\(\int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^3 \, dx\) [405]

Optimal result

Integrand size = 26, antiderivative size = 83 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^3 \, dx=-\frac {81\ 2^{\frac {1}{2}+m} c^3 \cos ^7(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {1}{2}-m,\frac {9}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2}-m} (3+3 \sin (e+f x))^{-4+m}}{7 f} \]

[Out]

-1/7*2^(1/2+m)*a^4*c^3*cos(f*x+e)^7*hypergeom([7/2, 1/2-m],[9/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(1/2-m)*(a
+a*sin(f*x+e))^(-4+m)/f

Rubi [A] (verified)

Time = 0.10 (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} \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^3 \, dx=-\frac {a^4 c^3 2^{m+\frac {1}{2}} \cos ^7(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-4} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {1}{2}-m,\frac {9}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{7 f} \]

[In]

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

[Out]

-1/7*(2^(1/2 + m)*a^4*c^3*Cos[e + f*x]^7*Hypergeometric2F1[7/2, 1/2 - m, 9/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e
+ f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^(-4 + 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^3 c^3\right ) \int \cos ^6(e+f x) (a+a \sin (e+f x))^{-3+m} \, dx \\ & = \frac {\left (a^5 c^3 \cos ^7(e+f x)\right ) \text {Subst}\left (\int (a-a x)^{5/2} (a+a x)^{-\frac {1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{7/2} (a+a \sin (e+f x))^{7/2}} \\ & = \frac {\left (2^{-\frac {1}{2}+m} a^5 c^3 \cos ^7(e+f x) (a+a \sin (e+f x))^{-4+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)^{5/2} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{7/2}} \\ & = -\frac {2^{\frac {1}{2}+m} a^4 c^3 \cos ^7(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {1}{2}-m,\frac {9}{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))^{-4+m}}{7 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 15.71 (sec) , antiderivative size = 326, normalized size of antiderivative = 3.93 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^3 \, dx=\frac {(1+i) 3^m (1+\sin (e+f x))^m (c-c \sin (e+f x))^3 \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (\left (\frac {1}{2}-\frac {i}{2}\right ) (7+2 m) \sec ^4\left (\frac {1}{2} (e+f x)\right ) \left (-115-58 m-12 m^2+\left (5+6 m+4 m^2\right ) \cos (2 (e+f x))+16 m (4+m) \sin (e+f x)\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (-i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^4+15 i \operatorname {Hypergeometric2F1}\left (4+m,7+2 m,2 (4+m),\frac {(1-i) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{-i+\tan \left (\frac {1}{2} (e+f x)\right )}\right ) (-i \cos (e+f x)-\sin (e+f x))^m \left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )^6\right )}{f (1+2 m) (3+2 m) (5+2 m) (7+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \left (-i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

[In]

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

[Out]

((1 + I)*3^m*(1 + Sin[e + f*x])^m*(c - c*Sin[e + f*x])^3*(1 + Tan[(e + f*x)/2])*((1/2 - I/2)*(7 + 2*m)*Sec[(e
+ f*x)/2]^4*(-115 - 58*m - 12*m^2 + (5 + 6*m + 4*m^2)*Cos[2*(e + f*x)] + 16*m*(4 + m)*Sin[e + f*x])*(-1 + Tan[
(e + f*x)/2])*(-I + Tan[(e + f*x)/2])^4 + (15*I)*Hypergeometric2F1[4 + m, 7 + 2*m, 2*(4 + m), ((1 - I)*(1 + Ta
n[(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])^3*(1 + T
an[(e + f*x)/2])^6))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(7 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*(-I +
Tan[(e + f*x)/2])^7*(I + Tan[(e + f*x)/2])^3)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(-(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e))*(a*sin(f*x + e) + a)^m, x
)

Sympy [F]

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

[In]

integrate((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**3,x)

[Out]

-c**3*(Integral(3*(a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral(-3*(a*sin(e + f*x) + a)**m*sin(e + f*x)*
*2, x) + Integral((a*sin(e + f*x) + a)**m*sin(e + f*x)**3, x) + Integral(-(a*sin(e + f*x) + a)**m, x))

Maxima [F]

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

[In]

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

[Out]

-integrate((c*sin(f*x + e) - c)^3*(a*sin(f*x + e) + a)^m, x)

Giac [F]

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

[In]

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

[Out]

integrate(-(c*sin(f*x + e) - c)^3*(a*sin(f*x + e) + a)^m, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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