Integrand size = 27, antiderivative size = 94 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=-\frac {2 a^2 (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (6-5 m+m^2\right )}-\frac {a (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (3-m)} \]
-2*a^2*(e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^(-2+m)/d/e/(2-m)/(3-m)-a*(e *cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^(-1+m)/d/e/(3-m)
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {e^3 \cos ^4(c+d x) (e \cos (c+d x))^{-2 m} (a (1+\sin (c+d x)))^m (-4+m+(-2+m) \sin (c+d x))}{d (-3+m) (-2+m) (1+\sin (c+d x))^2} \]
(e^3*Cos[c + d*x]^4*(a*(1 + Sin[c + d*x]))^m*(-4 + m + (-2 + m)*Sin[c + d* x]))/(d*(-3 + m)*(-2 + m)*(e*Cos[c + d*x])^(2*m)*(1 + Sin[c + d*x])^2)
Time = 0.42 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3153, 3042, 3152}
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.
\(\displaystyle \int (a \sin (c+d x)+a)^m (e \cos (c+d x))^{3-2 m} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (c+d x)+a)^m (e \cos (c+d x))^{3-2 m}dx\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {2 a \int (e \cos (c+d x))^{3-2 m} (\sin (c+d x) a+a)^{m-1}dx}{3-m}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \int (e \cos (c+d x))^{3-2 m} (\sin (c+d x) a+a)^{m-1}dx}{3-m}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)}\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle -\frac {2 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{4-2 m}}{d e (2-m) (3-m)}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)}\) |
(-2*a^2*(e*Cos[c + d*x])^(4 - 2*m)*(a + a*Sin[c + d*x])^(-2 + m))/(d*e*(2 - m)*(3 - m)) - (a*(e*Cos[c + d*x])^(4 - 2*m)*(a + a*Sin[c + d*x])^(-1 + m ))/(d*e*(3 - m))
3.4.68.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
\[\int \left (e \cos \left (d x +c \right )\right )^{3-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]
Time = 0.29 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.81 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {{\left ({\left (m - 2\right )} \cos \left (d x + c\right )^{2} + {\left (m - 4\right )} \cos \left (d x + c\right ) + {\left ({\left (m - 2\right )} \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - 2\right )} \left (e \cos \left (d x + c\right )\right )^{-2 \, m + 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{2 \, d m^{2} - {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right )^{2} - 10 \, d m + {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right ) + {\left (2 \, d m^{2} - 10 \, d m + {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right ) + 12 \, d\right )} \sin \left (d x + c\right ) + 12 \, d} \]
((m - 2)*cos(d*x + c)^2 + (m - 4)*cos(d*x + c) + ((m - 2)*cos(d*x + c) + 2 )*sin(d*x + c) - 2)*(e*cos(d*x + c))^(-2*m + 3)*(a*sin(d*x + c) + a)^m/(2* d*m^2 - (d*m^2 - 5*d*m + 6*d)*cos(d*x + c)^2 - 10*d*m + (d*m^2 - 5*d*m + 6 *d)*cos(d*x + c) + (2*d*m^2 - 10*d*m + (d*m^2 - 5*d*m + 6*d)*cos(d*x + c) + 12*d)*sin(d*x + c) + 12*d)
\[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{3 - 2 m}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (91) = 182\).
Time = 0.30 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.73 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {{\left (a^{m} e^{3} {\left (m - 4\right )} - \frac {2 \, a^{m} e^{3} {\left (m - 6\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {a^{m} e^{3} {\left (m + 12\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{m} e^{3} {\left (m + 2\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {a^{m} e^{3} {\left (m + 12\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2 \, a^{m} e^{3} {\left (m - 6\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{m} e^{3} {\left (m - 4\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} e^{\left (-2 \, m \log \left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + m \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left ({\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} + \frac {3 \, {\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, {\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \]
(a^m*e^3*(m - 4) - 2*a^m*e^3*(m - 6)*sin(d*x + c)/(cos(d*x + c) + 1) - a^m *e^3*(m + 12)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^m*e^3*(m + 2)*sin( d*x + c)^3/(cos(d*x + c) + 1)^3 - a^m*e^3*(m + 12)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2*a^m*e^3*(m - 6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a^m *e^3*(m - 4)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*e^(-2*m*log(-sin(d*x + c )/(cos(d*x + c) + 1) + 1) + m*log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1) )/(((m^2 - 5*m + 6)*e^(2*m) + 3*(m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^2/(co s(d*x + c) + 1)^2 + 3*(m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*d)
Leaf count of result is larger than twice the leaf count of optimal. 16008 vs. \(2 (91) = 182\).
Time = 13.03 (sec) , antiderivative size = 16008, normalized size of antiderivative = 170.30 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\text {Too large to display} \]
2*(m*e^(m*log(2) - 2*m*log(4*abs(e)*abs(tan(1/8*pi - 1/4*d*x - 1/4*c))/(ta n(1/8*pi - 1/4*d*x - 1/4*c)^2 + 1)) + m*log(abs(a)) - 3*log(2) + 3*log(4*a bs(e)*abs(tan(1/8*pi - 1/4*d*x - 1/4*c))/(tan(1/8*pi - 1/4*d*x - 1/4*c)^2 + 1)))*tan(-3/8*pi + 1/2*pi*m*sgn(e*tan(1/2*d*x + 1/2*c)^2 - 2*e*tan(1/2*d *x + 1/2*c) + e)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(e) - 3/4*pi*sgn(e*tan (1/2*d*x + 1/2*c)^2 - 2*e*tan(1/2*d*x + 1/2*c) + e)*sgn(tan(1/2*d*x + 1/2* c)^2 - 1)*sgn(e) + pi*m*floor(1/2*d*x/pi + 1/2*c/pi - floor(1/2*d*x/pi + 1 /2*c/pi + 1/2) + 1/4) + pi*m*floor(1/2*d*x/pi + 1/2*c/pi + 1/2) - pi*m*flo or(-1/4*sgn(a) + 1/2) + 1/2*pi*m*sgn(e*tan(1/2*d*x + 1/2*c)^2 - 2*e*tan(1/ 2*d*x + 1/2*c) + e) - 1/4*pi*m*sgn(a) + 1/4*pi*m - 3/4*d*x - 3/2*pi*floor( 1/2*d*x/pi + 1/2*c/pi - floor(1/2*d*x/pi + 1/2*c/pi + 1/2) + 1/4) - 3/2*pi *floor(1/2*d*x/pi + 1/2*c/pi + 1/2) - 3/4*pi*sgn(e*tan(1/2*d*x + 1/2*c)^2 - 2*e*tan(1/2*d*x + 1/2*c) + e) - 3/4*c)^2*tan(1/2*d*x + 1/2*c)^6 + 4*m*e^ (m*log(2) - 2*m*log(4*abs(e)*abs(tan(1/8*pi - 1/4*d*x - 1/4*c))/(tan(1/8*p i - 1/4*d*x - 1/4*c)^2 + 1)) + m*log(abs(a)) - 3*log(2) + 3*log(4*abs(e)*a bs(tan(1/8*pi - 1/4*d*x - 1/4*c))/(tan(1/8*pi - 1/4*d*x - 1/4*c)^2 + 1)))* tan(-3/8*pi + 1/2*pi*m*sgn(e*tan(1/2*d*x + 1/2*c)^2 - 2*e*tan(1/2*d*x + 1/ 2*c) + e)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(e) - 3/4*pi*sgn(e*tan(1/2*d* x + 1/2*c)^2 - 2*e*tan(1/2*d*x + 1/2*c) + e)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(e) + pi*m*floor(1/2*d*x/pi + 1/2*c/pi - floor(1/2*d*x/pi + 1/2*c...
Time = 8.60 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.56 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {e^3\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (14\,m-24\,\sin \left (c+d\,x\right )-36\,\sin \left (3\,c+3\,d\,x\right )-12\,\sin \left (5\,c+5\,d\,x\right )+24\,{\sin \left (2\,c+2\,d\,x\right )}^2-4\,{\sin \left (3\,c+3\,d\,x\right )}^2+8\,m\,\sin \left (c+d\,x\right )-17\,m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )+12\,m\,\sin \left (3\,c+3\,d\,x\right )+4\,m\,\sin \left (5\,c+5\,d\,x\right )-2\,m\,\left (2\,{\sin \left (2\,c+2\,d\,x\right )}^2-1\right )+m\,\left (2\,{\sin \left (3\,c+3\,d\,x\right )}^2-1\right )+132\,{\sin \left (c+d\,x\right )}^2-128\right )}{8\,d\,{\left (-e\,\left (2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\right )}^{2\,m}\,\left (m^2-5\,m+6\right )\,\left (12\,{\sin \left (c+d\,x\right )}^2+15\,\sin \left (c+d\,x\right )-\sin \left (3\,c+3\,d\,x\right )+4\right )} \]
(e^3*(a*(sin(c + d*x) + 1))^m*(14*m - 24*sin(c + d*x) - 36*sin(3*c + 3*d*x ) - 12*sin(5*c + 5*d*x) + 24*sin(2*c + 2*d*x)^2 - 4*sin(3*c + 3*d*x)^2 + 8 *m*sin(c + d*x) - 17*m*(2*sin(c + d*x)^2 - 1) + 12*m*sin(3*c + 3*d*x) + 4* m*sin(5*c + 5*d*x) - 2*m*(2*sin(2*c + 2*d*x)^2 - 1) + m*(2*sin(3*c + 3*d*x )^2 - 1) + 132*sin(c + d*x)^2 - 128))/(8*d*(-e*(2*sin(c/2 + (d*x)/2)^2 - 1 ))^(2*m)*(m^2 - 5*m + 6)*(15*sin(c + d*x) - sin(3*c + 3*d*x) + 12*sin(c + d*x)^2 + 4))