Integrand size = 25, antiderivative size = 122 \[ \int \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 \sqrt {a+b} f}+\frac {3 a \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 f} \]
3/8*a^2*arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/f/(a+b)^( 1/2)+1/4*sec(f*x+e)^3*(a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)/f+3/8*a*sec(f*x+ e)*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.52 \[ \int \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {(a+b) \sin ^2(e+f x)}{a+b \sin ^2(e+f x)}\right ) \sin (e+f x)}{f \sqrt {a+b \sin ^2(e+f x)}} \]
(a^2*Hypergeometric2F1[1/2, 3, 3/2, ((a + b)*Sin[e + f*x]^2)/(a + b*Sin[e + f*x]^2)]*Sin[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]^2])
Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3669, 292, 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.
| \(\displaystyle \int \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\cos (e+f x)^5}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{\left (1-\sin ^2(e+f x)\right )^3}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle \frac {\frac {3}{4} a \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\left (1-\sin ^2(e+f x)\right )^2}d\sin (e+f x)+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 \left (1-\sin ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 \left (1-\sin ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {(a+b) \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 \left (1-\sin ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 \sqrt {a+b}}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 \left (1-\sin ^2(e+f x)\right )^2}}{f}\) |
((Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2))/(4*(1 - Sin[e + f*x]^2)^2) + (3*a*((a*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/( 2*Sqrt[a + b]) + (Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(2*(1 - Sin[e + f*x]^2))))/4)/f
3.4.38.3.1 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. \(405\) vs. \(2(106)=212\).
Time = 1.39 (sec) , antiderivative size = 406, normalized size of antiderivative = 3.33
| method | result | size |
| default | \(\frac {3 a^{2} \left (\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{2}-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}-2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{2}\right ) \left (\cos ^{4}\left (f x +e \right )\right )+2 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}} \left (3 a -2 b \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {7}{2}} \sin \left (f x +e \right )}{16 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{4} f}\) | \(406\) |
1/16*(3*a^2*(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^2+2*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*b+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^2-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))*a^2-2*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))*a*b-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^2)*cos(f*x+e)^4+2 *(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*(3*a-2*b)*cos(f*x+e)^2*sin(f*x+e)+ 4*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(7/2)*sin(f*x+e))/(a+b)^(5/2)/cos(f*x+e )^4/f
Time = 0.67 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.39 \[ \int \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, \sqrt {a + b} a^{2} \cos \left (f x + e\right )^{4} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left ({\left (3 \, a^{2} + a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{32 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{4}}, -\frac {3 \, a^{2} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{4} - 2 \, {\left ({\left (3 \, a^{2} + a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{16 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{4}}\right ] \]
[1/32*(3*sqrt(a + b)*a^2*cos(f*x + e)^4*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*((3*a^2 + a*b - 2*b^2)*cos( f*x + e)^2 + 2*a^2 + 4*a*b + 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f* x + e))/((a + b)*f*cos(f*x + e)^4), -1/16*(3*a^2*sqrt(-a - b)*arctan(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)^4 - 2*((3*a^2 + a*b - 2*b^2)*cos(f*x + e)^2 + 2*a^2 + 4*a*b + 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a + b)*f*cos(f*x + e)^4)]
Timed out. \[ \int \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^5} \,d x \]