Optimal. Leaf size=207 \[ -\frac{\left (12 a^2-18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d (a-b)^{3/2}}+\frac{\left (12 a^2+18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d (a+b)^{3/2}}-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 d \left (a^2-b^2\right )}+\frac{\tan (c+d x) \sec ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{4 d} \]
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Rubi [A] time = 0.323333, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2668, 737, 823, 827, 1166, 206} \[ -\frac{\left (12 a^2-18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d (a-b)^{3/2}}+\frac{\left (12 a^2+18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d (a+b)^{3/2}}-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 d \left (a^2-b^2\right )}+\frac{\tan (c+d x) \sec ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{4 d} \]
Antiderivative was successfully verified.
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Rule 2668
Rule 737
Rule 823
Rule 827
Rule 1166
Rule 206
Rubi steps
\begin{align*} \int \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{-3 a-\frac{5 x}{2}}{\sqrt{a+x} \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{\frac{1}{4} a \left (12 a^2-13 b^2\right )+\frac{1}{4} \left (6 a^2-5 b^2\right ) x}{\sqrt{a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{\frac{1}{4} a \left (12 a^2-13 b^2\right )-\frac{1}{4} a \left (6 a^2-5 b^2\right )+\frac{1}{4} \left (6 a^2-5 b^2\right ) x^2}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}-\frac{\left (12 a^2-18 a b+5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 (a-b) d}+\frac{\left (12 a^2+18 a b+5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 (a+b) d}\\ &=-\frac{\left (12 a^2-18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 (a-b)^{3/2} d}+\frac{\left (12 a^2+18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 (a+b)^{3/2} d}-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 1.48296, size = 224, normalized size = 1.08 \[ \frac{-\sqrt{a-b} (a+b)^2 \left (12 a^2-18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )+(a-b)^2 \sqrt{a+b} \left (12 a^2+18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )+\frac{1}{2} \left (a^2-b^2\right ) \sec ^4(c+d x) \sqrt{a+b \sin (c+d x)} \left (\left (22 a^2-21 b^2\right ) \sin (c+d x)+6 a^2 \sin (3 (c+d x))-2 a b \cos (2 (c+d x))-2 a b-5 b^2 \sin (3 (c+d x))\right )}{32 d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.717, size = 509, normalized size = 2.5 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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 (\sqrt{b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{5}, 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 \sqrt{b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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