Integrand size = 21, antiderivative size = 127 \[ \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=a^2 x-\frac {3 b^2 x}{2}+\frac {3 a b \text {arctanh}(\cos (c+d x))}{d}-\frac {2 a b \cos (c+d x)}{d}+\frac {a^2 \cot (c+d x)}{d}-\frac {b^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a b \cot (c+d x) \csc (c+d x)}{d}-\frac {b^2 \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
a^2*x-3/2*b^2*x+3*a*b*arctanh(cos(d*x+c))/d-2*a*b*cos(d*x+c)/d+a^2*cot(d*x +c)/d-b^2*cot(d*x+c)/d-1/3*a^2*cot(d*x+c)^3/d-a*b*cot(d*x+c)*csc(d*x+c)/d- 1/2*b^2*cos(d*x+c)*sin(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(293\) vs. \(2(127)=254\).
Time = 6.13 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.31 \[ \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (2 a^2-3 b^2\right ) (c+d x)}{2 d}-\frac {2 a b \cos (c+d x)}{d}+\frac {\left (4 a^2 \cos \left (\frac {1}{2} (c+d x)\right )-3 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right ) \csc \left (\frac {1}{2} (c+d x)\right )}{6 d}-\frac {a b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{4 d}-\frac {a^2 \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}+\frac {3 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {3 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{4 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (-4 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+3 b^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d}-\frac {b^2 \sin (2 (c+d x))}{4 d}+\frac {a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 d} \] Input:
Integrate[Cot[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]
Output:
((2*a^2 - 3*b^2)*(c + d*x))/(2*d) - (2*a*b*Cos[c + d*x])/d + ((4*a^2*Cos[( c + d*x)/2] - 3*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*d) - (a*b*Csc[( c + d*x)/2]^2)/(4*d) - (a^2*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*d) + (3*a*b*Log[Cos[(c + d*x)/2]])/d - (3*a*b*Log[Sin[(c + d*x)/2]])/d + (a*b*S ec[(c + d*x)/2]^2)/(4*d) + (Sec[(c + d*x)/2]*(-4*a^2*Sin[(c + d*x)/2] + 3* b^2*Sin[(c + d*x)/2]))/(6*d) - (b^2*Sin[2*(c + d*x)])/(4*d) + (a^2*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(24*d)
Time = 0.40 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3201, 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.
| \(\displaystyle \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2}{\tan (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3201 |
| \(\displaystyle \int \left (a^2 \cot ^4(c+d x)+2 a b \cos (c+d x) \cot ^3(c+d x)+b^2 \cos ^2(c+d x) \cot ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x)}{d}+a^2 x+\frac {3 a b \text {arctanh}(\cos (c+d x))}{d}-\frac {3 a b \cos (c+d x)}{d}-\frac {a b \cos (c+d x) \cot ^2(c+d x)}{d}-\frac {3 b^2 \cot (c+d x)}{2 d}+\frac {b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {3 b^2 x}{2}\) |
Input:
Int[Cot[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]
Output:
a^2*x - (3*b^2*x)/2 + (3*a*b*ArcTanh[Cos[c + d*x]])/d - (3*a*b*Cos[c + d*x ])/d + (a^2*Cot[c + d*x])/d - (3*b^2*Cot[c + d*x])/(2*d) + (b^2*Cos[c + d* x]^2*Cot[c + d*x])/(2*d) - (a*b*Cos[c + d*x]*Cot[c + d*x]^2)/d - (a^2*Cot[ c + d*x]^3)/(3*d)
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*( x_)])^(p_.), x_Symbol] :> Int[ExpandIntegrand[(g*Tan[e + f*x])^p, (a + b*Si n[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Time = 1.38 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(145\) |
| default | \(\frac {a^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(145\) |
| risch | \(a^{2} x -\frac {3 b^{2} x}{2}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{8 d}-\frac {a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {4 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-2 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+2 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-4 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+\frac {8 i a^{2}}{3}-2 i b^{2}-2 a b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\) | \(236\) |
Input:
int(cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c)+2*a*b*(-1/2/sin(d*x+c)^2*cos (d*x+c)^5-1/2*cos(d*x+c)^3-3/2*cos(d*x+c)-3/2*ln(csc(d*x+c)-cot(d*x+c)))+b ^2*(-1/sin(d*x+c)*cos(d*x+c)^5-(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)-3/ 2*d*x-3/2*c))
Time = 0.10 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.72 \[ \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \, b^{2} \cos \left (d x + c\right )^{5} + 4 \, {\left (2 \, a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 9 \, {\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 9 \, {\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (2 \, a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right ) + 3 \, {\left ({\left (2 \, a^{2} - 3 \, b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right )^{3} - {\left (2 \, a^{2} - 3 \, b^{2}\right )} d x + 6 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/6*(3*b^2*cos(d*x + c)^5 + 4*(2*a^2 - 3*b^2)*cos(d*x + c)^3 + 9*(a*b*cos( d*x + c)^2 - a*b)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 9*(a*b*cos(d* x + c)^2 - a*b)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(2*a^2 - 3*b ^2)*cos(d*x + c) + 3*((2*a^2 - 3*b^2)*d*x*cos(d*x + c)^2 - 4*a*b*cos(d*x + c)^3 - (2*a^2 - 3*b^2)*d*x + 6*a*b*cos(d*x + c))*sin(d*x + c))/((d*cos(d* x + c)^2 - d)*sin(d*x + c))
\[ \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \cot ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)**4*(a+b*sin(d*x+c))**2,x)
Output:
Integral((a + b*sin(c + d*x))**2*cot(c + d*x)**4, x)
Time = 0.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.09 \[ \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {2 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} - 3 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} b^{2} + 3 \, a b {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \] Input:
integrate(cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
1/6*(2*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^2 - 3*(3*d* x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*b^2 + 3* a*b*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - 4*cos(d*x + c) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)))/d
Time = 0.17 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.90 \[ \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 72 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, {\left (2 \, a^{2} - 3 \, b^{2}\right )} {\left (d x + c\right )} + \frac {24 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac {132 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \] Input:
integrate(cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a*b*tan(1/2*d*x + 1/2*c)^2 - 72*a*b*l og(abs(tan(1/2*d*x + 1/2*c))) - 15*a^2*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1 /2*d*x + 1/2*c) + 12*(2*a^2 - 3*b^2)*(d*x + c) + 24*(b^2*tan(1/2*d*x + 1/2 *c)^3 - 4*a*b*tan(1/2*d*x + 1/2*c)^2 - b^2*tan(1/2*d*x + 1/2*c) - 4*a*b)/( tan(1/2*d*x + 1/2*c)^2 + 1)^2 + (132*a*b*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*t an(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*a*b*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^3)/d
Time = 35.41 (sec) , antiderivative size = 584, normalized size of antiderivative = 4.60 \[ \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx =\text {Too large to display} \] Input:
int(cot(c + d*x)^4*(a + b*sin(c + d*x))^2,x)
Output:
-((5*b^2*cos(c + d*x))/16 + (a^2*cos(3*c + 3*d*x))/3 - (11*b^2*cos(3*c + 3 *d*x))/32 + (b^2*cos(5*c + 5*d*x))/32 + (a^2*atan((3*b^2*cos(c/2 + (d*x)/2 ) - 2*a^2*cos(c/2 + (d*x)/2) + 6*a*b*sin(c/2 + (d*x)/2))/(2*a^2*sin(c/2 + (d*x)/2) - 3*b^2*sin(c/2 + (d*x)/2) + 6*a*b*cos(c/2 + (d*x)/2)))*sin(3*c + 3*d*x))/2 - (3*b^2*atan((3*b^2*cos(c/2 + (d*x)/2) - 2*a^2*cos(c/2 + (d*x) /2) + 6*a*b*sin(c/2 + (d*x)/2))/(2*a^2*sin(c/2 + (d*x)/2) - 3*b^2*sin(c/2 + (d*x)/2) + 6*a*b*cos(c/2 + (d*x)/2)))*sin(3*c + 3*d*x))/4 + (3*a*b*sin(c + d*x))/2 - (3*a^2*atan((3*b^2*cos(c/2 + (d*x)/2) - 2*a^2*cos(c/2 + (d*x) /2) + 6*a*b*sin(c/2 + (d*x)/2))/(2*a^2*sin(c/2 + (d*x)/2) - 3*b^2*sin(c/2 + (d*x)/2) + 6*a*b*cos(c/2 + (d*x)/2)))*sin(c + d*x))/2 + (9*b^2*atan((3*b ^2*cos(c/2 + (d*x)/2) - 2*a^2*cos(c/2 + (d*x)/2) + 6*a*b*sin(c/2 + (d*x)/2 ))/(2*a^2*sin(c/2 + (d*x)/2) - 3*b^2*sin(c/2 + (d*x)/2) + 6*a*b*cos(c/2 + (d*x)/2)))*sin(c + d*x))/4 + a*b*sin(2*c + 2*d*x) - (a*b*sin(3*c + 3*d*x)) /2 - (a*b*sin(4*c + 4*d*x))/4 + (9*a*b*sin(c + d*x)*log(sin(c/2 + (d*x)/2) /cos(c/2 + (d*x)/2)))/4 - (3*a*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2) )*sin(3*c + 3*d*x))/4)/(d*sin(c + d*x)^3)
Time = 0.18 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.42 \[ \int \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -2 \cos \left (d x +c \right ) a^{2}-18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a b +6 \sin \left (d x +c \right )^{3} a^{2} d x +15 \sin \left (d x +c \right )^{3} a b -9 \sin \left (d x +c \right )^{3} b^{2} d x}{6 \sin \left (d x +c \right )^{3} d} \] Input:
int(cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x)
Output:
( - 3*cos(c + d*x)*sin(c + d*x)**4*b**2 - 12*cos(c + d*x)*sin(c + d*x)**3* a*b + 8*cos(c + d*x)*sin(c + d*x)**2*a**2 - 6*cos(c + d*x)*sin(c + d*x)**2 *b**2 - 6*cos(c + d*x)*sin(c + d*x)*a*b - 2*cos(c + d*x)*a**2 - 18*log(tan ((c + d*x)/2))*sin(c + d*x)**3*a*b + 6*sin(c + d*x)**3*a**2*d*x + 15*sin(c + d*x)**3*a*b - 9*sin(c + d*x)**3*b**2*d*x)/(6*sin(c + d*x)**3*d)