\(\int \sin ^5(e+f x) (A+C \sin ^2(e+f x)) \, dx\) [16]

Optimal result

Integrand size = 21, antiderivative size = 73 \[ \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx=-\frac {(A+C) \cos (e+f x)}{f}+\frac {(2 A+3 C) \cos ^3(e+f x)}{3 f}-\frac {(A+3 C) \cos ^5(e+f x)}{5 f}+\frac {C \cos ^7(e+f x)}{7 f} \]

[Out]

-(A+C)*cos(f*x+e)/f+1/3*(2*A+3*C)*cos(f*x+e)^3/f-1/5*(A+3*C)*cos(f*x+e)^5/f+1/7*C*cos(f*x+e)^7/f

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3092, 380} \[ \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx=-\frac {(A+3 C) \cos ^5(e+f x)}{5 f}+\frac {(2 A+3 C) \cos ^3(e+f x)}{3 f}-\frac {(A+C) \cos (e+f x)}{f}+\frac {C \cos ^7(e+f x)}{7 f} \]

[In]

Int[Sin[e + f*x]^5*(A + C*Sin[e + f*x]^2),x]

[Out]

-(((A + C)*Cos[e + f*x])/f) + ((2*A + 3*C)*Cos[e + f*x]^3)/(3*f) - ((A + 3*C)*Cos[e + f*x]^5)/(5*f) + (C*Cos[e
+ f*x]^7)/(7*f)

Rule 380

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rule 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \left (1-x^2\right )^2 \left (A+C-C x^2\right ) \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \left (A \left (1+\frac {C}{A}\right )-(2 A+3 C) x^2+(A+3 C) x^4-C x^6\right ) \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {(A+C) \cos (e+f x)}{f}+\frac {(2 A+3 C) \cos ^3(e+f x)}{3 f}-\frac {(A+3 C) \cos ^5(e+f x)}{5 f}+\frac {C \cos ^7(e+f x)}{7 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.49 \[ \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx=-\frac {5 A \cos (e+f x)}{8 f}-\frac {35 C \cos (e+f x)}{64 f}+\frac {5 A \cos (3 (e+f x))}{48 f}+\frac {7 C \cos (3 (e+f x))}{64 f}-\frac {A \cos (5 (e+f x))}{80 f}-\frac {7 C \cos (5 (e+f x))}{320 f}+\frac {C \cos (7 (e+f x))}{448 f} \]

[In]

Integrate[Sin[e + f*x]^5*(A + C*Sin[e + f*x]^2),x]

[Out]

(-5*A*Cos[e + f*x])/(8*f) - (35*C*Cos[e + f*x])/(64*f) + (5*A*Cos[3*(e + f*x)])/(48*f) + (7*C*Cos[3*(e + f*x)]
)/(64*f) - (A*Cos[5*(e + f*x)])/(80*f) - (7*C*Cos[5*(e + f*x)])/(320*f) + (C*Cos[7*(e + f*x)])/(448*f)

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {\left (700 A +735 C \right ) \cos \left (3 f x +3 e \right )+\left (-84 A -147 C \right ) \cos \left (5 f x +5 e \right )+15 C \cos \left (7 f x +7 e \right )+\left (-4200 A -3675 C \right ) \cos \left (f x +e \right )-3584 A -3072 C}{6720 f}\)	\(73\)
derivativedivides	\(\frac {-\frac {C \left (\frac {16}{5}+\sin ^{6}\left (f x +e \right )+\frac {6 \left (\sin ^{4}\left (f x +e \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (f x +e \right )\right )}{5}\right ) \cos \left (f x +e \right )}{7}-\frac {A \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}}{f}\)	\(74\)
default	\(\frac {-\frac {C \left (\frac {16}{5}+\sin ^{6}\left (f x +e \right )+\frac {6 \left (\sin ^{4}\left (f x +e \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (f x +e \right )\right )}{5}\right ) \cos \left (f x +e \right )}{7}-\frac {A \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}}{f}\)	\(74\)
parts	\(-\frac {A \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5 f}-\frac {C \left (\frac {16}{5}+\sin ^{6}\left (f x +e \right )+\frac {6 \left (\sin ^{4}\left (f x +e \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (f x +e \right )\right )}{5}\right ) \cos \left (f x +e \right )}{7 f}\)	\(76\)
risch	\(-\frac {5 \cos \left (f x +e \right ) A}{8 f}-\frac {35 \cos \left (f x +e \right ) C}{64 f}+\frac {C \cos \left (7 f x +7 e \right )}{448 f}-\frac {\cos \left (5 f x +5 e \right ) A}{80 f}-\frac {7 \cos \left (5 f x +5 e \right ) C}{320 f}+\frac {5 \cos \left (3 f x +3 e \right ) A}{48 f}+\frac {7 \cos \left (3 f x +3 e \right ) C}{64 f}\)	\(101\)
norman	\(\frac {-\frac {112 A +96 C}{105 f}-\frac {32 A \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {\left (80 A +96 C \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {\left (112 A +96 C \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 f}-\frac {\left (112 A +96 C \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{7}}\)	\(116\)

[In]

int(sin(f*x+e)^5*(A+C*sin(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

1/6720*((700*A+735*C)*cos(3*f*x+3*e)+(-84*A-147*C)*cos(5*f*x+5*e)+15*C*cos(7*f*x+7*e)+(-4200*A-3675*C)*cos(f*x
+e)-3584*A-3072*C)/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx=\frac {15 \, C \cos \left (f x + e\right )^{7} - 21 \, {\left (A + 3 \, C\right )} \cos \left (f x + e\right )^{5} + 35 \, {\left (2 \, A + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 105 \, {\left (A + C\right )} \cos \left (f x + e\right )}{105 \, f} \]

[In]

integrate(sin(f*x+e)^5*(A+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

1/105*(15*C*cos(f*x + e)^7 - 21*(A + 3*C)*cos(f*x + e)^5 + 35*(2*A + 3*C)*cos(f*x + e)^3 - 105*(A + C)*cos(f*x
+ e))/f

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (61) = 122\).

Time = 0.44 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.10 \[ \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx=\begin {cases} - \frac {A \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 A \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {8 A \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac {C \sin ^{6}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 C \sin ^{4}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{f} - \frac {8 C \sin ^{2}{\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{5 f} - \frac {16 C \cos ^{7}{\left (e + f x \right )}}{35 f} & \text {for}\: f \neq 0 \\x \left (A + C \sin ^{2}{\left (e \right )}\right ) \sin ^{5}{\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(f*x+e)**5*(A+C*sin(f*x+e)**2),x)

[Out]

Piecewise((-A*sin(e + f*x)**4*cos(e + f*x)/f - 4*A*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - 8*A*cos(e + f*x)**5
/(15*f) - C*sin(e + f*x)**6*cos(e + f*x)/f - 2*C*sin(e + f*x)**4*cos(e + f*x)**3/f - 8*C*sin(e + f*x)**2*cos(e
+ f*x)**5/(5*f) - 16*C*cos(e + f*x)**7/(35*f), Ne(f, 0)), (x*(A + C*sin(e)**2)*sin(e)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx=\frac {15 \, C \cos \left (f x + e\right )^{7} - 21 \, {\left (A + 3 \, C\right )} \cos \left (f x + e\right )^{5} + 35 \, {\left (2 \, A + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 105 \, {\left (A + C\right )} \cos \left (f x + e\right )}{105 \, f} \]

[In]

integrate(sin(f*x+e)^5*(A+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

1/105*(15*C*cos(f*x + e)^7 - 21*(A + 3*C)*cos(f*x + e)^5 + 35*(2*A + 3*C)*cos(f*x + e)^3 - 105*(A + C)*cos(f*x
+ e))/f

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx=\frac {C \cos \left (7 \, f x + 7 \, e\right )}{448 \, f} - \frac {{\left (4 \, A + 7 \, C\right )} \cos \left (5 \, f x + 5 \, e\right )}{320 \, f} + \frac {{\left (20 \, A + 21 \, C\right )} \cos \left (3 \, f x + 3 \, e\right )}{192 \, f} - \frac {5 \, {\left (8 \, A + 7 \, C\right )} \cos \left (f x + e\right )}{64 \, f} \]

[In]

integrate(sin(f*x+e)^5*(A+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

1/448*C*cos(7*f*x + 7*e)/f - 1/320*(4*A + 7*C)*cos(5*f*x + 5*e)/f + 1/192*(20*A + 21*C)*cos(3*f*x + 3*e)/f - 5
/64*(8*A + 7*C)*cos(f*x + e)/f

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx=\frac {\frac {C\,{\cos \left (e+f\,x\right )}^7}{7}+\left (-\frac {A}{5}-\frac {3\,C}{5}\right )\,{\cos \left (e+f\,x\right )}^5+\left (\frac {2\,A}{3}+C\right )\,{\cos \left (e+f\,x\right )}^3+\left (-A-C\right )\,\cos \left (e+f\,x\right )}{f} \]

[In]

int(sin(e + f*x)^5*(A + C*sin(e + f*x)^2),x)

[Out]

(cos(e + f*x)^3*((2*A)/3 + C) + (C*cos(e + f*x)^7)/7 - cos(e + f*x)*(A + C) - cos(e + f*x)^5*(A/5 + (3*C)/5))/
f