3.325 \(\int \frac{(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=136 \[ \frac{256 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c f}+\frac{8 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 f}-\frac{64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f} \]

[Out]

(256*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*a^2*f) - (64*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(5/2
))/(a^2*f) + (8*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(a^2*f) + (2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9
/2))/(3*a^2*c*f)

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Rubi [A]  time = 0.332457, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ \frac{256 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c f}+\frac{8 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 f}-\frac{64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^2,x]

[Out]

(256*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*a^2*f) - (64*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(5/2
))/(a^2*f) + (8*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(a^2*f) + (2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9
/2))/(3*a^2*c*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 2674

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 - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), 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]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2673

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 - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin{align*} \int \frac{(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx &=\frac{\int \sec ^4(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{a^2 c^2}\\ &=\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c f}+\frac{4 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a^2 c}\\ &=\frac{8 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c f}+\frac{32 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a^2}\\ &=-\frac{64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac{8 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c f}-\frac{(128 c) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{a^2}\\ &=\frac{256 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac{64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac{8 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c f}\\ \end{align*}

Mathematica [A]  time = 1.18221, size = 112, normalized size = 0.82 \[ \frac{c^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (273 \sin (e+f x)+\sin (3 (e+f x))-30 \cos (2 (e+f x))+210)}{6 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^2,x]

[Out]

(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(210 - 30*Cos[2*(e + f*x)] + 273*Sin[e + f
*x] + Sin[3*(e + f*x)]))/(6*a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)

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Maple [A]  time = 0.581, size = 79, normalized size = 0.6 \begin{align*}{\frac{2\,{c}^{4} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}-15\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}-69\,\sin \left ( fx+e \right ) -45 \right ) }{3\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^2,x)

[Out]

2/3*c^4/a^2*(-1+sin(f*x+e))/(1+sin(f*x+e))*(sin(f*x+e)^3-15*sin(f*x+e)^2-69*sin(f*x+e)-45)/cos(f*x+e)/(c-c*sin
(f*x+e))^(1/2)/f

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Maxima [B]  time = 1.86055, size = 451, normalized size = 3.32 \begin{align*} -\frac{2 \,{\left (45 \, c^{\frac{7}{2}} + \frac{138 \, c^{\frac{7}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{285 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{544 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{630 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{812 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{630 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{544 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac{285 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac{138 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac{45 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}\right )}}{3 \,{\left (a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} f{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/3*(45*c^(7/2) + 138*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 285*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 + 544*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 630*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 812*c
^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 630*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 544*c^(7/2)*sin
(f*x + e)^7/(cos(f*x + e) + 1)^7 + 285*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 138*c^(7/2)*sin(f*x + e)^
9/(cos(f*x + e) + 1)^9 + 45*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)/((a^2 + 3*a^2*sin(f*x + e)/(cos(f*x
+ e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*f*(sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 1)^(7/2))

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Fricas [A]  time = 1.06585, size = 223, normalized size = 1.64 \begin{align*} -\frac{2 \,{\left (15 \, c^{3} \cos \left (f x + e\right )^{2} - 60 \, c^{3} -{\left (c^{3} \cos \left (f x + e\right )^{2} + 68 \, c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{3 \,{\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-2/3*(15*c^3*cos(f*x + e)^2 - 60*c^3 - (c^3*cos(f*x + e)^2 + 68*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(
a^2*f*cos(f*x + e)*sin(f*x + e) + a^2*f*cos(f*x + e))

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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((c-c*sin(f*x+e))**(7/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 2.67561, size = 802, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/3*(8*(14*sqrt(2)*a^8*sqrt(c) - 20*a^8*sqrt(c) - 7*sqrt(2)*c^(13/2) + 9*c^(13/2))*sgn(tan(1/2*f*x + 1/2*e) -
1)/(5*sqrt(2)*a^2*c^3 - 7*a^2*c^3) + (17*a^6*sgn(tan(1/2*f*x + 1/2*e) - 1)/c + (15*a^6*sgn(tan(1/2*f*x + 1/2*
e) - 1)/c + (17*a^6*sgn(tan(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x + 1/2*e)/c + 15*a^6*sgn(tan(1/2*f*x + 1/2*e) - 1
)/c)*tan(1/2*f*x + 1/2*e))*tan(1/2*f*x + 1/2*e))/(c*tan(1/2*f*x + 1/2*e)^2 + c)^(3/2) + 16*(3*(sqrt(c)*tan(1/2
*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) + 21*(sqrt(c)*tan(1/2*
f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 14*(sqrt(c)*tan(1
/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 42*(sqrt(c)*tan(1/
2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 39*(sqrt(c)*ta
n(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1) - 7*c^(13/2)*sgn(ta
n(1/2*f*x + 1/2*e) - 1))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(sqrt(c)*
tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)^3*a^2))/f