Integrand size = 25, antiderivative size = 143 \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {(3 a-b) (a+b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{8 a^{3/2} f}-\frac {(3 a-b) \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 a f}-\frac {\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 a f} \]
-1/8*(3*a-b)*(a+b)*arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/ a^(3/2)/f-1/4*(a+b-b*cos(f*x+e)^2)^(3/2)*cot(f*x+e)*csc(f*x+e)^3/a/f-1/8*( 3*a-b)*cot(f*x+e)*csc(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/a/f
Time = 0.52 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\left (-6 a^2-4 a b+2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )-\sqrt {2} \sqrt {a} \sqrt {2 a+b-b \cos (2 (e+f x))} \cot (e+f x) \csc (e+f x) \left (3 a+b+2 a \csc ^2(e+f x)\right )}{16 a^{3/2} f} \]
((-6*a^2 - 4*a*b + 2*b^2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] - Sqrt[2]*Sqrt[a]*Sqrt[2*a + b - b*Cos[2*(e + f *x)]]*Cot[e + f*x]*Csc[e + f*x]*(3*a + b + 2*a*Csc[e + f*x]^2))/(16*a^(3/2 )*f)
Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3665, 296, 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.
\(\displaystyle \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \sin (e+f x)^2}}{\sin (e+f x)^5}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {\sqrt {-b \cos ^2(e+f x)+a+b}}{\left (1-\cos ^2(e+f x)\right )^3}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 296 |
\(\displaystyle -\frac {\frac {(3 a-b) \int \frac {\sqrt {-b \cos ^2(e+f x)+a+b}}{\left (1-\cos ^2(e+f x)\right )^2}d\cos (e+f x)}{4 a}+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a \left (1-\cos ^2(e+f x)\right )^2}}{f}\) |
\(\Big \downarrow \) 292 |
\(\displaystyle -\frac {\frac {(3 a-b) \left (\frac {1}{2} (a+b) \int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )}{4 a}+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a \left (1-\cos ^2(e+f x)\right )^2}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {\frac {(3 a-b) \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {a \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )}{4 a}+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a \left (1-\cos ^2(e+f x)\right )^2}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {(3 a-b) \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {a}}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )}{4 a}+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a \left (1-\cos ^2(e+f x)\right )^2}}{f}\) |
-(((Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(3/2))/(4*a*(1 - Cos[e + f*x]^ 2)^2) + ((3*a - b)*(((a + b)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b *Cos[e + f*x]^2]])/(2*Sqrt[a]) + (Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x] ^2])/(2*(1 - Cos[e + f*x]^2))))/(4*a))/f)
3.2.26.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[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. \(378\) vs. \(2(127)=254\).
Time = 1.10 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.65
method | result | size |
default | \(-\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (3 a^{3} \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right )+2 b \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right ) a^{2}-\ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) b^{2} \left (\sin ^{4}\left (f x +e \right )\right ) a +6 \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, a^{\frac {5}{2}}+2 \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, a^{\frac {3}{2}} b +4 \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, a^{\frac {5}{2}}\right )}{16 \sin \left (f x +e \right )^{4} a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(379\) |
-1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(3*a^3*ln(((a-b)*cos(f*x+e)^ 2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)* sin(f*x+e)^4+2*b*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*c os(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^4*a^2-ln(((a-b)*cos(f*x+e )^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2 )*b^2*sin(f*x+e)^4*a+6*sin(f*x+e)^2*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2 )*a^(5/2)+2*sin(f*x+e)^2*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*a^(3/2)*b +4*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*a^(5/2))/sin(f*x+e)^4/a^(5/2)/c os(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.45 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.64 \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [-\frac {{\left ({\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 2 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left ({\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{32 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}, \frac {{\left ({\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 2 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right ) + 2 \, {\left ({\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}\right ] \]
[-1/32*(((3*a^2 + 2*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 2*a*b - b^2)*co s(f*x + e)^2 + 3*a^2 + 2*a*b - b^2)*sqrt(a)*log(2*((a^2 - 6*a*b + b^2)*cos (f*x + e)^4 + 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + (a + b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) + a ^2 + 2*a*b + b^2)/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 4*((3*a^2 + a *b)*cos(f*x + e)^3 - (5*a^2 + a*b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f), 1/16*(((3 *a^2 + 2*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^ 2 + 3*a^2 + 2*a*b - b^2)*sqrt(-a)*arctan(-1/2*((a - b)*cos(f*x + e)^2 + a + b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(a*b*cos(f*x + e)^3 - (a^2 + a*b)*cos(f*x + e))) + 2*((3*a^2 + a*b)*cos(f*x + e)^3 - (5*a^2 + a*b)*cos (f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^2*f*cos(f*x + e)^4 - 2*a^2* f*cos(f*x + e)^2 + a^2*f)]
\[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \csc ^{5}{\left (e + f x \right )}\, dx \]
\[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{5} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 914 vs. \(2 (127) = 254\).
Time = 0.57 (sec) , antiderivative size = 914, normalized size of antiderivative = 6.39 \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Too large to display} \]
1/64*(sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan (1/2*f*x + 1/2*e)^2 + a)*(tan(1/2*f*x + 1/2*e)^2 + (7*a + 2*b)/a) + 8*(3*a ^2 + 2*a*b - b^2)*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2 *f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))/(sqrt(-a)*a) - 4*(3*a^(5/2) + 2*a^(3/2)*b - sqrt(a)*b^2)*l og(abs(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a - a^(3/2) - 2*sqrt(a)*b))/a^2 + 4*(4*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/ 2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a^2 + 8*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/ 2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a *b + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2 *a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*b^2 + 5*(sq rt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2 *f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(5/2) + 4*(sqrt(a)* tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(3/2)*b - 2*(sqrt(a)*tan( 1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2 *e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^3 - 4*(sqrt(a)*tan(1/2*f*x + 1/ 2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4...
Timed out. \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^5} \,d x \]