3.286 \(\int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=256 \[ \frac {4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) (-9 A d+B c-8 B d) \cos (e+f x)}{315 d f \sqrt {a \sin (e+f x)+a}}+\frac {2 a (-9 A d+B c-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt {a \sin (e+f x)+a}}+\frac {4 d (c+d) (-9 A d+B c-8 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{105 a f}+\frac {8 (5 c-d) (c+d) (-9 A d+B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{315 f}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}} \]

[Out]

4/105*d*(c+d)*(-9*A*d+B*c-8*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/a/f+4/315*a*(c+d)*(-9*A*d+B*c-8*B*d)*(15*c^
2+10*c*d+7*d^2)*cos(f*x+e)/d/f/(a+a*sin(f*x+e))^(1/2)+2/63*a*(-9*A*d+B*c-8*B*d)*cos(f*x+e)*(c+d*sin(f*x+e))^3/
d/f/(a+a*sin(f*x+e))^(1/2)-2/9*a*B*cos(f*x+e)*(c+d*sin(f*x+e))^4/d/f/(a+a*sin(f*x+e))^(1/2)+8/315*(5*c-d)*(c+d
)*(-9*A*d+B*c-8*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.46, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {2981, 2770, 2761, 2751, 2646} \[ \frac {4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) (-9 A d+B c-8 B d) \cos (e+f x)}{315 d f \sqrt {a \sin (e+f x)+a}}+\frac {2 a (-9 A d+B c-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt {a \sin (e+f x)+a}}+\frac {4 d (c+d) (-9 A d+B c-8 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{105 a f}+\frac {8 (5 c-d) (c+d) (-9 A d+B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{315 f}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^3,x]

[Out]

(4*a*(c + d)*(B*c - 9*A*d - 8*B*d)*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(315*d*f*Sqrt[a + a*Sin[e + f*x]])
+ (8*(5*c - d)*(c + d)*(B*c - 9*A*d - 8*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(315*f) + (4*d*(c + d)*(B*
c - 9*A*d - 8*B*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(105*a*f) + (2*a*(B*c - 9*A*d - 8*B*d)*Cos[e + f*x
]*(c + d*Sin[e + f*x])^3)/(63*d*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^4)/(9*d
*f*Sqrt[a + a*Sin[e + f*x]])

Rule 2646

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

Rule 2751

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

Rule 2761

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

Rule 2770

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] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx &=-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a+a \sin (e+f x)}}+\frac {(9 a A d-B (a c-8 a d)) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3 \, dx}{9 a d}\\ &=\frac {2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a+a \sin (e+f x)}}+\frac {(2 (c+d) (9 a A d-B (a c-8 a d))) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx}{21 a d}\\ &=\frac {4 d (c+d) (B c-9 A d-8 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 a f}+\frac {2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a+a \sin (e+f x)}}+\frac {(4 (c+d) (9 a A d-B (a c-8 a d))) \int \sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a \left (5 c^2+3 d^2\right )+a (5 c-d) d \sin (e+f x)\right ) \, dx}{105 a^2 d}\\ &=\frac {8 (5 c-d) (c+d) (B c-9 A d-8 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{315 f}+\frac {4 d (c+d) (B c-9 A d-8 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 a f}+\frac {2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a+a \sin (e+f x)}}+\frac {\left (2 (c+d) \left (15 c^2+10 c d+7 d^2\right ) (9 a A d-B (a c-8 a d))\right ) \int \sqrt {a+a \sin (e+f x)} \, dx}{315 a d}\\ &=\frac {4 a (c+d) (B c-9 A d-8 B d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{315 d f \sqrt {a+a \sin (e+f x)}}+\frac {8 (5 c-d) (c+d) (B c-9 A d-8 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{315 f}+\frac {4 d (c+d) (B c-9 A d-8 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 a f}+\frac {2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 1.25, size = 305, normalized size = 1.19 \[ -\frac {\sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-4 d \left (27 A d (7 c+2 d)+B \left (189 c^2+162 c d+83 d^2\right )\right ) \cos (2 (e+f x))+2520 A c^3+2520 A c^2 d \sin (e+f x)+5040 A c^2 d+2016 A c d^2 \sin (e+f x)+4788 A c d^2+846 A d^3 \sin (e+f x)-90 A d^3 \sin (3 (e+f x))+1368 A d^3+840 B c^3 \sin (e+f x)+1680 B c^3+2016 B c^2 d \sin (e+f x)+4788 B c^2 d+2538 B c d^2 \sin (e+f x)-270 B c d^2 \sin (3 (e+f x))+4104 B c d^2+752 B d^3 \sin (e+f x)-80 B d^3 \sin (3 (e+f x))+35 B d^3 \cos (4 (e+f x))+1321 B d^3\right )}{1260 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^3,x]

[Out]

-1/1260*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(2520*A*c^3 + 1680*B*c^3 + 5040*A*c^
2*d + 4788*B*c^2*d + 4788*A*c*d^2 + 4104*B*c*d^2 + 1368*A*d^3 + 1321*B*d^3 - 4*d*(27*A*d*(7*c + 2*d) + B*(189*
c^2 + 162*c*d + 83*d^2))*Cos[2*(e + f*x)] + 35*B*d^3*Cos[4*(e + f*x)] + 840*B*c^3*Sin[e + f*x] + 2520*A*c^2*d*
Sin[e + f*x] + 2016*B*c^2*d*Sin[e + f*x] + 2016*A*c*d^2*Sin[e + f*x] + 2538*B*c*d^2*Sin[e + f*x] + 846*A*d^3*S
in[e + f*x] + 752*B*d^3*Sin[e + f*x] - 270*B*c*d^2*Sin[3*(e + f*x)] - 90*A*d^3*Sin[3*(e + f*x)] - 80*B*d^3*Sin
[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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fricas [A]  time = 0.45, size = 467, normalized size = 1.82 \[ -\frac {2 \, {\left (35 \, B d^{3} \cos \left (f x + e\right )^{5} - 5 \, {\left (27 \, B c d^{2} + {\left (9 \, A + B\right )} d^{3}\right )} \cos \left (f x + e\right )^{4} + 105 \, {\left (3 \, A + B\right )} c^{3} + 63 \, {\left (5 \, A + 7 \, B\right )} c^{2} d + 9 \, {\left (49 \, A + 27 \, B\right )} c d^{2} + {\left (81 \, A + 107 \, B\right )} d^{3} - {\left (189 \, B c^{2} d + 27 \, {\left (7 \, A + 6 \, B\right )} c d^{2} + 2 \, {\left (27 \, A + 59 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (105 \, B c^{3} + 63 \, {\left (5 \, A + B\right )} c^{2} d + 9 \, {\left (7 \, A + 36 \, B\right )} c d^{2} + 2 \, {\left (54 \, A + 13 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (105 \, {\left (3 \, A + 2 \, B\right )} c^{3} + 63 \, {\left (10 \, A + 11 \, B\right )} c^{2} d + 99 \, {\left (7 \, A + 6 \, B\right )} c d^{2} + {\left (198 \, A + 211 \, B\right )} d^{3}\right )} \cos \left (f x + e\right ) - {\left (35 \, B d^{3} \cos \left (f x + e\right )^{4} + 105 \, {\left (3 \, A + B\right )} c^{3} + 63 \, {\left (5 \, A + 7 \, B\right )} c^{2} d + 9 \, {\left (49 \, A + 27 \, B\right )} c d^{2} + {\left (81 \, A + 107 \, B\right )} d^{3} + 5 \, {\left (27 \, B c d^{2} + {\left (9 \, A + 8 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (63 \, B c^{2} d + 9 \, {\left (7 \, A + B\right )} c d^{2} + {\left (3 \, A + 26 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (105 \, B c^{3} + 63 \, {\left (5 \, A + 4 \, B\right )} c^{2} d + 9 \, {\left (28 \, A + 39 \, B\right )} c d^{2} + 13 \, {\left (9 \, A + 8 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{315 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-2/315*(35*B*d^3*cos(f*x + e)^5 - 5*(27*B*c*d^2 + (9*A + B)*d^3)*cos(f*x + e)^4 + 105*(3*A + B)*c^3 + 63*(5*A
+ 7*B)*c^2*d + 9*(49*A + 27*B)*c*d^2 + (81*A + 107*B)*d^3 - (189*B*c^2*d + 27*(7*A + 6*B)*c*d^2 + 2*(27*A + 59
*B)*d^3)*cos(f*x + e)^3 + (105*B*c^3 + 63*(5*A + B)*c^2*d + 9*(7*A + 36*B)*c*d^2 + 2*(54*A + 13*B)*d^3)*cos(f*
x + e)^2 + (105*(3*A + 2*B)*c^3 + 63*(10*A + 11*B)*c^2*d + 99*(7*A + 6*B)*c*d^2 + (198*A + 211*B)*d^3)*cos(f*x
+ e) - (35*B*d^3*cos(f*x + e)^4 + 105*(3*A + B)*c^3 + 63*(5*A + 7*B)*c^2*d + 9*(49*A + 27*B)*c*d^2 + (81*A +
107*B)*d^3 + 5*(27*B*c*d^2 + (9*A + 8*B)*d^3)*cos(f*x + e)^3 - 3*(63*B*c^2*d + 9*(7*A + B)*c*d^2 + (3*A + 26*B
)*d^3)*cos(f*x + e)^2 - (105*B*c^3 + 63*(5*A + 4*B)*c^2*d + 9*(28*A + 39*B)*c*d^2 + 13*(9*A + 8*B)*d^3)*cos(f*
x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(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+B*sin(f*x+e))*(c+d*sin(f*x+e))^3*(a+a*sin(f*x+e))^(1/2),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)sqrt(2*a)*(-40*f*(-2*A*d^3*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-6*B*c*d^2*sign(cos(1/2
*(f*x+exp(1))-1/4*pi)))*cos(1/4*(10*f*x+10*exp(1)+pi))/(40*f)^2-56*f*(-2*A*d^3*sign(cos(1/2*(f*x+exp(1))-1/4*p
i))-6*B*c*d^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(14*f*x+14*exp(1)-pi))/(56*f)^2+12*f*(-2*B*d^3*sign(
cos(1/2*(f*x+exp(1))-1/4*pi))-6*A*c*d^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-6*B*c^2*d*sign(cos(1/2*(f*x+exp(1))
-1/4*pi)))*sin(1/4*(6*f*x+6*exp(1)+pi))/(12*f)^2+20*f*(-2*B*d^3*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-6*A*c*d^2*s
ign(cos(1/2*(f*x+exp(1))-1/4*pi))-6*B*c^2*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(10*f*x+10*exp(1)-pi))
/(20*f)^2-8*f*(6*A*d^3*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+8*B*c^3*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+24*A*c^2*
d*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+18*B*c*d^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(2*f*x+2*exp(1)+pi
))/(8*f)^2-24*f*(6*A*d^3*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+8*B*c^3*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+24*A*c^
2*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+18*B*c*d^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(6*f*x+6*exp(1)-
pi))/(24*f)^2+8*f*(16*A*c^3*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+6*B*d^3*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+24*A
*c*d^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+24*B*c^2*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(2*f*x-pi)+1/
2*exp(1))/(8*f)^2+224*B*d^3*f*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(14*f*x+14*exp(1)+pi))/(112*f)^2+288*
B*d^3*f*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(18*f*x+18*exp(1)-pi))/(144*f)^2)

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maple [A]  time = 1.40, size = 242, normalized size = 0.95 \[ \frac {2 \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (\left (-45 A \,d^{3}-135 B c \,d^{2}-40 B \,d^{3}\right ) \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (315 A \,c^{2} d +252 A c \,d^{2}+117 A \,d^{3}+105 B \,c^{3}+252 B \,c^{2} d +351 B c \,d^{2}+104 B \,d^{3}\right ) \sin \left (f x +e \right )+35 B \left (\cos ^{4}\left (f x +e \right )\right ) d^{3}+\left (-189 A c \,d^{2}-54 A \,d^{3}-189 B \,c^{2} d -162 B c \,d^{2}-118 B \,d^{3}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+315 A \,c^{3}+630 A \,c^{2} d +693 A c \,d^{2}+198 A \,d^{3}+210 B \,c^{3}+693 B \,c^{2} d +594 B c \,d^{2}+211 B \,d^{3}\right )}{315 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3*(a+a*sin(f*x+e))^(1/2),x)

[Out]

2/315*(1+sin(f*x+e))*a*(sin(f*x+e)-1)*((-45*A*d^3-135*B*c*d^2-40*B*d^3)*sin(f*x+e)*cos(f*x+e)^2+(315*A*c^2*d+2
52*A*c*d^2+117*A*d^3+105*B*c^3+252*B*c^2*d+351*B*c*d^2+104*B*d^3)*sin(f*x+e)+35*B*cos(f*x+e)^4*d^3+(-189*A*c*d
^2-54*A*d^3-189*B*c^2*d-162*B*c*d^2-118*B*d^3)*cos(f*x+e)^2+315*A*c^3+630*A*c^2*d+693*A*c*d^2+198*A*d^3+210*B*
c^3+693*B*c^2*d+594*B*c*d^2+211*B*d^3)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3*(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^3,x)

[Out]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**3*(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))*(A + B*sin(e + f*x))*(c + d*sin(e + f*x))**3, x)

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