Optimal. Leaf size=192 \[ \frac {(2 a B+2 A b-b B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {\sqrt {2} (a-b) (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {b B \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \]
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Rubi [A] time = 0.70, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4221, 3045, 2982, 2782, 205, 2774, 216} \[ \frac {(2 a B+2 A b-b B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {\sqrt {2} (a-b) (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {b B \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 216
Rule 2774
Rule 2782
Rule 2982
Rule 3045
Rule 4221
Rubi steps
\begin {align*} \int \frac {\left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {b B \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (2 a A+b B)+\frac {1}{2} a (2 A b+2 a B-b B) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{a}\\ &=\frac {b B \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\left ((a-b) (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx+\frac {\left ((2 A b+2 a B-b B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{2 a}\\ &=\frac {b B \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left (2 a (a-b) (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{d}-\frac {\left ((2 A b+2 a B-b B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a d}\\ &=\frac {(2 A b+2 a B-b B) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {\sqrt {2} (a-b) (A-B) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {b B \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.46, size = 143, normalized size = 0.74 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\sqrt {2} (2 a B+2 A b-b B) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (a-b) (A-B) \tan ^{-1}\left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )+2 b B \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)}\right )}{d \sqrt {a (\cos (c+d x)+1)}} \]
Antiderivative was successfully verified.
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fricas [A] time = 24.98, size = 208, normalized size = 1.08 \[ \frac {\sqrt {a \cos \left (d x + c\right ) + a} B b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (2 \, B a + {\left (2 \, A - B\right )} b + {\left (2 \, B a + {\left (2 \, A - B\right )} b\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {\sqrt {2} {\left ({\left (A - B\right )} a^{2} - {\left (A - B\right )} a b + {\left ({\left (A - B\right )} a^{2} - {\left (A - B\right )} a b\right )} \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{a d \cos \left (d x + c\right ) + a d} \]
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 \[ \int \frac {{\left (B b \cos \left (d x + c\right )^{2} + A a + {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}}{\sqrt {a \cos \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.59, size = 317, normalized size = 1.65 \[ -\frac {\left (B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, b \sin \left (d x +c \right )+2 A \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) b +2 B \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) a -B \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) b -2 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) a +2 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) b +2 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) a -2 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) b \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right ) \sqrt {2}}{2 d \sin \left (d x +c \right )^{2} a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
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 \[ \int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (B\,b\,{\cos \left (c+d\,x\right )}^2+\left (A\,b+B\,a\right )\,\cos \left (c+d\,x\right )+A\,a\right )}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]
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 \[ \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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