Integrand size = 23, antiderivative size = 77 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {b^2 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a+b} d}-\frac {(a-b) \cot (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a d} \]
-(a-b)*cot(d*x+c)/a^2/d-1/3*cot(d*x+c)^3/a/d+b^2*arctan((a+b)^(1/2)*tan(d* x+c)/a^(1/2))/a^(5/2)/d/(a+b)^(1/2)
Time = 0.80 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.55 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {(2 a+b-b \cos (2 (c+d x))) \csc ^2(c+d x) \left (-3 b^2 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a} \sqrt {a+b} \cot (c+d x) \left (2 a-3 b+a \csc ^2(c+d x)\right )\right )}{6 a^{5/2} \sqrt {a+b} d \left (b+a \csc ^2(c+d x)\right )} \]
-1/6*((2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*(-3*b^2*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a]*Sqrt[a + b]*Cot[c + d*x]*(2*a - 3*b + a*Csc[c + d*x]^2)))/(a^(5/2)*Sqrt[a + b]*d*(b + a*Csc[c + d*x]^2))
Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94, 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.
\(\displaystyle \int \frac {\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^4 \left (a+b \sin (c+d x)^2\right )}dx\) |
\(\Big \downarrow \) 3666 |
\(\displaystyle \frac {\int \frac {\cot ^4(c+d x) \left (\tan ^2(c+d x)+1\right )^2}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \frac {\int \left (\frac {\cot ^4(c+d x)}{a}+\frac {(a-b) \cot ^2(c+d x)}{a^2}+\frac {b^2}{a^2 \left ((a+b) \tan ^2(c+d x)+a\right )}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^2 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a+b}}-\frac {(a-b) \cot (c+d x)}{a^2}-\frac {\cot ^3(c+d x)}{3 a}}{d}\) |
((b^2*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(5/2)*Sqrt[a + b]) - ((a - b)*Cot[c + d*x])/a^2 - Cot[c + d*x]^3/(3*a))/d
3.1.91.3.1 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 = 0.84 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 a \tan \left (d x +c \right )^{3}}-\frac {a -b}{a^{2} \tan \left (d x +c \right )}+\frac {b^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{2} \sqrt {a \left (a +b \right )}}}{d}\) | \(69\) |
default | \(\frac {-\frac {1}{3 a \tan \left (d x +c \right )^{3}}-\frac {a -b}{a^{2} \tan \left (d x +c \right )}+\frac {b^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{2} \sqrt {a \left (a +b \right )}}}{d}\) | \(69\) |
risch | \(\frac {2 i \left (3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a \,{\mathrm e}^{2 i \left (d x +c \right )}-6 b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a +3 b \right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d \,a^{2}}+\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d \,a^{2}}\) | \(251\) |
1/d*(-1/3/a/tan(d*x+c)^3-(a-b)/a^2/tan(d*x+c)+b^2/a^2/(a*(a+b))^(1/2)*arct an((a+b)*tan(d*x+c)/(a*(a+b))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (67) = 134\).
Time = 0.27 (sec) , antiderivative size = 451, normalized size of antiderivative = 5.86 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [-\frac {4 \, {\left (2 \, a^{3} - a^{2} b - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) - 12 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )}{12 \, {\left ({\left (a^{4} + a^{3} b\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + a^{3} b\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (2 \, a^{3} - a^{2} b - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 6 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )}{6 \, {\left ({\left (a^{4} + a^{3} b\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + a^{3} b\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
[-1/12*(4*(2*a^3 - a^2*b - 3*a*b^2)*cos(d*x + c)^3 + 3*(b^2*cos(d*x + c)^2 - b^2)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(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) - 12*(a^3 - a*b^2)*cos(d*x + c))/(((a^4 + a^3*b)*d*cos(d*x + c)^2 - (a^ 4 + a^3*b)*d)*sin(d*x + c)), -1/6*(2*(2*a^3 - a^2*b - 3*a*b^2)*cos(d*x + c )^3 + 3*(b^2*cos(d*x + c)^2 - b^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*c os(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c)))*sin(d* x + c) - 6*(a^3 - a*b^2)*cos(d*x + c))/(((a^4 + a^3*b)*d*cos(d*x + c)^2 - (a^4 + a^3*b)*d)*sin(d*x + c))]
\[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {3 \, b^{2} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{2}} - \frac {3 \, {\left (a - b\right )} \tan \left (d x + c\right )^{2} + a}{a^{2} \tan \left (d x + c\right )^{3}}}{3 \, d} \]
1/3*(3*b^2*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*a ^2) - (3*(a - b)*tan(d*x + c)^2 + a)/(a^2*tan(d*x + c)^3))/d
Time = 0.46 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.44 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{2}}{\sqrt {a^{2} + a b} a^{2}} - \frac {3 \, a \tan \left (d x + c\right )^{2} - 3 \, b \tan \left (d x + c\right )^{2} + a}{a^{2} \tan \left (d x + c\right )^{3}}}{3 \, d} \]
1/3*(3*(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^2/(sqrt(a^2 + a*b)*a^2) - (3*a*ta n(d*x + c)^2 - 3*b*tan(d*x + c)^2 + a)/(a^2*tan(d*x + c)^3))/d
Time = 13.89 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {b^2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a+b}}{\sqrt {a}}\right )}{a^{5/2}\,d\,\sqrt {a+b}}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a-b\right )}{a^2}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^3} \]