\(\int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx\) [608]

Optimal result

Integrand size = 25, antiderivative size = 192 \[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {d (d-2 c (2+m)) \cos (e+f x) (3+3 \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {2^{\frac {1}{2}+m} \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (3+3 \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {d^2 \cos (e+f x) (3+3 \sin (e+f x))^{1+m}}{3 f (2+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2840, 2830, 2731, 2730} \[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=-\frac {2^{m+\frac {1}{2}} \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac {d (d-2 c (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

[In]

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

[Out]

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

Rule 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[a^IntPart[n]*((a + b*Sin[c + d*x])^FracPart
[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n]), Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

Rule 2840

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

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 21.30 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.62 \[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=-\frac {3^m (1+\sin (e+f x))^m (\cos (e+f x)+i (1+\sin (e+f x))) \left (-\frac {2 \left (2 c^2+d^2\right ) \operatorname {Hypergeometric2F1}(1,1+m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}-\frac {4 i c d \operatorname {Hypergeometric2F1}(1,m,-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)-i \sin (e+f x))}{1+m}+\frac {4 i c d \operatorname {Hypergeometric2F1}(1,2+m,2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)+i \sin (e+f x))}{-1+m}+\frac {d^2 \operatorname {Hypergeometric2F1}(1,-1+m,-1-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{2+m}+\frac {d^2 \operatorname {Hypergeometric2F1}(1,3+m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}\right )}{4 f} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

\[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (c + d \sin {\left (e + f x \right )}\right )^{2}\, dx \]

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(c + d*sin(e + f*x))**2, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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