3.253 \(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx\)

Optimal. Leaf size=166 \[ -\frac {a^2 (12 A c+8 A d+8 B c+7 B d) \cos (e+f x)}{6 f}-\frac {a^2 (12 A c+8 A d+8 B c+7 B d) \sin (e+f x) \cos (e+f x)}{24 f}+\frac {1}{8} a^2 x (12 A c+8 A d+8 B c+7 B d)-\frac {(4 A d+4 B c-B d) \cos (e+f x) (a \sin (e+f x)+a)^2}{12 f}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^3}{4 a f} \]

[Out]

1/8*a^2*(12*A*c+8*A*d+8*B*c+7*B*d)*x-1/6*a^2*(12*A*c+8*A*d+8*B*c+7*B*d)*cos(f*x+e)/f-1/24*a^2*(12*A*c+8*A*d+8*
B*c+7*B*d)*cos(f*x+e)*sin(f*x+e)/f-1/12*(4*A*d+4*B*c-B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^2/f-1/4*B*d*cos(f*x+e)*(
a+a*sin(f*x+e))^3/a/f

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Rubi [A]  time = 0.27, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2968, 3023, 2751, 2644} \[ -\frac {a^2 (12 A c+8 A d+8 B c+7 B d) \cos (e+f x)}{6 f}-\frac {a^2 (12 A c+8 A d+8 B c+7 B d) \sin (e+f x) \cos (e+f x)}{24 f}+\frac {1}{8} a^2 x (12 A c+8 A d+8 B c+7 B d)-\frac {(4 A d+4 B c-B d) \cos (e+f x) (a \sin (e+f x)+a)^2}{12 f}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^3}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A*c + 8*B*c + 8*A*d + 7*B*d)*x)/8 - (a^2*(12*A*c + 8*B*c + 8*A*d + 7*B*d)*Cos[e + f*x])/(6*f) - (a^2*
(12*A*c + 8*B*c + 8*A*d + 7*B*d)*Cos[e + f*x]*Sin[e + f*x])/(24*f) - ((4*B*c + 4*A*d - B*d)*Cos[e + f*x]*(a +
a*Sin[e + f*x])^2)/(12*f) - (B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^3)/(4*a*f)

Rule 2644

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

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 2968

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*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[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx &=\int (a+a \sin (e+f x))^2 \left (A c+(B c+A d) \sin (e+f x)+B d \sin ^2(e+f x)\right ) \, dx\\ &=-\frac {B d \cos (e+f x) (a+a \sin (e+f x))^3}{4 a f}+\frac {\int (a+a \sin (e+f x))^2 (a (4 A c+3 B d)+a (4 B c+4 A d-B d) \sin (e+f x)) \, dx}{4 a}\\ &=-\frac {(4 B c+4 A d-B d) \cos (e+f x) (a+a \sin (e+f x))^2}{12 f}-\frac {B d \cos (e+f x) (a+a \sin (e+f x))^3}{4 a f}+\frac {1}{12} (12 A c+8 B c+8 A d+7 B d) \int (a+a \sin (e+f x))^2 \, dx\\ &=\frac {1}{8} a^2 (12 A c+8 B c+8 A d+7 B d) x-\frac {a^2 (12 A c+8 B c+8 A d+7 B d) \cos (e+f x)}{6 f}-\frac {a^2 (12 A c+8 B c+8 A d+7 B d) \cos (e+f x) \sin (e+f x)}{24 f}-\frac {(4 B c+4 A d-B d) \cos (e+f x) (a+a \sin (e+f x))^2}{12 f}-\frac {B d \cos (e+f x) (a+a \sin (e+f x))^3}{4 a f}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 160, normalized size = 0.96 \[ -\frac {a^2 \cos (e+f x) \left (6 (12 A c+8 A d+8 B c+7 B d) \sin ^{-1}\left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(e+f x)} \left (8 (A d+B (c+2 d)) \sin ^2(e+f x)+3 (4 A c+8 A d+8 B c+7 B d) \sin (e+f x)+8 (6 A c+5 A d+5 B c+4 B d)+6 B d \sin ^3(e+f x)\right )\right )}{24 f \sqrt {\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(a^2*Cos[e + f*x]*(6*(12*A*c + 8*B*c + 8*A*d + 7*B*d)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[
e + f*x]^2]*(8*(6*A*c + 5*B*c + 5*A*d + 4*B*d) + 3*(4*A*c + 8*B*c + 8*A*d + 7*B*d)*Sin[e + f*x] + 8*(A*d + B*(
c + 2*d))*Sin[e + f*x]^2 + 6*B*d*Sin[e + f*x]^3)))/(f*Sqrt[Cos[e + f*x]^2])

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fricas [A]  time = 0.45, size = 144, normalized size = 0.87 \[ \frac {8 \, {\left (B a^{2} c + {\left (A + 2 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (4 \, {\left (3 \, A + 2 \, B\right )} a^{2} c + {\left (8 \, A + 7 \, B\right )} a^{2} d\right )} f x - 48 \, {\left ({\left (A + B\right )} a^{2} c + {\left (A + B\right )} a^{2} d\right )} \cos \left (f x + e\right ) + 3 \, {\left (2 \, B a^{2} d \cos \left (f x + e\right )^{3} - {\left (4 \, {\left (A + 2 \, B\right )} a^{2} c + {\left (8 \, A + 9 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*(B*a^2*c + (A + 2*B)*a^2*d)*cos(f*x + e)^3 + 3*(4*(3*A + 2*B)*a^2*c + (8*A + 7*B)*a^2*d)*f*x - 48*((A
+ B)*a^2*c + (A + B)*a^2*d)*cos(f*x + e) + 3*(2*B*a^2*d*cos(f*x + e)^3 - (4*(A + 2*B)*a^2*c + (8*A + 9*B)*a^2*
d)*cos(f*x + e))*sin(f*x + e))/f

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giac [A]  time = 0.17, size = 172, normalized size = 1.04 \[ \frac {B a^{2} d \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {1}{8} \, {\left (12 \, A a^{2} c + 8 \, B a^{2} c + 8 \, A a^{2} d + 7 \, B a^{2} d\right )} x + \frac {{\left (B a^{2} c + A a^{2} d + 2 \, B a^{2} d\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {{\left (8 \, A a^{2} c + 7 \, B a^{2} c + 7 \, A a^{2} d + 6 \, B a^{2} d\right )} \cos \left (f x + e\right )}{4 \, f} - \frac {{\left (A a^{2} c + 2 \, B a^{2} c + 2 \, A a^{2} d + 2 \, B a^{2} d\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*B*a^2*d*sin(4*f*x + 4*e)/f + 1/8*(12*A*a^2*c + 8*B*a^2*c + 8*A*a^2*d + 7*B*a^2*d)*x + 1/12*(B*a^2*c + A*a
^2*d + 2*B*a^2*d)*cos(3*f*x + 3*e)/f - 1/4*(8*A*a^2*c + 7*B*a^2*c + 7*A*a^2*d + 6*B*a^2*d)*cos(f*x + e)/f - 1/
4*(A*a^2*c + 2*B*a^2*c + 2*A*a^2*d + 2*B*a^2*d)*sin(2*f*x + 2*e)/f

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maple [A]  time = 0.41, size = 278, normalized size = 1.67 \[ \frac {a^{2} A c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {a^{2} A d \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-\frac {B \,a^{2} c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+B \,a^{2} d \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-2 a^{2} A c \cos \left (f x +e \right )+2 a^{2} A d \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+2 B \,a^{2} c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {2 B \,a^{2} d \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+a^{2} A c \left (f x +e \right )-a^{2} A d \cos \left (f x +e \right )-B \,a^{2} c \cos \left (f x +e \right )+B \,a^{2} d \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 268, normalized size = 1.61 \[ \frac {24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c + 96 \, {\left (f x + e\right )} A a^{2} c + 32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c + 48 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c + 32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{2} d + 48 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} d + 64 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} d + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} d + 24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} d - 192 \, A a^{2} c \cos \left (f x + e\right ) - 96 \, B a^{2} c \cos \left (f x + e\right ) - 96 \, A a^{2} d \cos \left (f x + e\right )}{96 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(24*(2*f*x + 2*e - sin(2*f*x + 2*e))*A*a^2*c + 96*(f*x + e)*A*a^2*c + 32*(cos(f*x + e)^3 - 3*cos(f*x + e)
)*B*a^2*c + 48*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^2*c + 32*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^2*d + 48*(2
*f*x + 2*e - sin(2*f*x + 2*e))*A*a^2*d + 64*(cos(f*x + e)^3 - 3*cos(f*x + e))*B*a^2*d + 3*(12*f*x + 12*e + sin
(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B*a^2*d + 24*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^2*d - 192*A*a^2*c*cos(f*
x + e) - 96*B*a^2*c*cos(f*x + e) - 96*A*a^2*d*cos(f*x + e))/f

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mupad [B]  time = 14.60, size = 492, normalized size = 2.96 \[ \frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (12\,A\,c+8\,A\,d+8\,B\,c+7\,B\,d\right )}{4\,\left (3\,A\,a^2\,c+2\,A\,a^2\,d+2\,B\,a^2\,c+\frac {7\,B\,a^2\,d}{4}\right )}\right )\,\left (12\,A\,c+8\,A\,d+8\,B\,c+7\,B\,d\right )}{4\,f}-\frac {a^2\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )\,\left (12\,A\,c+8\,A\,d+8\,B\,c+7\,B\,d\right )}{4\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (A\,a^2\,c+2\,A\,a^2\,d+2\,B\,a^2\,c+\frac {15\,B\,a^2\,d}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (A\,a^2\,c+2\,A\,a^2\,d+2\,B\,a^2\,c+\frac {7\,B\,a^2\,d}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (A\,a^2\,c+2\,A\,a^2\,d+2\,B\,a^2\,c+\frac {15\,B\,a^2\,d}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (12\,A\,a^2\,c+10\,A\,a^2\,d+10\,B\,a^2\,c+8\,B\,a^2\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (12\,A\,a^2\,c+\frac {34\,A\,a^2\,d}{3}+\frac {34\,B\,a^2\,c}{3}+\frac {32\,B\,a^2\,d}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (4\,A\,a^2\,c+2\,A\,a^2\,d+2\,B\,a^2\,c\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,a^2\,c+2\,A\,a^2\,d+2\,B\,a^2\,c+\frac {7\,B\,a^2\,d}{4}\right )+4\,A\,a^2\,c+\frac {10\,A\,a^2\,d}{3}+\frac {10\,B\,a^2\,c}{3}+\frac {8\,B\,a^2\,d}{3}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*atan((a^2*tan(e/2 + (f*x)/2)*(12*A*c + 8*A*d + 8*B*c + 7*B*d))/(4*(3*A*a^2*c + 2*A*a^2*d + 2*B*a^2*c + (7
*B*a^2*d)/4)))*(12*A*c + 8*A*d + 8*B*c + 7*B*d))/(4*f) - (a^2*(atan(tan(e/2 + (f*x)/2)) - (f*x)/2)*(12*A*c + 8
*A*d + 8*B*c + 7*B*d))/(4*f) - (tan(e/2 + (f*x)/2)^3*(A*a^2*c + 2*A*a^2*d + 2*B*a^2*c + (15*B*a^2*d)/4) - tan(
e/2 + (f*x)/2)^7*(A*a^2*c + 2*A*a^2*d + 2*B*a^2*c + (7*B*a^2*d)/4) - tan(e/2 + (f*x)/2)^5*(A*a^2*c + 2*A*a^2*d
+ 2*B*a^2*c + (15*B*a^2*d)/4) + tan(e/2 + (f*x)/2)^4*(12*A*a^2*c + 10*A*a^2*d + 10*B*a^2*c + 8*B*a^2*d) + tan
(e/2 + (f*x)/2)^2*(12*A*a^2*c + (34*A*a^2*d)/3 + (34*B*a^2*c)/3 + (32*B*a^2*d)/3) + tan(e/2 + (f*x)/2)^6*(4*A*
a^2*c + 2*A*a^2*d + 2*B*a^2*c) + tan(e/2 + (f*x)/2)*(A*a^2*c + 2*A*a^2*d + 2*B*a^2*c + (7*B*a^2*d)/4) + 4*A*a^
2*c + (10*A*a^2*d)/3 + (10*B*a^2*c)/3 + (8*B*a^2*d)/3)/(f*(4*tan(e/2 + (f*x)/2)^2 + 6*tan(e/2 + (f*x)/2)^4 + 4
*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1))

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sympy [A]  time = 2.76, size = 571, normalized size = 3.44 \[ \begin {cases} \frac {A a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {A a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{2} c x - \frac {A a^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 A a^{2} c \cos {\left (e + f x \right )}}{f} + A a^{2} d x \sin ^{2}{\left (e + f x \right )} + A a^{2} d x \cos ^{2}{\left (e + f x \right )} - \frac {A a^{2} d \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {A a^{2} d \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 A a^{2} d \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {A a^{2} d \cos {\left (e + f x \right )}}{f} + B a^{2} c x \sin ^{2}{\left (e + f x \right )} + B a^{2} c x \cos ^{2}{\left (e + f x \right )} - \frac {B a^{2} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {B a^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 B a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {B a^{2} c \cos {\left (e + f x \right )}}{f} + \frac {3 B a^{2} d x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 B a^{2} d x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {B a^{2} d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 B a^{2} d x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {B a^{2} d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {5 B a^{2} d \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {2 B a^{2} d \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 B a^{2} d \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {B a^{2} d \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {4 B a^{2} d \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\relax (e )}\right ) \left (c + d \sin {\relax (e )}\right ) \left (a \sin {\relax (e )} + a\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*a**2*c*x*sin(e + f*x)**2/2 + A*a**2*c*x*cos(e + f*x)**2/2 + A*a**2*c*x - A*a**2*c*sin(e + f*x)*co
s(e + f*x)/(2*f) - 2*A*a**2*c*cos(e + f*x)/f + A*a**2*d*x*sin(e + f*x)**2 + A*a**2*d*x*cos(e + f*x)**2 - A*a**
2*d*sin(e + f*x)**2*cos(e + f*x)/f - A*a**2*d*sin(e + f*x)*cos(e + f*x)/f - 2*A*a**2*d*cos(e + f*x)**3/(3*f) -
A*a**2*d*cos(e + f*x)/f + B*a**2*c*x*sin(e + f*x)**2 + B*a**2*c*x*cos(e + f*x)**2 - B*a**2*c*sin(e + f*x)**2*
cos(e + f*x)/f - B*a**2*c*sin(e + f*x)*cos(e + f*x)/f - 2*B*a**2*c*cos(e + f*x)**3/(3*f) - B*a**2*c*cos(e + f*
x)/f + 3*B*a**2*d*x*sin(e + f*x)**4/8 + 3*B*a**2*d*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + B*a**2*d*x*sin(e + f*
x)**2/2 + 3*B*a**2*d*x*cos(e + f*x)**4/8 + B*a**2*d*x*cos(e + f*x)**2/2 - 5*B*a**2*d*sin(e + f*x)**3*cos(e + f
*x)/(8*f) - 2*B*a**2*d*sin(e + f*x)**2*cos(e + f*x)/f - 3*B*a**2*d*sin(e + f*x)*cos(e + f*x)**3/(8*f) - B*a**2
*d*sin(e + f*x)*cos(e + f*x)/(2*f) - 4*B*a**2*d*cos(e + f*x)**3/(3*f), Ne(f, 0)), (x*(A + B*sin(e))*(c + d*sin
(e))*(a*sin(e) + a)**2, True))

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