Integrand size = 21, antiderivative size = 139 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {3 a x}{b^4}+\frac {3 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2} d}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))} \]
-3*a*x/b^4-1/2*cos(d*x+c)^3/b/d/(a+b*sin(d*x+c))^2-3/2*cos(d*x+c)*(2*a+b*s in(d*x+c))/b^3/d/(a+b*sin(d*x+c))+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/d/(a^2-b^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(486\) vs. \(2(139)=278\).
Time = 6.30 (sec) , antiderivative size = 486, normalized size of antiderivative = 3.50 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\cos ^3(c+d x) \left (-6 \left (2 a^3-2 a^2 b-a b^2+b^3\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}}{\sqrt {a+b} \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}}}\right ) \sqrt {1-\sin (c+d x)} (a+b \sin (c+d x))^2+\sqrt {a+b} \left (6 \sqrt {a-b} \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {b (1+\sin (c+d x))}{-a+b}}}{\sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}}}\right ) \sqrt {1-\sin (c+d x)} (a+b \sin (c+d x))^2+(-a+b) \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \left (6 \sqrt {b} (2 a+b) \text {arcsinh}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}}{\sqrt {2} \sqrt {b}}\right ) (a+b \sin (c+d x))^2+\sqrt {a-b} (a+b) \sqrt {1-\sin (c+d x)} \sqrt {\frac {b (1+\sin (c+d x))}{-a+b}} \left (6 a^2+b^2+9 a b \sin (c+d x)+2 b^2 \sin ^2(c+d x)\right )\right )\right )\right )}{2 (a-b)^{5/2} b^2 (a+b)^{3/2} d (1-\sin (c+d x))^{3/2} \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \left (-\frac {b (1+\sin (c+d x))}{a-b}\right )^{3/2} (a+b \sin (c+d x))^2} \]
-1/2*(Cos[c + d*x]^3*(-6*(2*a^3 - 2*a^2*b - a*b^2 + b^3)*ArcTanh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])/(Sqrt[a + b]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))])]*Sqrt[1 - Sin[c + d*x]]*(a + b*Sin[c + d*x])^2 + Sqrt[a + b]*(6*Sqrt[a - b]*(2*a^2 - b^2)*ArcTanh[Sqrt[(b*(1 + Sin[c + d*x ]))/(-a + b)]/Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]]*Sqrt[1 - Sin[c + d *x]]*(a + b*Sin[c + d*x])^2 + (-a + b)*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*(6*Sqrt[b]*(2*a + b)*ArcSinh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[c + d*x ]))/(a - b))])/(Sqrt[2]*Sqrt[b])]*(a + b*Sin[c + d*x])^2 + Sqrt[a - b]*(a + b)*Sqrt[1 - Sin[c + d*x]]*Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)]*(6*a^2 + b^2 + 9*a*b*Sin[c + d*x] + 2*b^2*Sin[c + d*x]^2)))))/((a - b)^(5/2)*b^2*( a + b)^(3/2)*d*(1 - Sin[c + d*x])^(3/2)*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*(-((b*(1 + Sin[c + d*x]))/(a - b)))^(3/2)*(a + b*Sin[c + d*x])^2)
Time = 0.65 (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 |
| \(\displaystyle -\frac {3 \left (\frac {\frac {2 a x}{b}-\frac {2 \left (2 a^2-b^2\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}}{b^2}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\right )}{2 b}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
-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)
3.5.57.3.1 Defintions of rubi rules used
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 = 1.94 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.71
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {2 \left (-\frac {b^{2} \left (3 a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {b \left (4 a^{4}+9 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}-\frac {b^{2} \left (13 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-2 a^{2} b -\frac {b^{3}}{2}\right )}{{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{4}}-\frac {2 \left (\frac {b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+3 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{4}}}{d}\) | \(238\) |
| default | \(\frac {\frac {\frac {2 \left (-\frac {b^{2} \left (3 a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {b \left (4 a^{4}+9 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}-\frac {b^{2} \left (13 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-2 a^{2} b -\frac {b^{3}}{2}\right )}{{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{4}}-\frac {2 \left (\frac {b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+3 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{4}}}{d}\) | \(238\) |
| risch | \(-\frac {3 a x}{b^{4}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{3} d}+\frac {i \left (-6 i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+14 i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+i b^{3} {\mathrm e}^{i \left (d x +c \right )}+10 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+5 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 a \,b^{2}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+i b \right )^{2} d \,b^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\) | \(464\) |
1/d*(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)/(a*tan(1/2*d*x+1/2*c)^2+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)))-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))))
Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (130) = 260\).
Time = 0.32 (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=\left [-\frac {12 \, {\left (a^{3} b^{2} - a b^{4}\right )} d x \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} - 12 \, {\left (a^{5} - a b^{4}\right )} d x + 3 \, {\left (2 \, a^{4} + a^{2} b^{2} - b^{4} - {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 6 \, {\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right ) - 6 \, {\left (4 \, {\left (a^{4} b - a^{2} b^{3}\right )} d x + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) - {\left (a^{4} b^{4} - b^{8}\right )} d\right )}}, -\frac {6 \, {\left (a^{3} b^{2} - a b^{4}\right )} d x \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{5} - a b^{4}\right )} d x - 3 \, {\left (2 \, a^{4} + a^{2} b^{2} - b^{4} - {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 3 \, {\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right ) - 3 \, {\left (4 \, {\left (a^{4} b - a^{2} b^{3}\right )} d x + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) - {\left (a^{4} b^{4} - b^{8}\right )} d\right )}}\right ] \]
[-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. \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.37 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.96 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} a}{b^{4}} - \frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (2 \, a^{2} - b^{2}\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b^{3}} + \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{4} + a^{2} b^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2} a^{2} b^{3}}}{d} \]
-(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 = 7.18 (sec) , antiderivative size = 1360, normalized size of antiderivative = 9.78 \[ \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]
- ((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 ...