3.4.25 \(\int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx\) [325]

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

[Out]

256/3*c^2*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^2/f-64*c*sec(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/a^2/f+8*sec(f*x+e
)^3*(c-c*sin(f*x+e))^(7/2)/a^2/f+2/3*sec(f*x+e)^3*(c-c*sin(f*x+e))^(9/2)/a^2/c/f

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Rubi [A]
time = 0.24, antiderivative size = 136, normalized size of antiderivative = 1.00, 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 = {2815, 2753, 2752} \begin {gather*} \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} \end {gather*}

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 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))^{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*}

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Mathematica [A]
time = 0.75, size = 112, normalized size = 0.82 \begin {gather*} \frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \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 \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])^(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 = 2.29, size = 79, normalized size = 0.58

method	result	size
default	\(\frac {2 c^{4} \left (\sin \left (f x +e \right )-1\right ) \left (\sin ^{3}\left (f x +e \right )-15 \left (\sin ^{2}\left (f x +e \right )\right )-69 \sin \left (f x +e \right )-45\right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(79\)

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,method=_RETURNVERBOSE)

[Out]

2/3*c^4/a^2*(sin(f*x+e)-1)/(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] Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (132) = 264\).
time = 0.50, size = 362, normalized size = 2.66 \begin {gather*} -\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 {gather*}

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 = 0.34, size = 98, normalized size = 0.72 \begin {gather*} -\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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3061 deep

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Giac [A]
time = 0.58, size = 125, normalized size = 0.92 \begin {gather*} \frac {128 \, \sqrt {2} {\left (c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {3 \, c^{3} {\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}}\right )} \sqrt {c}}{3 \, a^{2} f {\left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - 1\right )}^{3}} \end {gather*}

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]

128/3*sqrt(2)*(c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*c^3*(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)*sqrt(c)/(a^2*f*((cos(-1/4*pi + 1/2*f*x + 1
/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 1)^3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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