\(\int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx\) [406]

Optimal result

Integrand size = 26, antiderivative size = 83 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=-\frac {27\ 2^{\frac {1}{2}+m} c^2 \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2}-m} (3+3 \sin (e+f x))^{-3+m}}{5 f} \]

[Out]

-1/5*2^(1/2+m)*a^3*c^2*cos(f*x+e)^5*hypergeom([5/2, 1/2-m],[7/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(1/2-m)*(a
+a*sin(f*x+e))^(-3+m)/f

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2815, 2768, 72, 71} \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=-\frac {a^3 c^2 2^{m+\frac {1}{2}} \cos ^5(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{5 f} \]

[In]

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

[Out]

-1/5*(2^(1/2 + m)*a^3*c^2*Cos[e + f*x]^5*Hypergeometric2F1[5/2, 1/2 - m, 7/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e
+ f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^(-3 + m))/f

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 2815

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^{-2+m} \, dx \\ & = \frac {\left (a^4 c^2 \cos ^5(e+f x)\right ) \text {Subst}\left (\int (a-a x)^{3/2} (a+a x)^{-\frac {1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}} \\ & = \frac {\left (2^{-\frac {1}{2}+m} a^4 c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-3+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {1}{2}+m} (a-a x)^{3/2} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2}} \\ & = -\frac {2^{\frac {1}{2}+m} a^3 c^2 \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-3+m}}{5 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.16 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.01 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\frac {(1+i) 3^m c^2 (1+\sin (e+f x))^m \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left ((2-2 i) (5+2 m) \sec ^2\left (\frac {1}{2} (e+f x)\right ) (-3-m+m \sin (e+f x)) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (-i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^3+3 \operatorname {Hypergeometric2F1}\left (3+m,5+2 m,2 (3+m),\frac {(1-i) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{-i+\tan \left (\frac {1}{2} (e+f x)\right )}\right ) (-i \cos (e+f x)-\sin (e+f x))^m \left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{f (1+2 m) (3+2 m) (5+2 m) \left (-i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^2} \]

[In]

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

[Out]

((1 + I)*3^m*c^2*(1 + Sin[e + f*x])^m*(1 + Tan[(e + f*x)/2])*((2 - 2*I)*(5 + 2*m)*Sec[(e + f*x)/2]^2*(-3 - m +
m*Sin[e + f*x])*(-1 + Tan[(e + f*x)/2])*(-I + Tan[(e + f*x)/2])^3 + 3*Hypergeometric2F1[3 + m, 5 + 2*m, 2*(3
+ m), ((1 - I)*(1 + Tan[(e + f*x)/2]))/(-I + Tan[(e + f*x)/2])]*((-I)*Cos[e + f*x] - Sin[e + f*x])^m*(I + Tan[
(e + f*x)/2])^2*(1 + Tan[(e + f*x)/2])^4))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(-I + Tan[(e + f*x)/2])^5*(I + Tan
[(e + f*x)/2])^2)

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{2}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int { {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

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

[Out]

integral(-(c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2)*(a*sin(f*x + e) + a)^m, x)

Sympy [F]

\[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=c^{2} \left (\int \left (- 2 \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\right )\, dx + \int \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}\, dx + \int \left (a \sin {\left (e + f x \right )} + a\right )^{m}\, dx\right ) \]

[In]

integrate((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**2,x)

[Out]

c**2*(Integral(-2*(a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral((a*sin(e + f*x) + a)**m*sin(e + f*x)**2,
x) + Integral((a*sin(e + f*x) + a)**m, x))

Maxima [F]

\[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int { {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

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

[Out]

integrate((c*sin(f*x + e) - c)^2*(a*sin(f*x + e) + a)^m, x)

Giac [F]

\[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int { {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

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

[Out]

integrate((c*sin(f*x + e) - c)^2*(a*sin(f*x + e) + a)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]

[In]

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

[Out]

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