Optimal. Leaf size=186 \[ \frac{10 \left (a^2+b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}} \text{EllipticF}\left (\frac{1}{2} \left (-\tan ^{-1}(a,b)+c+d x\right ),2\right )}{21 d \sqrt{a \cos (c+d x)+b \sin (c+d x)}}-\frac{10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt{a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac{2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d} \]
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Rubi [A] time = 0.0958722, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3073, 3078, 2641} \[ \frac{10 \left (a^2+b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}} F\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{21 d \sqrt{a \cos (c+d x)+b \sin (c+d x)}}-\frac{10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt{a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac{2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d} \]
Antiderivative was successfully verified.
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Rule 3073
Rule 3078
Rule 2641
Rubi steps
\begin{align*} \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx &=-\frac{2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac{1}{7} \left (5 \left (a^2+b^2\right )\right ) \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx\\ &=-\frac{10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt{a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac{2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac{1}{21} \left (5 \left (a^2+b^2\right )^2\right ) \int \frac{1}{\sqrt{a \cos (c+d x)+b \sin (c+d x)}} \, dx\\ &=-\frac{10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt{a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac{2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac{\left (5 \left (a^2+b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}\right ) \int \frac{1}{\sqrt{\cos \left (c+d x-\tan ^{-1}(a,b)\right )}} \, dx}{21 \sqrt{a \cos (c+d x)+b \sin (c+d x)}}\\ &=-\frac{10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt{a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac{2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac{10 \left (a^2+b^2\right )^2 F\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}}{21 d \sqrt{a \cos (c+d x)+b \sin (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.85859, size = 205, normalized size = 1.1 \[ \frac{\frac{20 \left (a^2+b^2\right )^2 \tan \left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d x\right ) \sqrt{\cos ^2\left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d x\right )} \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d x\right )\right )}{\sqrt{b \sqrt{\frac{a^2}{b^2}+1} \sin \left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d x\right )}}+\sqrt{a \cos (c+d x)+b \sin (c+d x)} \left (-23 b \left (a^2+b^2\right ) \cos (c+d x)+\left (3 b^3-9 a^2 b\right ) \cos (3 (c+d x))+2 a \sin (c+d x) \left (3 \left (a^2-3 b^2\right ) \cos (2 (c+d x))+13 a^2+7 b^2\right )\right )}{42 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.891, size = 183, normalized size = 1. \begin{align*}{\frac{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}{21\,\cos \left ( dx+c-\arctan \left ( -a,b \right ) \right ) d} \left ( 6\, \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{5}+5\,\sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) +2}\sqrt{-\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }{\it EllipticF} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },1/2\,\sqrt{2} \right ) +4\, \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{3}-10\,\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ){\frac{1}{\sqrt{\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \sqrt{{a}^{2}+{b}^{2}}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac{7}{2}}\,{d x} \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 (3 \, a b^{2} \cos \left (d x + c\right ) +{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (b^{3} +{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}, 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 (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac{7}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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