Optimal. Leaf size=271 \[ \frac {4 a^3 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {4 a^3 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a^3 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{15 a d}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}} \]
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Rubi [A] time = 0.63, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4086, 4018, 3997, 3787, 3771, 2639, 2641} \[ \frac {4 a^3 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}-\frac {2 (5 A-5 B-9 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {4 a^3 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a^3 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{15 a d}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 3771
Rule 3787
Rule 3997
Rule 4018
Rule 4086
Rubi steps
\begin {align*} \int \frac {(a+a \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+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \int \frac {(a+a \sec (c+d x))^3 \left (\frac {3}{2} a (2 A+B)-\frac {1}{2} a (5 A-3 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{3 a}\\ &=\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}+\frac {4 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{4} a^2 (35 A+15 B-3 C)-\frac {3}{4} a^2 (5 A-5 B-9 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{15 a}\\ &=\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {8 \int \frac {(a+a \sec (c+d x)) \left (\frac {3}{4} a^3 (20 A+5 B-6 C)+\frac {3}{4} a^3 (5 A+20 B+21 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{45 a}\\ &=\frac {4 a^3 (5 A+20 B+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {16 \int \frac {\frac {9}{8} a^4 (5 A-5 B-9 C)+\frac {15}{8} a^4 (5 A+5 B+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{45 a}\\ &=\frac {4 a^3 (5 A+20 B+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (2 a^3 (5 A-5 B-9 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^3 (5 A+5 B+3 C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {4 a^3 (5 A+20 B+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (2 a^3 (5 A-5 B-9 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^3 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^3 (5 A-5 B-9 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^3 (5 A+5 B+3 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 {4 a^3 (5 A+20 B+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}\\ \end {align*}
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Mathematica [C] time = 4.71, size = 316, normalized size = 1.17 \[ \frac {a^3 e^{-i d x} \sec ^{\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-4 i (5 A-5 B-9 C) e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+80 (5 A+5 B+3 C) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+30 A \sin (c+d x)+10 A \sin (2 (c+d x))+30 A \sin (3 (c+d x))+5 A \sin (4 (c+d x))+180 i A \cos (c+d x)+60 i A \cos (3 (c+d x))+90 B \sin (c+d x)+20 B \sin (2 (c+d x))+90 B \sin (3 (c+d x))-180 i B \cos (c+d x)-60 i B \cos (3 (c+d x))+132 C \sin (c+d x)+60 C \sin (2 (c+d x))+108 C \sin (3 (c+d x))-324 i C \cos (c+d x)-108 i C \cos (3 (c+d x))\right )}{60 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{3} \sec \left (d x + c\right )^{5} + {\left (B + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{4} + {\left (A + 3 \, B + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{3} + {\left (3 \, A + 3 \, B + C\right )} a^{3} \sec \left (d x + c\right )^{2} + {\left (3 \, A + B\right )} a^{3} \sec \left (d x + c\right ) + A a^{3}}{\sec \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]
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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 18.16, size = 1328, normalized size = 4.90 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
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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