Optimal. Leaf size=144 \[ -\frac{\sqrt{b} \left (15 a^2+20 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 f (a+b)^{5/2}}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{x}{a^3}-\frac{b \tan (e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rubi [A] time = 0.174835, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4128, 414, 527, 522, 203, 205} \[ -\frac{\sqrt{b} \left (15 a^2+20 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 f (a+b)^{5/2}}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{x}{a^3}-\frac{b \tan (e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
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Rule 4128
Rule 414
Rule 527
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{4 a+b-3 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a (a+b) f}\\ &=-\frac{b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2+9 a b+4 b^2-b (7 a+4 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a+b)^2 f}\\ &=-\frac{b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}-\frac{\left (b \left (15 a^2+20 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a+b)^2 f}\\ &=\frac{x}{a^3}-\frac{\sqrt{b} \left (15 a^2+20 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 (a+b)^{5/2} f}-\frac{b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [C] time = 5.76198, size = 332, normalized size = 2.31 \[ \frac{\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{b \left (\left (9 a^2+28 a b+16 b^2\right ) \sin (2 e)-3 a (3 a+2 b) \sin (2 f x)\right ) (a \cos (2 (e+f x))+a+2 b)}{f (a+b)^2 (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}+\frac{b \left (15 a^2+20 a b+8 b^2\right ) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{f (a+b)^{5/2} \sqrt{b (\cos (e)-i \sin (e))^4}}-\frac{4 b^2 ((a+2 b) \sin (2 e)-a \sin (2 f x))}{f (a+b) (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}+8 x (a \cos (2 (e+f x))+a+2 b)^2\right )}{64 a^3 \left (a+b \sec ^2(e+f x)\right )^3} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.086, size = 321, normalized size = 2.2 \begin{align*}{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{3}}}-{\frac{7\,{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,fa \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{2\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{9\,b\tan \left ( fx+e \right ) }{8\, \left ( a+b \right ) af \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{2}\tan \left ( fx+e \right ) }{2\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ( a+b \right ) }}-{\frac{15\,b}{8\,fa \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-{\frac{5\,{b}^{2}}{2\,f{a}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-{\frac{{b}^{3}}{f{a}^{3} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\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 = 0.716671, size = 1854, normalized size = 12.88 \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 = 1.27303, size = 277, normalized size = 1.92 \begin{align*} -\frac{\frac{{\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )}{\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 )}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt{a b + b^{2}}} + \frac{7 \, a b^{2} \tan \left (f x + e\right )^{3} + 4 \, b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{2} b \tan \left (f x + e\right ) + 13 \, a b^{2} \tan \left (f x + e\right ) + 4 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac{8 \,{\left (f x + e\right )}}{a^{3}}}{8 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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