3.4.33 \(\int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx\) [333]

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

[Out]

-256/5*c*sec(f*x+e)^5*(c-c*sin(f*x+e))^(5/2)/a^3/f+64*sec(f*x+e)^5*(c-c*sin(f*x+e))^(7/2)/a^3/f-24*sec(f*x+e)^
5*(c-c*sin(f*x+e))^(9/2)/a^3/c/f+2*sec(f*x+e)^5*(c-c*sin(f*x+e))^(11/2)/a^3/c^2/f

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Rubi [A]
time = 0.24, antiderivative size = 134, 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 {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} \end {gather*}

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 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))^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*}

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Mathematica [A]
time = 0.79, size = 114, normalized size = 0.85 \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)} (-182+90 \cos (2 (e+f x))-235 \sin (e+f x)+5 \sin (3 (e+f x)))}{10 a^3 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))^3} \end {gather*}

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 = 2.68, size = 81, normalized size = 0.60

method	result	size
default	\(\frac {2 c^{4} \left (\sin \left (f x +e \right )-1\right ) \left (5 \left (\sin ^{3}\left (f x +e \right )\right )+45 \left (\sin ^{2}\left (f x +e \right )\right )+55 \sin \left (f x +e \right )+23\right )}{5 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(81\)

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,method=_RETURNVERBOSE)

[Out]

2/5*c^4/a^3*(sin(f*x+e)-1)/(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] Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (132) = 264\).
time = 0.52, size = 462, normalized size = 3.45 \begin {gather*} \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 {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))^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 = 0.35, size = 114, normalized size = 0.85 \begin {gather*} -\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 {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))^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(-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))**3,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (132) = 264\).
time = 0.63, size = 343, normalized size = 2.56 \begin {gather*} -\frac {4 \, \sqrt {2} \sqrt {c} {\left (\frac {5 \, c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a^{3} {\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 {11 \, c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {50 \, c^{3} {\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 {80 \, 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}} + \frac {30 \, c^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3} \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 )}^{3}} + \frac {5 \, c^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4} \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 )}^{4}}}{a^{3} {\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 )}^{5}}\right )}}{5 \, f} \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))^3,x, algorithm="giac")

[Out]

-4/5*sqrt(2)*sqrt(c)*(5*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a^3*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(co
s(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)) - (11*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 50*c^3*(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) + 80*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 +
30*c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*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 + 5*c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1
/2*f*x + 1/2*e) + 1)^4)/(a^3*((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
))/f

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Mupad [B]
time = 13.87, size = 542, normalized size = 4.04 \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^3}{a^3\,f}-\frac {c^3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^3\,f}\right )}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}}-\frac {24\,c^3\,{\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^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 )}+\frac {c^3\,{\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}}{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 {288\,c^3\,{\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 )}^3}-\frac {c^3\,{\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 )}\,256{}\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 {128\,c^3\,{\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} \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))^3,x)

[Out]

(c^3*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)/(a^3*f*(e
xp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^2) - (24*c^3*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^3*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)) - ((c
- c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((2*c^3)/(a^3*f) - (c^3*exp(e*1i + f*x*1
i)*2i)/(a^3*f)))/(exp(e*1i + f*x*1i) - 1i) + (288*c^3*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)^3) - (c^3*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)*256i)/(5*a^3*f*(exp(e*
1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^4) - (128*c^3*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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