Integrand size = 29, antiderivative size = 137 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}-\frac {2 \cos ^5(c+d x)}{5 d (a+a \sin (c+d x))^{5/2}}-\frac {2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {4 \cos (c+d x)}{a^2 d \sqrt {a+a \sin (c+d x)}} \]
-2/5*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-2/3*cos(d*x+c)^3/a/d/(a+a*sin(d *x+c))^(3/2)+4*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/ 2))/a^(5/2)/d*2^(1/2)-4*cos(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 0.99 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {a (1+\sin (c+d x))} \left ((240+240 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )-180 \cos \left (\frac {1}{2} (c+d x)\right )+25 \cos \left (\frac {3}{2} (c+d x)\right )+3 \cos \left (\frac {5}{2} (c+d x)\right )+180 \sin \left (\frac {1}{2} (c+d x)\right )+25 \sin \left (\frac {3}{2} (c+d x)\right )-3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
(Sqrt[a*(1 + Sin[c + d*x])]*((240 + 240*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)* (-1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*c + d*x)/4] - Sin[(2*c + d*x)/4])] - 180*C os[(c + d*x)/2] + 25*Cos[(3*(c + d*x))/2] + 3*Cos[(5*(c + d*x))/2] + 180*S in[(c + d*x)/2] + 25*Sin[(3*(c + d*x))/2] - 3*Sin[(5*(c + d*x))/2]))/(30*a ^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
Time = 0.67 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3339, 3042, 3158, 3042, 3158, 3042, 3128, 219}
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.
| \(\displaystyle \int \frac {\sin (c+d x) \cos ^4(c+d x)}{(a \sin (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^4}{(a \sin (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3339 |
| \(\displaystyle -\int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{5/2}}dx-\frac {2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{5/2}}dx-\frac {2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle -\frac {2 \int \frac {\cos ^2(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx}{a}-\frac {2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \int \frac {\cos (c+d x)^2}{(\sin (c+d x) a+a)^{3/2}}dx}{a}-\frac {2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle -\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{a}+\frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\right )}{a}-\frac {2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{a}+\frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\right )}{a}-\frac {2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle -\frac {2 \left (\frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}-\frac {4 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{a d}\right )}{a}-\frac {2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2 \left (\frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\right )}{a}-\frac {2 \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
(-2*Cos[c + d*x]^5)/(5*d*(a + a*Sin[c + d*x])^(5/2)) - (2*Cos[c + d*x]^3)/ (3*a*d*(a + a*Sin[c + d*x])^(3/2)) - (2*((-2*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[ c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(3/2)*d) + (2*Cos[c + d* x])/(a*d*Sqrt[a + a*Sin[c + d*x]])))/a
3.5.84.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[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 + p))), x] + Simp[g^2*((p - 1)/(a*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In tegersQ[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && NeQ[m + p + 1, 0]
Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.95
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (30 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-3 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}-5 a \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-30 a^{2} \sqrt {a -a \sin \left (d x +c \right )}\right )}{15 a^{5} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(130\) |
2/15*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(30*a^(5/2)*2^(1/2)*arctanh( 1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-3*(a-a*sin(d*x+c))^(5/2)-5*a*( a-a*sin(d*x+c))^(3/2)-30*a^2*(a-a*sin(d*x+c))^(1/2))/a^5/cos(d*x+c)/(a+a*s in(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (118) = 236\).
Time = 0.28 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.74 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left (\frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} + {\left (3 \, \cos \left (d x + c\right )^{3} + 14 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) - 52\right )} \sin \left (d x + c\right ) - 41 \, \cos \left (d x + c\right ) - 52\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{15 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]
2/15*(15*sqrt(2)*(a*cos(d*x + c) + a*sin(d*x + c) + a)*log(-(cos(d*x + c)^ 2 - (cos(d*x + c) - 2)*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*( cos(d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2))/sqrt(a) + (3*c os(d*x + c)^3 + 14*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 11*cos(d*x + c) - 52)*sin(d*x + c) - 41*cos(d*x + c) - 52)*sqrt(a*sin(d*x + c) + a))/(a^3*d* cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.42 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (\frac {15 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {15 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (6 \, a^{\frac {25}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, a^{\frac {25}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{\frac {25}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{15} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{15 \, d} \]
-2/15*(15*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos (-1/4*pi + 1/2*d*x + 1/2*c))) - 15*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/ 2*c) + 1)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 2*sqrt(2)*(6*a^( 25/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 5*a^(25/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 15*a^(25/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^15*sgn(cos(-1/ 4*pi + 1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]