Integrand size = 40, antiderivative size = 96 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}+\frac {(A-5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}} \]
1/6*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(7/2)+1/24* (A-5*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(5/2)
Time = 9.78 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.30 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (A+4 B-3 B \cos (2 (e+f x))+3 (A-B) \sin (e+f x))}{6 c^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \]
-1/6*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*( A + 4*B - 3*B*Cos[2*(e + f*x)] + 3*(A - B)*Sin[e + f*x]))/(c^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x] ])
Time = 0.56 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3042, 3451, 3042, 3221}
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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}}dx\) |
\(\Big \downarrow \) 3451 |
\(\displaystyle \frac {(A-5 B) \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{5/2}}dx}{6 c}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-5 B) \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{5/2}}dx}{6 c}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3221 |
\(\displaystyle \frac {(A-5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\) |
((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(6*f*(c - c*Sin[e + f*x] )^(7/2)) + ((A - 5*B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(24*c*f*(c - c*Sin[e + f*x])^(5/2))
3.2.46.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && Ne Q[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] + Simp[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[ {a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] && !SumSimplerQ[n, 1])) && NeQ[2* m + 1, 0]
Time = 3.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08
method | result | size |
default | \(\frac {a \tan \left (f x +e \right ) \left (A \left (\cos ^{2}\left (f x +e \right )\right )-B \left (\sin ^{2}\left (f x +e \right )\right )+3 A \sin \left (f x +e \right )-3 B \sin \left (f x +e \right )-7 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{6 c^{3} f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\) | \(104\) |
parts | \(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )-3 \cos \left (f x +e \right )-7 \tan \left (f x +e \right )+3 \sec \left (f x +e \right )\right )}{6 f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}+\frac {B \sec \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )+3\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a}{6 f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}\) | \(180\) |
1/6*a/c^3/f*tan(f*x+e)*(A*cos(f*x+e)^2-B*sin(f*x+e)^2+3*A*sin(f*x+e)-3*B*s in(f*x+e)-7*A)*(a*(1+sin(f*x+e)))^(1/2)/(cos(f*x+e)^2+2*sin(f*x+e)-2)/(-c* (sin(f*x+e)-1))^(1/2)
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.30 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {{\left (6 \, B a \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} a \sin \left (f x + e\right ) - {\left (A + 7 \, B\right )} a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (3 \, c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right ) - {\left (c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x , algorithm="fricas")
1/6*(6*B*a*cos(f*x + e)^2 - 3*(A - B)*a*sin(f*x + e) - (A + 7*B)*a)*sqrt(a *sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(3*c^4*f*cos(f*x + e)^3 - 4*c ^4*f*cos(f*x + e) - (c^4*f*cos(f*x + e)^3 - 4*c^4*f*cos(f*x + e))*sin(f*x + e))
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x , algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (84) = 168\).
Time = 0.41 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.91 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {{\left (12 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3 \, A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{24 \, c^{4} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}} \]
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x , algorithm="giac")
-1/24*(12*B*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/ 2*f*x + 1/2*e)^4 - 3*A*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(- 1/4*pi + 1/2*f*x + 1/2*e)^2 - 9*B*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/ 2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 2*A*a*sqrt(c)*sgn(cos(-1/4*pi + 1 /2*f*x + 1/2*e)) + 2*B*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt (a)/(c^4*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2 *e)^6)
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]