Integrand size = 23, antiderivative size = 767 \[ \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {2 \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} \sqrt {a^{2/3}-b^{2/3}} d}+\frac {2 a^{2/3} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^{4/3} d}-\frac {4 \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^{2/3} d}+\frac {2 \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} d}-\frac {2 \sqrt [3]{-1} a^{2/3} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} b^{4/3} d}-\frac {2 \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} d}-\frac {2 (-1)^{2/3} a^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} b^{4/3} d}+\frac {4 \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}} b^{2/3} d}+\frac {4 \text {arctanh}\left (\frac {\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}} b^{2/3} d}-\frac {\cos (c+d x)}{b d} \] Output:
2/3*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/a ^(2/3)/(a^(2/3)-b^(2/3))^(1/2)/d+2/3*a^(2/3)*arctan((b^(1/3)+a^(1/3)*tan(1 /2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/(a^(2/3)-b^(2/3))^(1/2)/b^(4/3)/d- 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^(2/3)-b^(2/3))^(1/2)/b^(2/3)/d+2/3*arctan(((-1)^(2/3)*b^(1/3)+a^(1/3)*ta n(1/2*d*x+1/2*c))/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2))/a^(2/3)/(a^(2/3)+(-1 )^(1/3)*b^(2/3))^(1/2)/d-2/3*(-1)^(1/3)*a^(2/3)*arctan(((-1)^(2/3)*b^(1/3) +a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2))/(a^(2/3)+ (-1)^(1/3)*b^(2/3))^(1/2)/b^(4/3)/d-2/3*arctan((-1)^(1/3)*(b^(1/3)+(-1)^(2 /3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/a^(2/3 )/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2)/d-2/3*(-1)^(2/3)*a^(2/3)*arctan((-1)^ (1/3)*(b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-(-1)^(2/3)* b^(2/3))^(1/2))/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2)/b^(4/3)/d+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))/(-(-1)^(2/3)*a^(2/3)+b^(2/3))^(1/2)/b^(2/3)/d+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))/((-1)^(1/3)*a^(2/3)+b^(2/3))^(1/2)/b^(2/3)/d-cos(d*x+c)/b/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.64 (sec) , antiderivative size = 300, normalized size of antiderivative = 0.39 \[ \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {3 \cos (c+d x)+i \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-2 i a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}-a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}+2 i a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^3+a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3+2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]}{3 b d} \] Input:
Integrate[Cos[c + d*x]^4/(a + b*Sin[c + d*x]^3),x]
Output:
-1/3*(3*Cos[c + d*x] + I*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)* b*#1^4 + I*b*#1^6 & , (2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - I*b* Log[1 - 2*Cos[c + d*x]*#1 + #1^2] - (2*I)*a*ArcTan[Sin[c + d*x]/(Cos[c + d *x] - #1)]*#1 - a*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 + (2*I)*a*ArcTan[Si n[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 + a*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] *#1^3 + 2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*b*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) & ])/(b*d)
Time = 1.73 (sec) , antiderivative size = 767, 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, 3705, 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.
\(\displaystyle \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{a+b \sin (c+d x)^3}dx\) |
\(\Big \downarrow \) 3705 |
\(\displaystyle \int \left (\frac {1}{a+b \sin ^3(c+d x)}+\frac {\sin ^4(c+d x)}{a+b \sin ^3(c+d x)}-\frac {2 \sin ^2(c+d x)}{a+b \sin ^3(c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {4 \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 a^{2/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 \sqrt [3]{-1} a^{2/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {2 (-1)^{2/3} a^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {4 \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{2/3} d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac {4 \text {arctanh}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{2/3} d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac {\cos (c+d x)}{b d}\) |
Input:
Int[Cos[c + d*x]^4/(a + b*Sin[c + d*x]^3),x]
Output:
(2*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/( 3*a^(2/3)*Sqrt[a^(2/3) - b^(2/3)]*d) + (2*a^(2/3)*ArcTan[(b^(1/3) + a^(1/3 )*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)]*b ^(4/3)*d) - (4*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)]*b^(2/3)*d) + (2*ArcTan[((-1)^(2/3)*b ^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]])/(3 *a^(2/3)*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*d) - (2*(-1)^(1/3)*a^(2/3)*Arc Tan[((-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1 /3)*b^(2/3)]])/(3*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b^(4/3)*d) - (2*ArcTa n[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2]))/Sqrt[a^(2/3 ) - (-1)^(2/3)*b^(2/3)]])/(3*a^(2/3)*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*d) - (2*(-1)^(2/3)*a^(2/3)*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)* Tan[(c + d*x)/2]))/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^(4/3)*d) + (4*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)* Tan[(c + d*x)/2])/Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]])/(3*Sqrt[-((-1)^( 2/3)*a^(2/3)) + b^(2/3)]*b^(2/3)*d) + (4*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*Sqrt[(-1)^( 1/3)*a^(2/3) + b^(2/3)]*b^(2/3)*d) - Cos[c + d*x]/(b*d)
Int[cos[(e_.) + (f_.)*(x_)]^(m_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_) ), x_Symbol] :> Int[Expand[(1 - Sin[e + f*x]^2)^(m/2)/(a + b*Sin[e + f*x]^n ), x], x] /; FreeQ[{a, b, e, f}, x] && IGtQ[m/2, 0] && IntegerQ[(n - 1)/2]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.00 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.16
method | result | size |
derivativedivides | \(\frac {-\frac {2}{b \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4} b -2 \textit {\_R}^{3} a -6 \textit {\_R}^{2} b -2 a \textit {\_R} +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}}{3 b}}{d}\) | \(121\) |
default | \(\frac {-\frac {2}{b \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4} b -2 \textit {\_R}^{3} a -6 \textit {\_R}^{2} b -2 a \textit {\_R} +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}}{3 b}}{d}\) | \(121\) |
risch | \(-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (729 a^{4} b^{8} d^{6} \textit {\_Z}^{6}-729 a^{4} b^{6} d^{4} \textit {\_Z}^{4}+\left (162 a^{4} b^{4} d^{2}+81 a^{2} b^{6} d^{2}\right ) \textit {\_Z}^{2}+a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {243 d^{5} b^{7} a^{6}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}-\frac {972 d^{5} b^{9} a^{4}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}\right ) \textit {\_R}^{5}+\left (\frac {81 i d^{4} b^{5} a^{7}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}-\frac {162 i d^{4} b^{7} a^{5}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}+\frac {81 i d^{4} b^{9} a^{3}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}\right ) \textit {\_R}^{4}+\left (-\frac {189 d^{3} b^{5} a^{6}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}+\frac {945 d^{3} b^{7} a^{4}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}-\frac {27 d^{3} b^{9} a^{2}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}\right ) \textit {\_R}^{3}+\left (-\frac {54 i d^{2} b^{3} a^{7}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}+\frac {189 i d^{2} b^{5} a^{5}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}-\frac {135 i d^{2} b^{7} a^{3}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}\right ) \textit {\_R}^{2}+\left (\frac {3 d b \,a^{8}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}+\frac {9 d \,b^{3} a^{6}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}-\frac {189 d \,b^{5} a^{4}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}-\frac {66 d \,b^{7} a^{2}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}\right ) \textit {\_R} +\frac {3 i b \,a^{7}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}-\frac {27 i b^{3} a^{5}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}+\frac {18 i b^{5} a^{3}}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}+\frac {6 i b^{7} a}{a^{8}-4 a^{6} b^{2}-21 a^{4} b^{4}+23 a^{2} b^{6}+b^{8}}\right )\right )\) | \(976\) |
Input:
int(cos(d*x+c)^4/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/d*(-2/b/(1+tan(1/2*d*x+1/2*c)^2)+1/3/b*sum((_R^4*b-2*_R^3*a-6*_R^2*b-2*_ R*a+b)/(_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)))
Result contains complex when optimal does not.
Time = 3.46 (sec) , antiderivative size = 23437, normalized size of antiderivative = 30.56 \[ \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)**4/(a+b*sin(d*x+c)**3),x)
Output:
Integral(cos(c + d*x)**4/(a + b*sin(c + d*x)**3), x)
\[ \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\cos \left (d x + c\right )^{4}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(cos(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
-(b*d*integrate(4*(3*a*b*cos(4*d*x + 4*c)^2 + 3*a*b*cos(2*d*x + 2*c)^2 + 3 *a*b*sin(4*d*x + 4*c)^2 + 3*b^2*cos(d*x + c)*sin(2*d*x + 2*c) + 3*a*b*sin( 2*d*x + 2*c)^2 + b^2*sin(d*x + c) - (a*b*cos(4*d*x + 4*c) - a*b*cos(2*d*x + 2*c) + b^2*sin(5*d*x + 5*c) + b^2*sin(d*x + c))*cos(6*d*x + 6*c) - (8*a* b*cos(3*d*x + 3*c) + 3*b^2*sin(4*d*x + 4*c) - 3*b^2*sin(2*d*x + 2*c))*cos( 5*d*x + 5*c) - (6*a*b*cos(2*d*x + 2*c) + 8*a^2*sin(3*d*x + 3*c) - 3*b^2*si n(d*x + c) - a*b)*cos(4*d*x + 4*c) - 8*(a*b*cos(d*x + c) + a^2*sin(2*d*x + 2*c))*cos(3*d*x + 3*c) - (3*b^2*sin(d*x + c) + a*b)*cos(2*d*x + 2*c) + (b ^2*cos(5*d*x + 5*c) + b^2*cos(d*x + c) - a*b*sin(4*d*x + 4*c) + a*b*sin(2* d*x + 2*c))*sin(6*d*x + 6*c) + (3*b^2*cos(4*d*x + 4*c) - 3*b^2*cos(2*d*x + 2*c) - 8*a*b*sin(3*d*x + 3*c) + b^2)*sin(5*d*x + 5*c) + (8*a^2*cos(3*d*x + 3*c) - 3*b^2*cos(d*x + c) - 6*a*b*sin(2*d*x + 2*c))*sin(4*d*x + 4*c) + 8 *(a^2*cos(2*d*x + 2*c) - a*b*sin(d*x + c))*sin(3*d*x + 3*c))/(b^3*cos(6*d* x + 6*c)^2 + 9*b^3*cos(4*d*x + 4*c)^2 + 64*a^2*b*cos(3*d*x + 3*c)^2 + 9*b^ 3*cos(2*d*x + 2*c)^2 + b^3*sin(6*d*x + 6*c)^2 + 9*b^3*sin(4*d*x + 4*c)^2 + 64*a^2*b*sin(3*d*x + 3*c)^2 - 48*a*b^2*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 9*b^3*sin(2*d*x + 2*c)^2 - 6*b^3*cos(2*d*x + 2*c) + b^3 - 2*(3*b^3*cos(4 *d*x + 4*c) - 3*b^3*cos(2*d*x + 2*c) - 8*a*b^2*sin(3*d*x + 3*c) + b^3)*cos (6*d*x + 6*c) - 6*(3*b^3*cos(2*d*x + 2*c) + 8*a*b^2*sin(3*d*x + 3*c) - b^3 )*cos(4*d*x + 4*c) - 2*(8*a*b^2*cos(3*d*x + 3*c) + 3*b^3*sin(4*d*x + 4*...
\[ \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\cos \left (d x + c\right )^{4}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(cos(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
sage0*x
Time = 38.76 (sec) , antiderivative size = 2338, normalized size of antiderivative = 3.05 \[ \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
int(cos(c + d*x)^4/(a + b*sin(c + d*x)^3),x)
Output:
symsum(log(172032*a^4*b^9 - 81920*a^2*b^11 - 98304*a^6*b^7 + 8192*a^8*b^5 - 319488*root(729*a^4*b^8*d^6 - 729*a^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2 *b^6*d^2 - 3*a^4*b^2 + 3*a^2*b^4 + a^6 - b^6, d, k)*a^2*b^12 - 688128*root (729*a^4*b^8*d^6 - 729*a^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2*b^6*d^2 - 3* a^4*b^2 + 3*a^2*b^4 + a^6 - b^6, d, k)*a^4*b^10 + 344064*root(729*a^4*b^8* d^6 - 729*a^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2*b^6*d^2 - 3*a^4*b^2 + 3*a ^2*b^4 + a^6 - b^6, d, k)*a^6*b^8 - 98304*a*b^12*tan(c/2 + (d*x)/2) - 2949 12*root(729*a^4*b^8*d^6 - 729*a^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2*b^6*d ^2 - 3*a^4*b^2 + 3*a^2*b^4 + a^6 - b^6, d, k)^2*a^2*b^13 - 1400832*root(72 9*a^4*b^8*d^6 - 729*a^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2*b^6*d^2 - 3*a^4 *b^2 + 3*a^2*b^4 + a^6 - b^6, d, k)^2*a^4*b^11 + 1695744*root(729*a^4*b^8* d^6 - 729*a^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2*b^6*d^2 - 3*a^4*b^2 + 3*a ^2*b^4 + a^6 - b^6, d, k)^2*a^6*b^9 + 5750784*root(729*a^4*b^8*d^6 - 729*a ^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2*b^6*d^2 - 3*a^4*b^2 + 3*a^2*b^4 + a^ 6 - b^6, d, k)^3*a^4*b^12 - 3760128*root(729*a^4*b^8*d^6 - 729*a^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2*b^6*d^2 - 3*a^4*b^2 + 3*a^2*b^4 + a^6 - b^6, d , k)^3*a^6*b^10 + 3317760*root(729*a^4*b^8*d^6 - 729*a^4*b^6*d^4 + 162*a^4 *b^4*d^2 + 81*a^2*b^6*d^2 - 3*a^4*b^2 + 3*a^2*b^4 + a^6 - b^6, d, k)^4*a^4 *b^13 - 3317760*root(729*a^4*b^8*d^6 - 729*a^4*b^6*d^4 + 162*a^4*b^4*d^2 + 81*a^2*b^6*d^2 - 3*a^4*b^2 + 3*a^2*b^4 + a^6 - b^6, d, k)^4*a^6*b^11 -...
\[ \int \frac {\cos ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{4}}{\sin \left (d x +c \right )^{3} b +a}d x \] Input:
int(cos(d*x+c)^4/(a+b*sin(d*x+c)^3),x)
Output:
int(cos(c + d*x)**4/(sin(c + d*x)**3*b + a),x)