Integrand size = 25, antiderivative size = 143 \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {a (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 (a+b)^{3/2} f}+\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 (a+b) f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f} \]
[Out]
Time = 0.24 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3269, 390, 386, 385, 212} \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {a (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 f (a+b)^{3/2}}+\frac {\tan (e+f x) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f (a+b)}+\frac {(3 a+4 b) \tan (e+f x) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f (a+b)} \]
[In]
[Out]
Rule 212
Rule 385
Rule 386
Rule 390
Rule 3269
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f}+\frac {(3 a+4 b) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 (a+b) f} \\ & = \frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 (a+b) f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f}+\frac {(a (3 a+4 b)) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{8 (a+b) f} \\ & = \frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 (a+b) f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f}+\frac {(a (3 a+4 b)) \text {Subst}\left (\int \frac {1}{1-(a+b) x^2} \, dx,x,\frac {\sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 (a+b) f} \\ & = \frac {a (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 (a+b)^{3/2} f}+\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 (a+b) f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 14.16 (sec) , antiderivative size = 669, normalized size of antiderivative = 4.68 \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {\sec ^3(e+f x) \left (1+\frac {b \sin ^2(e+f x)}{a}\right ) \tan (e+f x) \left (-15 a \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right )-10 b \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x)-30 a \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}-20 b \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}-32 a \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}-32 b \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}+32 a \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}+32 b \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}+15 a \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}+10 b \sin ^2(e+f x) \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}\right )}{40 f \sqrt {a+b \sin ^2(e+f x)} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. \(569\) vs. \(2(127)=254\).
Time = 1.59 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.99
method | result | size |
default | \(\frac {2 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (3 a +4 b \right ) \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+a \left (3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{3}+10 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b +11 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a \,b^{2}+4 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{3}-3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{3}-10 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b -11 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a \,b^{2}-4 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{3}\right ) \left (\cos ^{4}\left (f x +e \right )\right )+2 \left (a +b \right )^{\frac {3}{2}} {\left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}} \left (3 a +4 b \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 \left (a +b \right )^{\frac {5}{2}} {\left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}} \sin \left (f x +e \right )}{16 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{4} \left (a^{2}+2 a b +b^{2}\right ) f}\) | \(570\) |
[In]
[Out]
none
Time = 0.55 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.10 \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [\frac {{\left (3 \, a^{2} + 4 \, a b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left ({\left (3 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{32 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}, -\frac {{\left (3 \, a^{2} + 4 \, a b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{4} - 2 \, {\left ({\left (3 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{16 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}\right ] \]
[In]
[Out]
\[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \sec ^{5}{\left (e + f x \right )}\, dx \]
[In]
[Out]
\[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5} \,d x } \]
[In]
[Out]
\[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5} \,d x } \]
[In]
[Out]
Timed out. \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\cos \left (e+f\,x\right )}^5} \,d x \]
[In]
[Out]