Integrand size = 37, antiderivative size = 35 \[ \int \sin ^{-2-m}(c+d x) (a+a \sin (c+d x))^m (1+m-m \sin (c+d x)) \, dx=-\frac {\cos (c+d x) \sin ^{-1-m}(c+d x) (a+a \sin (c+d x))^m}{d} \] Output:
-cos(d*x+c)*sin(d*x+c)^(-1-m)*(a+a*sin(d*x+c))^m/d
Time = 2.63 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \sin ^{-2-m}(c+d x) (a+a \sin (c+d x))^m (1+m-m \sin (c+d x)) \, dx=-\frac {\cos (c+d x) \sin ^{-1-m}(c+d x) (a (1+\sin (c+d x)))^m}{d} \] Input:
Integrate[Sin[c + d*x]^(-2 - m)*(a + a*Sin[c + d*x])^m*(1 + m - m*Sin[c + d*x]),x]
Output:
-((Cos[c + d*x]*Sin[c + d*x]^(-1 - m)*(a*(1 + Sin[c + d*x]))^m)/d)
Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {3042, 3453}
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.
\(\displaystyle \int \sin ^{-m-2}(c+d x) (m (-\sin (c+d x))+m+1) (a \sin (c+d x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^{-m-2} (m (-\sin (c+d x))+m+1) (a \sin (c+d x)+a)^mdx\) |
\(\Big \downarrow \) 3453 |
\(\displaystyle -\frac {\cos (c+d x) \sin ^{-m-1}(c+d x) (a \sin (c+d x)+a)^m}{d}\) |
Input:
Int[Sin[c + d*x]^(-2 - m)*(a + a*Sin[c + d*x])^m*(1 + m - m*Sin[c + d*x]), x]
Output:
-((Cos[c + d*x]*Sin[c + d*x]^(-1 - m)*(a + a*Sin[c + d*x])^m)/d)
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[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && EqQ[m + n + 2, 0] && EqQ[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)), 0]
\[\int \sin \left (d x +c \right )^{-2-m} \left (a +a \sin \left (d x +c \right )\right )^{m} \left (1+m -m \sin \left (d x +c \right )\right )d x\]
Input:
int(sin(d*x+c)^(-2-m)*(a+a*sin(d*x+c))^m*(1+m-m*sin(d*x+c)),x)
Output:
int(sin(d*x+c)^(-2-m)*(a+a*sin(d*x+c))^m*(1+m-m*sin(d*x+c)),x)
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int \sin ^{-2-m}(c+d x) (a+a \sin (c+d x))^m (1+m-m \sin (c+d x)) \, dx=-\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sin \left (d x + c\right )^{-m - 2} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{d} \] Input:
integrate(sin(d*x+c)^(-2-m)*(a+a*sin(d*x+c))^m*(1+m-m*sin(d*x+c)),x, algor ithm="fricas")
Output:
-(a*sin(d*x + c) + a)^m*sin(d*x + c)^(-m - 2)*cos(d*x + c)*sin(d*x + c)/d
\[ \int \sin ^{-2-m}(c+d x) (a+a \sin (c+d x))^m (1+m-m \sin (c+d x)) \, dx=- \int \left (- \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{- m - 2}{\left (c + d x \right )}\right )\, dx - \int \left (- m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{- m - 2}{\left (c + d x \right )}\right )\, dx - \int m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )} \sin ^{- m - 2}{\left (c + d x \right )}\, dx \] Input:
integrate(sin(d*x+c)**(-2-m)*(a+a*sin(d*x+c))**m*(1+m-m*sin(d*x+c)),x)
Output:
-Integral(-(a*sin(c + d*x) + a)**m*sin(c + d*x)**(-m - 2), x) - Integral(- m*(a*sin(c + d*x) + a)**m*sin(c + d*x)**(-m - 2), x) - Integral(m*(a*sin(c + d*x) + a)**m*sin(c + d*x)*sin(c + d*x)**(-m - 2), x)
\[ \int \sin ^{-2-m}(c+d x) (a+a \sin (c+d x))^m (1+m-m \sin (c+d x)) \, dx=\int { -{\left (m \sin \left (d x + c\right ) - m - 1\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sin \left (d x + c\right )^{-m - 2} \,d x } \] Input:
integrate(sin(d*x+c)^(-2-m)*(a+a*sin(d*x+c))^m*(1+m-m*sin(d*x+c)),x, algor ithm="maxima")
Output:
-integrate((m*sin(d*x + c) - m - 1)*(a*sin(d*x + c) + a)^m*sin(d*x + c)^(- m - 2), x)
Leaf count of result is larger than twice the leaf count of optimal. 5502 vs. \(2 (35) = 70\).
Time = 34.36 (sec) , antiderivative size = 5502, normalized size of antiderivative = 157.20 \[ \int \sin ^{-2-m}(c+d x) (a+a \sin (c+d x))^m (1+m-m \sin (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)^(-2-m)*(a+a*sin(d*x+c))^m*(1+m-m*sin(d*x+c)),x, algor ithm="giac")
Output:
-8*(cos(2*pi*m*floor(-1/8*sgn(4*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c)^2 + 8* tan(d*x + c)^2*tan(1/2*d*x + 1/2*c) + 4*tan(d*x + c)^2 + 2*tan(1/2*d*x + 1 /2*c)^2 + 8*tan(1/2*d*x + 1/2*c) + 2) + 5/8) + 1/4*pi*m*sgn(4*tan(d*x + c) ^2*tan(1/2*d*x + 1/2*c)^2 + 8*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c) + 4*tan( d*x + c)^2 + 2*tan(1/2*d*x + 1/2*c)^2 + 8*tan(1/2*d*x + 1/2*c) + 2) - 1/4* pi*m)*e^(m*log(sqrt(2)*sqrt(abs(4*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c)^2 + 8*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c) + 4*tan(d*x + c)^2 + 2*tan(1/2*d*x + 1/2*c)^2 + 8*tan(1/2*d*x + 1/2*c) + 2)*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c )^2 + abs(4*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c)^2 + 8*tan(d*x + c)^2*tan(1 /2*d*x + 1/2*c) + 4*tan(d*x + c)^2 + 2*tan(1/2*d*x + 1/2*c)^2 + 8*tan(1/2* d*x + 1/2*c) + 2)*tan(d*x + c)^2 + abs(4*tan(d*x + c)^2*tan(1/2*d*x + 1/2* c)^2 + 8*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c) + 4*tan(d*x + c)^2 + 2*tan(1/ 2*d*x + 1/2*c)^2 + 8*tan(1/2*d*x + 1/2*c) + 2)*tan(1/2*d*x + 1/2*c)^2 + ab s(4*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c)^2 + 8*tan(d*x + c)^2*tan(1/2*d*x + 1/2*c) + 4*tan(d*x + c)^2 + 2*tan(1/2*d*x + 1/2*c)^2 + 8*tan(1/2*d*x + 1/ 2*c) + 2))*abs(a)/(tan(d*x + c)^2*tan(1/2*d*x + 1/2*c)^2 + tan(d*x + c)^2 + tan(1/2*d*x + 1/2*c)^2 + 1)) - m*log(4*abs(tan(1/2*d*x + 1/2*c))/(tan(1/ 2*d*x + 1/2*c)^2 + 1)) - 2*log(4*abs(tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)))*tan(-1/2*pi + 1/4*pi*m*sgn(2*a*tan(1/2*d*x + 1/2*c)^4 + 4* a*tan(1/2*d*x + 1/2*c)^3 - 4*a*tan(1/2*d*x + 1/2*c) - 2*a)*sgn(4*a*tan(...
Time = 35.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \sin ^{-2-m}(c+d x) (a+a \sin (c+d x))^m (1+m-m \sin (c+d x)) \, dx=-\frac {\sin \left (2\,c+2\,d\,x\right )\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m}{2\,d\,{\sin \left (c+d\,x\right )}^{m+2}} \] Input:
int(((a + a*sin(c + d*x))^m*(m - m*sin(c + d*x) + 1))/sin(c + d*x)^(m + 2) ,x)
Output:
-(sin(2*c + 2*d*x)*(a*(sin(c + d*x) + 1))^m)/(2*d*sin(c + d*x)^(m + 2))
\[ \int \sin ^{-2-m}(c+d x) (a+a \sin (c+d x))^m (1+m-m \sin (c+d x)) \, dx=\left (\int \frac {\left (\sin \left (d x +c \right ) a +a \right )^{m}}{\sin \left (d x +c \right )^{m} \sin \left (d x +c \right )^{2}}d x \right ) m +\int \frac {\left (\sin \left (d x +c \right ) a +a \right )^{m}}{\sin \left (d x +c \right )^{m} \sin \left (d x +c \right )^{2}}d x -\left (\int \frac {\left (\sin \left (d x +c \right ) a +a \right )^{m}}{\sin \left (d x +c \right )^{m} \sin \left (d x +c \right )}d x \right ) m \] Input:
int(sin(d*x+c)^(-2-m)*(a+a*sin(d*x+c))^m*(1+m-m*sin(d*x+c)),x)
Output:
int((sin(c + d*x)*a + a)**m/(sin(c + d*x)**m*sin(c + d*x)**2),x)*m + int(( sin(c + d*x)*a + a)**m/(sin(c + d*x)**m*sin(c + d*x)**2),x) - int((sin(c + d*x)*a + a)**m/(sin(c + d*x)**m*sin(c + d*x)),x)*m