3.30 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx\)

Optimal. Leaf size=236 \[ -\frac {4 a^4 \cos (e+f x) (c-c \sin (e+f x))^{11/2}}{315 c f \sqrt {a \sin (e+f x)+a}}-\frac {4 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{11/2}}{105 c f}-\frac {a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}-\frac {4 a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f} \]

[Out]

-1/15*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(11/2)/c/f-4/45*a*cos(f*x+e)*(a+a*sin(f*x+e))^(5/
2)*(c-c*sin(f*x+e))^(11/2)/c/f-1/10*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(11/2)/c/f-4/315*a^4*co
s(f*x+e)*(c-c*sin(f*x+e))^(11/2)/c/f/(a+a*sin(f*x+e))^(1/2)-4/105*a^3*cos(f*x+e)*(c-c*sin(f*x+e))^(11/2)*(a+a*
sin(f*x+e))^(1/2)/c/f

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Rubi [A]  time = 0.72, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac {a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac {4 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{11/2}}{105 c f}-\frac {4 a^4 \cos (e+f x) (c-c \sin (e+f x))^{11/2}}{315 c f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}-\frac {4 a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(-4*a^4*Cos[e + f*x]*(c - c*Sin[e + f*x])^(11/2))/(315*c*f*Sqrt[a + a*Sin[e + f*x]]) - (4*a^3*Cos[e + f*x]*Sqr
t[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(11/2))/(105*c*f) - (a^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c
- c*Sin[e + f*x])^(11/2))/(15*c*f) - (4*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(11/2)
)/(45*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(11/2))/(10*c*f)

Rule 2738

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2841

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rubi steps

\begin {align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx &=\frac {\int (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{11/2} \, dx}{a c}\\ &=-\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}+\frac {4 \int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2} \, dx}{5 c}\\ &=-\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}+\frac {(8 a) \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2} \, dx}{15 c}\\ &=-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}+\frac {\left (4 a^2\right ) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{11/2} \, dx}{15 c}\\ &=-\frac {4 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{11/2}}{105 c f}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}+\frac {\left (8 a^3\right ) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{11/2} \, dx}{105 c}\\ &=-\frac {4 a^4 \cos (e+f x) (c-c \sin (e+f x))^{11/2}}{315 c f \sqrt {a+a \sin (e+f x)}}-\frac {4 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{11/2}}{105 c f}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}\\ \end {align*}

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Mathematica [A]  time = 5.63, size = 209, normalized size = 0.89 \[ \frac {a^3 c^4 (\sin (e+f x)-1)^4 (\sin (e+f x)+1)^3 \sqrt {a (\sin (e+f x)+1)} \sqrt {c-c \sin (e+f x)} (158760 \sin (e+f x)+35280 \sin (3 (e+f x))+9072 \sin (5 (e+f x))+1620 \sin (7 (e+f x))+140 \sin (9 (e+f x))+13230 \cos (2 (e+f x))+7560 \cos (4 (e+f x))+2835 \cos (6 (e+f x))+630 \cos (8 (e+f x))+63 \cos (10 (e+f x)))}{322560 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a^3*c^4*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x])^3*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(13230
*Cos[2*(e + f*x)] + 7560*Cos[4*(e + f*x)] + 2835*Cos[6*(e + f*x)] + 630*Cos[8*(e + f*x)] + 63*Cos[10*(e + f*x)
] + 158760*Sin[e + f*x] + 35280*Sin[3*(e + f*x)] + 9072*Sin[5*(e + f*x)] + 1620*Sin[7*(e + f*x)] + 140*Sin[9*(
e + f*x)]))/(322560*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7)

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fricas [A]  time = 1.23, size = 144, normalized size = 0.61 \[ \frac {{\left (63 \, a^{3} c^{4} \cos \left (f x + e\right )^{10} - 63 \, a^{3} c^{4} + 2 \, {\left (35 \, a^{3} c^{4} \cos \left (f x + e\right )^{8} + 40 \, a^{3} c^{4} \cos \left (f x + e\right )^{6} + 48 \, a^{3} c^{4} \cos \left (f x + e\right )^{4} + 64 \, a^{3} c^{4} \cos \left (f x + e\right )^{2} + 128 \, a^{3} c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{630 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

1/630*(63*a^3*c^4*cos(f*x + e)^10 - 63*a^3*c^4 + 2*(35*a^3*c^4*cos(f*x + e)^8 + 40*a^3*c^4*cos(f*x + e)^6 + 48
*a^3*c^4*cos(f*x + e)^4 + 64*a^3*c^4*cos(f*x + e)^2 + 128*a^3*c^4)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt
(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)sqrt(2*a)*sqrt(2*c
)*(-16128*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(f*x+exp(1))/(256
*f)^2-8064*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(3*f*x+3*exp(1))
/(384*f)^2-5760*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(5*f*x+5*ex
p(1))/(640*f)^2-32256*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(7*f*
x+7*exp(1))/(3584*f)^2-4608*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*si
n(9*f*x+9*exp(1))/(4608*f)^2-7168*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*p
i))*cos(2*f*x+2*exp(1))/(1024*f)^2-7168*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))
-1/4*pi))*cos(4*f*x+4*exp(1))/(1024*f)^2-7680*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+e
xp(1))-1/4*pi))*cos(6*f*x+6*exp(1))/(1536*f)^2-57344*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2
*(f*x+exp(1))-1/4*pi))*cos(8*f*x+8*exp(1))/(8192*f)^2-10240*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(
cos(1/2*(f*x+exp(1))-1/4*pi))*cos(10*f*x+10*exp(1))/(10240*f)^2-3584*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1))-1/4*p
i))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(-2*f*x-2*exp(1))/(-512*f)^2-5120*a^3*c^4*f*sign(sin(1/2*(f*x+exp(1)
)-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(-4*f*x-4*exp(1))/(-1024*f)^2-43008*a^3*c^4*f*sign(sin(1/2*(f
*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(-6*f*x-6*exp(1))/(-6144*f)^2-8192*a^3*c^4*f*sign(si
n(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(-8*f*x-8*exp(1))/(-8192*f)^2)

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maple [A]  time = 0.51, size = 169, normalized size = 0.72 \[ \frac {\left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {9}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {7}{2}} \left (63 \left (\cos ^{10}\left (f x +e \right )\right )+7 \sin \left (f x +e \right ) \left (\cos ^{8}\left (f x +e \right )\right )+70 \left (\cos ^{8}\left (f x +e \right )\right )+17 \left (\cos ^{6}\left (f x +e \right )\right ) \sin \left (f x +e \right )+80 \left (\cos ^{6}\left (f x +e \right )\right )+33 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+96 \left (\cos ^{4}\left (f x +e \right )\right )+65 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+128 \left (\cos ^{2}\left (f x +e \right )\right )+193 \sin \left (f x +e \right )+193\right )}{630 f \cos \left (f x +e \right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(9/2),x)

[Out]

1/630/f*(-c*(sin(f*x+e)-1))^(9/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7/2)*(63*cos(f*x+e)^10+7*sin(f*x+e)*cos(f*x+e
)^8+70*cos(f*x+e)^8+17*cos(f*x+e)^6*sin(f*x+e)+80*cos(f*x+e)^6+33*sin(f*x+e)*cos(f*x+e)^4+96*cos(f*x+e)^4+65*c
os(f*x+e)^2*sin(f*x+e)+128*cos(f*x+e)^2+193*sin(f*x+e)+193)/cos(f*x+e)^9

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}} \cos \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(7/2)*(-c*sin(f*x + e) + c)^(9/2)*cos(f*x + e)^2, x)

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mupad [B]  time = 13.53, size = 462, normalized size = 1.96 \[ \frac {{\mathrm {e}}^{-e\,10{}\mathrm {i}-f\,x\,10{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {63\,a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}+\frac {21\,a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{256\,f}+\frac {3\,a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}+\frac {9\,a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{512\,f}+\frac {a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{256\,f}+\frac {a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\cos \left (10\,e+10\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{2560\,f}+\frac {7\,a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {9\,a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {9\,a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{896\,f}+\frac {a^3\,c^4\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\sin \left (9\,e+9\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{1152\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^2*(a + a*sin(e + f*x))^(7/2)*(c - c*sin(e + f*x))^(9/2),x)

[Out]

(exp(- e*10i - f*x*10i)*(c - c*sin(e + f*x))^(1/2)*((63*a^3*c^4*exp(e*10i + f*x*10i)*sin(e + f*x)*(a + a*sin(e
+ f*x))^(1/2))/(64*f) + (21*a^3*c^4*exp(e*10i + f*x*10i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2))/(256*f)
+ (3*a^3*c^4*exp(e*10i + f*x*10i)*cos(4*e + 4*f*x)*(a + a*sin(e + f*x))^(1/2))/(64*f) + (9*a^3*c^4*exp(e*10i
+ f*x*10i)*cos(6*e + 6*f*x)*(a + a*sin(e + f*x))^(1/2))/(512*f) + (a^3*c^4*exp(e*10i + f*x*10i)*cos(8*e + 8*f*
x)*(a + a*sin(e + f*x))^(1/2))/(256*f) + (a^3*c^4*exp(e*10i + f*x*10i)*cos(10*e + 10*f*x)*(a + a*sin(e + f*x))
^(1/2))/(2560*f) + (7*a^3*c^4*exp(e*10i + f*x*10i)*sin(3*e + 3*f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f) + (9*a^
3*c^4*exp(e*10i + f*x*10i)*sin(5*e + 5*f*x)*(a + a*sin(e + f*x))^(1/2))/(160*f) + (9*a^3*c^4*exp(e*10i + f*x*1
0i)*sin(7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2))/(896*f) + (a^3*c^4*exp(e*10i + f*x*10i)*sin(9*e + 9*f*x)*(a +
a*sin(e + f*x))^(1/2))/(1152*f)))/(2*cos(e + f*x))

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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(cos(f*x+e)**2*(a+a*sin(f*x+e))**(7/2)*(c-c*sin(f*x+e))**(9/2),x)

[Out]

Timed out

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