3.3.77 \(\int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx\) [277]

Optimal. Leaf size=111 \[ \frac {\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {4 \tan (e+f x)}{7 a^2 c^4 f}+\frac {4 \tan ^3(e+f x)}{21 a^2 c^4 f} \]

[Out]

1/7*sec(f*x+e)^3/a^2/f/(c^2-c^2*sin(f*x+e))^2+1/7*sec(f*x+e)^3/a^2/f/(c^4-c^4*sin(f*x+e))+4/7*tan(f*x+e)/a^2/c
^4/f+4/21*tan(f*x+e)^3/a^2/c^4/f

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Rubi [A]
time = 0.11, antiderivative size = 111, normalized size of antiderivative = 1.00, 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 = {2815, 2751, 3852} \begin {gather*} \frac {4 \tan ^3(e+f x)}{21 a^2 c^4 f}+\frac {4 \tan (e+f x)}{7 a^2 c^4 f}+\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^4),x]

[Out]

Sec[e + f*x]^3/(7*a^2*f*(c^2 - c^2*Sin[e + f*x])^2) + Sec[e + f*x]^3/(7*a^2*f*(c^4 - c^4*Sin[e + f*x])) + (4*T
an[e + f*x])/(7*a^2*c^4*f) + (4*Tan[e + f*x]^3)/(21*a^2*c^4*f)

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]))

Rule 3852

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))^2 (c-c \sin (e+f x))^4} \, dx &=\frac {\int \frac {\sec ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{a^2 c^2}\\ &=\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {5 \int \frac {\sec ^4(e+f x)}{c-c \sin (e+f x)} \, dx}{7 a^2 c^3}\\ &=\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {4 \int \sec ^4(e+f x) \, dx}{7 a^2 c^4}\\ &=\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac {4 \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{7 a^2 c^4 f}\\ &=\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {4 \tan (e+f x)}{7 a^2 c^4 f}+\frac {4 \tan ^3(e+f x)}{21 a^2 c^4 f}\\ \end {align*}

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Mathematica [A]
time = 0.64, size = 151, normalized size = 1.36 \begin {gather*} \frac {-210+512 \cos (e+f x)-255 \cos (2 (e+f x))+768 \cos (3 (e+f x))-30 \cos (4 (e+f x))+256 \cos (5 (e+f x))+15 \cos (6 (e+f x))+120 \sin (e+f x)+1088 \sin (2 (e+f x))+180 \sin (3 (e+f x))+128 \sin (4 (e+f x))+60 \sin (5 (e+f x))-64 \sin (6 (e+f x))}{5376 a^2 c^4 f (-1+\sin (e+f x))^4 (1+\sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^4),x]

[Out]

(-210 + 512*Cos[e + f*x] - 255*Cos[2*(e + f*x)] + 768*Cos[3*(e + f*x)] - 30*Cos[4*(e + f*x)] + 256*Cos[5*(e +
f*x)] + 15*Cos[6*(e + f*x)] + 120*Sin[e + f*x] + 1088*Sin[2*(e + f*x)] + 180*Sin[3*(e + f*x)] + 128*Sin[4*(e +
f*x)] + 60*Sin[5*(e + f*x)] - 64*Sin[6*(e + f*x)])/(5376*a^2*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x])^2
)

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Maple [A]
time = 0.38, size = 163, normalized size = 1.47

method	result	size
risch	\(-\frac {16 i \left (-8 i {\mathrm e}^{3 i \left (f x +e \right )}+14 \,{\mathrm e}^{4 i \left (f x +e \right )}-4 i {\mathrm e}^{i \left (f x +e \right )}+3 \,{\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{21 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3} f \,c^{4} a^{2}}\)	\(89\)
derivativedivides	\(\frac {-\frac {4}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {5}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {55}{12 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {23}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {13}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {1}{12 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{2} f \,c^{4}}\)	\(163\)
default	\(\frac {-\frac {4}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {5}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {55}{12 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {23}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {13}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {1}{12 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{2} f \,c^{4}}\)	\(163\)
norman	\(\frac {\frac {4 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {1}{14 a c f}-\frac {\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a c f}+\frac {5 \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a c f}-\frac {20 \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}-\frac {12 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{7 a c f}-\frac {68 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{21 a c f}+\frac {5 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}-\frac {13 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}+\frac {53 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{14 a c f}}{a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3} c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

2/f/a^2/c^4*(-2/7/(tan(1/2*f*x+1/2*e)-1)^7-1/(tan(1/2*f*x+1/2*e)-1)^6-2/(tan(1/2*f*x+1/2*e)-1)^5-5/2/(tan(1/2*
f*x+1/2*e)-1)^4-55/24/(tan(1/2*f*x+1/2*e)-1)^3-23/16/(tan(1/2*f*x+1/2*e)-1)^2-13/16/(tan(1/2*f*x+1/2*e)-1)-1/2
4/(tan(1/2*f*x+1/2*e)+1)^3+1/16/(tan(1/2*f*x+1/2*e)+1)^2-3/16/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 463 vs. \(2 (111) = 222\).
time = 0.33, size = 463, normalized size = 4.17 \begin {gather*} -\frac {2 \, {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {24 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {76 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {28 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {42 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {56 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {28 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {42 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - \frac {21 \, \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} - 6\right )}}{21 \, {\left (a^{2} c^{4} - \frac {4 \, a^{2} c^{4} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} c^{4} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {8 \, a^{2} c^{4} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {14 \, a^{2} c^{4} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {14 \, a^{2} c^{4} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {8 \, a^{2} c^{4} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {3 \, a^{2} c^{4} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac {4 \, a^{2} c^{4} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} - \frac {a^{2} c^{4} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

-2/21*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 24*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 76*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 + 28*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 42*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 56*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 - 28*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 42*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 -
21*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 6)/((a^2*c^4 - 4*a^2*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*c^4*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^2*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*a^2*c^4*sin(f*x + e)^
4/(cos(f*x + e) + 1)^4 + 14*a^2*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 8*a^2*c^4*sin(f*x + e)^7/(cos(f*x +
e) + 1)^7 - 3*a^2*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 4*a^2*c^4*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - a^
2*c^4*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*f)

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Fricas [A]
time = 0.31, size = 122, normalized size = 1.10 \begin {gather*} -\frac {16 \, \cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - {\left (8 \, \cos \left (f x + e\right )^{4} - 12 \, \cos \left (f x + e\right )^{2} - 5\right )} \sin \left (f x + e\right ) - 2}{21 \, {\left (a^{2} c^{4} f \cos \left (f x + e\right )^{5} + 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

-1/21*(16*cos(f*x + e)^4 - 8*cos(f*x + e)^2 - (8*cos(f*x + e)^4 - 12*cos(f*x + e)^2 - 5)*sin(f*x + e) - 2)/(a^
2*c^4*f*cos(f*x + e)^5 + 2*a^2*c^4*f*cos(f*x + e)^3*sin(f*x + e) - 2*a^2*c^4*f*cos(f*x + e)^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 2213 vs. \(2 (97) = 194\).
time = 12.96, size = 2213, normalized size = 19.94 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**4,x)

[Out]

Piecewise((-42*tan(e/2 + f*x/2)**9/(21*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 84*a**2*c**4*f*tan(e/2 + f*x/2)**9 +
63*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 168*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 294*a**2*c**4*f*tan(e/2 + f*x/2)**
6 + 294*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 168*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 63*a**2*c**4*f*tan(e/2 + f*x/2
)**2 + 84*a**2*c**4*f*tan(e/2 + f*x/2) - 21*a**2*c**4*f) + 84*tan(e/2 + f*x/2)**8/(21*a**2*c**4*f*tan(e/2 + f*
x/2)**10 - 84*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 63*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 168*a**2*c**4*f*tan(e/2 +
f*x/2)**7 - 294*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 294*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 168*a**2*c**4*f*tan(e
/2 + f*x/2)**3 - 63*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 84*a**2*c**4*f*tan(e/2 + f*x/2) - 21*a**2*c**4*f) - 56*t
an(e/2 + f*x/2)**7/(21*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 84*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 63*a**2*c**4*f*
tan(e/2 + f*x/2)**8 + 168*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 294*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 294*a**2*c**
4*f*tan(e/2 + f*x/2)**4 - 168*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 63*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 84*a**2*c
**4*f*tan(e/2 + f*x/2) - 21*a**2*c**4*f) - 112*tan(e/2 + f*x/2)**6/(21*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 84*a
**2*c**4*f*tan(e/2 + f*x/2)**9 + 63*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 168*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 29
4*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 294*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 168*a**2*c**4*f*tan(e/2 + f*x/2)**3
- 63*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 84*a**2*c**4*f*tan(e/2 + f*x/2) - 21*a**2*c**4*f) + 84*tan(e/2 + f*x/2)
**5/(21*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 84*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 63*a**2*c**4*f*tan(e/2 + f*x/2
)**8 + 168*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 294*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 294*a**2*c**4*f*tan(e/2 + f
*x/2)**4 - 168*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 63*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 84*a**2*c**4*f*tan(e/2 +
f*x/2) - 21*a**2*c**4*f) + 56*tan(e/2 + f*x/2)**4/(21*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 84*a**2*c**4*f*tan(e
/2 + f*x/2)**9 + 63*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 168*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 294*a**2*c**4*f*ta
n(e/2 + f*x/2)**6 + 294*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 168*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 63*a**2*c**4*f
*tan(e/2 + f*x/2)**2 + 84*a**2*c**4*f*tan(e/2 + f*x/2) - 21*a**2*c**4*f) - 152*tan(e/2 + f*x/2)**3/(21*a**2*c*
*4*f*tan(e/2 + f*x/2)**10 - 84*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 63*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 168*a**2
*c**4*f*tan(e/2 + f*x/2)**7 - 294*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 294*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 168*
a**2*c**4*f*tan(e/2 + f*x/2)**3 - 63*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 84*a**2*c**4*f*tan(e/2 + f*x/2) - 21*a*
*2*c**4*f) + 48*tan(e/2 + f*x/2)**2/(21*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 84*a**2*c**4*f*tan(e/2 + f*x/2)**9
+ 63*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 168*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 294*a**2*c**4*f*tan(e/2 + f*x/2)*
*6 + 294*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 168*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 63*a**2*c**4*f*tan(e/2 + f*x/
2)**2 + 84*a**2*c**4*f*tan(e/2 + f*x/2) - 21*a**2*c**4*f) + 6*tan(e/2 + f*x/2)/(21*a**2*c**4*f*tan(e/2 + f*x/2
)**10 - 84*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 63*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 168*a**2*c**4*f*tan(e/2 + f*
x/2)**7 - 294*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 294*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 168*a**2*c**4*f*tan(e/2
+ f*x/2)**3 - 63*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 84*a**2*c**4*f*tan(e/2 + f*x/2) - 21*a**2*c**4*f) - 12/(21*
a**2*c**4*f*tan(e/2 + f*x/2)**10 - 84*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 63*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 1
68*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 294*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 294*a**2*c**4*f*tan(e/2 + f*x/2)**4
- 168*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 63*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 84*a**2*c**4*f*tan(e/2 + f*x/2)
- 21*a**2*c**4*f), Ne(f, 0)), (x/((a*sin(e) + a)**2*(-c*sin(e) + c)**4), True))

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Giac [A]
time = 0.47, size = 161, normalized size = 1.45 \begin {gather*} -\frac {\frac {7 \, {\left (9 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8\right )}}{a^{2} c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {273 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1155 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 2450 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2870 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2037 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 791 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 152}{a^{2} c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}}}{168 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="giac")

[Out]

-1/168*(7*(9*tan(1/2*f*x + 1/2*e)^2 + 15*tan(1/2*f*x + 1/2*e) + 8)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) + 1)^3) + (2
73*tan(1/2*f*x + 1/2*e)^6 - 1155*tan(1/2*f*x + 1/2*e)^5 + 2450*tan(1/2*f*x + 1/2*e)^4 - 2870*tan(1/2*f*x + 1/2
*e)^3 + 2037*tan(1/2*f*x + 1/2*e)^2 - 791*tan(1/2*f*x + 1/2*e) + 152)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/
f

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Mupad [B]
time = 6.98, size = 119, normalized size = 1.07 \begin {gather*} -\frac {\frac {\sin \left (e+f\,x\right )}{3}+\frac {4\,\cos \left (2\,e+2\,f\,x\right )}{21}+\frac {2\,\cos \left (4\,e+4\,f\,x\right )}{21}+\frac {\sin \left (3\,e+3\,f\,x\right )}{14}-\frac {\sin \left (5\,e+5\,f\,x\right )}{42}}{a^2\,c^4\,f\,\left (\frac {\cos \left (5\,e+5\,f\,x\right )}{16}-\frac {3\,\cos \left (3\,e+3\,f\,x\right )}{16}-\frac {7\,\cos \left (e+f\,x\right )}{8}+\frac {\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {\sin \left (4\,e+4\,f\,x\right )}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^4),x)

[Out]

-(sin(e + f*x)/3 + (4*cos(2*e + 2*f*x))/21 + (2*cos(4*e + 4*f*x))/21 + sin(3*e + 3*f*x)/14 - sin(5*e + 5*f*x)/
42)/(a^2*c^4*f*(cos(5*e + 5*f*x)/16 - (3*cos(3*e + 3*f*x))/16 - (7*cos(e + f*x))/8 + sin(2*e + 2*f*x)/2 + sin(
4*e + 4*f*x)/4))

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