3.333 \(\int \frac{(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=134 \[ \frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{a^3 c^2 f}-\frac{24 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c f}+\frac{64 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{a^3 f}-\frac{256 c \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f} \]

[Out]

(-256*c*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*f) + (64*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(7/2))/
(a^3*f) - (24*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/(a^3*c*f) + (2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(1
1/2))/(a^3*c^2*f)

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Rubi [A]  time = 0.330242, antiderivative size = 134, 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} \[ \frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{a^3 c^2 f}-\frac{24 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c f}+\frac{64 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{a^3 f}-\frac{256 c \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^3,x]

[Out]

(-256*c*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*f) + (64*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(7/2))/
(a^3*f) - (24*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/(a^3*c*f) + (2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(1
1/2))/(a^3*c^2*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))^3} \, dx &=\frac{\int \sec ^6(e+f x) (c-c \sin (e+f x))^{13/2} \, dx}{a^3 c^3}\\ &=\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{a^3 c^2 f}+\frac{12 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{a^3 c^2}\\ &=-\frac{24 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c f}+\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{a^3 c^2 f}-\frac{96 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a^3 c}\\ &=\frac{64 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{a^3 f}-\frac{24 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c f}+\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{a^3 c^2 f}+\frac{128 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a^3}\\ &=-\frac{256 c \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}+\frac{64 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{a^3 f}-\frac{24 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c f}+\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{a^3 c^2 f}\\ \end{align*}

Mathematica [A]  time = 1.24318, size = 114, normalized size = 0.85 \[ \frac{c^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (-235 \sin (e+f x)+5 \sin (3 (e+f x))+90 \cos (2 (e+f x))-182)}{10 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^3,x]

[Out]

(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-182 + 90*Cos[2*(e + f*x)] - 235*Sin[e +
f*x] + 5*Sin[3*(e + f*x)]))/(10*a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3)

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Maple [A]  time = 0.622, size = 81, normalized size = 0.6 \begin{align*}{\frac{2\,{c}^{4} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}+45\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+55\,\sin \left ( fx+e \right ) +23 \right ) }{5\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\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))^3,x)

[Out]

2/5*c^4/a^3*(-1+sin(f*x+e))/(1+sin(f*x+e))^2*(5*sin(f*x+e)^3+45*sin(f*x+e)^2+55*sin(f*x+e)+23)/cos(f*x+e)/(c-c
*sin(f*x+e))^(1/2)/f

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Maxima [B]  time = 1.83459, size = 575, normalized size = 4.29 \begin{align*} \frac{2 \,{\left (23 \, c^{\frac{7}{2}} + \frac{110 \, c^{\frac{7}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{318 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{590 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{1065 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{1220 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{1540 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{1220 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac{1065 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac{590 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac{318 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}} + \frac{110 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{11}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{11}} + \frac{23 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{12}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{12}}\right )}}{5 \,{\left (a^{3} + \frac{5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\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))^3,x, algorithm="maxima")

[Out]

2/5*(23*c^(7/2) + 110*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 318*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^
2 + 590*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1065*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 1220*
c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 1540*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1220*c^(7/2)*
sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 1065*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 590*c^(7/2)*sin(f*x +
e)^9/(cos(f*x + e) + 1)^9 + 318*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 110*c^(7/2)*sin(f*x + e)^11/(
cos(f*x + e) + 1)^11 + 23*c^(7/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12)/((a^3 + 5*a^3*sin(f*x + e)/(cos(f*x +
e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin
(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*f*(sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 1)^(7/2))

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Fricas [A]  time = 1.1173, size = 262, normalized size = 1.96 \begin{align*} -\frac{2 \,{\left (45 \, c^{3} \cos \left (f x + e\right )^{2} - 68 \, c^{3} + 5 \,{\left (c^{3} \cos \left (f x + e\right )^{2} - 12 \, c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{5 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

-2/5*(45*c^3*cos(f*x + e)^2 - 68*c^3 + 5*(c^3*cos(f*x + e)^2 - 12*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)
/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*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))**3,x)

[Out]

Timed out

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Giac [B]  time = 2.17722, size = 975, normalized size = 7.28 \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))^3,x, algorithm="giac")

[Out]

2/5*(2*(67*sqrt(2)*c^(7/2) - 95*c^(7/2))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(29*sqrt(2)*a^3 - 41*a^3) + 5*(c^4*sgn(
tan(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x + 1/2*e)/a^3 + c^4*sgn(tan(1/2*f*x + 1/2*e) - 1)/a^3)/sqrt(c*tan(1/2*f*x
+ 1/2*e)^2 + c) + 4*(5*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^9*c^4*sgn(tan(1/2*
f*x + 1/2*e) - 1) + 15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*c^(9/2)*sgn(tan(1
/2*f*x + 1/2*e) - 1) + 100*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*c^5*sgn(tan(1
/2*f*x + 1/2*e) - 1) - 60*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*c^(11/2)*sgn(t
an(1/2*f*x + 1/2*e) - 1) - 306*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*c^6*sgn(t
an(1/2*f*x + 1/2*e) - 1) + 210*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*c^(13/2)*
sgn(tan(1/2*f*x + 1/2*e) - 1) + 260*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c^7*
sgn(tan(1/2*f*x + 1/2*e) - 1) - 300*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*c^(1
5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 85*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*c^
8*sgn(tan(1/2*f*x + 1/2*e) - 1) - 9*c^(17/2)*sgn(tan(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)^5*a^3))/f