Optimal. Leaf size=148 \[ \frac{a^{3/2} (4 a+5 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 b^3 d (a+b)^{3/2}}-\frac{a (2 a+b) \tan (c+d x)}{2 b^2 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac{x (4 a-b)}{2 b^3}-\frac{\sin ^2(c+d x) \tan (c+d x)}{2 b d \left ((a+b) \tan ^2(c+d x)+a\right )} \]
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Rubi [A] time = 0.291697, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3187, 470, 578, 522, 203, 205} \[ \frac{a^{3/2} (4 a+5 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 b^3 d (a+b)^{3/2}}-\frac{a (2 a+b) \tan (c+d x)}{2 b^2 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac{x (4 a-b)}{2 b^3}-\frac{\sin ^2(c+d x) \tan (c+d x)}{2 b d \left ((a+b) \tan ^2(c+d x)+a\right )} \]
Antiderivative was successfully verified.
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Rule 3187
Rule 470
Rule 578
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a+(-a+b) x^2\right )}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=-\frac{a (2 a+b) \tan (c+d x)}{2 b^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac{\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 a (2 a+b)-2 \left (2 a^2+2 a b-b^2\right ) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{4 b^2 (a+b) d}\\ &=-\frac{a (2 a+b) \tan (c+d x)}{2 b^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac{\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac{(4 a-b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 b^3 d}+\frac{\left (a^2 (4 a+5 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{2 b^3 (a+b) d}\\ &=-\frac{(4 a-b) x}{2 b^3}+\frac{a^{3/2} (4 a+5 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 b^3 (a+b)^{3/2} d}-\frac{a (2 a+b) \tan (c+d x)}{2 b^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac{\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 1.47569, size = 106, normalized size = 0.72 \[ \frac{\frac{2 a^{3/2} (4 a+5 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{(a+b)^{3/2}}+b \sin (2 (c+d x)) \left (-\frac{2 a^2}{(a+b) (2 a-b \cos (2 (c+d x))+b)}-1\right )-2 (4 a-b) (c+d x)}{4 b^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.099, size = 187, normalized size = 1.3 \begin{align*} -{\frac{\tan \left ( dx+c \right ) }{2\,{b}^{2}d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{2\,{b}^{2}d}}-2\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d{b}^{3}}}-{\frac{{a}^{2}\tan \left ( dx+c \right ) }{2\,{b}^{2}d \left ( a+b \right ) \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) }}+2\,{\frac{{a}^{3}}{d{b}^{3} \left ( a+b \right ) \sqrt{a \left ( a+b \right ) }}\arctan \left ({\frac{ \left ( a+b \right ) \tan \left ( dx+c \right ) }{\sqrt{a \left ( a+b \right ) }}} \right ) }+{\frac{5\,{a}^{2}}{2\,{b}^{2}d \left ( a+b \right ) }\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \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 [A] time = 2.03464, size = 1413, normalized size = 9.55 \begin{align*} \text{result too large to display} \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 [A] time = 1.16718, size = 301, normalized size = 2.03 \begin{align*} \frac{\frac{{\left (4 \, a^{3} + 5 \, a^{2} b\right )}{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )}}{{\left (a b^{3} + b^{4}\right )} \sqrt{a^{2} + a b}} - \frac{2 \, a^{2} \tan \left (d x + c\right )^{3} + 2 \, a b \tan \left (d x + c\right )^{3} + b^{2} \tan \left (d x + c\right )^{3} + 2 \, a^{2} \tan \left (d x + c\right ) + a b \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{4} + b \tan \left (d x + c\right )^{4} + 2 \, a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}{\left (a b^{2} + b^{3}\right )}} - \frac{{\left (d x + c\right )}{\left (4 \, a - b\right )}}{b^{3}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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