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

Optimal. Leaf size=132 \[ \frac{64 c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac{256 c^3 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{5 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac{8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.344804, antiderivative size = 132, 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} \[ \frac{64 c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac{256 c^3 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{5 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac{8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*c^3*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(5*a*f) + (64*c^2*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(5
*a*f) + (8*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(5*a*f) + (2*Sec[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(5
*a*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)} \, dx &=\frac{\int \sec ^2(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a c}\\ &=\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac{12 \int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a}\\ &=\frac{8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac{(32 c) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a}\\ &=\frac{64 c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}+\frac{8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac{\left (128 c^2\right ) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{5 a}\\ &=-\frac{256 c^3 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{5 a f}+\frac{64 c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}+\frac{8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 2.12381, size = 112, normalized size = 0.85 \[ \frac{c^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (-175 \sin (e+f x)+\sin (3 (e+f x))-14 \cos (2 (e+f x))-350)}{10 a f (\sin (e+f x)+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-350 - 14*Cos[2*(e + f*x)] - 175*Sin[e +
f*x] + Sin[3*(e + f*x)]))/(10*a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x]))

________________________________________________________________________________________

Maple [A]  time = 0.479, size = 69, normalized size = 0.5 \begin{align*}{\frac{2\,{c}^{4} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}-7\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+43\,\sin \left ( fx+e \right ) +91 \right ) }{5\,af\cos \left ( fx+e \right ) }{\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)),x)

[Out]

2/5*c^4/a*(-1+sin(f*x+e))*(sin(f*x+e)^3-7*sin(f*x+e)^2+43*sin(f*x+e)+91)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [B]  time = 1.89519, size = 321, normalized size = 2.43 \begin{align*} \frac{2 \,{\left (91 \, c^{\frac{7}{2}} + \frac{86 \, c^{\frac{7}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{336 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{266 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{490 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{266 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{336 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{86 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac{91 \, c^{\frac{7}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )}}{5 \,{\left (a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\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)),x, algorithm="maxima")

[Out]

2/5*(91*c^(7/2) + 86*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 336*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 266*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 490*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 266*c^(
7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 336*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 86*c^(7/2)*sin(f*
x + e)^7/(cos(f*x + e) + 1)^7 + 91*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)/((a + a*sin(f*x + e)/(cos(f*x
+ e) + 1))*f*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(7/2))

________________________________________________________________________________________

Fricas [A]  time = 1.03254, size = 173, normalized size = 1.31 \begin{align*} -\frac{2 \,{\left (7 \, c^{3} \cos \left (f x + e\right )^{2} + 84 \, c^{3} -{\left (c^{3} \cos \left (f x + e\right )^{2} - 44 \, c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{5 \, a f \cos \left (f x + e\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)),x, algorithm="fricas")

[Out]

-2/5*(7*c^3*cos(f*x + e)^2 + 84*c^3 - (c^3*cos(f*x + e)^2 - 44*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a
*f*cos(f*x + e))

________________________________________________________________________________________

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)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.78945, size = 575, normalized size = 4.36 \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)),x, algorithm="giac")

[Out]

-1/20*(4*(40*sqrt(2)*c^10 - 11*sqrt(2)*a^6*c + 22*a^6*c)*sgn(tan(1/2*f*x + 1/2*e) - 1)/(sqrt(2)*a*c^(13/2) - a
*c^(13/2)) - 640*((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*c^4*sgn(tan(1/2*f*x + 1/
2*e) - 1) - c^(9/2)*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)*a) - (51*a^5
*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^3 + (35*a^5*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^3 + (90*a^5*sgn(tan(1/2*f*x + 1/2
*e) - 1)/c^3 + (90*a^5*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^3 + (51*a^5*sgn(tan(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x +
1/2*e)/c^3 + 35*a^5*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^3)*tan(1/2*f*x + 1/2*e))*tan(1/2*f*x + 1/2*e))*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)^(5/2))/f