3.440 \(\int \frac {\cos ^4(e+f x) \sin (e+f x)}{(a+a \sin (e+f x))^6} \, dx\)

Optimal. Leaf size=58 \[ \frac {\cos ^5(e+f x)}{7 f (a \sin (e+f x)+a)^6}-\frac {6 \cos ^5(e+f x)}{35 a f (a \sin (e+f x)+a)^5} \]

[Out]

1/7*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^6-6/35*cos(f*x+e)^5/a/f/(a+a*sin(f*x+e))^5

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Rubi [A]  time = 0.11, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2859, 2671} \[ \frac {\cos ^5(e+f x)}{7 f (a \sin (e+f x)+a)^6}-\frac {6 \cos ^5(e+f x)}{35 a f (a \sin (e+f x)+a)^5} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^4*Sin[e + f*x])/(a + a*Sin[e + f*x])^6,x]

[Out]

Cos[e + f*x]^5/(7*f*(a + a*Sin[e + f*x])^6) - (6*Cos[e + f*x]^5)/(35*a*f*(a + a*Sin[e + f*x])^5)

Rule 2671

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

Rule 2859

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(e+f x) \sin (e+f x)}{(a+a \sin (e+f x))^6} \, dx &=\frac {\cos ^5(e+f x)}{7 f (a+a \sin (e+f x))^6}+\frac {6 \int \frac {\cos ^4(e+f x)}{(a+a \sin (e+f x))^5} \, dx}{7 a}\\ &=\frac {\cos ^5(e+f x)}{7 f (a+a \sin (e+f x))^6}-\frac {6 \cos ^5(e+f x)}{35 a f (a+a \sin (e+f x))^5}\\ \end {align*}

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Mathematica [B]  time = 1.26, size = 143, normalized size = 2.47 \[ \frac {1134 \sin \left (2 e+\frac {3 f x}{2}\right )-224 \sin \left (2 e+\frac {5 f x}{2}\right )+\sin \left (4 e+\frac {7 f x}{2}\right )+4585 \cos \left (e+\frac {f x}{2}\right )-2982 \cos \left (e+\frac {3 f x}{2}\right )-1148 \cos \left (3 e+\frac {5 f x}{2}\right )+197 \cos \left (3 e+\frac {7 f x}{2}\right )+2275 \sin \left (\frac {f x}{2}\right )}{4620 a^6 f \left (\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[e + f*x]^4*Sin[e + f*x])/(a + a*Sin[e + f*x])^6,x]

[Out]

(4585*Cos[e + (f*x)/2] - 2982*Cos[e + (3*f*x)/2] - 1148*Cos[3*e + (5*f*x)/2] + 197*Cos[3*e + (7*f*x)/2] + 2275
*Sin[(f*x)/2] + 1134*Sin[2*e + (3*f*x)/2] - 224*Sin[2*e + (5*f*x)/2] + Sin[4*e + (7*f*x)/2])/(4620*a^6*f*(Cos[
e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7)

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fricas [B]  time = 0.44, size = 195, normalized size = 3.36 \[ \frac {6 \, \cos \left (f x + e\right )^{4} - 11 \, \cos \left (f x + e\right )^{3} - 27 \, \cos \left (f x + e\right )^{2} + {\left (6 \, \cos \left (f x + e\right )^{3} + 17 \, \cos \left (f x + e\right )^{2} - 10 \, \cos \left (f x + e\right ) - 20\right )} \sin \left (f x + e\right ) + 10 \, \cos \left (f x + e\right ) + 20}{35 \, {\left (a^{6} f \cos \left (f x + e\right )^{4} - 3 \, a^{6} f \cos \left (f x + e\right )^{3} - 8 \, a^{6} f \cos \left (f x + e\right )^{2} + 4 \, a^{6} f \cos \left (f x + e\right ) + 8 \, a^{6} f - {\left (a^{6} f \cos \left (f x + e\right )^{3} + 4 \, a^{6} f \cos \left (f x + e\right )^{2} - 4 \, a^{6} f \cos \left (f x + e\right ) - 8 \, a^{6} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*sin(f*x+e)/(a+a*sin(f*x+e))^6,x, algorithm="fricas")

[Out]

1/35*(6*cos(f*x + e)^4 - 11*cos(f*x + e)^3 - 27*cos(f*x + e)^2 + (6*cos(f*x + e)^3 + 17*cos(f*x + e)^2 - 10*co
s(f*x + e) - 20)*sin(f*x + e) + 10*cos(f*x + e) + 20)/(a^6*f*cos(f*x + e)^4 - 3*a^6*f*cos(f*x + e)^3 - 8*a^6*f
*cos(f*x + e)^2 + 4*a^6*f*cos(f*x + e) + 8*a^6*f - (a^6*f*cos(f*x + e)^3 + 4*a^6*f*cos(f*x + e)^2 - 4*a^6*f*co
s(f*x + e) - 8*a^6*f)*sin(f*x + e))

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giac [A]  time = 0.29, size = 92, normalized size = 1.59 \[ -\frac {2 \, {\left (35 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 35 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 70 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 14 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}{35 \, a^{6} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*sin(f*x+e)/(a+a*sin(f*x+e))^6,x, algorithm="giac")

[Out]

-2/35*(35*tan(1/2*f*x + 1/2*e)^5 - 35*tan(1/2*f*x + 1/2*e)^4 + 70*tan(1/2*f*x + 1/2*e)^3 - 14*tan(1/2*f*x + 1/
2*e)^2 + 7*tan(1/2*f*x + 1/2*e) + 1)/(a^6*f*(tan(1/2*f*x + 1/2*e) + 1)^7)

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maple [A]  time = 0.46, size = 100, normalized size = 1.72 \[ \frac {\frac {224}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {64}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{7}}-\frac {32}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{6}}-\frac {32}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {12}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{f \,a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^4*sin(f*x+e)/(a+a*sin(f*x+e))^6,x)

[Out]

4/f/a^6*(56/5/(tan(1/2*f*x+1/2*e)+1)^5-1/2/(tan(1/2*f*x+1/2*e)+1)^2+16/7/(tan(1/2*f*x+1/2*e)+1)^7-8/(tan(1/2*f
*x+1/2*e)+1)^6-8/(tan(1/2*f*x+1/2*e)+1)^4+3/(tan(1/2*f*x+1/2*e)+1)^3)

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maxima [B]  time = 0.52, size = 269, normalized size = 4.64 \[ -\frac {2 \, {\left (\frac {7 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {14 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {70 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {35 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + 1\right )}}{35 \, {\left (a^{6} + \frac {7 \, a^{6} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {21 \, a^{6} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {35 \, a^{6} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {35 \, a^{6} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {21 \, a^{6} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {7 \, a^{6} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {a^{6} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*sin(f*x+e)/(a+a*sin(f*x+e))^6,x, algorithm="maxima")

[Out]

-2/35*(7*sin(f*x + e)/(cos(f*x + e) + 1) - 14*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 70*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 35*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 1)/((a^6 + 7*
a^6*sin(f*x + e)/(cos(f*x + e) + 1) + 21*a^6*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*a^6*sin(f*x + e)^3/(cos(
f*x + e) + 1)^3 + 35*a^6*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 21*a^6*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*
a^6*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^6*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)*f)

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mupad [B]  time = 8.89, size = 157, normalized size = 2.71 \[ -\frac {2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+7\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-14\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+70\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-35\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+35\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\right )}{35\,a^6\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(e + f*x)^4*sin(e + f*x))/(a + a*sin(e + f*x))^6,x)

[Out]

-(2*cos(e/2 + (f*x)/2)^2*(cos(e/2 + (f*x)/2)^5 + 35*sin(e/2 + (f*x)/2)^5 - 35*cos(e/2 + (f*x)/2)*sin(e/2 + (f*
x)/2)^4 + 7*cos(e/2 + (f*x)/2)^4*sin(e/2 + (f*x)/2) + 70*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^3 - 14*cos(e/
2 + (f*x)/2)^3*sin(e/2 + (f*x)/2)^2))/(35*a^6*f*(cos(e/2 + (f*x)/2) + sin(e/2 + (f*x)/2))^7)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**4*sin(f*x+e)/(a+a*sin(f*x+e))**6,x)

[Out]

Timed out

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