3.4.26 \(\int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx\) [326]

Optimal. Leaf size=100 \[ \frac {64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac {16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f} \]

[Out]

64/3*c*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^2/f-16*sec(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/a^2/f+2*sec(f*x+e)^3*(
c-c*sin(f*x+e))^(7/2)/a^2/c/f

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Rubi [A]
time = 0.18, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2815, 2753, 2752} \begin {gather*} \frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}-\frac {16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac {64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^2,x]

[Out]

(64*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*a^2*f) - (16*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(5/2))/(a
^2*f) + (2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(a^2*c*f)

Rule 2752

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 - 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 2753

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 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]))

Rubi steps

\begin {align*} \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx &=\frac {\int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a^2 c^2}\\ &=\frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}+\frac {8 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a^2 c}\\ &=-\frac {16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}-\frac {32 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{a^2}\\ &=\frac {64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac {16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 104, normalized size = 1.04 \begin {gather*} \frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (25-3 \cos (2 (e+f x))+36 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{3 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^2,x]

[Out]

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(25 - 3*Cos[2*(e + f*x)] + 36*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]
])/(3*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 = 2.26, size = 71, normalized size = 0.71

method	result	size
default	\(-\frac {2 c^{3} \left (\sin \left (f x +e \right )-1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )+18 \sin \left (f x +e \right )+11\right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

-2/3*c^3/a^2*(sin(f*x+e)-1)/(1+sin(f*x+e))*(3*sin(f*x+e)^2+18*sin(f*x+e)+11)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)
/f

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (98) = 196\).
time = 0.51, size = 312, normalized size = 3.12 \begin {gather*} -\frac {2 \, {\left (11 \, c^{\frac {5}{2}} + \frac {36 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {56 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {108 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {90 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {108 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {56 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {36 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {11 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\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 {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/3*(11*c^(5/2) + 36*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 56*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 108*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 90*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 108*c^(5
/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 56*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 36*c^(5/2)*sin(f*x
+ e)^7/(cos(f*x + e) + 1)^7 + 11*c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)/((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)^(5/2))

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Fricas [A]
time = 0.34, size = 82, normalized size = 0.82 \begin {gather*} -\frac {2 \, {\left (3 \, c^{2} \cos \left (f x + e\right )^{2} - 18 \, c^{2} \sin \left (f x + e\right ) - 14 \, c^{2}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-2/3*(3*c^2*cos(f*x + e)^2 - 18*c^2*sin(f*x + e) - 14*c^2)*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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (98) = 196\).
time = 0.57, size = 237, normalized size = 2.37 \begin {gather*} \frac {4 \, \sqrt {2} \sqrt {c} {\left (\frac {3 \, c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}} - \frac {5 \, c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {12 \, c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {3 \, c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}}{a^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{3}}\right )}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

4/3*sqrt(2)*sqrt(c)*(3*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a^2*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos
(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)) - (5*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*c^2*(cos(-1/4*pi + 1/
2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 3*c^2*(cos(-1/4
*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(a^2
*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)^3))/f

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Mupad [B]
time = 11.68, size = 360, normalized size = 3.60 \begin {gather*} \frac {\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 )}\,\left (\frac {2\,c^2}{a^2\,f}-\frac {c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^2\,f}\right )}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}}+\frac {16\,c^2\,{\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 )}}{a^2\,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 )}-\frac {c^2\,{\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 )}\,32{}\mathrm {i}}{3\,a^2\,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 {32\,c^2\,{\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 )}}{3\,a^2\,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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^2,x)

[Out]

((c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((2*c^2)/(a^2*f) - (c^2*exp(e*1i + f*
x*1i)*2i)/(a^2*f)))/(exp(e*1i + f*x*1i) - 1i) + (16*c^2*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))/(a^2*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)) - (c^2*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)*32i)/(3*a^2*f*(exp(e*1i
+ f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^2) - (32*c^2*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))/(3*a^2*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^3)

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