Integrand size = 25, antiderivative size = 128 \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {3 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{8 \sqrt {a} f}-\frac {3 (a+b) \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 f}-\frac {\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 f} \]
-1/4*(a+b-b*cos(f*x+e)^2)^(3/2)*cot(f*x+e)*csc(f*x+e)^3/f-3/8*(a+b)^2*arct anh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/f/a^(1/2)-3/8*(a+b)*cot (f*x+e)*csc(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/f
Time = 0.64 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89 \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {\frac {6 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )}{\sqrt {a}}+\sqrt {2} \sqrt {2 a+b-b \cos (2 (e+f x))} \cot (e+f x) \csc (e+f x) \left (3 a+5 b+2 a \csc ^2(e+f x)\right )}{16 f} \]
-1/16*((6*(a + b)^2*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/Sqrt[a] + Sqrt[2]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]] *Cot[e + f*x]*Csc[e + f*x]*(3*a + 5*b + 2*a*Csc[e + f*x]^2))/f
Time = 0.31 (sec) , antiderivative size = 141, 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, 3665, 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.
| \(\displaystyle \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\sin (e+f x)^5}dx\) |
| \(\Big \downarrow \) 3665 |
| \(\displaystyle -\frac {\int \frac {\left (-b \cos ^2(e+f x)+a+b\right )^{3/2}}{\left (1-\cos ^2(e+f x)\right )^3}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle -\frac {\frac {3}{4} (a+b) \int \frac {\sqrt {-b \cos ^2(e+f x)+a+b}}{\left (1-\cos ^2(e+f x)\right )^2}d\cos (e+f x)+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 \left (1-\cos ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle -\frac {\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 \left (1-\cos ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle -\frac {\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {a \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 \left (1-\cos ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {3}{4} (a+b) \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {a}}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 \left (1-\cos ^2(e+f x)\right )^2}}{f}\) |
-(((Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(3/2))/(4*(1 - Cos[e + f*x]^2) ^2) + (3*(a + b)*(((a + b)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*C os[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)/f)
3.2.36.3.1 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[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. \(375\) vs. \(2(112)=224\).
Time = 1.10 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.94
| method | result | size |
| default | \(-\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (3 a^{2} \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right )+6 a b \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right )+3 b^{2} \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right )+6 a^{\frac {3}{2}} \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (\sin ^{2}\left (f x +e \right )\right )+10 b \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (f x +e \right )\right )+4 a^{\frac {3}{2}} \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\right )}{16 \sqrt {a}\, \sin \left (f x +e \right )^{4} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(376\) |
-1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(3*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)* sin(f*x+e)^4+6*a*b*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+3*b^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)*sin(f*x+e)^4+6*a^(3/2)*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*sin(f *x+e)^2+10*b*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*a^(1/2)*sin(f*x+e)^2+ 4*a^(3/2)*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2))/a^(1/2)/sin(f*x+e)^4/co s(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.43 (sec) , antiderivative size = 484, normalized size of antiderivative = 3.78 \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} - 5 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{32 \, {\left (a f \cos \left (f x + e\right )^{4} - 2 \, a f \cos \left (f x + e\right )^{2} + a f\right )}}, \frac {3 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right ) + 2 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} - 5 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a f \cos \left (f x + e\right )^{4} - 2 \, a f \cos \left (f x + e\right )^{2} + a f\right )}}\right ] \]
[1/32*(3*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f *x + e)^2 + 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 + 5*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*f*cos(f*x + e)^4 - 2*a*f*cos(f*x + e)^2 + a*f), 1/16*(3*((a^2 + 2* a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + 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 + 5*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*f*cos(f*x + e)^4 - 2*a*f*cos(f*x + e)^ 2 + a*f)]
Timed out. \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{5} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 912 vs. \(2 (112) = 224\).
Time = 0.72 (sec) , antiderivative size = 912, normalized size of antiderivative = 7.12 \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
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)*(a*tan(1/2*f*x + 1/2*e)^2 + (7*a^2 + 10*a*b)/a) + 24*(a^2 + 2*a*b + b^2)*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*t an(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) - 12*(a^(5/2) + 2*a^(3/2)*b + sqrt(a)*b^2)* log(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 + 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 + 12*(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 + 10*(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*(s qrt(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) + 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))^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 - 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 + ...
Timed out. \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^5} \,d x \]