3.3.33 \(\int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx\) [233]

Optimal. Leaf size=60 \[ \frac {a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}+\frac {a \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^3} \]

[Out]

1/5*a*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^4+1/15*a*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^3

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Rubi [A]
time = 0.08, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2815, 2751, 2750} \begin {gather*} \frac {a \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^3}+\frac {a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c*Cos[e + f*x]^3)/(5*f*(c - c*Sin[e + f*x])^4) + (a*Cos[e + f*x]^3)/(15*f*(c - c*Sin[e + f*x])^3)

Rule 2750

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2751

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

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*} \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx &=(a c) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac {a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}+\frac {1}{5} a \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^3} \, dx\\ &=\frac {a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}+\frac {a \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^3}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 96, normalized size = 1.60 \begin {gather*} \frac {a \left (15 \cos \left (e+\frac {f x}{2}\right )-5 \cos \left (e+\frac {3 f x}{2}\right )+5 \sin \left (\frac {f x}{2}\right )+\sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 c^3 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(15*Cos[e + (f*x)/2] - 5*Cos[e + (3*f*x)/2] + 5*Sin[(f*x)/2] + Sin[2*e + (5*f*x)/2]))/(30*c^3*f*(Cos[e/2] -
Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5)

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Maple [A]
time = 0.32, size = 86, normalized size = 1.43

method	result	size
risch	\(\frac {2 i a \left (5 i {\mathrm e}^{2 i \left (f x +e \right )}+15 \,{\mathrm e}^{3 i \left (f x +e \right )}+i-5 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{15 f \,c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5}}\)	\(62\)
derivativedivides	\(\frac {2 a \left (-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {14}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\right )}{f \,c^{3}}\)	\(86\)
default	\(\frac {2 a \left (-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {14}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\right )}{f \,c^{3}}\)	\(86\)
norman	\(\frac {\frac {2 a \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {8 a}{15 c f}-\frac {2 a \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 c f}+\frac {8 a \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {16 a \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {58 a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/f*a/c^3*(-1/(tan(1/2*f*x+1/2*e)-1)-8/5/(tan(1/2*f*x+1/2*e)-1)^5-3/(tan(1/2*f*x+1/2*e)-1)^2-4/(tan(1/2*f*x+1/
2*e)-1)^4-14/3/(tan(1/2*f*x+1/2*e)-1)^3)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (62) = 124\).
time = 0.31, size = 423, normalized size = 7.05 \begin {gather*} -\frac {2 \, {\left (\frac {a {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 7\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {3 \, a {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(a*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos
(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 7)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) +
10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4
/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*a*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 5*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(co
s(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*
c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (62) = 124\).
time = 0.33, size = 166, normalized size = 2.77 \begin {gather*} -\frac {a \cos \left (f x + e\right )^{3} - 2 \, a \cos \left (f x + e\right )^{2} + 3 \, a \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} + 3 \, a \cos \left (f x + e\right ) + 6 \, a\right )} \sin \left (f x + e\right ) + 6 \, a}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(a*cos(f*x + e)^3 - 2*a*cos(f*x + e)^2 + 3*a*cos(f*x + e) + (a*cos(f*x + e)^2 + 3*a*cos(f*x + e) + 6*a)*
sin(f*x + e) + 6*a)/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*c
os(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (51) = 102\).
time = 2.90, size = 571, normalized size = 9.52 \begin {gather*} \begin {cases} - \frac {30 a \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} + \frac {30 a \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} - \frac {50 a \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} + \frac {10 a \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} - \frac {8 a}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )}{\left (- c \sin {\left (e \right )} + c\right )^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-30*a*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3
*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 30*a*tan(e
/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3
- 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 50*a*tan(e/2 + f*x/2)**2/(15*c**
3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2
+ f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 10*a*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2 + f*x/2)**5 -
75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*t
an(e/2 + f*x/2) - 15*c**3*f) - 8*a/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f
*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f), Ne(f, 0)), (x
*(a*sin(e) + a)/(-c*sin(e) + c)**3, True))

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Giac [A]
time = 0.43, size = 84, normalized size = 1.40 \begin {gather*} -\frac {2 \, {\left (15 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 25 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a\right )}}{15 \, c^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*a*tan(1/2*f*x + 1/2*e)^4 - 15*a*tan(1/2*f*x + 1/2*e)^3 + 25*a*tan(1/2*f*x + 1/2*e)^2 - 5*a*tan(1/2*f
*x + 1/2*e) + 4*a)/(c^3*f*(tan(1/2*f*x + 1/2*e) - 1)^5)

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Mupad [B]
time = 7.10, size = 136, normalized size = 2.27 \begin {gather*} \frac {2\,a\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-5\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+25\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-15\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+15\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{15\,c^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*cos(e/2 + (f*x)/2)*(4*cos(e/2 + (f*x)/2)^4 + 15*sin(e/2 + (f*x)/2)^4 - 15*cos(e/2 + (f*x)/2)*sin(e/2 + (f
*x)/2)^3 - 5*cos(e/2 + (f*x)/2)^3*sin(e/2 + (f*x)/2) + 25*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^2))/(15*c^3*
f*(cos(e/2 + (f*x)/2) - sin(e/2 + (f*x)/2))^5)

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