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

Optimal. Leaf size=73 \[ -\frac {(A+3 C) \cos ^5(e+f x)}{5 f}+\frac {(2 A+3 C) \cos ^3(e+f x)}{3 f}-\frac {(A+C) \cos (e+f x)}{f}+\frac {C \cos ^7(e+f x)}{7 f} \]

[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

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Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3013, 373} \[ -\frac {(A+3 C) \cos ^5(e+f x)}{5 f}+\frac {(2 A+3 C) \cos ^3(e+f x)}{3 f}-\frac {(A+C) \cos (e+f x)}{f}+\frac {C \cos ^7(e+f x)}{7 f} \]

Antiderivative was successfully verified.

[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 373

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 3013

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*} \int \sin ^5(e+f x) \left (A+C \sin ^2(e+f x)\right ) \, dx &=-\frac {\operatorname {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 {\operatorname {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*}

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Mathematica [A]  time = 0.03, size = 109, normalized size = 1.49 \[ -\frac {5 A \cos (e+f x)}{8 f}+\frac {5 A \cos (3 (e+f x))}{48 f}-\frac {A \cos (5 (e+f x))}{80 f}-\frac {35 C \cos (e+f x)}{64 f}+\frac {7 C \cos (3 (e+f x))}{64 f}-\frac {7 C \cos (5 (e+f x))}{320 f}+\frac {C \cos (7 (e+f x))}{448 f} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.45, size = 60, normalized size = 0.82 \[ \frac {15 \, C \cos \left (f x + e\right )^{7} - 21 \, {\left (A + 3 \, C\right )} \cos \left (f x + e\right )^{5} + 35 \, {\left (2 \, A + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 105 \, {\left (A + C\right )} \cos \left (f x + e\right )}{105 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.17, size = 97, normalized size = 1.33 \[ \frac {C \cos \left (7 \, f x + 7 \, e\right )}{448 \, f} - \frac {{\left (4 \, A + 7 \, C\right )} \cos \left (5 \, f x + 5 \, e\right )}{320 \, f} + \frac {{\left (20 \, A + 21 \, C\right )} \cos \left (3 \, f x + 3 \, e\right )}{192 \, f} - \frac {{\left (16 \, A + 23 \, C\right )} \cos \left (f x + e\right )}{64 \, f} - \frac {3 \, {\left (2 \, A + C\right )} \cos \left (f x + e\right )}{16 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 - 1
/64*(16*A + 23*C)*cos(f*x + e)/f - 3/16*(2*A + C)*cos(f*x + e)/f

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maple [A]  time = 0.44, size = 74, normalized size = 1.01 \[ \frac {-\frac {C \left (\frac {16}{5}+\sin ^{6}\left (f x +e \right )+\frac {6 \left (\sin ^{4}\left (f x +e \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (f x +e \right )\right )}{5}\right ) \cos \left (f x +e \right )}{7}-\frac {A \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1/7*C*(16/5+sin(f*x+e)^6+6/5*sin(f*x+e)^4+8/5*sin(f*x+e)^2)*cos(f*x+e)-1/5*A*(8/3+sin(f*x+e)^4+4/3*sin(f
*x+e)^2)*cos(f*x+e))

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maxima [A]  time = 0.61, size = 60, normalized size = 0.82 \[ \frac {15 \, C \cos \left (f x + e\right )^{7} - 21 \, {\left (A + 3 \, C\right )} \cos \left (f x + e\right )^{5} + 35 \, {\left (2 \, A + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 105 \, {\left (A + C\right )} \cos \left (f x + e\right )}{105 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 13.15, size = 58, normalized size = 0.79 \[ \frac {\frac {C\,{\cos \left (e+f\,x\right )}^7}{7}+\left (-\frac {A}{5}-\frac {3\,C}{5}\right )\,{\cos \left (e+f\,x\right )}^5+\left (\frac {2\,A}{3}+C\right )\,{\cos \left (e+f\,x\right )}^3+\left (-A-C\right )\,\cos \left (e+f\,x\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 7.13, size = 153, normalized size = 2.10 \[ \begin {cases} - \frac {A \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 A \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {8 A \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac {C \sin ^{6}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 C \sin ^{4}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{f} - \frac {8 C \sin ^{2}{\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{5 f} - \frac {16 C \cos ^{7}{\left (e + f x \right )}}{35 f} & \text {for}\: f \neq 0 \\x \left (A + C \sin ^{2}{\relax (e )}\right ) \sin ^{5}{\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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