Optimal. Leaf size=215 \[ -\frac {4 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {4 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d} \]
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Rubi [A] time = 0.26, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2695, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {4 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {4 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2695
Rule 2752
Rule 2753
Rubi steps
\begin {align*} \int \cos ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx &=\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {2 \int (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)} \, dx}{5 b}\\ &=-\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {4 \int \frac {2 a b+\frac {1}{2} \left (a^2+3 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{15 b}\\ &=-\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\left (2 a \left (a^2-b^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{15 b^2}+\frac {\left (2 \left (a^2+3 b^2\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{15 b^2}\\ &=-\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {\left (2 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{15 b^2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (2 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}} \, dx}{15 b^2 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {4 \left (a^2+3 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{15 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{15 b^2 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.83, size = 185, normalized size = 0.86 \[ \frac {b \cos (c+d x) \left (2 a^2+8 a b \sin (c+d x)-3 b^2 \cos (2 (c+d x))+3 b^2\right )+4 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )-4 \left (a^3+a^2 b+3 a b^2+3 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2}, x\right ) \]
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 \[ \int \sqrt {b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.72, size = 792, normalized size = 3.68 \[ \frac {\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b}{15}+\frac {4 a^{2} \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2}}{5}-\frac {4 a \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}}{15}-\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{4}}{5}-\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{4}}{15}-\frac {8 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b^{2}}{15}+\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{4}}{5}-\frac {2 b^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{5}-\frac {8 a \,b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{15}-\frac {2 a^{2} b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{15}+\frac {2 b^{4} \left (\sin ^{2}\left (d x +c \right )\right )}{5}+\frac {8 a \,b^{3} \sin \left (d x +c \right )}{15}+\frac {2 a^{2} b^{2}}{15}}{b^{3} \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d} \]
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 \[ \int \sqrt {b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^2\,\sqrt {a+b\,\sin \left (c+d\,x\right )} \,d x \]
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 \[ \int \sqrt {a + b \sin {\left (c + d x \right )}} \cos ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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