Optimal. Leaf size=144 \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} d (a+b)^{5/2}}+\frac{3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{8 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )}+\frac{b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2} \]
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Rubi [A] time = 0.146423, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3184, 3173, 12, 3181, 205} \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} d (a+b)^{5/2}}+\frac{3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{8 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )}+\frac{b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2} \]
Antiderivative was successfully verified.
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Rule 3184
Rule 3173
Rule 12
Rule 3181
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx &=\frac{b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}-\frac{\int \frac{-4 a-3 b+2 b \sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx}{4 a (a+b)}\\ &=\frac{b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}-\frac{\int \frac{-8 a^2-8 a b-3 b^2}{a+b \sin ^2(c+d x)} \, dx}{8 a^2 (a+b)^2}\\ &=\frac{b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}+\frac{\left (8 a^2+8 a b+3 b^2\right ) \int \frac{1}{a+b \sin ^2(c+d x)} \, dx}{8 a^2 (a+b)^2}\\ &=\frac{b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}+\frac{\left (8 a^2+8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{8 a^2 (a+b)^2 d}\\ &=\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} (a+b)^{5/2} d}+\frac{b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 1.24192, size = 125, normalized size = 0.87 \[ \frac{\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{(a+b)^{5/2}}+\frac{\sqrt{a} b \sin (2 (c+d x)) \left (16 a^2-3 b (2 a+b) \cos (2 (c+d x))+16 a b+3 b^2\right )}{(a+b)^2 (2 a-b \cos (2 (c+d x))+b)^2}}{8 a^{5/2} d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.093, size = 334, normalized size = 2.3 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}b}{d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2}a \left ( a+b \right ) }}+{\frac{3\,{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2}{a}^{2} \left ( a+b \right ) }}+{\frac{\tan \left ( dx+c \right ) b}{d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{5\,{b}^{2}\tan \left ( dx+c \right ) }{8\,d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2}a \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{1}{d \left ({a}^{2}+2\,ab+{b}^{2} \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 ) }}}}+{\frac{b}{d \left ({a}^{2}+2\,ab+{b}^{2} \right ) a}\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 ) }}}}+{\frac{3\,{b}^{2}}{8\,{a}^{2}d \left ({a}^{2}+2\,ab+{b}^{2} \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 [B] time = 2.3178, size = 1883, normalized size = 13.08 \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 = 1.16285, size = 285, normalized size = 1.98 \begin{align*} \frac{\frac{{\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 (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt{a^{2} + a b}} + \frac{8 \, a^{2} b \tan \left (d x + c\right )^{3} + 11 \, a b^{2} \tan \left (d x + c\right )^{3} + 3 \, b^{3} \tan \left (d x + c\right )^{3} + 8 \, a^{2} b \tan \left (d x + c\right ) + 5 \, a b^{2} \tan \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )}{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}}}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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