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

Optimal. Leaf size=73 \[ \frac {8 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 c f}-\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c^2 f} \]

[Out]

8/15*sec(f*x+e)^5*(c-c*sin(f*x+e))^(5/2)/a^3/c/f-2/3*sec(f*x+e)^5*(c-c*sin(f*x+e))^(7/2)/a^3/c^2/f

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Rubi [A]  time = 0.20, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ \frac {8 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 c f}-\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(15*a^3*c*f) - (2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(7/2))/(3*
a^3*c^2*f)

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]

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 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]))

Rubi steps

\begin {align*} \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx &=\frac {\int \sec ^6(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a^3 c^3}\\ &=-\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c^2 f}-\frac {4 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{3 a^3 c^2}\\ &=\frac {8 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 c f}-\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c^2 f}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 92, normalized size = 1.26 \[ \frac {2 c (5 \sin (e+f x)-1) \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{15 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + 5*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(15*a^3*f*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3)

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fricas [A]  time = 0.46, size = 74, normalized size = 1.01 \[ -\frac {2 \, {\left (5 \, c \sin \left (f x + e\right ) - c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/15*(5*c*sin(f*x + e) - c)*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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.82, size = 61, normalized size = 0.84 \[ -\frac {2 c^{2} \left (\sin \left (f x +e \right )-1\right ) \left (5 \sin \left (f x +e \right )-1\right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^3,x)

[Out]

-2/15*c^2/a^3*(sin(f*x+e)-1)/(1+sin(f*x+e))^2*(5*sin(f*x+e)-1)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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maxima [B]  time = 0.85, size = 331, normalized size = 4.53 \[ \frac {2 \, {\left (c^{\frac {3}{2}} - \frac {10 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {4 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {30 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {6 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {30 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {4 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {10 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {c^{\frac {3}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )}}{15 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/15*(c^(3/2) - 10*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 4*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 3
0*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 6*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 30*c^(3/2)*sin
(f*x + e)^5/(cos(f*x + e) + 1)^5 + 4*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 10*c^(3/2)*sin(f*x + e)^7/(
cos(f*x + e) + 1)^7 + c^(3/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)/((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)^(3/2))

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mupad [B]  time = 10.91, size = 355, normalized size = 4.86 \[ \frac {c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,8{}\mathrm {i}}{3\,a^3\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^2}+\frac {136\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}}{15\,a^3\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^3}-\frac {c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,64{}\mathrm {i}}{5\,a^3\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^4}-\frac {32\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}}{5\,a^3\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^3,x)

[Out]

(c*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*8i)/(3*a^3*f*(ex
p(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^2) + (136*c*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*
1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2))/(15*a^3*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^3) -
(c*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*64i)/(5*a^3*f*(e
xp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^4) - (32*c*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*
1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2))/(5*a^3*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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