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

Optimal. Leaf size=103 \[ \frac {64 a c^4 \cos ^3(e+f x)}{105 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a c^3 \cos ^3(e+f x)}{35 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f} \]

[Out]

64/105*a*c^4*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+16/35*a*c^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(1/2)+2/7*a*c^2
*cos(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.22, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2736, 2674, 2673} \[ \frac {64 a c^4 \cos ^3(e+f x)}{105 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a c^3 \cos ^3(e+f x)}{35 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a*c^4*Cos[e + f*x]^3)/(105*f*(c - c*Sin[e + f*x])^(3/2)) + (16*a*c^3*Cos[e + f*x]^3)/(35*f*Sqrt[c - c*Sin[
e + f*x]]) + (2*a*c^2*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(7*f)

Rule 2673

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 - 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 2674

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 2736

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 (a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx &=(a c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac {2 a c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}+\frac {1}{7} \left (8 a c^2\right ) \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx\\ &=\frac {16 a c^3 \cos ^3(e+f x)}{35 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}+\frac {1}{35} \left (32 a c^3\right ) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=\frac {64 a c^4 \cos ^3(e+f x)}{105 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a c^3 \cos ^3(e+f x)}{35 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 94, normalized size = 0.91 \[ -\frac {a c^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 (108 \sin (e+f x)+15 \cos (2 (e+f x))-157)}{105 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/105*(a*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-157 + 15*Cos[2*(e + f*x)] + 108*Sin[e + f*x])*Sqrt[c -
c*Sin[e + f*x]])/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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fricas [A]  time = 0.43, size = 152, normalized size = 1.48 \[ \frac {2 \, {\left (15 \, a c^{2} \cos \left (f x + e\right )^{4} + 39 \, a c^{2} \cos \left (f x + e\right )^{3} - 8 \, a c^{2} \cos \left (f x + e\right )^{2} + 32 \, a c^{2} \cos \left (f x + e\right ) + 64 \, a c^{2} - {\left (15 \, a c^{2} \cos \left (f x + e\right )^{3} - 24 \, a c^{2} \cos \left (f x + e\right )^{2} - 32 \, a c^{2} \cos \left (f x + e\right ) - 64 \, a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{105 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a*c^2*cos(f*x + e)^4 + 39*a*c^2*cos(f*x + e)^3 - 8*a*c^2*cos(f*x + e)^2 + 32*a*c^2*cos(f*x + e) + 64
*a*c^2 - (15*a*c^2*cos(f*x + e)^3 - 24*a*c^2*cos(f*x + e)^2 - 32*a*c^2*cos(f*x + e) - 64*a*c^2)*sin(f*x + e))*
sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)

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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((a+a*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),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)sqrt(2*c)*(-2*a*c^
2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(2*f*x-pi)+1/2*exp(1))/f-16*a*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4
*pi))*sin(1/4*(2*f*x+2*exp(1)+pi))/(8*f)^2+48*a*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(6*f*x+6*exp(
1)-pi))/(24*f)^2-80*a*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(10*f*x+10*exp(1)+pi))/(40*f)^2+112*a*c
^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(14*f*x+14*exp(1)-pi))/(56*f)^2+24*a*c^2*f*sign(sin(1/2*(f*x+e
xp(1))-1/4*pi))*cos(1/4*(6*f*x+6*exp(1)+pi))/(12*f)^2-40*a*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(1
0*f*x+10*exp(1)-pi))/(20*f)^2+4*a*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(2*f*x-pi)+1/2*exp(1))/(2*f
)^2)

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maple [A]  time = 0.88, size = 69, normalized size = 0.67 \[ -\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (15 \left (\sin ^{2}\left (f x +e \right )\right )-54 \sin \left (f x +e \right )+71\right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(sin(f*x+e)-1)*c^3*(1+sin(f*x+e))^2*a*(15*sin(f*x+e)^2-54*sin(f*x+e)+71)/cos(f*x+e)/(c-c*sin(f*x+e))^(1
/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int \left (- c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\right )\, dx + \int \left (- c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(c**2*sqrt(-c*sin(e + f*x) + c), x) + Integral(-c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + I
ntegral(-c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e +
f*x)**3, x))

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