Integrand size = 25, antiderivative size = 82 \[ \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 \sqrt {a+b} f}+\frac {\sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{2 f} \]
1/2*a*arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/f/(a+b)^(1/ 2)+1/2*sec(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/f
Time = 2.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.00 \[ \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\sin (e+f x) \left (\sqrt {2} a \text {arctanh}\left (\frac {\sqrt {\frac {(a+b) \sin ^2(e+f x)}{a}}}{\sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}}+(2 a+b-b \cos (2 (e+f x))) \sec ^2(e+f x) \sqrt {\frac {(a+b) \sin ^2(e+f x)}{a}}\right )}{4 f \sqrt {\frac {(a+b) \sin ^2(e+f x)}{a}} \sqrt {a+b \sin ^2(e+f x)}} \]
(Sin[e + f*x]*(Sqrt[2]*a*ArcTanh[Sqrt[((a + b)*Sin[e + f*x]^2)/a]/Sqrt[1 + (b*Sin[e + f*x]^2)/a]]*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a] + (2*a + b - b*Cos[2*(e + f*x)])*Sec[e + f*x]^2*Sqrt[((a + b)*Sin[e + f*x]^2)/a]))/(4 *f*Sqrt[((a + b)*Sin[e + f*x]^2)/a]*Sqrt[a + b*Sin[e + f*x]^2])
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3669, 292, 291, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \sin (e+f x)^2}}{\cos (e+f x)^3}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\left (1-\sin ^2(e+f x)\right )^2}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 292 |
\(\displaystyle \frac {\frac {1}{2} a \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {1}{2} a \int \frac {1}{1-\frac {(a+b) \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 \sqrt {a+b}}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}}{f}\) |
((a*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(2*Sqr t[a + b]) + (Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(2*(1 - Sin[e + f*x] ^2)))/f
3.4.27.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( a*(p + 1))) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ {a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt Q[q, 0] && NeQ[p, -1]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(289\) vs. \(2(70)=140\).
Time = 1.34 (sec) , antiderivative size = 290, normalized size of antiderivative = 3.54
method | result | size |
default | \(\frac {2 b \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {a +b}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+a \left (\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 {\left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}} \sqrt {a +b}\, \sin \left (f x +e \right )}{4 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{2} f}\) | \(290\) |
1/4*(2*b*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)^2*sin(f*x+e)+a* (ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+ a))*a+ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f* x+e)+a))*b-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*s in(f*x+e)+a))*a-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2 )-b*sin(f*x+e)+a))*b)*cos(f*x+e)^2+2*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(1/2 )*sin(f*x+e))/(a+b)^(3/2)/cos(f*x+e)^2/f
Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (70) = 140\).
Time = 0.33 (sec) , antiderivative size = 337, normalized size of antiderivative = 4.11 \[ \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [\frac {\sqrt {a + b} a \cos \left (f x + e\right )^{2} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \sin \left (f x + e\right )}{8 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{2}}, -\frac {a \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \sin \left (f x + e\right )}{4 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{2}}\right ] \]
[1/8*(sqrt(a + b)*a*cos(f*x + e)^2*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e) ^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x + e) + 8* a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) + 4*sqrt(-b*cos(f*x + e)^2 + a + b)* (a + b)*sin(f*x + e))/((a + b)*f*cos(f*x + e)^2), -1/4*(a*sqrt(-a - b)*arc tan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f* x + e)))*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)*sin(f* x + e))/((a + b)*f*cos(f*x + e)^2)]
\[ \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \sec ^{3}{\left (e + f x \right )}\, dx \]
\[ \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{3} \,d x } \]
\[ \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\cos \left (e+f\,x\right )}^3} \,d x \]