Optimal. Leaf size=241 \[ -\frac {2 \left (10 a b B-5 a^2 (A-C)+b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^2 B+b^2 B+2 a b (3 A+C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (5 A b^2+10 a b B+4 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 b B+4 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d} \]
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Rubi [A]
time = 0.35, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4181, 4161,
4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (4 a^2 C+10 a b B+5 A b^2+3 b^2 C\right )}{5 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 (3 a^2 B+2 a b (3 A+C)+b^2 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^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right )}{5 d}+\frac {2 b (4 a C+5 b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 3856
Rule 4131
Rule 4132
Rule 4161
Rule 4181
Rubi steps
\begin {align*} \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx &=\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2}{5} \int \frac {(a+b \sec (c+d x)) \left (\frac {1}{2} a (5 A-C)+\frac {1}{2} (5 A b+5 a B+3 b C) \sec (c+d x)+\frac {1}{2} (5 b B+4 a C) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b (5 b B+4 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {3}{4} a^2 (5 A-C)+\frac {5}{4} \left (3 a^2 B+b^2 B+2 a b (3 A+C)\right ) \sec (c+d x)+\frac {3}{4} \left (5 A b^2+10 a b B+4 a^2 C+3 b^2 C\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b (5 b B+4 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {3}{4} a^2 (5 A-C)+\frac {3}{4} \left (5 A b^2+10 a b B+4 a^2 C+3 b^2 C\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (3 a^2 B+b^2 B+2 a b (3 A+C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 \left (5 A b^2+10 a b B+4 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 b B+4 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (-10 a b B+5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (3 a^2 B+b^2 B+2 a b (3 A+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 (3 a^2 B+b^2 B+2 a b (3 A+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 \left (5 A b^2+10 a b B+4 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 b B+4 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (\left (-10 a b B+5 a^2 (A-C)-b^2 (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 (10 a b B-5 a^2 (A-C)+b^2 (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 (3 a^2 B+b^2 B+2 a b (3 A+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 \left (5 A b^2+10 a b B+4 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 b B+4 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}\\ \end {align*}
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Mathematica [A]
time = 2.25, size = 271, normalized size = 1.12 \begin {gather*} \frac {4 (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (3 \left (-10 a b B+5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (3 a^2 B+b^2 B+2 a b (3 A+C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+15 A b^2 \sin (c+d x)+30 a b B \sin (c+d x)+15 a^2 C \sin (c+d x)+9 b^2 C \sin (c+d x)+5 b^2 B \tan (c+d x)+10 a b C \tan (c+d x)+3 b^2 C \sec (c+d x) \tan (c+d x)\right )}{15 d (b+a \cos (c+d x))^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {7}{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. \(972\) vs.
\(2(269)=538\).
time = 0.22, size = 973, normalized size = 4.04
method | result | size |
default | \(\text {Expression too large to display}\) | \(973\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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 = 1.01, size = 319, normalized size = 1.32 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (3 i \, B a^{2} + 2 i \, {\left (3 \, A + C\right )} a b + i \, B b^{2}\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 (-3 i \, B a^{2} - 2 i \, {\left (3 \, A + C\right )} a b - i \, B b^{2}\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 \, {\left (A - C\right )} a^{2} + 10 i \, B a b + i \, {\left (5 \, A + 3 \, C\right )} b^{2}\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 \, {\left (A - C\right )} a^{2} - 10 i \, B a b - i \, {\left (5 \, A + 3 \, C\right )} b^{2}\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 (3 \, C b^{2} + 3 \, {\left (5 \, C a^{2} + 10 \, B a b + {\left (5 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, C a b + B b^{2}\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 )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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