3.2.41 \(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\) [141]

Optimal. Leaf size=134 \[ -\frac {a^2 A \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{3 f \sqrt {a+a \sin (e+f x)}}-\frac {a A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f} \]

[Out]

-1/4*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2)/f-1/3*a^2*A*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/
f/(a+a*sin(f*x+e))^(1/2)-1/3*a*A*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.23, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3052, 2819, 2817} \begin {gather*} -\frac {a^2 A \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {a A \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

-1/3*(a^2*A*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(f*Sqrt[a + a*Sin[e + f*x]]) - (a*A*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2))/(3*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e +
f*x])^(3/2))/(4*f)

Rule 2817

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 2819

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-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] &&
EqQ[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 3052

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

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx &=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f}+A \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx\\ &=-\frac {a A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f}+\frac {1}{3} (2 a A) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx\\ &=-\frac {a^2 A \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{3 f \sqrt {a+a \sin (e+f x)}}-\frac {a A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f}\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 96, normalized size = 0.72 \begin {gather*} -\frac {c \sec ^3(e+f x) (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)} (-12 B \cos (2 (e+f x))-3 B \cos (4 (e+f x))+8 A (9 \sin (e+f x)+\sin (3 (e+f x))))}{96 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

-1/96*(c*Sec[e + f*x]^3*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(-12*B*Cos[2
*(e + f*x)] - 3*B*Cos[4*(e + f*x)] + 8*A*(9*Sin[e + f*x] + Sin[3*(e + f*x)])))/f

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Maple [A]
time = 0.35, size = 86, normalized size = 0.64

method	result	size
default	\(\frac {\left (3 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 A \left (\cos ^{2}\left (f x +e \right )\right )+3 B \sin \left (f x +e \right )+8 A \right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}{12 f \cos \left (f x +e \right )^{3}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/12/f*(3*B*cos(f*x+e)^2*sin(f*x+e)+4*A*cos(f*x+e)^2+3*B*sin(f*x+e)+8*A)*(-c*(sin(f*x+e)-1))^(3/2)*sin(f*x+e)*
(a*(1+sin(f*x+e)))^(3/2)/cos(f*x+e)^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(3/2), x)

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Fricas [A]
time = 0.38, size = 89, normalized size = 0.66 \begin {gather*} -\frac {{\left (3 \, B a c \cos \left (f x + e\right )^{4} - 3 \, B a c - 4 \, {\left (A a c \cos \left (f x + e\right )^{2} + 2 \, A a c\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}}{12 \, f \cos \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*B*a*c*cos(f*x + e)^4 - 3*B*a*c - 4*(A*a*c*cos(f*x + e)^2 + 2*A*a*c)*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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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((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3003 deep

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (125) = 250\).
time = 0.53, size = 252, normalized size = 1.88 \begin {gather*} \frac {4 \, {\left (3 \, B a c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, A a c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 6 \, B a c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, A a c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

4/3*(3*B*a*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x +
1/2*e)) + 2*A*a*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f
*x + 1/2*e)) - 6*B*a*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi +
1/2*f*x + 1/2*e)) - 3*A*a*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*
pi + 1/2*f*x + 1/2*e)) + 3*B*a*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(
-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)*sqrt(c)/f

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Mupad [B]
time = 1.84, size = 103, normalized size = 0.77 \begin {gather*} -\frac {a\,c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,B\,\cos \left (e+f\,x\right )+15\,B\,\cos \left (3\,e+3\,f\,x\right )+3\,B\,\cos \left (5\,e+5\,f\,x\right )-80\,A\,\sin \left (2\,e+2\,f\,x\right )-8\,A\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(3/2),x)

[Out]

-(a*c*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(12*B*cos(e + f*x) + 15*B*cos(3*e + 3*f*x) +
3*B*cos(5*e + 5*f*x) - 80*A*sin(2*e + 2*f*x) - 8*A*sin(4*e + 4*f*x)))/(96*f*(cos(2*e + 2*f*x) + 1))

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