )} a^{3} - 9 i \, B a^{2} b - 3 i \, {\left (3 \, A + C\right )} a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{3} - 15 i \, {\left (A - C\right )} a^{2} b + 15 i \, B a b^{2} + i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{3} + 15 i \, {\left (A - C\right )} a^{2} b - 15 i \, B a b^{2} - i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (5 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, C b^{3} + 3 \, {\left (15 \, C a^{2} b + 15 \, B a b^{2} + {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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