Optimal. Leaf size=392 \[ -\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (-(2 A-3 C))-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right )}-\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b (4 A-C)+2 a^3 B-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right )}-\frac{\left (-a^2 b^2 (7 A-C)+5 a^3 b B-3 a^4 C-3 a b^3 B+5 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac{\sin (c+d x) \left (a^2 (-(2 A-3 C))-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}+\frac{\sin (c+d x) \left (-a^2 b (4 A-C)+2 a^3 B-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]
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Rubi [A] time = 1.56235, antiderivative size = 392, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (-(2 A-3 C))-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right )}-\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b (4 A-C)+2 a^3 B-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right )}-\frac{\left (-a^2 b^2 (7 A-C)+5 a^3 b B-3 a^4 C-3 a b^3 B+5 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac{\sin (c+d x) \left (a^2 (-(2 A-3 C))-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}+\frac{\sin (c+d x) \left (-a^2 b (4 A-C)+2 a^3 B-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3055
Rule 3059
Rule 2639
Rule 3002
Rule 2641
Rule 2805
Rubi steps
\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} \left (-5 A b^2+3 a b B+a^2 (2 A-3 C)\right )-a (A b-a B+b C) \cos (c+d x)+\frac{3}{2} \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{2 \int \frac{\frac{3}{4} \left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right )+\frac{1}{2} a \left (2 A b^2-3 a b B+a^2 (A+3 C)\right ) \cos (c+d x)-\frac{1}{4} b \left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{4 \int \frac{\frac{1}{8} \left (-15 A b^4-12 a^3 b B+9 a b^3 B+a^2 b^2 (16 A-3 C)+2 a^4 (A+3 C)\right )-\frac{1}{4} a \left (10 A b^3+3 a^3 B-6 a b^2 B-a^2 (7 A b-3 b C)\right ) \cos (c+d x)-\frac{3}{8} b \left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac{4 \int \frac{\frac{1}{8} b \left (15 A b^4+12 a^3 b B-9 a b^3 B-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)\right )+\frac{1}{8} a b^2 \left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^3 b \left (a^2-b^2\right )}-\frac{\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac{\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2 \left (a^2-b^2\right )}-\frac{\left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 \left (a^2-b^2\right ) d}-\frac{\left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}-\frac{\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 7.08061, size = 428, normalized size = 1.09 \[ \frac{\frac{6 b \sin (2 (c+d x)) \left (a^2 b (C-4 A)+2 a^3 B-3 a b^2 B+5 A b^3\right )+8 a \left (a^2-b^2\right ) (3 a B-5 A b) \sin (c+d x)+8 a^2 A \left (a^2-b^2\right ) \tan (c+d x)}{\left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))}+\frac{\frac{2 \left (a^2 b^2 (44 A-9 C)+4 a^4 (A+3 C)-30 a^3 b B+27 a b^3 B-45 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}-\frac{8 a \left (a^2 (3 b C-7 A b)+3 a^3 B-6 a b^2 B+10 A b^3\right ) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}-\frac{6 \sin (c+d x) \left (a^2 b (C-4 A)+2 a^3 B-3 a b^2 B+5 A b^3\right ) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}}{(a-b) (a+b)}}{12 a^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 5.498, size = 1038, normalized size = 2.7 \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 \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \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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