Optimal. Leaf size=154 \[ -\frac {3 (5 A+2 C) \sin (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )}{5 b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )}{2 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 b^2 d} \]
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Rubi [A] time = 0.15, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {16, 4047, 3772, 2643, 4046} \[ -\frac {3 (5 A+2 C) \sin (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )}{5 b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )}{2 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 b^2 d} \]
Antiderivative was successfully verified.
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Rule 16
Rule 2643
Rule 3772
Rule 4046
Rule 4047
Rubi steps
\begin {align*} \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx &=\frac {\int (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac {\int (b \sec (c+d x))^{2/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}+\frac {B \int (b \sec (c+d x))^{5/3} \, dx}{b^3}\\ &=\frac {3 C (b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d}+\frac {(5 A+2 C) \int (b \sec (c+d x))^{2/3} \, dx}{5 b^2}+\frac {\left (B \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{5/3}} \, dx}{b^3}\\ &=\frac {3 B \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{2 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d}+\frac {\left ((5 A+2 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{2/3}} \, dx}{5 b^2}\\ &=\frac {3 B \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{2 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 (5 A+2 C) \cos (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{5 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d}\\ \end {align*}
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Mathematica [C] time = 1.96, size = 299, normalized size = 1.94 \[ \frac {\left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {3 (5 B \csc (c) \cos (d x)+2 C \tan (c+d x))}{d}-\frac {3 i 2^{2/3} e^{-i (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (\left (-1+e^{2 i c}\right ) (5 A+2 C) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-e^{2 i (c+d x)}\right )+5 B \left (-1+e^{2 i c}\right ) \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (-\frac {1}{6},\frac {2}{3};\frac {5}{6};-e^{2 i (c+d x)}\right )+5 B \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sec ^{\frac {2}{3}}(c+d x)}\right )}{5 (b \sec (c+d x))^{4/3} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}}{b^{2}}, x\right ) \]
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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.77, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{2}\left (d x +c \right )\right ) \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]
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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]
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 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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