Optimal. Leaf size=142 \[ \frac {\left (3 a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 b^{5/2} f (a+b)^{5/2}}-\frac {3 a (a+2 b) \tan (e+f x)}{8 b^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {a \tan (e+f x) \sec ^2(e+f x)}{4 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.16, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4146, 413, 385, 205} \[ \frac {\left (3 a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 b^{5/2} f (a+b)^{5/2}}-\frac {3 a (a+2 b) \tan (e+f x)}{8 b^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {a \tan (e+f x) \sec ^2(e+f x)}{4 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 205
Rule 385
Rule 413
Rule 4146
Rubi steps
\begin {align*} \int \frac {\sec ^6(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 )^2}{\left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a \sec ^2(e+f x) \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {a+4 b+(3 a+4 b) x^2}{\left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 b (a+b) f}\\ &=-\frac {a \sec ^2(e+f x) \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {3 a (a+2 b) \tan (e+f x)}{8 b^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\left (3 a^2+8 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 b^2 (a+b)^2 f}\\ &=\frac {\left (3 a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 b^{5/2} (a+b)^{5/2} f}-\frac {a \sec ^2(e+f x) \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {3 a (a+2 b) \tan (e+f x)}{8 b^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.90, size = 125, normalized size = 0.88 \[ \frac {\frac {\left (3 a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}-\frac {a \sqrt {b} \sin (2 (e+f x)) \left (3 a^2+3 a (a+2 b) \cos (2 (e+f x))+16 a b+16 b^2\right )}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^2}}{8 b^{5/2} f} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 1.45, size = 722, normalized size = 5.08 \[ \left [-\frac {{\left ({\left (3 \, a^{4} + 8 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} + 8 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b + 8 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, {\left (3 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 2 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} b^{2} + 13 \, a^{2} b^{3} + 8 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{5} b^{3} + 3 \, a^{4} b^{4} + 3 \, a^{3} b^{5} + a^{2} b^{6}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{4} + 3 \, a^{3} b^{5} + 3 \, a^{2} b^{6} + a b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{5} + 3 \, a^{2} b^{6} + 3 \, a b^{7} + b^{8}\right )} f\right )}}, -\frac {{\left ({\left (3 \, a^{4} + 8 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} + 8 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b + 8 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left (3 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 2 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} b^{2} + 13 \, a^{2} b^{3} + 8 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{5} b^{3} + 3 \, a^{4} b^{4} + 3 \, a^{3} b^{5} + a^{2} b^{6}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{4} + 3 \, a^{3} b^{5} + 3 \, a^{2} b^{6} + a b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{5} + 3 \, a^{2} b^{6} + 3 \, a b^{7} + b^{8}\right )} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.36, size = 193, normalized size = 1.36 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a^{2} + 8 \, a b + 8 \, b^{2}\right )}}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \sqrt {a b + b^{2}}} - \frac {5 \, a^{2} b \tan \left (f x + e\right )^{3} + 8 \, a b^{2} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 11 \, a^{2} b \tan \left (f x + e\right ) + 8 \, a b^{2} \tan \left (f x + e\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.64, size = 294, normalized size = 2.07 \[ -\frac {5 a^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} b \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \left (\tan ^{3}\left (f x +e \right )\right )}{f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} \left (a^{2}+2 a b +b^{2}\right )}-\frac {3 a^{2} \tan \left (f x +e \right )}{8 f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} b^{2} \left (a +b \right )}-\frac {a \tan \left (f x +e \right )}{b \left (a +b \right ) f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {3 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a^{2}}{8 f \left (a^{2}+2 a b +b^{2}\right ) b^{2} \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a}{f \left (a^{2}+2 a b +b^{2}\right ) b \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.45, size = 210, normalized size = 1.48 \[ \frac {\frac {{\left (3 \, a^{2} + 8 \, a b + 8 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} b}} - \frac {{\left (5 \, a^{2} b + 8 \, a b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{3} + 11 \, a^{2} b + 8 \, a b^{2}\right )} \tan \left (f x + e\right )}{a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6} + {\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{2}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 5.21, size = 149, normalized size = 1.05 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{8\,b^{5/2}\,f\,{\left (a+b\right )}^{5/2}}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (5\,a^2+8\,b\,a\right )}{8\,b\,{\left (a+b\right )}^2}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (3\,a^2+8\,b\,a\right )}{8\,b^2\,\left (a+b\right )}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________