3.287 \(\int \frac{1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=97 \[ \frac{6 \tan ^5(e+f x)}{35 a^3 c^4 f}+\frac{4 \tan ^3(e+f x)}{7 a^3 c^4 f}+\frac{6 \tan (e+f x)}{7 a^3 c^4 f}+\frac{\sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )} \]

[Out]

Sec[e + f*x]^5/(7*a^3*f*(c^4 - c^4*Sin[e + f*x])) + (6*Tan[e + f*x])/(7*a^3*c^4*f) + (4*Tan[e + f*x]^3)/(7*a^3
*c^4*f) + (6*Tan[e + f*x]^5)/(35*a^3*c^4*f)

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Rubi [A]  time = 0.119633, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2672, 3767} \[ \frac{6 \tan ^5(e+f x)}{35 a^3 c^4 f}+\frac{4 \tan ^3(e+f x)}{7 a^3 c^4 f}+\frac{6 \tan (e+f x)}{7 a^3 c^4 f}+\frac{\sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^4),x]

[Out]

Sec[e + f*x]^5/(7*a^3*f*(c^4 - c^4*Sin[e + f*x])) + (6*Tan[e + f*x])/(7*a^3*c^4*f) + (4*Tan[e + f*x]^3)/(7*a^3
*c^4*f) + (6*Tan[e + f*x]^5)/(35*a^3*c^4*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))^4} \, dx &=\frac{\int \frac{\sec ^6(e+f x)}{c-c \sin (e+f x)} \, dx}{a^3 c^3}\\ &=\frac{\sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{6 \int \sec ^6(e+f x) \, dx}{7 a^3 c^4}\\ &=\frac{\sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{6 \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{7 a^3 c^4 f}\\ &=\frac{\sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{6 \tan (e+f x)}{7 a^3 c^4 f}+\frac{4 \tan ^3(e+f x)}{7 a^3 c^4 f}+\frac{6 \tan ^5(e+f x)}{35 a^3 c^4 f}\\ \end{align*}

Mathematica [A]  time = 1.07088, size = 193, normalized size = 1.99 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (5120 \sin (e+f x)+125 \sin (2 (e+f x))+2560 \sin (3 (e+f x))+100 \sin (4 (e+f x))+512 \sin (5 (e+f x))+25 \sin (6 (e+f x))-500 \cos (e+f x)+1280 \cos (2 (e+f x))-250 \cos (3 (e+f x))+1024 \cos (4 (e+f x))-50 \cos (5 (e+f x))+256 \cos (6 (e+f x)))}{17920 f (a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^4),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-500*Cos[e + f*x] + 1280*Cos[2*(
e + f*x)] - 250*Cos[3*(e + f*x)] + 1024*Cos[4*(e + f*x)] - 50*Cos[5*(e + f*x)] + 256*Cos[6*(e + f*x)] + 5120*S
in[e + f*x] + 125*Sin[2*(e + f*x)] + 2560*Sin[3*(e + f*x)] + 100*Sin[4*(e + f*x)] + 512*Sin[5*(e + f*x)] + 25*
Sin[6*(e + f*x)]))/(17920*f*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^4)

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Maple [B]  time = 0.069, size = 193, normalized size = 2. \begin{align*} 2\,{\frac{1}{f{a}^{3}{c}^{4}} \left ( -1/7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-7}-1/2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-6}-{\frac{21}{20\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-{\frac{11}{8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-{\frac{11}{8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-{\frac{15}{16\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{21}{32\,\tan \left ( 1/2\,fx+e/2 \right ) -32}}-1/20\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-5}+1/8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-4}-1/4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3}+1/4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}-{\frac{11}{32\,\tan \left ( 1/2\,fx+e/2 \right ) +32}} \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))^4,x)

[Out]

2/f/a^3/c^4*(-1/7/(tan(1/2*f*x+1/2*e)-1)^7-1/2/(tan(1/2*f*x+1/2*e)-1)^6-21/20/(tan(1/2*f*x+1/2*e)-1)^5-11/8/(t
an(1/2*f*x+1/2*e)-1)^4-11/8/(tan(1/2*f*x+1/2*e)-1)^3-15/16/(tan(1/2*f*x+1/2*e)-1)^2-21/32/(tan(1/2*f*x+1/2*e)-
1)-1/20/(tan(1/2*f*x+1/2*e)+1)^5+1/8/(tan(1/2*f*x+1/2*e)+1)^4-1/4/(tan(1/2*f*x+1/2*e)+1)^3+1/4/(tan(1/2*f*x+1/
2*e)+1)^2-11/32/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.29869, size = 701, normalized size = 7.23 \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))^4,x, algorithm="maxima")

[Out]

2/35*(25*sin(f*x + e)/(cos(f*x + e) + 1) - 55*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 + 130*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 26*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 182*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 + 126*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 105*sin(f*x + e)^8/(cos(f*x + e) + 1)^
8 - 35*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 35*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 35*sin(f*x + e)^11/(co
s(f*x + e) + 1)^11 + 5)/((a^3*c^4 - 2*a^3*c^4*sin(f*x + e)/(cos(f*x + e) + 1) - 4*a^3*c^4*sin(f*x + e)^2/(cos(
f*x + e) + 1)^2 + 10*a^3*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)
^4 - 20*a^3*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 20*a^3*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 5*a^3*c
^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 10*a^3*c^4*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 4*a^3*c^4*sin(f*x +
e)^10/(cos(f*x + e) + 1)^10 + 2*a^3*c^4*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 - a^3*c^4*sin(f*x + e)^12/(cos(f
*x + e) + 1)^12)*f)

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Fricas [A]  time = 1.66689, size = 263, normalized size = 2.71 \begin{align*} -\frac{16 \, \cos \left (f x + e\right )^{6} - 8 \, \cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 2 \,{\left (8 \, \cos \left (f x + e\right )^{4} + 4 \, \cos \left (f x + e\right )^{2} + 3\right )} \sin \left (f x + e\right ) - 1}{35 \,{\left (a^{3} c^{4} f \cos \left (f x + e\right )^{5} \sin \left (f x + e\right ) - a^{3} c^{4} 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))^4,x, algorithm="fricas")

[Out]

-1/35*(16*cos(f*x + e)^6 - 8*cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 2*(8*cos(f*x + e)^4 + 4*cos(f*x + e)^2 + 3)*s
in(f*x + e) - 1)/(a^3*c^4*f*cos(f*x + e)^5*sin(f*x + e) - a^3*c^4*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))**4,x)

[Out]

Timed out

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Giac [B]  time = 2.27087, size = 255, normalized size = 2.63 \begin{align*} -\frac{\frac{7 \,{\left (55 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 180 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 250 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 160 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 43\right )}}{a^{3} c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}} + \frac{735 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 3360 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 7315 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 8820 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 6321 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2492 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 461}{a^{3} c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}}}{560 \, 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))^4,x, algorithm="giac")

[Out]

-1/560*(7*(55*tan(1/2*f*x + 1/2*e)^4 + 180*tan(1/2*f*x + 1/2*e)^3 + 250*tan(1/2*f*x + 1/2*e)^2 + 160*tan(1/2*f
*x + 1/2*e) + 43)/(a^3*c^4*(tan(1/2*f*x + 1/2*e) + 1)^5) + (735*tan(1/2*f*x + 1/2*e)^6 - 3360*tan(1/2*f*x + 1/
2*e)^5 + 7315*tan(1/2*f*x + 1/2*e)^4 - 8820*tan(1/2*f*x + 1/2*e)^3 + 6321*tan(1/2*f*x + 1/2*e)^2 - 2492*tan(1/
2*f*x + 1/2*e) + 461)/(a^3*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f