Integrand size = 23, antiderivative size = 296 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {2 b^{4/3} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^2 \sqrt {a^{2/3}-b^{2/3}} d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {2 b^{4/3} \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 a^2 \sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}} d}-\frac {2 b^{4/3} \text {arctanh}\left (\frac {\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a^2 \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}} d}-\frac {\cot (c+d x)}{a d}-\frac {\cot ^3(c+d x)}{3 a d} \] Output:
2/3*b^(4/3)*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^ (1/2))/a^2/(a^(2/3)-b^(2/3))^(1/2)/d+b*arctanh(cos(d*x+c))/a^2/d-2/3*b^(4/ 3)*arctanh((b^(1/3)-(-1)^(1/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(-(-1)^(2/3)*a^ (2/3)+b^(2/3))^(1/2))/a^2/(-(-1)^(2/3)*a^(2/3)+b^(2/3))^(1/2)/d-2/3*b^(4/3 )*arctanh((b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/((-1)^(1/3)*a^(2 /3)+b^(2/3))^(1/2))/a^2/((-1)^(1/3)*a^(2/3)+b^(2/3))^(1/2)/d-cot(d*x+c)/a/ d-1/3*cot(d*x+c)^3/a/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 2.94 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {-8 a \cot \left (\frac {1}{2} (c+d x)\right )+24 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-24 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 i b^2 \text {RootSum}\left [-b+3 b \text {$\#$1}^2-8 i a \text {$\#$1}^3-3 b \text {$\#$1}^4+b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-4 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+2 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]+8 a \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-\frac {1}{2} a \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+8 a \tan \left (\frac {1}{2} (c+d x)\right )}{24 a^2 d} \] Input:
Integrate[Csc[c + d*x]^4/(a + b*Sin[c + d*x]^3),x]
Output:
(-8*a*Cot[(c + d*x)/2] + 24*b*Log[Cos[(c + d*x)/2]] - 24*b*Log[Sin[(c + d* x)/2]] + (4*I)*b^2*RootSum[-b + 3*b*#1^2 - (8*I)*a*#1^3 - 3*b*#1^4 + b*#1^ 6 & , (2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - I*Log[1 - 2*Cos[c + d* x]*#1 + #1^2] - 4*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (2*I)*Lo g[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)*a*# 1^2 - 2*b*#1^3 + b*#1^5) & ] + 8*a*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - (a* Csc[(c + d*x)/2]^4*Sin[c + d*x])/2 + 8*a*Tan[(c + d*x)/2])/(24*a^2*d)
Time = 0.68 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3699, 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.
| \(\displaystyle \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^4 \left (a+b \sin (c+d x)^3\right )}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {b^2 \sin ^2(c+d x)}{a^2 \left (a+b \sin ^3(c+d x)\right )}-\frac {b \csc (c+d x)}{a^2}+\frac {\csc ^4(c+d x)}{a}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {2 b^{4/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^2 d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 b^{4/3} \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 a^2 d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}-\frac {2 b^{4/3} \text {arctanh}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a^2 d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}\) |
Input:
Int[Csc[c + d*x]^4/(a + b*Sin[c + d*x]^3),x]
Output:
(2*b^(4/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2 /3)]])/(3*a^2*Sqrt[a^(2/3) - b^(2/3)]*d) + (b*ArcTanh[Cos[c + d*x]])/(a^2* d) - (2*b^(4/3)*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/Sq rt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]])/(3*a^2*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*d) - (2*b^(4/3)*ArcTanh[(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d* x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*a^2*Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]*d) - Cot[c + d*x]/(a*d) - Cot[c + d*x]^3/(3*a*d)
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n) ^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4] || Gt Q[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.46 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.55
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}+\frac {4 b^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 a^{2}}-\frac {1}{24 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\) | \(162\) |
| default | \(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}+\frac {4 b^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 a^{2}}-\frac {1}{24 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\) | \(162\) |
| risch | \(\frac {4 i \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+16 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (12230590464 a^{14} d^{6}-12230590464 a^{12} b^{2} d^{6}\right ) \textit {\_Z}^{6}+15925248 a^{8} b^{4} d^{4} \textit {\_Z}^{4}-6912 a^{4} b^{6} d^{2} \textit {\_Z}^{2}+b^{8}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {254803968 a^{12} d^{5}}{b^{7}}-\frac {254803968 a^{10} d^{5}}{b^{5}}\right ) \textit {\_R}^{5}+\left (-\frac {5308416 i d^{4} a^{9}}{b^{5}}+\frac {5308416 i d^{4} a^{7}}{b^{3}}\right ) \textit {\_R}^{4}+\frac {331776 a^{6} d^{3} \textit {\_R}^{3}}{b^{3}}+\left (-\frac {2304 i a^{5} d^{2}}{b^{3}}-\frac {4608 i a^{3} d^{2}}{b}\right ) \textit {\_R}^{2}-\frac {144 d \,a^{2} \textit {\_R}}{b}+\frac {i b}{a}\right )\right )+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{a^{2} d}\) | \(269\) |
Input:
int(csc(d*x+c)^4/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/d*(1/8/a*(1/3*tan(1/2*d*x+1/2*c)^3+3*tan(1/2*d*x+1/2*c))+4/3*b^2/a^2*sum (_R^2/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf( _Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))-1/24/a/tan(1/2*d*x+1/2*c)^3-3/8/a/ta n(1/2*d*x+1/2*c)-b/a^2*ln(tan(1/2*d*x+1/2*c)))
Result contains complex when optimal does not.
Time = 3.55 (sec) , antiderivative size = 29423, normalized size of antiderivative = 99.40 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(csc(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)**4/(a+b*sin(d*x+c)**3),x)
Output:
Integral(csc(c + d*x)**4/(a + b*sin(c + d*x)**3), x)
Exception generated. \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(csc(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
\[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )^{4}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(csc(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
sage0*x
Time = 41.31 (sec) , antiderivative size = 1503, normalized size of antiderivative = 5.08 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(sin(c + d*x)^4*(a + b*sin(c + d*x)^3)),x)
Output:
symsum(log((98304*b^11 + 589824*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243 *a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)*a^2*b^10 - 98304*root(729*a^12* b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^2*a ^4*b^9 - 5898240*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^3*a^6*b^8 - 7962624*root(729*a^12*b^2*z^6 - 72 9*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^4*a^8*b^7 - 663 552*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^ 2 - b^8, z, k)^4*a^10*b^5 + 5308416*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^5*a^10*b^6 - 10616832*root( 729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^5*a^12*b^4 + 7962624*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8* b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^6*a^12*b^5 - 9953280*root(729*a^12*b ^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^6*a^ 14*b^3 + 24576*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27 *a^4*b^6*z^2 - b^8, z, k)*a^3*b^9*tan(c/2 + (d*x)/2) - 3145728*root(729*a^ 12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^ 2*a^3*b^10*tan(c/2 + (d*x)/2) + 466944*root(729*a^12*b^2*z^6 - 729*a^14*z^ 6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^2*a^5*b^8*tan(c/2 + (d*x )/2) + 18874368*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 2 7*a^4*b^6*z^2 - b^8, z, k)^3*a^5*b^9*tan(c/2 + (d*x)/2) + 3981312*root(...
\[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)^4/(a+b*sin(d*x+c)^3),x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)**2*a**2 + 64*cos(c + d*x)*sin(c + d*x)**2* b**2 - cos(c + d*x)*a**2 - 64*cos(c + d*x)*b**2 - 24*int(1/(tan((c + d*x)/ 2)**10*a + 3*tan((c + d*x)/2)**8*a + 8*tan((c + d*x)/2)**7*b + 3*tan((c + d*x)/2)**6*a + tan((c + d*x)/2)**4*a),x)*sin(c + d*x)**3*a*b**2*d - 72*int (1/(tan((c + d*x)/2)**8*a + 3*tan((c + d*x)/2)**6*a + 8*tan((c + d*x)/2)** 5*b + 3*tan((c + d*x)/2)**4*a + tan((c + d*x)/2)**2*a),x)*sin(c + d*x)**3* a*b**2*d - 192*int(1/(tan((c + d*x)/2)**7*a + 3*tan((c + d*x)/2)**5*a + 8* tan((c + d*x)/2)**4*b + 3*tan((c + d*x)/2)**3*a + tan((c + d*x)/2)*a),x)*s in(c + d*x)**3*b**3*d - 72*int(1/(tan((c + d*x)/2)**6*a + 3*tan((c + d*x)/ 2)**4*a + 8*tan((c + d*x)/2)**3*b + 3*tan((c + d*x)/2)**2*a + a),x)*sin(c + d*x)**3*a*b**2*d + 3*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3*a*b - log(tan((c + d*x)/2)**6*a + 3*tan((c + d*x)/2)**4*a + 8*tan((c + d*x)/2)** 3*b + 3*tan((c + d*x)/2)**2*a + a)*sin(c + d*x)**3*a*b - 3*log(tan((c + d* x)/2))*sin(c + d*x)**3*a*b + 24*sin(c + d*x)**3*b**2*d*x + 96*sin(c + d*x) **2*b**2 - 64*b**2)/(3*sin(c + d*x)**3*a**3*d)