Optimal. Leaf size=181 \[ \frac{b^4 \sin (e+f x)}{4 a^4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}-\frac{b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{b^2 \left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{9/2} f (a+b)^{5/2}}+\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f} \]
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Rubi [A] time = 0.242773, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4147, 390, 1157, 385, 208} \[ \frac{b^4 \sin (e+f x)}{4 a^4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}-\frac{b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{b^2 \left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{9/2} f (a+b)^{5/2}}+\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f} \]
Antiderivative was successfully verified.
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Rule 4147
Rule 390
Rule 1157
Rule 385
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a-3 b}{a^4}-\frac{x^2}{a^3}+\frac{b^2 \left (6 a^2+8 a b+3 b^2\right )-4 a b^2 (3 a+2 b) x^2+6 a^2 b^2 x^4}{a^4 \left (a+b-a x^2\right )^3}\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f}+\frac{\operatorname{Subst}\left (\int \frac{b^2 \left (6 a^2+8 a b+3 b^2\right )-4 a b^2 (3 a+2 b) x^2+6 a^2 b^2 x^4}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{a^4 f}\\ &=\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f}+\frac{b^4 \sin (e+f x)}{4 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{-b^2 \left (24 a^2+32 a b+11 b^2\right )+24 a b^2 (a+b) x^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 a^4 (a+b) f}\\ &=\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f}+\frac{b^4 \sin (e+f x)}{4 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}-\frac{b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}+\frac{\left (b^2 \left (48 a^2+80 a b+35 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{8 a^4 (a+b)^2 f}\\ &=\frac{b^2 \left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{9/2} (a+b)^{5/2} f}+\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f}+\frac{b^4 \sin (e+f x)}{4 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}-\frac{b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 4.45507, size = 194, normalized size = 1.07 \[ \frac{-\frac{3 b^2 \left (48 a^2+80 a b+35 b^2\right ) \left (\log \left (\sqrt{a+b}-\sqrt{a} \sin (e+f x)\right )-\log \left (\sqrt{a+b}+\sqrt{a} \sin (e+f x)\right )\right )}{(a+b)^{5/2}}+4 a^{3/2} \sin (3 (e+f x))+12 \sqrt{a} \sin (e+f x) \left (-\frac{b^4 (13 a \cos (2 (e+f x))+9 a+22 b)}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^2}+a \left (3-\frac{16 b^3}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)}\right )-12 b\right )}{48 a^{9/2} f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.102, size = 177, normalized size = 1. \begin{align*}{\frac{1}{f} \left ( -{\frac{1}{{a}^{4}} \left ({\frac{a \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{3}}-\sin \left ( fx+e \right ) a+3\,\sin \left ( fx+e \right ) b \right ) }-{\frac{{b}^{2}}{{a}^{4}} \left ({\frac{1}{ \left ( -a-b+a \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}} \left ( -{\frac{ab \left ( 16\,a+13\,b \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{8\,{a}^{2}+16\,ab+8\,{b}^{2}}}+{\frac{ \left ( 16\,a+11\,b \right ) b\sin \left ( fx+e \right ) }{8\,a+8\,b}} \right ) }-{\frac{48\,{a}^{2}+80\,ab+35\,{b}^{2}}{8\,{a}^{2}+16\,ab+8\,{b}^{2}}{\it Artanh} \left ({\sin \left ( fx+e \right ) a{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ) } \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 = 0.778927, size = 1908, normalized size = 10.54 \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 = 1.29686, size = 323, normalized size = 1.78 \begin{align*} -\frac{\frac{3 \,{\left (48 \, a^{2} b^{2} + 80 \, a b^{3} + 35 \, b^{4}\right )} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \sqrt{-a^{2} - a b}} - \frac{3 \,{\left (16 \, a^{2} b^{3} \sin \left (f x + e\right )^{3} + 13 \, a b^{4} \sin \left (f x + e\right )^{3} - 16 \, a^{2} b^{3} \sin \left (f x + e\right ) - 27 \, a b^{4} \sin \left (f x + e\right ) - 11 \, b^{5} \sin \left (f x + e\right )\right )}}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )}{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}} + \frac{8 \,{\left (a^{6} \sin \left (f x + e\right )^{3} - 3 \, a^{6} \sin \left (f x + e\right ) + 9 \, a^{5} b \sin \left (f x + e\right )\right )}}{a^{9}}}{24 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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