Optimal. Leaf size=236 \[ \frac{2 (b B-a C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 b^2 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}+\frac{2 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d \sqrt{\cos (c+d x)}}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac{2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 C \sin (c+d x)}{5 b d \cos ^{\frac{5}{2}}(c+d x)} \]
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Rubi [A] time = 1.2782, antiderivative size = 236, normalized size of antiderivative = 1., 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 = {4112, 3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}+\frac{2 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d \sqrt{\cos (c+d x)}}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac{2 (b B-a C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^2 d}+\frac{2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 C \sin (c+d x)}{5 b d \cos ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 4112
Rule 3055
Rule 3059
Rule 2639
Rule 3002
Rule 2641
Rule 2805
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx &=\int \frac{C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (b+a \cos (c+d x))} \, dx\\ &=\frac{2 C \sin (c+d x)}{5 b d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{\frac{5}{2} (b B-a C)+\frac{1}{2} b (5 A+3 C) \cos (c+d x)+\frac{3}{2} a C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{5 b}\\ &=\frac{2 C \sin (c+d x)}{5 b d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\frac{3}{4} \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )+\frac{1}{4} b (5 b B+4 a C) \cos (c+d x)+\frac{5}{4} a (b B-a C) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{15 b^2}\\ &=\frac{2 C \sin (c+d x)}{5 b d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt{\cos (c+d x)}}+\frac{8 \int \frac{\frac{5}{8} \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right )-\frac{1}{8} b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right ) \cos (c+d x)-\frac{3}{8} a \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{15 b^3}\\ &=\frac{2 C \sin (c+d x)}{5 b d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt{\cos (c+d x)}}-\frac{8 \int \frac{-\frac{5}{8} a \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right )-\frac{5}{8} a^2 b (b B-a C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{15 a b^3}-\frac{\left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 b^3}\\ &=-\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac{2 C \sin (c+d x)}{5 b d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt{\cos (c+d x)}}+\frac{(b B-a C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^2}-\frac{\left (a \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^3}\\ &=-\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac{2 (b B-a C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^2 d}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 (a+b) d}+\frac{2 C \sin (c+d x)}{5 b d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 4.82478, size = 334, normalized size = 1.42 \[ \frac{-\frac{2 b \left (20 a^2 C-20 a b B+15 A b^2+9 b^2 C\right ) \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac{6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}-\frac{2 \left (-45 a^2 b B+45 a^3 C+a b^2 (45 A+19 C)-10 b^3 B\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{2 \left (3 \sin (2 (c+d x)) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )+10 b (b B-a C) \sin (c+d x)+6 b^2 C \tan (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)}}{30 b^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 10.045, size = 800, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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