Integrand size = 23, antiderivative size = 219 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {\left (a^{4/3}-b^{4/3}\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{5/3} d}+\frac {\left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} b^{5/3} d}-\frac {\left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} b^{5/3} d}-\frac {2 \log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\sin ^2(c+d x)}{2 b d} \] Output:
1/3*(a^(4/3)-b^(4/3))*arctan(1/3*(a^(1/3)-2*b^(1/3)*sin(d*x+c))*3^(1/2)/a^ (1/3))*3^(1/2)/a^(2/3)/b^(5/3)/d+1/3*(a^(4/3)+b^(4/3))*ln(a^(1/3)+b^(1/3)* sin(d*x+c))/a^(2/3)/b^(5/3)/d-1/6*(a^(4/3)+b^(4/3))*ln(a^(2/3)-a^(1/3)*b^( 1/3)*sin(d*x+c)+b^(2/3)*sin(d*x+c)^2)/a^(2/3)/b^(5/3)/d-2/3*ln(a+b*sin(d*x +c)^3)/b/d+1/2*sin(d*x+c)^2/b/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.31 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {-2 \sqrt {3} b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )+2 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )-b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )-4 a^{2/3} \log \left (a+b \sin ^3(c+d x)\right )+3 a^{2/3} \sin ^2(c+d x)-3 a^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b \sin ^3(c+d x)}{a}\right ) \sin ^2(c+d x)}{6 a^{2/3} b d} \] Input:
Integrate[Cos[c + d*x]^5/(a + b*Sin[c + d*x]^3),x]
Output:
(-2*Sqrt[3]*b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^( 1/3))] + 2*b^(2/3)*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]] - b^(2/3)*Log[a^(2/ 3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d*x]^2] - 4*a^(2/3)*Lo g[a + b*Sin[c + d*x]^3] + 3*a^(2/3)*Sin[c + d*x]^2 - 3*a^(2/3)*Hypergeomet ric2F1[2/3, 1, 5/3, -((b*Sin[c + d*x]^3)/a)]*Sin[c + d*x]^2)/(6*a^(2/3)*b* d)
Time = 0.48 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3702, 2426, 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.
| \(\displaystyle \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{a+b \sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3702 |
| \(\displaystyle \frac {\int \frac {\left (1-\sin ^2(c+d x)\right )^2}{b \sin ^3(c+d x)+a}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 2426 |
| \(\displaystyle \frac {\int \left (\frac {\sin (c+d x)}{b}+\frac {-2 b \sin ^2(c+d x)-a \sin (c+d x)+b}{b \left (b \sin ^3(c+d x)+a\right )}\right )d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\left (a^{4/3}-b^{4/3}\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{5/3}}-\frac {\left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} b^{5/3}}+\frac {\left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} b^{5/3}}-\frac {2 \log \left (a+b \sin ^3(c+d x)\right )}{3 b}+\frac {\sin ^2(c+d x)}{2 b}}{d}\) |
Input:
Int[Cos[c + d*x]^5/(a + b*Sin[c + d*x]^3),x]
Output:
(((a^(4/3) - b^(4/3))*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a ^(1/3))])/(Sqrt[3]*a^(2/3)*b^(5/3)) + ((a^(4/3) + b^(4/3))*Log[a^(1/3) + b ^(1/3)*Sin[c + d*x]])/(3*a^(2/3)*b^(5/3)) - ((a^(4/3) + b^(4/3))*Log[a^(2/ 3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d*x]^2])/(6*a^(2/3)*b^ (5/3)) - (2*Log[a + b*Sin[c + d*x]^3])/(3*b) + Sin[c + d*x]^2/(2*b))/d
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IntegerQ[n]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x _)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
Time = 1.50 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{2}}{2 b}+\frac {b \left (\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )-a \left (-\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\sin \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )-\frac {2 \ln \left (a +b \sin \left (d x +c \right )^{3}\right )}{3}}{b}}{d}\) | \(264\) |
| default | \(\frac {\frac {\sin \left (d x +c \right )^{2}}{2 b}+\frac {b \left (\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )-a \left (-\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\sin \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )-\frac {2 \ln \left (a +b \sin \left (d x +c \right )^{3}\right )}{3}}{b}}{d}\) | \(264\) |
| risch | \(\frac {2 i x}{b}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}+\frac {4 i c}{b d}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (27 a^{2} b^{5} d^{3} \textit {\_Z}^{3}+54 a^{2} b^{4} d^{2} \textit {\_Z}^{2}+27 a^{2} b^{3} d \textit {\_Z} -a^{4}+2 a^{2} b^{2}-b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {18 i a^{3} b^{3} d^{2} \textit {\_R}^{2}}{a^{4}-b^{4}}+\left (\frac {24 i a^{3} b^{2} d}{a^{4}-b^{4}}-\frac {6 i a \,b^{4} d}{a^{4}-b^{4}}\right ) \textit {\_R} +\frac {4 i a^{3} b}{a^{4}-b^{4}}-\frac {4 i a \,b^{3}}{a^{4}-b^{4}}\right ) {\mathrm e}^{i \left (d x +c \right )}-\frac {a^{4}}{a^{4}-b^{4}}+\frac {b^{4}}{a^{4}-b^{4}}\right )\right )\) | \(274\) |
Input:
int(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/d*(1/2*sin(d*x+c)^2/b+(b*(1/3/b/(1/b*a)^(2/3)*ln(sin(d*x+c)+(1/b*a)^(1/3 ))-1/6/b/(1/b*a)^(2/3)*ln(sin(d*x+c)^2-(1/b*a)^(1/3)*sin(d*x+c)+(1/b*a)^(2 /3))+1/3/b/(1/b*a)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*sin(d *x+c)-1)))-a*(-1/3/b/(1/b*a)^(1/3)*ln(sin(d*x+c)+(1/b*a)^(1/3))+1/6/b/(1/b *a)^(1/3)*ln(sin(d*x+c)^2-(1/b*a)^(1/3)*sin(d*x+c)+(1/b*a)^(2/3))+1/3*3^(1 /2)/b/(1/b*a)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*sin(d*x+c)-1)))-2/ 3*ln(a+b*sin(d*x+c)^3))/b)
Result contains complex when optimal does not.
Time = 0.80 (sec) , antiderivative size = 3216, normalized size of antiderivative = 14.68 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5/(a+b*sin(d*x+c)**3),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {\frac {9 \, \sin \left (d x + c\right )^{2}}{b} - \frac {2 \, \sqrt {3} {\left (a {\left (3 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} - 4\right )} - b {\left (3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}} - \frac {4 \, a}{b}\right )}\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {3 \, {\left (b {\left (4 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} + 1\right )} + a \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \log \left (\sin \left (d x + c\right )^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {6 \, {\left (b {\left (2 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} - 1\right )} - a \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right )\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{18 \, d} \] Input:
integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
1/18*(9*sin(d*x + c)^2/b - 2*sqrt(3)*(a*(3*(a/b)^(2/3) - 4) - b*(3*(a/b)^( 1/3) - 4*a/b))*arctan(-1/3*sqrt(3)*((a/b)^(1/3) - 2*sin(d*x + c))/(a/b)^(1 /3))/(a*b) - 3*(b*(4*(a/b)^(2/3) + 1) + a*(a/b)^(1/3))*log(sin(d*x + c)^2 - (a/b)^(1/3)*sin(d*x + c) + (a/b)^(2/3))/(b^2*(a/b)^(2/3)) - 6*(b*(2*(a/b )^(2/3) - 1) - a*(a/b)^(1/3))*log((a/b)^(1/3) + sin(d*x + c))/(b^2*(a/b)^( 2/3)))/d
Time = 0.44 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {\sin \left (d x + c\right )^{2}}{2 \, b d} - \frac {2 \, \log \left ({\left | b \sin \left (d x + c\right )^{3} + a \right |}\right )}{3 \, b d} + \frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} + \left (-a b^{2}\right )^{\frac {2}{3}} a\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a b^{3} d} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} - \left (-a b^{2}\right )^{\frac {2}{3}} a\right )} \log \left (\sin \left (d x + c\right )^{2} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a b^{3} d} + \frac {{\left (a b^{4} d^{5} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - b^{5} d^{5}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right ) \right |}\right )}{3 \, a b^{5} d^{6}} \] Input:
integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
1/2*sin(d*x + c)^2/(b*d) - 2/3*log(abs(b*sin(d*x + c)^3 + a))/(b*d) + 1/3* sqrt(3)*((-a*b^2)^(1/3)*b^2 + (-a*b^2)^(2/3)*a)*arctan(1/3*sqrt(3)*((-a/b) ^(1/3) + 2*sin(d*x + c))/(-a/b)^(1/3))/(a*b^3*d) + 1/6*((-a*b^2)^(1/3)*b^2 - (-a*b^2)^(2/3)*a)*log(sin(d*x + c)^2 + (-a/b)^(1/3)*sin(d*x + c) + (-a/ b)^(2/3))/(a*b^3*d) + 1/3*(a*b^4*d^5*(-a/b)^(1/3) - b^5*d^5)*(-a/b)^(1/3)* log(abs(-(-a/b)^(1/3) + sin(d*x + c)))/(a*b^5*d^6)
Time = 35.73 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {\left (\sum _{k=1}^3\ln \left (3\,a+\mathrm {root}\left (27\,a^2\,b^5\,d^3+54\,a^2\,b^4\,d^2+27\,a^2\,b^3\,d+2\,a^2\,b^2-b^4-a^4,d,k\right )\,\left (12\,a\,b+3\,b^2\,\sin \left (c+d\,x\right )+\mathrm {root}\left (27\,a^2\,b^5\,d^3+54\,a^2\,b^4\,d^2+27\,a^2\,b^3\,d+2\,a^2\,b^2-b^4-a^4,d,k\right )\,a\,b^2\,9\right )+\frac {\sin \left (c+d\,x\right )\,\left (a^2+2\,b^2\right )}{b}\right )\,\mathrm {root}\left (27\,a^2\,b^5\,d^3+54\,a^2\,b^4\,d^2+27\,a^2\,b^3\,d+2\,a^2\,b^2-b^4-a^4,d,k\right )\right )+\frac {{\sin \left (c+d\,x\right )}^2}{2\,b}}{d} \] Input:
int(cos(c + d*x)^5/(a + b*sin(c + d*x)^3),x)
Output:
(symsum(log(3*a + root(27*a^2*b^5*d^3 + 54*a^2*b^4*d^2 + 27*a^2*b^3*d + 2* a^2*b^2 - b^4 - a^4, d, k)*(12*a*b + 3*b^2*sin(c + d*x) + 9*root(27*a^2*b^ 5*d^3 + 54*a^2*b^4*d^2 + 27*a^2*b^3*d + 2*a^2*b^2 - b^4 - a^4, d, k)*a*b^2 ) + (sin(c + d*x)*(a^2 + 2*b^2))/b)*root(27*a^2*b^5*d^3 + 54*a^2*b^4*d^2 + 27*a^2*b^3*d + 2*a^2*b^2 - b^4 - a^4, d, k), k, 1, 3) + sin(c + d*x)^2/(2 *b))/d
\[ \int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )^{3} b +a}d x \] Input:
int(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x)
Output:
int(cos(c + d*x)**5/(sin(c + d*x)**3*b + a),x)