Integrand size = 26, antiderivative size = 113 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}+\frac {a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}} \]
1/2*a*cos(f*x+e)/f/(c-c*sin(f*x+e))^(5/2)-1/8*a*cos(f*x+e)/c/f/(c-c*sin(f* x+e))^(3/2)-1/16*a*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e)) ^(1/2))/c^(5/2)/f*2^(1/2)
Time = 2.12 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.56 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a \left (-2 \sqrt {c} (-7+\cos (2 (e+f x))-8 \sin (e+f x))+2 \sqrt {2} \arctan \left (\frac {\sqrt {-c (1+\sin (e+f x))}}{\sqrt {2} \sqrt {c}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sqrt {-c (1+\sin (e+f x))}\right )}{32 c^{5/2} f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)}} \]
(a*(-2*Sqrt[c]*(-7 + Cos[2*(e + f*x)] - 8*Sin[e + f*x]) + 2*Sqrt[2]*ArcTan [Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[2]*Sqrt[c])]*(Cos[(e + f*x)/2] - Sin[ (e + f*x)/2])^4*Sqrt[-(c*(1 + Sin[e + f*x]))]))/(32*c^(5/2)*f*(Cos[(e + f* x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]])
Time = 0.52 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3215, 3042, 3159, 3042, 3129, 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.
| \(\displaystyle \int \frac {a \sin (e+f x)+a}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a \sin (e+f x)+a}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a c \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a c \left (\frac {\cos (e+f x)}{2 c f (c-c \sin (e+f x))^{5/2}}-\frac {\int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx}{4 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {\cos (e+f x)}{2 c f (c-c \sin (e+f x))^{5/2}}-\frac {\int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx}{4 c^2}\right )\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle a c \left (\frac {\cos (e+f x)}{2 c f (c-c \sin (e+f x))^{5/2}}-\frac {\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}}{4 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {\cos (e+f x)}{2 c f (c-c \sin (e+f x))^{5/2}}-\frac {\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}}{4 c^2}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a c \left (\frac {\cos (e+f x)}{2 c f (c-c \sin (e+f x))^{5/2}}-\frac {\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}}{4 c^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a c \left (\frac {\cos (e+f x)}{2 c f (c-c \sin (e+f x))^{5/2}}-\frac {\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}}{4 c^2}\right )\) |
a*c*(Cos[e + f*x]/(2*c*f*(c - c*Sin[e + f*x])^(5/2)) - (ArcTanh[(Sqrt[c]*C os[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(2*Sqrt[2]*c^(3/2)*f) + C os[e + f*x]/(2*f*(c - c*Sin[e + f*x])^(3/2)))/(4*c^2))
3.3.96.3.1 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[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 2.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.65
| method | result | size |
| default | \(\frac {a \left (\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {2}\, c^{3}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) \sqrt {2}\, c^{3}-2 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-4 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {5}{2}}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{16 c^{\frac {11}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(187\) |
| parts | \(\frac {a \left (-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}+6 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}} \sin \left (f x +e \right )+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )-14 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{32 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a \left (5 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {2}\, c^{3}+12 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {5}{2}}-10 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-10 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) \sqrt {2}\, c^{3}+5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{32 c^{\frac {11}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(384\) |
1/16*a*(arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2 *2^(1/2)*c^3-2*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*sin(f *x+e)*2^(1/2)*c^3-2*(c*(sin(f*x+e)+1))^(3/2)*c^(3/2)-4*(c*(sin(f*x+e)+1))^ (1/2)*c^(5/2)+2^(1/2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2) )*c^3)*(c*(sin(f*x+e)+1))^(1/2)/c^(11/2)/(sin(f*x+e)-1)/cos(f*x+e)/(c-c*si n(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (94) = 188\).
Time = 0.28 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.97 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (a \cos \left (f x + e\right )^{3} + 3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) - 4 \, a\right )} \sin \left (f x + e\right ) - 4 \, a\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (a \cos \left (f x + e\right )^{2} - 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) + 4 \, a\right )} \sin \left (f x + e\right ) - 4 \, a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{32 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
1/32*(sqrt(2)*(a*cos(f*x + e)^3 + 3*a*cos(f*x + e)^2 - 2*a*cos(f*x + e) - (a*cos(f*x + e)^2 - 2*a*cos(f*x + e) - 4*a)*sin(f*x + e) - 4*a)*sqrt(c)*lo g(-(c*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f* x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin (f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f *x + e) - 2)) + 4*(a*cos(f*x + e)^2 - 3*a*cos(f*x + e) - (a*cos(f*x + e) + 4*a)*sin(f*x + e) - 4*a)*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f* x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))
\[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=a \left (\int \frac {\sin {\left (e + f x \right )}}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {1}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \]
a*(Integral(sin(e + f*x)/(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2 - 2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c**2*sqrt(-c*sin(e + f*x) + c)), x) + Integral(1/(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2 - 2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c**2*sqrt(-c*sin(e + f*x) + c)), x))
\[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {a \sin \left (f x + e\right ) + a}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (94) = 188\).
Time = 0.40 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.93 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\frac {2 \, \sqrt {2} a \log \left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}{c^{\frac {5}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} a {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{c^{\frac {5}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (a \sqrt {c} - \frac {2 \, a \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}{c^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{128 \, f} \]
-1/128*(2*sqrt(2)*a*log((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*p i + 1/2*f*x + 1/2*e) + 1)^2)/(c^(5/2)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*a*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(c^(5/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)*( a*sqrt(c) - 2*a*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*p i + 1/2*f*x + 1/2*e) + 1)^2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2/(c^3*( cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) )/f
Timed out. \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {a+a\,\sin \left (e+f\,x\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]