Optimal. Leaf size=521 \[ \frac{2 \sqrt{a+b} \cot (c+d x) \left (a^2 b (315 A-161 B+135 C)+15 a^3 (7 B-C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{105 b d}+\frac{2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{105 d}-\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (161 a^2 b B+15 a^3 C+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b^2 d}-\frac{2 a^2 A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d}+\frac{2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d} \]
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Rubi [A] time = 0.971456, antiderivative size = 521, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4056, 4058, 3921, 3784, 3832, 4004} \[ \frac{2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{105 d}+\frac{2 \sqrt{a+b} \cot (c+d x) \left (a^2 b (315 A-161 B+135 C)+15 a^3 (7 B-C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b d}-\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (161 a^2 b B+15 a^3 C+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b^2 d}-\frac{2 a^2 A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d}+\frac{2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d} \]
Antiderivative was successfully verified.
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Rule 4056
Rule 4058
Rule 3921
Rule 3784
Rule 3832
Rule 4004
Rubi steps
\begin{align*} \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{2}{7} \int (a+b \sec (c+d x))^{3/2} \left (\frac{7 a A}{2}+\frac{1}{2} (7 A b+7 a B+5 b C) \sec (c+d x)+\frac{1}{2} (7 b B+5 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{4}{35} \int \sqrt{a+b \sec (c+d x)} \left (\frac{35 a^2 A}{4}+\frac{1}{4} \left (70 a A b+35 a^2 B+21 b^2 B+40 a b C\right ) \sec (c+d x)+\frac{1}{4} \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{8}{105} \int \frac{\frac{105 a^3 A}{8}+\frac{1}{8} \left (105 a^3 B+119 a b^2 B+45 a^2 b (7 A+3 C)+5 b^3 (7 A+5 C)\right ) \sec (c+d x)+\frac{1}{8} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{8}{105} \int \frac{\frac{105 a^3 A}{8}+\left (\frac{1}{8} \left (105 a^3 B+119 a b^2 B+45 a^2 b (7 A+3 C)+5 b^3 (7 A+5 C)\right )+\frac{1}{8} \left (-161 a^2 b B-63 b^3 B-15 a^3 C-5 a b^2 (49 A+29 C)\right )\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{105} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{2 (a-b) \sqrt{a+b} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}+\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\left (a^3 A\right ) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{105} \left (15 a^3 (7 B-C)+b^3 (35 A-63 B+25 C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{2 (a-b) \sqrt{a+b} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}+\frac{2 \sqrt{a+b} \left (15 a^3 (7 B-C)+b^3 (35 A-63 B+25 C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{105 b d}-\frac{2 a^2 A \sqrt{a+b} \cot (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{d}+\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end{align*}
Mathematica [B] time = 21.2756, size = 1405, normalized size = 2.7 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 1.259, size = 5138, normalized size = 9.9 \begin{align*} \text{output 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] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \sec \left (d x + c\right )^{4} +{\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \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{\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 ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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