3.19 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=188 \[ -\frac {a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt {a \sin (e+f x)+a}}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}-\frac {3 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f} \]

[Out]

-3/28*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(9/2)/c/f-1/8*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-
c*sin(f*x+e))^(9/2)/c/f-1/35*a^3*cos(f*x+e)*(c-c*sin(f*x+e))^(9/2)/c/f/(a+a*sin(f*x+e))^(1/2)-1/14*a^2*cos(f*x
+e)*(c-c*sin(f*x+e))^(9/2)*(a+a*sin(f*x+e))^(1/2)/c/f

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Rubi [A]  time = 0.62, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac {a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}-\frac {3 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(a^3*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(35*c*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(9/2))/(14*c*f) - (3*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin
[e + f*x])^(9/2))/(28*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(9/2))/(8*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))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx &=\frac {\int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx}{a c}\\ &=-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac {3 \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2} \, dx}{4 c}\\ &=-\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac {(3 a) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2} \, dx}{7 c}\\ &=-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac {a^2 \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2} \, dx}{7 c}\\ &=-\frac {a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}\\ \end {align*}

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Mathematica [A]  time = 3.11, size = 176, normalized size = 0.94 \[ -\frac {c^3 (\sin (e+f x)-1)^3 (a (\sin (e+f x)+1))^{5/2} \sqrt {c-c \sin (e+f x)} (19600 \sin (e+f x)+3920 \sin (3 (e+f x))+784 \sin (5 (e+f x))+80 \sin (7 (e+f x))+1960 \cos (2 (e+f x))+980 \cos (4 (e+f x))+280 \cos (6 (e+f x))+35 \cos (8 (e+f x)))}{35840 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-1/35840*(c^3*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[e + f*x]]*(1960*Cos[2*(e + f*x
)] + 980*Cos[4*(e + f*x)] + 280*Cos[6*(e + f*x)] + 35*Cos[8*(e + f*x)] + 19600*Sin[e + f*x] + 3920*Sin[3*(e +
f*x)] + 784*Sin[5*(e + f*x)] + 80*Sin[7*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])^5)

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fricas [A]  time = 0.49, size = 128, normalized size = 0.68 \[ \frac {{\left (35 \, a^{2} c^{3} \cos \left (f x + e\right )^{8} - 35 \, a^{2} c^{3} + 8 \, {\left (5 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} + 6 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 8 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{280 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/280*(35*a^2*c^3*cos(f*x + e)^8 - 35*a^2*c^3 + 8*(5*a^2*c^3*cos(f*x + e)^6 + 6*a^2*c^3*cos(f*x + e)^4 + 8*a^2
*c^3*cos(f*x + e)^2 + 16*a^2*c^3)*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))^(5/2)*(c-c*sin(f*x+e))^(7/2),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)sqrt(2*a)*sqrt(2*c)*(-4480*a^2*c^3*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))/(128*f)^2-8064*a^2*c^3*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)^2-4480*a^2*c^3*f*sign(sin(1/2*(f*x+e
xp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(5*f*x+5*exp(1))/(640*f)^2-896*a^2*c^3*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))/(896*f)^2-2560*a^2*c^3*f*sign(sin(1
/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(2*f*x+2*exp(1))/(512*f)^2-9216*a^2*c^3*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^2*c^3*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))/(1536*f)^2-2048*a^2
*c^3*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))/(2048*f)^2-46
08*a^2*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(-2*f*x-2*exp(1))/(-512*
f)^2-5120*a^2*c^3*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)^2-1536*a^2*c^3*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))/(-1536*f)^2)

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maple [A]  time = 0.44, size = 143, normalized size = 0.76 \[ \frac {\left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {7}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (35 \left (\cos ^{8}\left (f x +e \right )\right )+5 \left (\cos ^{6}\left (f x +e \right )\right ) \sin \left (f x +e \right )+40 \left (\cos ^{6}\left (f x +e \right )\right )+13 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+48 \left (\cos ^{4}\left (f x +e \right )\right )+29 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+64 \left (\cos ^{2}\left (f x +e \right )\right )+93 \sin \left (f x +e \right )+93\right )}{280 f \cos \left (f x +e \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2),x)

[Out]

1/280/f*(-c*(sin(f*x+e)-1))^(7/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)*(35*cos(f*x+e)^8+5*cos(f*x+e)^6*sin(f*x+
e)+40*cos(f*x+e)^6+13*sin(f*x+e)*cos(f*x+e)^4+48*cos(f*x+e)^4+29*cos(f*x+e)^2*sin(f*x+e)+64*cos(f*x+e)^2+93*si
n(f*x+e)+93)/cos(f*x+e)^7

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \cos \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(7/2)*cos(f*x + e)^2, x)

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mupad [B]  time = 12.62, size = 376, normalized size = 2.00 \[ \frac {{\mathrm {e}}^{-e\,8{}\mathrm {i}-f\,x\,8{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {35\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}+\frac {7\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{128\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{512\,f}+\frac {7\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^2*(a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(7/2),x)

[Out]

(exp(- e*8i - f*x*8i)*(c - c*sin(e + f*x))^(1/2)*((35*a^2*c^3*exp(e*8i + f*x*8i)*sin(e + f*x)*(a + a*sin(e + f
*x))^(1/2))/(32*f) + (7*a^2*c^3*exp(e*8i + f*x*8i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2))/(64*f) + (7*a^
2*c^3*exp(e*8i + f*x*8i)*cos(4*e + 4*f*x)*(a + a*sin(e + f*x))^(1/2))/(128*f) + (a^2*c^3*exp(e*8i + f*x*8i)*co
s(6*e + 6*f*x)*(a + a*sin(e + f*x))^(1/2))/(64*f) + (a^2*c^3*exp(e*8i + f*x*8i)*cos(8*e + 8*f*x)*(a + a*sin(e
+ f*x))^(1/2))/(512*f) + (7*a^2*c^3*exp(e*8i + f*x*8i)*sin(3*e + 3*f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f) + (
7*a^2*c^3*exp(e*8i + f*x*8i)*sin(5*e + 5*f*x)*(a + a*sin(e + f*x))^(1/2))/(160*f) + (a^2*c^3*exp(e*8i + f*x*8i
)*sin(7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2))/(224*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))**(5/2)*(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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