3.288 \(\int \frac{1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx\)

Optimal. Leaf size=131 \[ \frac{2 \tan ^5(e+f x)}{15 a^3 c^5 f}+\frac{4 \tan ^3(e+f x)}{9 a^3 c^5 f}+\frac{2 \tan (e+f x)}{3 a^3 c^5 f}+\frac{\sec ^5(e+f x)}{9 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{\sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2} \]

[Out]

Sec[e + f*x]^5/(9*a^3*c^3*f*(c - c*Sin[e + f*x])^2) + Sec[e + f*x]^5/(9*a^3*f*(c^5 - c^5*Sin[e + f*x])) + (2*T
an[e + f*x])/(3*a^3*c^5*f) + (4*Tan[e + f*x]^3)/(9*a^3*c^5*f) + (2*Tan[e + f*x]^5)/(15*a^3*c^5*f)

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Rubi [A]  time = 0.170823, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2672, 3767} \[ \frac{2 \tan ^5(e+f x)}{15 a^3 c^5 f}+\frac{4 \tan ^3(e+f x)}{9 a^3 c^5 f}+\frac{2 \tan (e+f x)}{3 a^3 c^5 f}+\frac{\sec ^5(e+f x)}{9 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{\sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^5),x]

[Out]

Sec[e + f*x]^5/(9*a^3*c^3*f*(c - c*Sin[e + f*x])^2) + Sec[e + f*x]^5/(9*a^3*f*(c^5 - c^5*Sin[e + f*x])) + (2*T
an[e + f*x])/(3*a^3*c^5*f) + (4*Tan[e + f*x]^3)/(9*a^3*c^5*f) + (2*Tan[e + f*x]^5)/(15*a^3*c^5*f)

Rule 2736

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]))

Rule 2672

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*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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx &=\frac{\int \frac{\sec ^6(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{a^3 c^3}\\ &=\frac{\sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac{7 \int \frac{\sec ^6(e+f x)}{c-c \sin (e+f x)} \, dx}{9 a^3 c^4}\\ &=\frac{\sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac{\sec ^5(e+f x)}{9 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{2 \int \sec ^6(e+f x) \, dx}{3 a^3 c^5}\\ &=\frac{\sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac{\sec ^5(e+f x)}{9 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}-\frac{2 \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{3 a^3 c^5 f}\\ &=\frac{\sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac{\sec ^5(e+f x)}{9 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{2 \tan (e+f x)}{3 a^3 c^5 f}+\frac{4 \tan ^3(e+f x)}{9 a^3 c^5 f}+\frac{2 \tan ^5(e+f x)}{15 a^3 c^5 f}\\ \end{align*}

Mathematica [A]  time = 1.32107, size = 213, normalized size = 1.63 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (46080 \sin (e+f x)+3500 \sin (2 (e+f x))+19456 \sin (3 (e+f x))+2800 \sin (4 (e+f x))+1024 \sin (5 (e+f x))+700 \sin (6 (e+f x))-1024 \sin (7 (e+f x))-7875 \cos (e+f x)+20480 \cos (2 (e+f x))-3325 \cos (3 (e+f x))+16384 \cos (4 (e+f x))-175 \cos (5 (e+f x))+4096 \cos (6 (e+f x))+175 \cos (7 (e+f x)))}{184320 f (a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^5),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-7875*Cos[e + f*x] + 20480*Cos[2
*(e + f*x)] - 3325*Cos[3*(e + f*x)] + 16384*Cos[4*(e + f*x)] - 175*Cos[5*(e + f*x)] + 4096*Cos[6*(e + f*x)] +
175*Cos[7*(e + f*x)] + 46080*Sin[e + f*x] + 3500*Sin[2*(e + f*x)] + 19456*Sin[3*(e + f*x)] + 2800*Sin[4*(e + f
*x)] + 1024*Sin[5*(e + f*x)] + 700*Sin[6*(e + f*x)] - 1024*Sin[7*(e + f*x)]))/(184320*f*(a + a*Sin[e + f*x])^3
*(c - c*Sin[e + f*x])^5)

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Maple [A]  time = 0.081, size = 223, normalized size = 1.7 \begin{align*} 2\,{\frac{1}{f{c}^{5}{a}^{3}} \left ( -2/9\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-9}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-8}-5/2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-7}-{\frac{49}{12\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-{\frac{49}{10\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-{\frac{35}{8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-{\frac{49}{16\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-{\frac{51}{32\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{99}{128\,\tan \left ( 1/2\,fx+e/2 \right ) -128}}-1/40\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-5}+1/16\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-4}-{\frac{13}{96\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}+{\frac{9}{64\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-{\frac{29}{128\,\tan \left ( 1/2\,fx+e/2 \right ) +128}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x)

[Out]

2/f/c^5/a^3*(-2/9/(tan(1/2*f*x+1/2*e)-1)^9-1/(tan(1/2*f*x+1/2*e)-1)^8-5/2/(tan(1/2*f*x+1/2*e)-1)^7-49/12/(tan(
1/2*f*x+1/2*e)-1)^6-49/10/(tan(1/2*f*x+1/2*e)-1)^5-35/8/(tan(1/2*f*x+1/2*e)-1)^4-49/16/(tan(1/2*f*x+1/2*e)-1)^
3-51/32/(tan(1/2*f*x+1/2*e)-1)^2-99/128/(tan(1/2*f*x+1/2*e)-1)-1/40/(tan(1/2*f*x+1/2*e)+1)^5+1/16/(tan(1/2*f*x
+1/2*e)+1)^4-13/96/(tan(1/2*f*x+1/2*e)+1)^3+9/64/(tan(1/2*f*x+1/2*e)+1)^2-29/128/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.28406, size = 824, normalized size = 6.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="maxima")

[Out]

2/45*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 80*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 190*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 + 50*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 269*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 96*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 + 516*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 354*sin(f*x + e)^8/(cos(f*x + e) + 1)^8
- 69*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 240*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 30*sin(f*x + e)^11/(co
s(f*x + e) + 1)^11 - 90*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 45*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 + 10)
/((a^3*c^5 - 4*a^3*c^5*sin(f*x + e)/(cos(f*x + e) + 1) + a^3*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 16*a^3*
c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 19*a^3*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 20*a^3*c^5*sin(f*x
+ e)^5/(cos(f*x + e) + 1)^5 + 45*a^3*c^5*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 45*a^3*c^5*sin(f*x + e)^8/(cos(
f*x + e) + 1)^8 + 20*a^3*c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 19*a^3*c^5*sin(f*x + e)^10/(cos(f*x + e) +
1)^10 - 16*a^3*c^5*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 - a^3*c^5*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 4*a
^3*c^5*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 - a^3*c^5*sin(f*x + e)^14/(cos(f*x + e) + 1)^14)*f)

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Fricas [A]  time = 1.61598, size = 333, normalized size = 2.54 \begin{align*} -\frac{32 \, \cos \left (f x + e\right )^{6} - 16 \, \cos \left (f x + e\right )^{4} - 4 \, \cos \left (f x + e\right )^{2} -{\left (16 \, \cos \left (f x + e\right )^{6} - 24 \, \cos \left (f x + e\right )^{4} - 10 \, \cos \left (f x + e\right )^{2} - 7\right )} \sin \left (f x + e\right ) - 2}{45 \,{\left (a^{3} c^{5} f \cos \left (f x + e\right )^{7} + 2 \, a^{3} c^{5} f \cos \left (f x + e\right )^{5} \sin \left (f x + e\right ) - 2 \, a^{3} c^{5} f \cos \left (f x + e\right )^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="fricas")

[Out]

-1/45*(32*cos(f*x + e)^6 - 16*cos(f*x + e)^4 - 4*cos(f*x + e)^2 - (16*cos(f*x + e)^6 - 24*cos(f*x + e)^4 - 10*
cos(f*x + e)^2 - 7)*sin(f*x + e) - 2)/(a^3*c^5*f*cos(f*x + e)^7 + 2*a^3*c^5*f*cos(f*x + e)^5*sin(f*x + e) - 2*
a^3*c^5*f*cos(f*x + e)^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**5,x)

[Out]

Timed out

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Giac [A]  time = 2.02414, size = 293, normalized size = 2.24 \begin{align*} -\frac{\frac{3 \,{\left (435 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 1470 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2060 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1330 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 353\right )}}{a^{3} c^{5}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}} + \frac{4455 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 26460 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 78120 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 137340 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 157374 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 118356 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 57744 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 16596 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2339}{a^{3} c^{5}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{9}}}{2880 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x, algorithm="giac")

[Out]

-1/2880*(3*(435*tan(1/2*f*x + 1/2*e)^4 + 1470*tan(1/2*f*x + 1/2*e)^3 + 2060*tan(1/2*f*x + 1/2*e)^2 + 1330*tan(
1/2*f*x + 1/2*e) + 353)/(a^3*c^5*(tan(1/2*f*x + 1/2*e) + 1)^5) + (4455*tan(1/2*f*x + 1/2*e)^8 - 26460*tan(1/2*
f*x + 1/2*e)^7 + 78120*tan(1/2*f*x + 1/2*e)^6 - 137340*tan(1/2*f*x + 1/2*e)^5 + 157374*tan(1/2*f*x + 1/2*e)^4
- 118356*tan(1/2*f*x + 1/2*e)^3 + 57744*tan(1/2*f*x + 1/2*e)^2 - 16596*tan(1/2*f*x + 1/2*e) + 2339)/(a^3*c^5*(
tan(1/2*f*x + 1/2*e) - 1)^9))/f