Optimal. Leaf size=124 \[ -\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 f (a+b)^{7/2}}-\frac{15 \cot (e+f x)}{8 f (a+b)^3}+\frac{5 \cot (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\cot (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rubi [A] time = 0.111021, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4132, 290, 325, 205} \[ -\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 f (a+b)^{7/2}}-\frac{15 \cot (e+f x)}{8 f (a+b)^3}+\frac{5 \cot (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\cot (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
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Rule 4132
Rule 290
Rule 325
Rule 205
Rubi steps
\begin{align*} \int \frac{\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 (a+b) f}\\ &=\frac{\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^2 f}\\ &=-\frac{15 \cot (e+f x)}{8 (a+b)^3 f}+\frac{\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^3 f}\\ &=-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 (a+b)^{7/2} f}-\frac{15 \cot (e+f x)}{8 (a+b)^3 f}+\frac{\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [C] time = 6.79444, size = 987, normalized size = 7.96 \[ \frac{(\cos (2 e+2 f x) a+a+2 b)^3 \left (\frac{15 b \tan ^{-1}\left (\sec (f x) \left (\frac{\cos (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{i \sin (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \cos (2 e)}{64 \sqrt{a+b} f \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{15 i b \tan ^{-1}\left (\sec (f x) \left (\frac{\cos (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{i \sin (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \sin (2 e)}{64 \sqrt{a+b} f \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right ) \sec ^6(e+f x)}{(a+b)^3 \left (b \sec ^2(e+f x)+a\right )^3}+\frac{(\cos (2 e+2 f x) a+a+2 b) \csc (e) \csc (e+f x) \sec (2 e) \left (-32 \sin (f x) a^4+32 \sin (3 f x) a^4-48 \sin (2 e-f x) a^4+48 \sin (2 e+f x) a^4-32 \sin (4 e+f x) a^4-8 \sin (2 e+3 f x) a^4+32 \sin (4 e+3 f x) a^4-8 \sin (6 e+3 f x) a^4+8 \sin (2 e+5 f x) a^4+8 \sin (6 e+5 f x) a^4-64 b \sin (f x) a^3+46 b \sin (3 f x) a^3-128 b \sin (2 e-f x) a^3+146 b \sin (2 e+f x) a^3-82 b \sin (4 e+f x) a^3+18 b \sin (2 e+3 f x) a^3+73 b \sin (4 e+3 f x) a^3-9 b \sin (6 e+3 f x) a^3-9 b \sin (2 e+5 f x) a^3+9 b \sin (4 e+5 f x) a^3+22 b^2 \sin (f x) a^2-54 b^2 \sin (3 f x) a^2-106 b^2 \sin (2 e-f x) a^2+182 b^2 \sin (2 e+f x) a^2-54 b^2 \sin (4 e+f x) a^2+54 b^2 \sin (2 e+3 f x) a^2+24 b^2 \sin (4 e+3 f x) a^2-24 b^2 \sin (6 e+3 f x) a^2-2 b^2 \sin (2 e+5 f x) a^2+2 b^2 \sin (4 e+5 f x) a^2+80 b^3 \sin (f x) a-8 b^3 \sin (3 f x) a+80 b^3 \sin (2 e-f x) a+80 b^3 \sin (2 e+f x) a-80 b^3 \sin (4 e+f x) a+8 b^3 \sin (2 e+3 f x) a+8 b^3 \sin (4 e+3 f x) a-8 b^3 \sin (6 e+3 f x) a+16 b^4 \sin (f x)+16 b^4 \sin (2 e-f x)+16 b^4 \sin (2 e+f x)-16 b^4 \sin (4 e+f x)\right ) \sec ^6(e+f x)}{512 a^2 (a+b)^3 f \left (b \sec ^2(e+f x)+a\right )^3} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.104, size = 157, normalized size = 1.3 \begin{align*} -{\frac{1}{f \left ( a+b \right ) ^{3}\tan \left ( fx+e \right ) }}-{\frac{7\,{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a+b \right ) ^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{9\,ab\tan \left ( fx+e \right ) }{8\,f \left ( a+b \right ) ^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{9\,{b}^{2}\tan \left ( fx+e \right ) }{8\,f \left ( a+b \right ) ^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,b}{8\,f \left ( a+b \right ) ^{3}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \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 = 0.701863, size = 1430, normalized size = 11.53 \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 = 1.28838, size = 248, normalized size = 2. \begin{align*} -\frac{\frac{15 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )} b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b + b^{2}}} + \frac{7 \, b^{2} \tan \left (f x + e\right )^{3} + 9 \, a b \tan \left (f x + e\right ) + 9 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac{8}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )}}{8 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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