Integrand size = 22, antiderivative size = 71 \[ \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d} \] Output:
1/2*arctan(b^(1/4)*sin(d*x+c)/a^(1/4))/a^(3/4)/b^(1/4)/d+1/2*arctanh(b^(1/ 4)*sin(d*x+c)/a^(1/4))/a^(3/4)/b^(1/4)/d
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76 \[ \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d} \] Input:
Integrate[Cos[c + d*x]/(a - b*Sin[c + d*x]^4),x]
Output:
(ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)] + ArcTanh[(b^(1/4)*Sin[c + d*x])/a ^(1/4)])/(2*a^(3/4)*b^(1/4)*d)
Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {3042, 3702, 756, 218, 221}
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.
\(\displaystyle \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)}{a-b \sin (c+d x)^4}dx\) |
\(\Big \downarrow \) 3702 |
\(\displaystyle \frac {\int \frac {1}{a-b \sin ^4(c+d x)}d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {\frac {\int \frac {1}{\sqrt {a}-\sqrt {b} \sin ^2(c+d x)}d\sin (c+d x)}{2 \sqrt {a}}+\frac {\int \frac {1}{\sqrt {b} \sin ^2(c+d x)+\sqrt {a}}d\sin (c+d x)}{2 \sqrt {a}}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\int \frac {1}{\sqrt {a}-\sqrt {b} \sin ^2(c+d x)}d\sin (c+d x)}{2 \sqrt {a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b}}}{d}\) |
Input:
Int[Cos[c + d*x]/(a - b*Sin[c + d*x]^4),x]
Output:
(ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)]/(2*a^(3/4)*b^(1/4)) + ArcTanh[(b^( 1/4)*Sin[c + d*x])/a^(1/4)]/(2*a^(3/4)*b^(1/4)))/d
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
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])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.53 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.68
method | result | size |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (256 a^{3} b \,d^{4} \textit {\_Z}^{4}-1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+8 i a d \textit {\_R} \,{\mathrm e}^{i \left (d x +c \right )}-1\right )\) | \(48\) |
derivativedivides | \(\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 d a}\) | \(68\) |
default | \(\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 d a}\) | \(68\) |
Input:
int(cos(d*x+c)/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
sum(_R*ln(exp(2*I*(d*x+c))+8*I*a*d*_R*exp(I*(d*x+c))-1),_R=RootOf(256*_Z^4 *a^3*b*d^4-1))
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.27 \[ \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {1}{4} \, \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {1}{2} \, a d \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} + \frac {1}{2} \, \sin \left (d x + c\right )\right ) - \frac {1}{4} \, \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {1}{2} \, a d \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} - \frac {1}{2} \, \sin \left (d x + c\right )\right ) + \frac {1}{4} i \, \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {1}{2} i \, a d \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} + \frac {1}{2} \, \sin \left (d x + c\right )\right ) - \frac {1}{4} i \, \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {1}{2} i \, a d \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} - \frac {1}{2} \, \sin \left (d x + c\right )\right ) \] Input:
integrate(cos(d*x+c)/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
Output:
1/4*(1/(a^3*b*d^4))^(1/4)*log(1/2*a*d*(1/(a^3*b*d^4))^(1/4) + 1/2*sin(d*x + c)) - 1/4*(1/(a^3*b*d^4))^(1/4)*log(1/2*a*d*(1/(a^3*b*d^4))^(1/4) - 1/2* sin(d*x + c)) + 1/4*I*(1/(a^3*b*d^4))^(1/4)*log(1/2*I*a*d*(1/(a^3*b*d^4))^ (1/4) + 1/2*sin(d*x + c)) - 1/4*I*(1/(a^3*b*d^4))^(1/4)*log(1/2*I*a*d*(1/( a^3*b*d^4))^(1/4) - 1/2*sin(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (63) = 126\).
Time = 2.34 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.82 \[ \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx=\begin {cases} \frac {\tilde {\infty } x \cos {\left (c \right )}}{\sin ^{4}{\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {1}{3 b d \sin ^{3}{\left (c + d x \right )}} & \text {for}\: a = 0 \\\frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: b = 0 \\\frac {x \cos {\left (c \right )}}{a - b \sin ^{4}{\left (c \right )}} & \text {for}\: d = 0 \\- \frac {\sqrt [4]{\frac {a}{b}} \log {\left (- \sqrt [4]{\frac {a}{b}} + \sin {\left (c + d x \right )} \right )}}{4 a d} + \frac {\sqrt [4]{\frac {a}{b}} \log {\left (\sqrt [4]{\frac {a}{b}} + \sin {\left (c + d x \right )} \right )}}{4 a d} + \frac {\sqrt [4]{\frac {a}{b}} \operatorname {atan}{\left (\frac {\sin {\left (c + d x \right )}}{\sqrt [4]{\frac {a}{b}}} \right )}}{2 a d} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)/(a-b*sin(d*x+c)**4),x)
Output:
Piecewise((zoo*x*cos(c)/sin(c)**4, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (1/(3* b*d*sin(c + d*x)**3), Eq(a, 0)), (sin(c + d*x)/(a*d), Eq(b, 0)), (x*cos(c) /(a - b*sin(c)**4), Eq(d, 0)), (-(a/b)**(1/4)*log(-(a/b)**(1/4) + sin(c + d*x))/(4*a*d) + (a/b)**(1/4)*log((a/b)**(1/4) + sin(c + d*x))/(4*a*d) + (a /b)**(1/4)*atan(sin(c + d*x)/(a/b)**(1/4))/(2*a*d), True))
Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {2 \, \arctan \left (\frac {\sqrt {b} \sin \left (d x + c\right )}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} - \frac {\log \left (\frac {\sqrt {b} \sin \left (d x + c\right ) - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} \sin \left (d x + c\right ) + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}}}{4 \, d} \] Input:
integrate(cos(d*x+c)/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
Output:
1/4*(2*arctan(sqrt(b)*sin(d*x + c)/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sq rt(a)*sqrt(b))) - log((sqrt(b)*sin(d*x + c) - sqrt(sqrt(a)*sqrt(b)))/(sqrt (b)*sin(d*x + c) + sqrt(sqrt(a)*sqrt(b))))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) )/d
Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (51) = 102\).
Time = 0.36 (sec) , antiderivative size = 232, normalized size of antiderivative = 3.27 \[ \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b d} + \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b d} + \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} \log \left (\sin \left (d x + c\right )^{2} + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{8 \, a b d} - \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} \log \left (\sin \left (d x + c\right )^{2} - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{8 \, a b d} \] Input:
integrate(cos(d*x+c)/(a-b*sin(d*x+c)^4),x, algorithm="giac")
Output:
1/4*sqrt(2)*(-a*b^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) + 2*si n(d*x + c))/(-a/b)^(1/4))/(a*b*d) + 1/4*sqrt(2)*(-a*b^3)^(1/4)*arctan(-1/2 *sqrt(2)*(sqrt(2)*(-a/b)^(1/4) - 2*sin(d*x + c))/(-a/b)^(1/4))/(a*b*d) + 1 /8*sqrt(2)*(-a*b^3)^(1/4)*log(sin(d*x + c)^2 + sqrt(2)*(-a/b)^(1/4)*sin(d* x + c) + sqrt(-a/b))/(a*b*d) - 1/8*sqrt(2)*(-a*b^3)^(1/4)*log(sin(d*x + c) ^2 - sqrt(2)*(-a/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/(a*b*d)
Time = 35.52 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.56 \[ \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sin \left (c+d\,x\right )}{a^{1/4}}\right )+\mathrm {atanh}\left (\frac {b^{1/4}\,\sin \left (c+d\,x\right )}{a^{1/4}}\right )}{2\,a^{3/4}\,b^{1/4}\,d} \] Input:
int(cos(c + d*x)/(a - b*sin(c + d*x)^4),x)
Output:
(atan((b^(1/4)*sin(c + d*x))/a^(1/4)) + atanh((b^(1/4)*sin(c + d*x))/a^(1/ 4)))/(2*a^(3/4)*b^(1/4)*d)
Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {2 \mathit {atan} \left (\frac {\sqrt {b}\, \sin \left (d x +c \right )}{b^{\frac {1}{4}} a^{\frac {1}{4}}}\right )+\mathrm {log}\left (a^{\frac {1}{4}}+b^{\frac {1}{4}} \sin \left (d x +c \right )\right )-\mathrm {log}\left (-a^{\frac {1}{4}}+b^{\frac {1}{4}} \sin \left (d x +c \right )\right )}{4 b^{\frac {1}{4}} a^{\frac {3}{4}} d} \] Input:
int(cos(d*x+c)/(a-b*sin(d*x+c)^4),x)
Output:
(b**(3/4)*a**(1/4)*(2*atan((sqrt(b)*sin(c + d*x))/(b**(1/4)*a**(1/4))) + l og(a**(1/4) + b**(1/4)*sin(c + d*x)) - log( - a**(1/4) + b**(1/4)*sin(c + d*x))))/(4*a*b*d)