Integrand size = 30, antiderivative size = 88 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}} \] Output:
1/6*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(7/2)+1/24*cos(f* x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(5/2)
Time = 3.60 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.20 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (1+3 \sin (e+f x))}{6 c^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[(a + a*Sin[e + f*x])^(3/2)/(c - c*Sin[e + f*x])^(7/2),x]
Output:
-1/6*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*( 1 + 3*Sin[e + f*x]))/(c^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Si n[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]])
Time = 0.49 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3222, 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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c-c \sin (e+f x))^{7/2}}dx\) |
\(\Big \downarrow \) 3222 |
\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{5/2}}dx}{6 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{5/2}}dx}{6 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3221 |
\(\displaystyle \frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\) |
Input:
Int[(a + a*Sin[e + f*x])^(3/2)/(c - c*Sin[e + f*x])^(7/2),x]
Output:
(Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(6*f*(c - c*Sin[e + f*x])^(7/2)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(24*c*f*(c - c*Sin[e + f*x])^ (5/2))
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_)*((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] + Simp[(m + n + 1)/(a*(2*m + 1) ) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
Time = 1.81 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {\sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a \left (\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3}+2 \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \sqrt {4}}{48 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{3}}\) | \(94\) |
Input:
int((a+sin(f*x+e)*a)^(3/2)/(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)
Output:
1/48/f*(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*a/(c*cos(1/4*Pi+1/2*f*x+1/2*e )^2)^(1/2)/c^3*(tan(1/4*Pi+1/2*f*x+1/2*e)^3+2*tan(1/4*Pi+1/2*f*x+1/2*e)^3* sec(1/4*Pi+1/2*f*x+1/2*e)^2)*4^(1/2)
Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.15 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {{\left (3 \, a \sin \left (f x + e\right ) + a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (3 \, c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right ) - {\left (c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="fric as")
Output:
-1/6*(3*a*sin(f*x + e) + 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. \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(7/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxi ma")
Output:
integrate((a*sin(f*x + e) + a)^(3/2)/(-c*sin(f*x + e) + c)^(7/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac ")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 21.42 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.41 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {a\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}+3\,a\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{\frac {9\,c^4\,f\,\cos \left (3\,e+3\,f\,x\right )}{2}+\frac {21\,c^4\,f\,\sin \left (2\,e+2\,f\,x\right )}{2}-\frac {3\,c^4\,f\,\sin \left (4\,e+4\,f\,x\right )}{4}-\frac {21\,c^4\,f\,\cos \left (e+f\,x\right )}{2}} \] Input:
int((a + a*sin(e + f*x))^(3/2)/(c - c*sin(e + f*x))^(7/2),x)
Output:
-(a*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2) + 3*a*sin(e + f* x)*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/((9*c^4*f*cos(3* e + 3*f*x))/2 + (21*c^4*f*sin(2*e + 2*f*x))/2 - (3*c^4*f*sin(4*e + 4*f*x)) /4 - (21*c^4*f*cos(e + f*x))/2)
\[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{4}-4 \sin \left (f x +e \right )^{3}+6 \sin \left (f x +e \right )^{2}-4 \sin \left (f x +e \right )+1}d x +\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4}-4 \sin \left (f x +e \right )^{3}+6 \sin \left (f x +e \right )^{2}-4 \sin \left (f x +e \right )+1}d x \right )}{c^{4}} \] Input:
int((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(7/2),x)
Output:
(sqrt(c)*sqrt(a)*a*(int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)* sin(e + f*x))/(sin(e + f*x)**4 - 4*sin(e + f*x)**3 + 6*sin(e + f*x)**2 - 4 *sin(e + f*x) + 1),x) + int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x)**4 - 4*sin(e + f*x)**3 + 6*sin(e + f*x)**2 - 4*sin(e + f*x) + 1),x)))/c**4