3.176 \(\int (g \cos (e+f x))^{3-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx\)

Optimal. Leaf size=127 \[ -\frac {2 a^2 (a \sin (e+f x)+a)^{m-2} (c-c \sin (e+f x))^n (g \cos (e+f x))^{4-2 m}}{f g (-m+n+2) (-m+n+3)}-\frac {a (a \sin (e+f x)+a)^{m-1} (c-c \sin (e+f x))^n (g \cos (e+f x))^{4-2 m}}{f g (-m+n+3)} \]

[Out]

-2*a^2*(g*cos(f*x+e))^(4-2*m)*(a+a*sin(f*x+e))^(-2+m)*(c-c*sin(f*x+e))^n/f/g/(2-m+n)/(3-m+n)-a*(g*cos(f*x+e))^
(4-2*m)*(a+a*sin(f*x+e))^(-1+m)*(c-c*sin(f*x+e))^n/f/g/(3-m+n)

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Rubi [A]  time = 0.41, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2846, 2844} \[ -\frac {2 a^2 (a \sin (e+f x)+a)^{m-2} (c-c \sin (e+f x))^n (g \cos (e+f x))^{4-2 m}}{f g (-m+n+2) (-m+n+3)}-\frac {a (a \sin (e+f x)+a)^{m-1} (c-c \sin (e+f x))^n (g \cos (e+f x))^{4-2 m}}{f g (-m+n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^(3 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n,x]

[Out]

(-2*a^2*(g*Cos[e + f*x])^(4 - 2*m)*(a + a*Sin[e + f*x])^(-2 + m)*(c - c*Sin[e + f*x])^n)/(f*g*(2 - m + n)*(3 -
m + n)) - (a*(g*Cos[e + f*x])^(4 - 2*m)*(a + a*Sin[e + f*x])^(-1 + m)*(c - c*Sin[e + f*x])^n)/(f*g*(3 - m + n
))

Rule 2844

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e +
f*x])^n)/(f*g*(m - n - 1)), x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b
^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m - n - 1, 0]

Rule 2846

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e +
f*x])^n)/(f*g*(m + n + p)), x] + Dist[(a*(2*m + p - 1))/(m + n + p), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && Eq
Q[a^2 - b^2, 0] && IGtQ[Simplify[m + p/2 - 1/2], 0] &&  !LtQ[n, -1] &&  !(IGtQ[Simplify[n + p/2 - 1/2], 0] &&
GtQ[m - n, 0]) &&  !(ILtQ[Simplify[m + n + p], 0] && GtQ[Simplify[2*m + n + (3*p)/2 + 1], 0])

Rubi steps

\begin {align*} \int (g \cos (e+f x))^{3-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx &=-\frac {a (g \cos (e+f x))^{4-2 m} (a+a \sin (e+f x))^{-1+m} (c-c \sin (e+f x))^n}{f g (3-m+n)}+\frac {(2 a) \int (g \cos (e+f x))^{3-2 m} (a+a \sin (e+f x))^{-1+m} (c-c \sin (e+f x))^n \, dx}{3-m+n}\\ &=-\frac {2 a^2 (g \cos (e+f x))^{4-2 m} (a+a \sin (e+f x))^{-2+m} (c-c \sin (e+f x))^n}{f g (2-m+n) (3-m+n)}-\frac {a (g \cos (e+f x))^{4-2 m} (a+a \sin (e+f x))^{-1+m} (c-c \sin (e+f x))^n}{f g (3-m+n)}\\ \end {align*}

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Mathematica [A]  time = 1.38, size = 143, normalized size = 1.13 \[ -\frac {g^3 \cos ^{2 n}(e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (g \cos (e+f x))^{-2 m} ((-m+n+2) \sin (e+f x)-m+n+4) (a (\sin (e+f x)+1))^{m-n} \exp (n (\log (a (\sin (e+f x)+1))+\log (c-c \sin (e+f x))-2 \log (\cos (e+f x))))}{f (-m+n+2) (-m+n+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Cos[e + f*x])^(3 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n,x]

[Out]

-((E^(n*(-2*Log[Cos[e + f*x]] + Log[a*(1 + Sin[e + f*x])] + Log[c - c*Sin[e + f*x]]))*g^3*Cos[e + f*x]^(2*n)*(
Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(a*(1 + Sin[e + f*x]))^(m - n)*(4 - m + n + (2 - m + n)*Sin[e + f*x]))/
(f*(2 - m + n)*(3 - m + n)*(g*Cos[e + f*x])^(2*m)))

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fricas [B]  time = 0.52, size = 298, normalized size = 2.35 \[ \frac {{\left ({\left (m - n - 2\right )} \cos \left (f x + e\right )^{2} + {\left (m - n - 4\right )} \cos \left (f x + e\right ) + {\left ({\left (m - n - 2\right )} \cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - 2\right )} \left (g \cos \left (f x + e\right )\right )^{-2 \, m + 3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} e^{\left (2 \, n \log \left (g \cos \left (f x + e\right )\right ) - n \log \left (a \sin \left (f x + e\right ) + a\right ) + n \log \left (\frac {a c}{g^{2}}\right )\right )}}{2 \, f m^{2} + 2 \, f n^{2} - {\left (f m^{2} + f n^{2} - 5 \, f m - {\left (2 \, f m - 5 \, f\right )} n + 6 \, f\right )} \cos \left (f x + e\right )^{2} - 10 \, f m - 2 \, {\left (2 \, f m - 5 \, f\right )} n + {\left (f m^{2} + f n^{2} - 5 \, f m - {\left (2 \, f m - 5 \, f\right )} n + 6 \, f\right )} \cos \left (f x + e\right ) + {\left (2 \, f m^{2} + 2 \, f n^{2} - 10 \, f m - 2 \, {\left (2 \, f m - 5 \, f\right )} n + {\left (f m^{2} + f n^{2} - 5 \, f m - {\left (2 \, f m - 5 \, f\right )} n + 6 \, f\right )} \cos \left (f x + e\right ) + 12 \, f\right )} \sin \left (f x + e\right ) + 12 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm="fricas")

[Out]

((m - n - 2)*cos(f*x + e)^2 + (m - n - 4)*cos(f*x + e) + ((m - n - 2)*cos(f*x + e) + 2)*sin(f*x + e) - 2)*(g*c
os(f*x + e))^(-2*m + 3)*(a*sin(f*x + e) + a)^m*e^(2*n*log(g*cos(f*x + e)) - n*log(a*sin(f*x + e) + a) + n*log(
a*c/g^2))/(2*f*m^2 + 2*f*n^2 - (f*m^2 + f*n^2 - 5*f*m - (2*f*m - 5*f)*n + 6*f)*cos(f*x + e)^2 - 10*f*m - 2*(2*
f*m - 5*f)*n + (f*m^2 + f*n^2 - 5*f*m - (2*f*m - 5*f)*n + 6*f)*cos(f*x + e) + (2*f*m^2 + 2*f*n^2 - 10*f*m - 2*
(2*f*m - 5*f)*n + (f*m^2 + f*n^2 - 5*f*m - (2*f*m - 5*f)*n + 6*f)*cos(f*x + e) + 12*f)*sin(f*x + e) + 12*f)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 26.10, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{3-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(3-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)

[Out]

int((g*cos(f*x+e))^(3-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)

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maxima [B]  time = 2.80, size = 485, normalized size = 3.82 \[ \frac {{\left (a^{m} c^{n} g^{3} {\left (m - n - 4\right )} - \frac {2 \, a^{m} c^{n} g^{3} {\left (m - n - 6\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {a^{m} c^{n} g^{3} {\left (m - n + 12\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {4 \, a^{m} c^{n} g^{3} {\left (m - n + 2\right )} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {a^{m} c^{n} g^{3} {\left (m - n + 12\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {2 \, a^{m} c^{n} g^{3} {\left (m - n - 6\right )} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {a^{m} c^{n} g^{3} {\left (m - n - 4\right )} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} e^{\left (2 \, n \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right ) - 2 \, m \log \left (-\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) + m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right ) - n \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left ({\left (m^{2} - m {\left (2 \, n + 5\right )} + n^{2} + 5 \, n + 6\right )} g^{2 \, m} + \frac {3 \, {\left (m^{2} - m {\left (2 \, n + 5\right )} + n^{2} + 5 \, n + 6\right )} g^{2 \, m} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, {\left (m^{2} - m {\left (2 \, n + 5\right )} + n^{2} + 5 \, n + 6\right )} g^{2 \, m} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {{\left (m^{2} - m {\left (2 \, n + 5\right )} + n^{2} + 5 \, n + 6\right )} g^{2 \, m} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm="maxima")

[Out]

(a^m*c^n*g^3*(m - n - 4) - 2*a^m*c^n*g^3*(m - n - 6)*sin(f*x + e)/(cos(f*x + e) + 1) - a^m*c^n*g^3*(m - n + 12
)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^m*c^n*g^3*(m - n + 2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - a^m*c^
n*g^3*(m - n + 12)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2*a^m*c^n*g^3*(m - n - 6)*sin(f*x + e)^5/(cos(f*x + e
) + 1)^5 + a^m*c^n*g^3*(m - n - 4)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*e^(2*n*log(sin(f*x + e)/(cos(f*x + e)
+ 1) - 1) - 2*m*log(-sin(f*x + e)/(cos(f*x + e) + 1) + 1) + m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1) - n
*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/(((m^2 - m*(2*n + 5) + n^2 + 5*n + 6)*g^(2*m) + 3*(m^2 - m*(2*n
+ 5) + n^2 + 5*n + 6)*g^(2*m)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*(m^2 - m*(2*n + 5) + n^2 + 5*n + 6)*g^(
2*m)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + (m^2 - m*(2*n + 5) + n^2 + 5*n + 6)*g^(2*m)*sin(f*x + e)^6/(cos(f*x
+ e) + 1)^6)*f)

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mupad [B]  time = 18.13, size = 476, normalized size = 3.75 \[ -\frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n\,\left (\frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3-2\,m}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (n-m+2\right )}{f\,\left (m^2-2\,m\,n-5\,m+n^2+5\,n+6\right )}-\frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3-2\,m}\,\left (\cos \left (3\,e+3\,f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\,1{}\mathrm {i}\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (-m\,1{}\mathrm {i}+n\,1{}\mathrm {i}+2{}\mathrm {i}\right )}{f\,\left (m^2-2\,m\,n-5\,m+n^2+5\,n+6\right )}-\frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3-2\,m}\,\left (\cos \left (e+f\,x\right )+\sin \left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (-m\,1{}\mathrm {i}+n\,1{}\mathrm {i}+6{}\mathrm {i}\right )}{f\,\left (m^2-2\,m\,n-5\,m+n^2+5\,n+6\right )}+\frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3-2\,m}\,\left (\cos \left (2\,e+2\,f\,x\right )+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (n-m+6\right )}{f\,\left (m^2-2\,m\,n-5\,m+n^2+5\,n+6\right )}\right )}{3\,\cos \left (e+f\,x\right )+\sin \left (e+f\,x\right )\,3{}\mathrm {i}-\cos \left (3\,e+3\,f\,x\right )-\sin \left (3\,e+3\,f\,x\right )\,1{}\mathrm {i}+\frac {m^2\,1{}\mathrm {i}-m\,n\,2{}\mathrm {i}-m\,5{}\mathrm {i}+n^2\,1{}\mathrm {i}+n\,5{}\mathrm {i}+6{}\mathrm {i}}{m^2-2\,m\,n-5\,m+n^2+5\,n+6}-\frac {3\,\left (\cos \left (2\,e+2\,f\,x\right )+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )\,\left (m^2\,1{}\mathrm {i}-m\,n\,2{}\mathrm {i}-m\,5{}\mathrm {i}+n^2\,1{}\mathrm {i}+n\,5{}\mathrm {i}+6{}\mathrm {i}\right )}{m^2-2\,m\,n-5\,m+n^2+5\,n+6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(e + f*x))^(3 - 2*m)*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n,x)

[Out]

-((c - c*sin(e + f*x))^n*(((g*cos(e + f*x))^(3 - 2*m)*(a + a*sin(e + f*x))^m*(n - m + 2))/(f*(5*n - 5*m - 2*m*
n + m^2 + n^2 + 6)) - ((g*cos(e + f*x))^(3 - 2*m)*(cos(3*e + 3*f*x) + sin(3*e + 3*f*x)*1i)*(a + a*sin(e + f*x)
)^m*(n*1i - m*1i + 2i))/(f*(5*n - 5*m - 2*m*n + m^2 + n^2 + 6)) - ((g*cos(e + f*x))^(3 - 2*m)*(cos(e + f*x) +
sin(e + f*x)*1i)*(a + a*sin(e + f*x))^m*(n*1i - m*1i + 6i))/(f*(5*n - 5*m - 2*m*n + m^2 + n^2 + 6)) + ((g*cos(
e + f*x))^(3 - 2*m)*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i)*(a + a*sin(e + f*x))^m*(n - m + 6))/(f*(5*n - 5*m
- 2*m*n + m^2 + n^2 + 6))))/(3*cos(e + f*x) + sin(e + f*x)*3i - cos(3*e + 3*f*x) - sin(3*e + 3*f*x)*1i + (n*5
i - m*5i - m*n*2i + m^2*1i + n^2*1i + 6i)/(5*n - 5*m - 2*m*n + m^2 + n^2 + 6) - (3*(cos(2*e + 2*f*x) + sin(2*e
+ 2*f*x)*1i)*(n*5i - m*5i - m*n*2i + m^2*1i + n^2*1i + 6i))/(5*n - 5*m - 2*m*n + m^2 + n^2 + 6))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))**(3-2*m)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**n,x)

[Out]

Timed out

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