Optimal. Leaf size=196 \[ \frac {2 a^2 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{15 d}-\frac {16 a^2 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d} \]
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Rubi [A] time = 0.39, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4089, 4018, 3997, 3787, 3771, 2639, 2641} \[ \frac {2 a^2 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{15 d}-\frac {16 a^2 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 3771
Rule 3787
Rule 3997
Rule 4018
Rule 4089
Rubi steps
\begin {align*} \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx &=\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{2} a (5 A-C)+2 a C \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{5 a}\\ &=\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {8 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {4 \int \frac {(a+a \sec (c+d x)) \left (\frac {1}{4} a^2 (15 A-7 C)+\frac {1}{4} a^2 (15 A+17 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{15 a}\\ &=\frac {2 a^2 (15 A+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {8 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {8 \int \frac {-3 a^3 C+\frac {5}{4} a^3 (3 A+C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a}\\ &=\frac {2 a^2 (15 A+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {8 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {1}{5} \left (8 a^2 C\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^2 (3 A+C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 a^2 (15 A+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {8 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {1}{5} \left (8 a^2 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^2 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {16 a^2 C \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 {4 a^2 (3 A+C) \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 a^2 (15 A+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {8 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}\\ \end {align*}
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Mathematica [C] time = 6.01, size = 312, normalized size = 1.59 \[ \frac {a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 \left (A+C \sec ^2(c+d x)\right ) \left (\frac {-3 \csc (c) \cos (d x) (5 A \cos (2 c)-5 A-16 C)+30 A \cos (c) \sin (d x)+2 C \tan (c+d x) (3 \sec (c+d x)+10)}{2 d \sec ^{\frac {7}{2}}(c+d x)}-\frac {2 i \sqrt {2} \cos ^4(c+d x) \left (5 \left (-1+e^{2 i c}\right ) (3 A+C) e^{i (c+d x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )+12 \left (-1+e^{2 i c}\right ) C \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )+12 C \sqrt {1+e^{2 i (c+d x)}}\right )}{\left (-1+e^{2 i c}\right ) d \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}}}\right )}{15 (A \cos (2 (c+d x))+A+2 C)} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{2} \sec \left (d x + c\right )^{4} + 2 \, C a^{2} \sec \left (d x + c\right )^{3} + {\left (A + C\right )} a^{2} \sec \left (d x + c\right )^{2} + 2 \, A a^{2} \sec \left (d x + c\right ) + A a^{2}}{\sqrt {\sec \left (d x + c\right )}}, x\right ) \]
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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 13.44, size = 756, normalized size = 3.86 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
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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