Integrand size = 23, antiderivative size = 85 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {(a-2 b) \text {arctanh}(\cos (c+d x))}{2 a^2 d}-\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a^2 \sqrt {a+b} d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]
-1/2*(a-2*b)*arctanh(cos(d*x+c))/a^2/d-1/2*cot(d*x+c)*csc(d*x+c)/a/d-b^(3/ 2)*arctanh(cos(d*x+c)*b^(1/2)/(a+b)^(1/2))/a^2/d/(a+b)^(1/2)
Result contains complex when optimal does not.
Time = 1.82 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.64 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {(2 a+b-b \cos (2 (c+d x))) \csc ^2(c+d x) \left (-8 b^{3/2} \arctan \left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )-8 b^{3/2} \arctan \left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )+\sqrt {-a-b} \left (a \csc ^2\left (\frac {1}{2} (c+d x)\right )+4 (a-2 b) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-a \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )}{16 a^2 \sqrt {-a-b} d \left (b+a \csc ^2(c+d x)\right )} \]
-1/16*((2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*(-8*b^(3/2)*ArcTan[(S qrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]] - 8*b^(3/2)*ArcTan[(Sqr t[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]] + Sqrt[-a - b]*(a*Csc[(c + d*x)/2]^2 + 4*(a - 2*b)*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) - a*Sec[(c + d*x)/2]^2)))/(a^2*Sqrt[-a - b]*d*(b + a*Csc[c + d*x]^2))
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3665, 316, 397, 219, 221}
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.
| \(\displaystyle \int \frac {\csc ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^3 \left (a+b \sin (c+d x)^2\right )}dx\) |
| \(\Big \downarrow \) 3665 |
| \(\displaystyle -\frac {\int \frac {1}{\left (1-\cos ^2(c+d x)\right )^2 \left (-b \cos ^2(c+d x)+a+b\right )}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle -\frac {\frac {\int \frac {-b \cos ^2(c+d x)+a-b}{\left (1-\cos ^2(c+d x)\right ) \left (-b \cos ^2(c+d x)+a+b\right )}d\cos (c+d x)}{2 a}+\frac {\cos (c+d x)}{2 a \left (1-\cos ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle -\frac {\frac {\frac {2 b^2 \int \frac {1}{-b \cos ^2(c+d x)+a+b}d\cos (c+d x)}{a}+\frac {(a-2 b) \int \frac {1}{1-\cos ^2(c+d x)}d\cos (c+d x)}{a}}{2 a}+\frac {\cos (c+d x)}{2 a \left (1-\cos ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {\frac {2 b^2 \int \frac {1}{-b \cos ^2(c+d x)+a+b}d\cos (c+d x)}{a}+\frac {(a-2 b) \text {arctanh}(\cos (c+d x))}{a}}{2 a}+\frac {\cos (c+d x)}{2 a \left (1-\cos ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\frac {\frac {2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}+\frac {(a-2 b) \text {arctanh}(\cos (c+d x))}{a}}{2 a}+\frac {\cos (c+d x)}{2 a \left (1-\cos ^2(c+d x)\right )}}{d}\) |
-(((((a - 2*b)*ArcTanh[Cos[c + d*x]])/a + (2*b^(3/2)*ArcTanh[(Sqrt[b]*Cos[ c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(2*a) + Cos[c + d*x]/(2*a*(1 - Co s[c + d*x]^2)))/d)
3.1.83.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
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]
Time = 0.85 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.26
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{4 a \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-a +2 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 a^{2}}+\frac {1}{4 a \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (a -2 b \right ) \ln \left (\cos \left (d x +c \right )-1\right )}{4 a^{2}}-\frac {b^{2} \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{2} \sqrt {\left (a +b \right ) b}}}{d}\) | \(107\) |
| default | \(\frac {\frac {1}{4 a \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-a +2 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 a^{2}}+\frac {1}{4 a \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (a -2 b \right ) \ln \left (\cos \left (d x +c \right )-1\right )}{4 a^{2}}-\frac {b^{2} \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{2} \sqrt {\left (a +b \right ) b}}}{d}\) | \(107\) |
| risch | \(\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}-\frac {i \sqrt {-\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{i \left (d x +c \right )}}{b}+1\right )}{2 \left (a +b \right ) d \,a^{2}}+\frac {i \sqrt {-\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{i \left (d x +c \right )}}{b}+1\right )}{2 \left (a +b \right ) d \,a^{2}}\) | \(238\) |
1/d*(1/4/a/(1+cos(d*x+c))+1/4/a^2*(-a+2*b)*ln(1+cos(d*x+c))+1/4/a/(cos(d*x +c)-1)+1/4*(a-2*b)/a^2*ln(cos(d*x+c)-1)-b^2/a^2/((a+b)*b)^(1/2)*arctanh(b* cos(d*x+c)/((a+b)*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 327, normalized size of antiderivative = 3.85 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [\frac {2 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}} \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) + 2 \, a \cos \left (d x + c\right ) - {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}}, \frac {4 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \cos \left (d x + c\right )\right ) + 2 \, a \cos \left (d x + c\right ) - {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}}\right ] \]
[1/4*(2*(b*cos(d*x + c)^2 - b)*sqrt(b/(a + b))*log(-(b*cos(d*x + c)^2 - 2* (a + b)*sqrt(b/(a + b))*cos(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a - b)) + 2*a*cos(d*x + c) - ((a - 2*b)*cos(d*x + c)^2 - a + 2*b)*log(1/2*cos(d*x + c) + 1/2) + ((a - 2*b)*cos(d*x + c)^2 - a + 2*b)*log(-1/2*cos(d*x + c) + 1/2))/(a^2*d*cos(d*x + c)^2 - a^2*d), 1/4*(4*(b*cos(d*x + c)^2 - b)*sqrt( -b/(a + b))*arctan(sqrt(-b/(a + b))*cos(d*x + c)) + 2*a*cos(d*x + c) - ((a - 2*b)*cos(d*x + c)^2 - a + 2*b)*log(1/2*cos(d*x + c) + 1/2) + ((a - 2*b) *cos(d*x + c)^2 - a + 2*b)*log(-1/2*cos(d*x + c) + 1/2))/(a^2*d*cos(d*x + c)^2 - a^2*d)]
\[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\csc ^{3}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]
Time = 0.45 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.41 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {2 \, b^{2} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2}} + \frac {2 \, \cos \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a} - \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{4 \, d} \]
1/4*(2*b^2*log((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x + c) + sqrt(( a + b)*b)))/(sqrt((a + b)*b)*a^2) + 2*cos(d*x + c)/(a*cos(d*x + c)^2 - a) - (a - 2*b)*log(cos(d*x + c) + 1)/a^2 + (a - 2*b)*log(cos(d*x + c) - 1)/a^ 2)/d
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (73) = 146\).
Time = 0.41 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.31 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {8 \, b^{2} \arctan \left (\frac {b \cos \left (d x + c\right ) + a + b}{\sqrt {-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{2}} + \frac {2 \, {\left (a - 2 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac {{\left (a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {\cos \left (d x + c\right ) - 1}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \]
1/8*(8*b^2*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/(sqrt(-a*b - b^2)*a^2) + 2*(a - 2*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + (a - 2*a*(cos(d*x + c) - 1)/(cos(d *x + c) + 1) + 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(a^2*(cos(d*x + c) - 1)) - (cos(d*x + c) - 1)/(a*(cos(d*x + c) + 1)))/d
Time = 13.81 (sec) , antiderivative size = 592, normalized size of antiderivative = 6.96 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a\,\left (b\,\cos \left (c+d\,x\right )-b\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )+b\,{\cos \left (c+d\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )\right )+a^2\,\left (\cos \left (c+d\,x\right )+\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )-{\cos \left (c+d\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )\right )-2\,b^2\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )+\mathrm {atan}\left (\frac {-a\,\cos \left (c+d\,x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,4{}\mathrm {i}-b\,\cos \left (c+d\,x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+b^5\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,8{}\mathrm {i}+a^2\,b^3\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}+a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}}{-a^5\,b^2+a^4\,b^3+5\,a^3\,b^4+3\,a^2\,b^5}\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}+2\,b^2\,{\cos \left (c+d\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )-{\cos \left (c+d\,x\right )}^2\,\mathrm {atan}\left (\frac {-a\,\cos \left (c+d\,x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,4{}\mathrm {i}-b\,\cos \left (c+d\,x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+b^5\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,8{}\mathrm {i}+a^2\,b^3\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}+a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}}{-a^5\,b^2+a^4\,b^3+5\,a^3\,b^4+3\,a^2\,b^5}\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}}{d\,\left (-2\,a^3\,{\cos \left (c+d\,x\right )}^2+2\,a^3-2\,b\,a^2\,{\cos \left (c+d\,x\right )}^2+2\,b\,a^2\right )} \]
-(a*(b*cos(c + d*x) - b*atanh(cos(c + d*x)) + b*cos(c + d*x)^2*atanh(cos(c + d*x))) + a^2*(cos(c + d*x) + atanh(cos(c + d*x)) - cos(c + d*x)^2*atanh (cos(c + d*x))) - 2*b^2*atanh(cos(c + d*x)) + atan((b^5*cos(c + d*x)*(a*b^ 3 + b^4)^(1/2)*8i - b*cos(c + d*x)*(a*b^3 + b^4)^(3/2)*8i - a*cos(c + d*x) *(a*b^3 + b^4)^(3/2)*4i + a^2*b^3*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*1i - a^ 3*b^2*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*2i + a*b^4*cos(c + d*x)*(a*b^3 + b^ 4)^(1/2)*12i + a^4*b*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*1i)/(3*a^2*b^5 + 5*a ^3*b^4 + a^4*b^3 - a^5*b^2))*(a*b^3 + b^4)^(1/2)*2i + 2*b^2*cos(c + d*x)^2 *atanh(cos(c + d*x)) - cos(c + d*x)^2*atan((b^5*cos(c + d*x)*(a*b^3 + b^4) ^(1/2)*8i - b*cos(c + d*x)*(a*b^3 + b^4)^(3/2)*8i - a*cos(c + d*x)*(a*b^3 + b^4)^(3/2)*4i + a^2*b^3*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*1i - a^3*b^2*co s(c + d*x)*(a*b^3 + b^4)^(1/2)*2i + a*b^4*cos(c + d*x)*(a*b^3 + b^4)^(1/2) *12i + a^4*b*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*1i)/(3*a^2*b^5 + 5*a^3*b^4 + a^4*b^3 - a^5*b^2))*(a*b^3 + b^4)^(1/2)*2i)/(d*(2*a^2*b + 2*a^3 - 2*a^3*c os(c + d*x)^2 - 2*a^2*b*cos(c + d*x)^2))