Integrand size = 23, antiderivative size = 109 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {b^3 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a+b} d}-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d} \]
-(a^2-a*b+b^2)*cot(d*x+c)/a^3/d-1/3*(2*a-b)*cot(d*x+c)^3/a^2/d-1/5*cot(d*x +c)^5/a/d-b^3*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(7/2)/d/(a+b)^(1/2)
Time = 2.31 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.35 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {(2 a+b-b \cos (2 (c+d x))) \csc ^2(c+d x) \left (15 b^3 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a} \sqrt {a+b} \cot (c+d x) \left (8 a^2-10 a b+15 b^2+a (4 a-5 b) \csc ^2(c+d x)+3 a^2 \csc ^4(c+d x)\right )\right )}{30 a^{7/2} \sqrt {a+b} d \left (b+a \csc ^2(c+d x)\right )} \]
-1/30*((2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*(15*b^3*ArcTan[(Sqrt[ a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a]*Sqrt[a + b]*Cot[c + d*x]*(8*a^2 - 10*a*b + 15*b^2 + a*(4*a - 5*b)*Csc[c + d*x]^2 + 3*a^2*Csc[c + d*x]^4)))/( a^(7/2)*Sqrt[a + b]*d*(b + a*Csc[c + d*x]^2))
Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3666, 364, 2009}
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.
\(\displaystyle \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^6 \left (a+b \sin (c+d x)^2\right )}dx\) |
\(\Big \downarrow \) 3666 |
\(\displaystyle \frac {\int \frac {\cot ^6(c+d x) \left (\tan ^2(c+d x)+1\right )^3}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \frac {\int \left (\frac {\cot ^6(c+d x)}{a}+\frac {(2 a-b) \cot ^4(c+d x)}{a^2}+\frac {\left (a^2-b a+b^2\right ) \cot ^2(c+d x)}{a^3}+\frac {b^3}{a^3 \left (-\left ((a+b) \tan ^2(c+d x)\right )-a\right )}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {b^3 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a+b}}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2}-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3}-\frac {\cot ^5(c+d x)}{5 a}}{d}\) |
(-((b^3*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(7/2)*Sqrt[a + b])) - ((a^2 - a*b + b^2)*Cot[c + d*x])/a^3 - ((2*a - b)*Cot[c + d*x]^3)/(3*a^ 2) - Cot[c + d*x]^5/(5*a))/d
3.1.92.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1)) , x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] & & IntegerQ[p]
Time = 1.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {-\frac {b^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {2 a -b}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {a^{2}-a b +b^{2}}{a^{3} \tan \left (d x +c \right )}}{d}\) | \(96\) |
default | \(\frac {-\frac {b^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {2 a -b}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {a^{2}-a b +b^{2}}{a^{3} \tan \left (d x +c \right )}}{d}\) | \(96\) |
risch | \(-\frac {2 i \left (15 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+30 a b \,{\mathrm e}^{6 i \left (d x +c \right )}-60 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+80 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-70 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+90 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-40 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+50 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-60 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+8 a^{2}-10 a b +15 b^{2}\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d \,a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d \,a^{3}}\) | \(346\) |
1/d*(-b^3/a^3/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-1/5 /a/tan(d*x+c)^5-1/3*(2*a-b)/a^2/tan(d*x+c)^3-(a^2-a*b+b^2)/a^3/tan(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (97) = 194\).
Time = 0.29 (sec) , antiderivative size = 595, normalized size of antiderivative = 5.46 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [-\frac {4 \, {\left (8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} - 20 \, {\left (4 \, a^{4} - a^{3} b + a^{2} b^{2} + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (b^{3} \cos \left (d x + c\right )^{4} - 2 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) + 60 \, {\left (a^{4} + a b^{3}\right )} \cos \left (d x + c\right )}{60 \, {\left ({\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} + a^{4} b\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (4 \, a^{4} - a^{3} b + a^{2} b^{2} + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (b^{3} \cos \left (d x + c\right )^{4} - 2 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 30 \, {\left (a^{4} + a b^{3}\right )} \cos \left (d x + c\right )}{30 \, {\left ({\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} + a^{4} b\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
[-1/60*(4*(8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3)*cos(d*x + c)^5 - 20*(4* a^4 - a^3*b + a^2*b^2 + 6*a*b^3)*cos(d*x + c)^3 + 15*(b^3*cos(d*x + c)^4 - 2*b^3*cos(d*x + c)^2 + b^3)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*c os(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 - 4*((2*a + b)*cos( d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(-a^2 - a*b)*sin(d*x + c) + a^2 + 2 *a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a *b + b^2))*sin(d*x + c) + 60*(a^4 + a*b^3)*cos(d*x + c))/(((a^5 + a^4*b)*d *cos(d*x + c)^4 - 2*(a^5 + a^4*b)*d*cos(d*x + c)^2 + (a^5 + a^4*b)*d)*sin( d*x + c)), -1/30*(2*(8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3)*cos(d*x + c)^ 5 - 10*(4*a^4 - a^3*b + a^2*b^2 + 6*a*b^3)*cos(d*x + c)^3 - 15*(b^3*cos(d* x + c)^4 - 2*b^3*cos(d*x + c)^2 + b^3)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c)))*si n(d*x + c) + 30*(a^4 + a*b^3)*cos(d*x + c))/(((a^5 + a^4*b)*d*cos(d*x + c) ^4 - 2*(a^5 + a^4*b)*d*cos(d*x + c)^2 + (a^5 + a^4*b)*d)*sin(d*x + c))]
\[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\csc ^{6}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]
Time = 0.36 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {15 \, b^{3} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{3}} + \frac {15 \, {\left (a^{2} - a b + b^{2}\right )} \tan \left (d x + c\right )^{4} + 5 \, {\left (2 \, a^{2} - a b\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \]
-1/15*(15*b^3*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a )*a^3) + (15*(a^2 - a*b + b^2)*tan(d*x + c)^4 + 5*(2*a^2 - a*b)*tan(d*x + c)^2 + 3*a^2)/(a^3*tan(d*x + c)^5))/d
Time = 0.47 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.42 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{3}}{\sqrt {a^{2} + a b} a^{3}} + \frac {15 \, a^{2} \tan \left (d x + c\right )^{4} - 15 \, a b \tan \left (d x + c\right )^{4} + 15 \, b^{2} \tan \left (d x + c\right )^{4} + 10 \, a^{2} \tan \left (d x + c\right )^{2} - 5 \, a b \tan \left (d x + c\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \]
-1/15*(15*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)))*b^3/(sqrt(a^2 + a*b)*a^3) + (15* a^2*tan(d*x + c)^4 - 15*a*b*tan(d*x + c)^4 + 15*b^2*tan(d*x + c)^4 + 10*a^ 2*tan(d*x + c)^2 - 5*a*b*tan(d*x + c)^2 + 3*a^2)/(a^3*tan(d*x + c)^5))/d
Time = 13.77 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.87 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2-a\,b+b^2\right )+\frac {a^2}{5}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a\,b}{3}-\frac {2\,a^2}{3}\right )}{a^3\,d\,{\mathrm {tan}\left (c+d\,x\right )}^5}-\frac {b^3\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a+b}}{\sqrt {a}}\right )}{a^{7/2}\,d\,\sqrt {a+b}} \]