Integrand size = 21, antiderivative size = 139 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {3 a x}{b^4}+\frac {3 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2} d}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))} \] Output:
-3*a*x/b^4+3*(2*a^2-b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/ b^4/(a^2-b^2)^(1/2)/d-1/2*cos(d*x+c)^3/b/d/(a+b*sin(d*x+c))^2-3/2*cos(d*x+ c)*(2*a+b*sin(d*x+c))/b^3/d/(a+b*sin(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(2641\) vs. \(2(139)=278\).
Time = 6.92 (sec) , antiderivative size = 2641, normalized size of antiderivative = 19.00 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \] Input:
Integrate[Cos[c + d*x]^4/(a + b*Sin[c + d*x])^3,x]
Output:
(Cos[c + d*x]^3*(-1/2*(b*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^(5/2)*( b/(a + b) - (b*Sin[c + d*x])/(a + b))^(5/2))/(((a*b)/(a - b) - b^2/(a - b) )*((a*b)/(a + b) + b^2/(a + b))*(a + b*Sin[c + d*x])^2) - (-((a*b^3*(-(b/( a - b)) - (b*Sin[c + d*x])/(a - b))^(5/2)*(b/(a + b) - (b*Sin[c + d*x])/(a + b))^(5/2))/((a^2 - b^2)*((a*b)/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b))*(a + b*Sin[c + d*x]))) - ((16*Sqrt[2]*a*b^4*(-(b/(a - b)) - ( b*Sin[c + d*x])/(a - b))^(5/2)*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]* (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(5/2)*((5* (1/(2*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(-1)))/8 - (15*b^3*(((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/b - ((a - b )^2*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^2)/(3*b^2) - (Sqrt[2]*Sqrt[a - b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/ (Sqrt[2]*Sqrt[b])]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[b] *Sqrt[1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b)])))/(3 2*(a - b)^3*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^3*(1 + ((a - b)*(-(b /(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^2)))/(5*(a - b)*(a + b)^3*Sq rt[((a + b)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)))/b]) + (((4*a^2*b^5)/(( a - b)^2*(a + b)^2) + (b^5*(2*a^2 - 3*b^2))/((a - b)^2*(a + b)^2))*((4*Sqr t[2]*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^(3/2)*Sqrt[b/(a + b) - (...
Time = 0.69 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3172, 3042, 3342, 25, 3042, 3214, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4}{(a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3172 |
| \(\displaystyle -\frac {3 \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2}dx}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \int \frac {\cos (c+d x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2}dx}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3342 |
| \(\displaystyle -\frac {3 \left (\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}-\frac {\int -\frac {b+2 a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b^2}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {3 \left (\frac {\int \frac {b+2 a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \left (\frac {\int \frac {b+2 a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -\frac {3 \left (\frac {\frac {2 a x}{b}-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b^2}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \left (\frac {\frac {2 a x}{b}-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b^2}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {3 \left (\frac {\frac {2 a x}{b}-\frac {2 \left (2 a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b^2}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {3 \left (\frac {\frac {4 \left (2 a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {2 a x}{b}}{b^2}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {3 \left (\frac {\frac {2 a x}{b}-\frac {2 \left (2 a^2-b^2\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}}{b^2}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
Input:
Int[Cos[c + d*x]^4/(a + b*Sin[c + d*x])^3,x]
Output:
-1/2*Cos[c + d*x]^3/(b*d*(a + b*Sin[c + d*x])^2) - (3*(((2*a*x)/b - (2*(2* a^2 - b^2)*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(b*Sq rt[a^2 - b^2]*d))/b^2 + (Cos[c + d*x]*(2*a + b*Sin[c + d*x]))/(b^2*d*(a + b*Sin[c + d*x]))))/(2*b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*C os[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d*p + b*d*(m + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Simp[g^2*(( p - 1)/(b^2*(m + 1)*(m + p + 1))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin [e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Sin[e + f*x ], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && Lt Q[m, -1] && GtQ[p, 1] && NeQ[m + p + 1, 0] && IntegerQ[2*m]
Time = 1.16 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.71
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+3 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{4}}+\frac {\frac {2 \left (-\frac {b^{2} \left (3 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {b \left (4 a^{4}+9 b^{2} a^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a^{2}}-\frac {b^{2} \left (13 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-2 a^{2} b -\frac {b^{3}}{2}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {3 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{4}}}{d}\) | \(238\) |
| default | \(\frac {-\frac {2 \left (\frac {b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+3 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{4}}+\frac {\frac {2 \left (-\frac {b^{2} \left (3 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {b \left (4 a^{4}+9 b^{2} a^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a^{2}}-\frac {b^{2} \left (13 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-2 a^{2} b -\frac {b^{3}}{2}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {3 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{4}}}{d}\) | \(238\) |
| risch | \(-\frac {3 a x}{b^{4}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{3} d}+\frac {i \left (-6 i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+i {\mathrm e}^{3 i \left (d x +c \right )} b^{3}+14 i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+i {\mathrm e}^{i \left (d x +c \right )} b^{3}+10 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{2 i \left (d x +c \right )} a \,b^{2}-5 a \,b^{2}\right )}{\left (-i {\mathrm e}^{2 i \left (d x +c \right )} b +i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d \,b^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\) | \(464\) |
Input:
int(cos(d*x+c)^4/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-2/b^4*(b/(1+tan(1/2*d*x+1/2*c)^2)+3*a*arctan(tan(1/2*d*x+1/2*c)))+2/ b^4*((-1/2*b^2*(3*a^2+2*b^2)/a*tan(1/2*d*x+1/2*c)^3-1/2*b*(4*a^4+9*a^2*b^2 +2*b^4)/a^2*tan(1/2*d*x+1/2*c)^2-1/2*b^2*(13*a^2+2*b^2)/a*tan(1/2*d*x+1/2* c)-2*a^2*b-1/2*b^3)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)^2+3/ 2*(2*a^2-b^2)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2 -b^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (130) = 260\).
Time = 0.13 (sec) , antiderivative size = 716, normalized size of antiderivative = 5.15 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
Output:
[-1/4*(12*(a^3*b^2 - a*b^4)*d*x*cos(d*x + c)^2 + 4*(a^2*b^3 - b^5)*cos(d*x + c)^3 - 12*(a^5 - a*b^4)*d*x + 3*(2*a^4 + a^2*b^2 - b^4 - (2*a^2*b^2 - b ^4)*cos(d*x + c)^2 + 2*(2*a^3*b - a*b^3)*sin(d*x + c))*sqrt(-a^2 + b^2)*lo g(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*c os(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 6*(2*a^4*b - a^2*b^3 - b^5)*co s(d*x + c) - 6*(4*(a^4*b - a^2*b^3)*d*x + 3*(a^3*b^2 - a*b^4)*cos(d*x + c) )*sin(d*x + c))/((a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(a^3*b^5 - a*b^7)*d* sin(d*x + c) - (a^4*b^4 - b^8)*d), -1/2*(6*(a^3*b^2 - a*b^4)*d*x*cos(d*x + c)^2 + 2*(a^2*b^3 - b^5)*cos(d*x + c)^3 - 6*(a^5 - a*b^4)*d*x - 3*(2*a^4 + a^2*b^2 - b^4 - (2*a^2*b^2 - b^4)*cos(d*x + c)^2 + 2*(2*a^3*b - a*b^3)*s in(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2) *cos(d*x + c))) - 3*(2*a^4*b - a^2*b^3 - b^5)*cos(d*x + c) - 3*(4*(a^4*b - a^2*b^3)*d*x + 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(a^3*b^5 - a*b^7)*d*sin(d*x + c) - (a^4*b^4 - b^8)*d)]
Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4/(a+b*sin(d*x+c))**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (130) = 260\).
Time = 0.16 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.96 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} a}{b^{4}} - \frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (2 \, a^{2} - b^{2}\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b^{3}} + \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{4} + a^{2} b^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2} a^{2} b^{3}}}{d} \] Input:
integrate(cos(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="giac")
Output:
-(3*(d*x + c)*a/b^4 - 3*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan( (a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*(2*a^2 - b^2)/(sqrt(a^2 - b ^2)*b^4) + 2/((tan(1/2*d*x + 1/2*c)^2 + 1)*b^3) + (3*a^3*b*tan(1/2*d*x + 1 /2*c)^3 + 2*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*tan(1/2*d*x + 1/2*c)^2 + 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 + 2*b^4*tan(1/2*d*x + 1/2*c)^2 + 13*a^3*b *tan(1/2*d*x + 1/2*c) + 2*a*b^3*tan(1/2*d*x + 1/2*c) + 4*a^4 + a^2*b^2)/(( a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a^2*b^3))/d
Time = 17.82 (sec) , antiderivative size = 1360, normalized size of antiderivative = 9.78 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
int(cos(c + d*x)^4/(a + b*sin(c + d*x))^3,x)
Output:
- ((6*a^2 + b^2)/b^3 + (2*tan(c/2 + (d*x)/2)^2*(6*a^4 + b^4 + 9*a^2*b^2))/ (a^2*b^3) + (tan(c/2 + (d*x)/2)*(21*a^2 + 2*b^2))/(a*b^2) + (4*tan(c/2 + ( d*x)/2)^3*(6*a^2 + b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)^4*(6*a^4 + 2*b^4 + 9*a^2*b^2))/(a^2*b^3) + (tan(c/2 + (d*x)/2)^5*(3*a^2 + 2*b^2))/(a*b^2))/(d *(tan(c/2 + (d*x)/2)^2*(3*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(3*a^2 + 4*b ^2) + a^2*tan(c/2 + (d*x)/2)^6 + a^2 + 8*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b* tan(c/2 + (d*x)/2)^5 + 4*a*b*tan(c/2 + (d*x)/2))) - (3*a*x)/b^4 - (atan((( (-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*((288*a^4)/b^5 - (8*tan(c/2 + (d*x) /2)*(9*a*b^7 - 108*a^3*b^5 + 72*a^5*b^3))/b^9 + (3*(-(a + b)*(a - b))^(1/2 )*(2*a^2 - b^2)*((8*tan(c/2 + (d*x)/2)*(12*a*b^10 - 24*a^3*b^8))/b^9 - 48* a^2 + (3*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 - 8*a^3*b^11))/b^9))/(2*(b^6 - a^2*b^4))))/(2*(b^6 - a^2*b^4)))*3i)/(2*(b^6 - a^2*b^4)) + ((-(a + b)*(a - b))^(1/2)*(2*a^2 - b^ 2)*((288*a^4)/b^5 - (8*tan(c/2 + (d*x)/2)*(9*a*b^7 - 108*a^3*b^5 + 72*a^5* b^3))/b^9 + (3*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*(48*a^2 - (8*tan(c/2 + (d*x)/2)*(12*a*b^10 - 24*a^3*b^8))/b^9 + (3*(-(a + b)*(a - b))^(1/2)*(2 *a^2 - b^2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 - 8*a^3*b^11))/ b^9))/(2*(b^6 - a^2*b^4))))/(2*(b^6 - a^2*b^4)))*3i)/(2*(b^6 - a^2*b^4)))/ ((16*(54*a^4 - 27*a^2*b^2))/b^8 + (16*tan(c/2 + (d*x)/2)*(216*a^5 - 108*a^ 3*b^2))/b^9 - (3*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*((288*a^4)/b^5 ...
Time = 0.26 (sec) , antiderivative size = 629, normalized size of antiderivative = 4.53 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {2 \cos \left (d x +c \right ) b^{5}-24 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a \,b^{3}-24 \sin \left (d x +c \right ) a^{4} b d x +24 \sin \left (d x +c \right ) a^{2} b^{3} d x +9 \sin \left (d x +c \right )^{2} b^{5}-9 a^{4} b +9 a^{2} b^{3}-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} b^{4}+24 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-12 \sin \left (d x +c \right )^{2} a^{3} b^{2} d x +12 \sin \left (d x +c \right )^{2} a \,b^{4} d x -12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} b^{2}-4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b^{3}-18 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}+18 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{4}+12 a^{3} b^{2} d x +48 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a^{3} b +24 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4}+4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{5}-12 \cos \left (d x +c \right ) a^{4} b +10 \cos \left (d x +c \right ) a^{2} b^{3}-12 a^{5} d x -9 \sin \left (d x +c \right )^{2} a^{2} b^{3}-18 \sin \left (d x +c \right ) a^{3} b^{2}+18 \sin \left (d x +c \right ) a \,b^{4}}{4 b^{4} d \left (\sin \left (d x +c \right )^{2} a^{2} b^{2}-\sin \left (d x +c \right )^{2} b^{4}+2 \sin \left (d x +c \right ) a^{3} b -2 \sin \left (d x +c \right ) a \,b^{3}+a^{4}-a^{2} b^{2}\right )} \] Input:
int(cos(d*x+c)^4/(a+b*sin(d*x+c))^3,x)
Output:
(24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin (c + d*x)**2*a**2*b**2 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b )/sqrt(a**2 - b**2))*sin(c + d*x)**2*b**4 + 48*sqrt(a**2 - b**2)*atan((tan ((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**3*b - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a*b **3 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2) )*a**4 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b* *2))*a**2*b**2 - 4*cos(c + d*x)*sin(c + d*x)**2*a**2*b**3 + 4*cos(c + d*x) *sin(c + d*x)**2*b**5 - 18*cos(c + d*x)*sin(c + d*x)*a**3*b**2 + 18*cos(c + d*x)*sin(c + d*x)*a*b**4 - 12*cos(c + d*x)*a**4*b + 10*cos(c + d*x)*a**2 *b**3 + 2*cos(c + d*x)*b**5 - 12*sin(c + d*x)**2*a**3*b**2*d*x - 9*sin(c + d*x)**2*a**2*b**3 + 12*sin(c + d*x)**2*a*b**4*d*x + 9*sin(c + d*x)**2*b** 5 - 24*sin(c + d*x)*a**4*b*d*x - 18*sin(c + d*x)*a**3*b**2 + 24*sin(c + d* x)*a**2*b**3*d*x + 18*sin(c + d*x)*a*b**4 - 12*a**5*d*x - 9*a**4*b + 12*a* *3*b**2*d*x + 9*a**2*b**3)/(4*b**4*d*(sin(c + d*x)**2*a**2*b**2 - sin(c + d*x)**2*b**4 + 2*sin(c + d*x)*a**3*b - 2*sin(c + d*x)*a*b**3 + a**4 - a**2 *b**2))