Integrand size = 23, antiderivative size = 335 \[ \int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {3 x}{8 b}+\frac {2 (-1)^{2/3} a^{5/3} \arctan \left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} b^{7/3} d}-\frac {2 a^{5/3} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^{7/3} d}+\frac {2 \sqrt [3]{-1} a^{5/3} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} b^{7/3} d}+\frac {a \cos (c+d x)}{b^2 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{8 b d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d} \]
3/8*x/b+a*cos(d*x+c)/b^2/d-3/8*cos(d*x+c)*sin(d*x+c)/b/d-1/4*cos(d*x+c)*si n(d*x+c)^3/b/d-2/3*a^(5/3)*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^ (2/3)-b^(2/3))^(1/2))/b^(7/3)/d/(a^(2/3)-b^(2/3))^(1/2)+2/3*(-1)^(1/3)*a^( 5/3)*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))/b^(7/3)/d/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2)+2/3*(-1 )^(2/3)*a^(5/3)*arctan(((-1)^(1/3)*b^(1/3)-a^(1/3)*tan(1/2*d*x+1/2*c))/(a^ (2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/b^(7/3)/d/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1 /2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.56 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.65 \[ \int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {96 a \cos (c+d x)-32 a^2 \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]+3 b (12 (c+d x)-8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{96 b^2 d} \]
(96*a*Cos[c + d*x] - 32*a^2*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3* I)*b*#1^4 + I*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] + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2)/(b - (4*I)*a*#1 - 2* b*#1^2 + b*#1^4) & ] + 3*b*(12*(c + d*x) - 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(96*b^2*d)
Time = 0.85 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^7}{a+b \sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {a^2 \sin (c+d x)}{b^2 \left (a+b \sin ^3(c+d x)\right )}-\frac {a \sin (c+d x)}{b^2}+\frac {\sin ^4(c+d x)}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^{5/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \sqrt [3]{-1} a^{5/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {2 (-1)^{2/3} a^{5/3} \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{7/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac {3 \sin (c+d x) \cos (c+d x)}{8 b d}+\frac {3 x}{8 b}\) |
(3*x)/(8*b) - (2*a^(5/3)*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^(7/3)*d) + (2*(-1)^(1/3) *a^(5/3)*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*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b^(7/3)*d ) + (2*(-1)^(2/3)*a^(5/3)*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^(7/3)*d) + (a*Cos[c + d*x])/(b^2*d) - (3*Cos[c + d* x]*Sin[c + d*x])/(8*b*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(4*b*d)
3.2.82.3.1 Defintions of rubi rules used
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 = 1.92 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.66
| method | result | size |
| derivativedivides | \(\frac {-\frac {4 \left (\frac {-\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{16}-\frac {a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {11 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{16}-\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{2}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{16}-\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{2}+\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {a}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 b \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}\right )}{b^{2}}+\frac {2 a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b^{2}}}{d}\) | \(221\) |
| default | \(\frac {-\frac {4 \left (\frac {-\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{16}-\frac {a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {11 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{16}-\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{2}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{16}-\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{2}+\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {a}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 b \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}\right )}{b^{2}}+\frac {2 a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b^{2}}}{d}\) | \(221\) |
| risch | \(\frac {3 x}{8 b}+\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d \,b^{2}}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (729 a^{2} b^{14} d^{6}-729 b^{16} d^{6}\right ) \textit {\_Z}^{6}-3981312 a^{4} b^{10} d^{4} \textit {\_Z}^{4}-4398046511104 a^{10}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {243 i d^{5} b^{12} a^{2}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}-\frac {243 i d^{5} b^{14}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}\right ) \textit {\_R}^{5}+\left (\frac {10368 i d^{4} b^{9} a^{4}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}-\frac {10368 i d^{4} b^{11} a^{2}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}\right ) \textit {\_R}^{4}+\left (-\frac {884736 i d^{3} b^{8} a^{4}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}-\frac {442368 i d^{3} b^{10} a^{2}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}\right ) \textit {\_R}^{3}+\left (-\frac {37748736 i d^{2} b^{5} a^{6}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}-\frac {18874368 i d^{2} b^{7} a^{4}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}\right ) \textit {\_R}^{2}+\left (-\frac {805306368 i d \,b^{2} a^{8}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}-\frac {1610612736 i d \,b^{4} a^{6}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}\right ) \textit {\_R} -\frac {34359738368 i b \,a^{8}}{34359738368 a^{9}+34359738368 a^{7} b^{2}}\right )\right )}{128}+\frac {\sin \left (4 d x +4 c \right )}{32 d b}-\frac {\sin \left (2 d x +2 c \right )}{4 d b}\) | \(461\) |
1/d*(-4/b^2*((-3/16*tan(1/2*d*x+1/2*c)^7*b-1/2*a*tan(1/2*d*x+1/2*c)^6-11/1 6*tan(1/2*d*x+1/2*c)^5*b-3/2*tan(1/2*d*x+1/2*c)^4*a+11/16*tan(1/2*d*x+1/2* c)^3*b-3/2*tan(1/2*d*x+1/2*c)^2*a+3/16*b*tan(1/2*d*x+1/2*c)-1/2*a)/(1+tan( 1/2*d*x+1/2*c)^2)^4-3/16*b*arctan(tan(1/2*d*x+1/2*c)))+2/3*a^2/b^2*sum((_R ^3+_R)/(_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 = 1.43 (sec) , antiderivative size = 21338, normalized size of antiderivative = 63.70 \[ \int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{7}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
-1/32*(32*b^2*d*integrate(-4*(3*a^2*b*cos(4*d*x + 4*c)^2 + 3*a^2*b*cos(2*d *x + 2*c)^2 + 3*a^2*b*sin(4*d*x + 4*c)^2 + 8*a^3*cos(2*d*x + 2*c)*sin(3*d* x + 3*c) - 8*a^3*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 3*a^2*b*sin(2*d*x + 2 *c)^2 - a^2*b*cos(2*d*x + 2*c) - (a^2*b*cos(4*d*x + 4*c) - a^2*b*cos(2*d*x + 2*c))*cos(6*d*x + 6*c) - (6*a^2*b*cos(2*d*x + 2*c) + 8*a^3*sin(3*d*x + 3*c) - a^2*b)*cos(4*d*x + 4*c) - (a^2*b*sin(4*d*x + 4*c) - a^2*b*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 2*(4*a^3*cos(3*d*x + 3*c) - 3*a^2*b*sin(2*d*x + 2*c))*sin(4*d*x + 4*c))/(b^4*cos(6*d*x + 6*c)^2 + 9*b^4*cos(4*d*x + 4*c) ^2 + 64*a^2*b^2*cos(3*d*x + 3*c)^2 + 9*b^4*cos(2*d*x + 2*c)^2 + b^4*sin(6* d*x + 6*c)^2 + 9*b^4*sin(4*d*x + 4*c)^2 + 64*a^2*b^2*sin(3*d*x + 3*c)^2 - 48*a*b^3*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 9*b^4*sin(2*d*x + 2*c)^2 - 6* b^4*cos(2*d*x + 2*c) + b^4 - 2*(3*b^4*cos(4*d*x + 4*c) - 3*b^4*cos(2*d*x + 2*c) - 8*a*b^3*sin(3*d*x + 3*c) + b^4)*cos(6*d*x + 6*c) - 6*(3*b^4*cos(2* d*x + 2*c) + 8*a*b^3*sin(3*d*x + 3*c) - b^4)*cos(4*d*x + 4*c) - 2*(8*a*b^3 *cos(3*d*x + 3*c) + 3*b^4*sin(4*d*x + 4*c) - 3*b^4*sin(2*d*x + 2*c))*sin(6 *d*x + 6*c) + 6*(8*a*b^3*cos(3*d*x + 3*c) - 3*b^4*sin(2*d*x + 2*c))*sin(4* d*x + 4*c) + 16*(3*a*b^3*cos(2*d*x + 2*c) - a*b^3)*sin(3*d*x + 3*c)), x) - 12*b*d*x - 32*a*cos(d*x + c) - b*sin(4*d*x + 4*c) + 8*b*sin(2*d*x + 2*c)) /(b^2*d)
\[ \int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{7}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
Time = 15.45 (sec) , antiderivative size = 1978, normalized size of antiderivative = 5.90 \[ \int \frac {\sin ^7(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
symsum(log((150994944*a^12*b^3*sin(c/2 + (d*x)/2) - 56623104*a^13*b^2*cos( c/2 + (d*x)/2) - 12582912*a^15*cos(c/2 + (d*x)/2) + 679477248*root(729*a^2 *b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)*a^11*b^5*sin(c/2 + (d*x)/2) + 679477248*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^1 0*z^4 + a^10, z, k)^2*a^9*b^8*cos(c/2 + (d*x)/2) - 42467328*root(729*a^2*b ^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^2*a^11*b^6*cos(c/2 + (d*x)/2) - 402653184*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^1 0*z^4 + a^10, z, k)^2*a^13*b^4*cos(c/2 + (d*x)/2) + 4586471424*root(729*a^ 2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^3*a^8*b^10*cos( c/2 + (d*x)/2) - 503316480*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4* b^10*z^4 + a^10, z, k)^3*a^12*b^6*cos(c/2 + (d*x)/2) + 1911029760*root(729 *a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^4*a^7*b^12*c os(c/2 + (d*x)/2) + 1774190592*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243* a^4*b^10*z^4 + a^10, z, k)^4*a^9*b^10*cos(c/2 + (d*x)/2) - 301989888*root( 729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^4*a^11*b^ 8*cos(c/2 + (d*x)/2) - 18345885696*root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^5*a^4*b^16*cos(c/2 + (d*x)/2) + 17199267840 *root(729*a^2*b^14*z^6 - 729*b^16*z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^5*a ^6*b^14*cos(c/2 + (d*x)/2) + 32614907904*root(729*a^2*b^14*z^6 - 729*b^16* z^6 + 243*a^4*b^10*z^4 + a^10, z, k)^5*a^8*b^12*cos(c/2 + (d*x)/2) + 91...