Integrand size = 31, antiderivative size = 169 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {4 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}+\frac {2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{7 a d (a+a \sin (c+d x))^{3/2}}+\frac {4 \cos (c+d x)}{a^2 d \sqrt {a+a \sin (c+d x)}} \]
4/7*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)-2/7*cos(d*x+c)^5/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 = 4.02 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {\sqrt {a (1+\sin (c+d x))} \left ((672+672 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )-525 \cos \left (\frac {1}{2} (c+d x)\right )+91 \cos \left (\frac {3}{2} (c+d x)\right )+21 \cos \left (\frac {5}{2} (c+d x)\right )-3 \cos \left (\frac {7}{2} (c+d x)\right )+525 \sin \left (\frac {1}{2} (c+d x)\right )+91 \sin \left (\frac {3}{2} (c+d x)\right )-21 \sin \left (\frac {5}{2} (c+d x)\right )-3 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{84 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
-1/84*(Sqrt[a*(1 + Sin[c + d*x])]*((672 + 672*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])] - 525*Cos[(c + d*x)/2] + 91*Cos[(3*(c + d*x))/2] + 21*Cos[(5*(c + d*x))/2] - 3*Cos[(7*(c + d*x))/2] + 525*Sin[(c + d*x)/2] + 91*Sin[(3*(c + d*x))/2] - 21*Sin[(5*(c + d*x))/2] - 3*Sin[(7*(c + d*x))/2]))/(a^3*d*(Cos[(c + d*x) /2] + Sin[(c + d*x)/2]))
Time = 0.95 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.10, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3357, 27, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^4(c+d x)}{(a \sin (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^4}{(a \sin (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3357 |
| \(\displaystyle \frac {2 \int -\frac {\cos ^4(c+d x) (10 \sin (c+d x) a+3 a)}{2 (\sin (c+d x) a+a)^{5/2}}dx}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\cos ^4(c+d x) (10 \sin (c+d x) a+3 a)}{(\sin (c+d x) a+a)^{5/2}}dx}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\cos (c+d x)^4 (10 \sin (c+d x) a+3 a)}{(\sin (c+d x) a+a)^{5/2}}dx}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3339 |
| \(\displaystyle -\frac {-7 a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{5/2}}dx-\frac {4 a \cos ^5(c+d x)}{d (a \sin (c+d x)+a)^{5/2}}}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-7 a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{5/2}}dx-\frac {4 a \cos ^5(c+d x)}{d (a \sin (c+d x)+a)^{5/2}}}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle -\frac {-7 a \left (\frac {2 \int \frac {\cos ^2(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx}{a}+\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {4 a \cos ^5(c+d x)}{d (a \sin (c+d x)+a)^{5/2}}}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-7 a \left (\frac {2 \int \frac {\cos (c+d x)^2}{(\sin (c+d x) a+a)^{3/2}}dx}{a}+\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {4 a \cos ^5(c+d x)}{d (a \sin (c+d x)+a)^{5/2}}}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle -\frac {-7 a \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{a}+\frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\right )}{a}+\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {4 a \cos ^5(c+d x)}{d (a \sin (c+d x)+a)^{5/2}}}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-7 a \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{a}+\frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\right )}{a}+\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {4 a \cos ^5(c+d x)}{d (a \sin (c+d x)+a)^{5/2}}}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle -\frac {-7 a \left (\frac {2 \left (\frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}-\frac {4 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{a d}\right )}{a}+\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {4 a \cos ^5(c+d x)}{d (a \sin (c+d x)+a)^{5/2}}}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {-7 a \left (\frac {2 \left (\frac {2 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\right )}{a}+\frac {2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {4 a \cos ^5(c+d x)}{d (a \sin (c+d x)+a)^{5/2}}}{7 a}-\frac {2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}\) |
(-2*Cos[c + d*x]^5)/(7*a*d*(a + a*Sin[c + d*x])^(3/2)) - ((-4*a*Cos[c + d* x]^5)/(d*(a + a*Sin[c + d*x])^(5/2)) - 7*a*((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*Sq rt[a + a*Sin[c + d*x]])))/a))/(7*a)
3.5.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^( p + 1))*((a + b*Sin[e + f*x])^(m + 1)/(b*f*g*(m + p + 2))), x] + Simp[1/(b* (m + p + 2)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a *(p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^ 2 - b^2, 0] && NeQ[m + p + 2, 0]
Time = 0.13 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.78
| method | result | size |
| default | \(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (42 a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-3 \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}-7 a^{2} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-42 a^{3} \sqrt {a -a \sin \left (d x +c \right )}\right )}{21 a^{6} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(132\) |
-2/21*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(42*a^(7/2)*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))^(7/2)-7*a^ 2*(a-a*sin(d*x+c))^(3/2)-42*a^3*(a-a*sin(d*x+c))^(1/2))/a^6/cos(d*x+c)/(a+ a*sin(d*x+c))^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.53 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left (\frac {21 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} + {\left (3 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{3} - 31 \, \cos \left (d x + c\right )^{2} + {\left (3 \, \cos \left (d x + c\right )^{3} + 12 \, \cos \left (d x + c\right )^{2} - 19 \, \cos \left (d x + c\right ) - 80\right )} \sin \left (d x + c\right ) + 61 \, \cos \left (d x + c\right ) + 80\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{21 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]
2/21*(21*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)^4 - 9*cos(d*x + c)^3 - 31*cos(d*x + c)^2 + (3*cos(d*x + c)^3 + 12*cos(d*x + c)^2 - 19*cos(d*x + c) - 80)*sin(d*x + c) + 61*cos(d*x + c) + 80)*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. \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.47 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left (\frac {21 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {21 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (12 \, a^{\frac {37}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7 \, a^{\frac {37}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, a^{\frac {37}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{21} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{21 \, d} \]
2/21*(21*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))) - 21*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)*(12*a^( 37/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 + 7*a^(37/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 21*a^(37/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^21*sgn(cos(-1/ 4*pi + 1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]