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

Optimal. Leaf size=156 \[ -\frac {a^2 \left (12 c^2+16 c d+7 d^2\right ) \cos (e+f x)}{6 f}-\frac {a^2 \left (12 c^2+16 c d+7 d^2\right ) \sin (e+f x) \cos (e+f x)}{24 f}+\frac {1}{8} a^2 x \left (12 c^2+16 c d+7 d^2\right )-\frac {d (8 c-d) \cos (e+f x) (a \sin (e+f x)+a)^2}{12 f}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^3}{4 a f} \]

[Out]

1/8*a^2*(12*c^2+16*c*d+7*d^2)*x-1/6*a^2*(12*c^2+16*c*d+7*d^2)*cos(f*x+e)/f-1/24*a^2*(12*c^2+16*c*d+7*d^2)*cos(
f*x+e)*sin(f*x+e)/f-1/12*(8*c-d)*d*cos(f*x+e)*(a+a*sin(f*x+e))^2/f-1/4*d^2*cos(f*x+e)*(a+a*sin(f*x+e))^3/a/f

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Rubi [A]  time = 0.20, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2761, 2751, 2644} \[ -\frac {a^2 \left (12 c^2+16 c d+7 d^2\right ) \cos (e+f x)}{6 f}-\frac {a^2 \left (12 c^2+16 c d+7 d^2\right ) \sin (e+f x) \cos (e+f x)}{24 f}+\frac {1}{8} a^2 x \left (12 c^2+16 c d+7 d^2\right )-\frac {d (8 c-d) \cos (e+f x) (a \sin (e+f x)+a)^2}{12 f}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^3}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*c^2 + 16*c*d + 7*d^2)*x)/8 - (a^2*(12*c^2 + 16*c*d + 7*d^2)*Cos[e + f*x])/(6*f) - (a^2*(12*c^2 + 16*c
*d + 7*d^2)*Cos[e + f*x]*Sin[e + f*x])/(24*f) - ((8*c - d)*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(12*f) - (d^
2*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 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]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2 \, dx &=-\frac {d^2 \cos (e+f x) (a+a \sin (e+f x))^3}{4 a f}+\frac {\int (a+a \sin (e+f x))^2 \left (a \left (4 c^2+3 d^2\right )+a (8 c-d) d \sin (e+f x)\right ) \, dx}{4 a}\\ &=-\frac {(8 c-d) d \cos (e+f x) (a+a \sin (e+f x))^2}{12 f}-\frac {d^2 \cos (e+f x) (a+a \sin (e+f x))^3}{4 a f}+\frac {1}{12} \left (12 c^2+16 c d+7 d^2\right ) \int (a+a \sin (e+f x))^2 \, dx\\ &=\frac {1}{8} a^2 \left (12 c^2+16 c d+7 d^2\right ) x-\frac {a^2 \left (12 c^2+16 c d+7 d^2\right ) \cos (e+f x)}{6 f}-\frac {a^2 \left (12 c^2+16 c d+7 d^2\right ) \cos (e+f x) \sin (e+f x)}{24 f}-\frac {(8 c-d) d \cos (e+f x) (a+a \sin (e+f x))^2}{12 f}-\frac {d^2 \cos (e+f x) (a+a \sin (e+f x))^3}{4 a f}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 148, normalized size = 0.95 \[ -\frac {a^2 \cos (e+f x) \left (6 \left (12 c^2+16 c d+7 d^2\right ) \sin ^{-1}\left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(e+f x)} \left (3 \left (4 c^2+16 c d+7 d^2\right ) \sin (e+f x)+16 \left (3 c^2+5 c d+2 d^2\right )+16 d (c+d) \sin ^2(e+f x)+6 d^2 \sin ^3(e+f x)\right )\right )}{24 f \sqrt {\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(a^2*Cos[e + f*x]*(6*(12*c^2 + 16*c*d + 7*d^2)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[e + f*x
]^2]*(16*(3*c^2 + 5*c*d + 2*d^2) + 3*(4*c^2 + 16*c*d + 7*d^2)*Sin[e + f*x] + 16*d*(c + d)*Sin[e + f*x]^2 + 6*d
^2*Sin[e + f*x]^3)))/(f*Sqrt[Cos[e + f*x]^2])

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fricas [A]  time = 0.45, size = 145, normalized size = 0.93 \[ \frac {16 \, {\left (a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (12 \, a^{2} c^{2} + 16 \, a^{2} c d + 7 \, a^{2} d^{2}\right )} f x - 48 \, {\left (a^{2} c^{2} + 2 \, a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right ) + 3 \, {\left (2 \, a^{2} d^{2} \cos \left (f x + e\right )^{3} - {\left (4 \, a^{2} c^{2} + 16 \, a^{2} c d + 9 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(16*(a^2*c*d + a^2*d^2)*cos(f*x + e)^3 + 3*(12*a^2*c^2 + 16*a^2*c*d + 7*a^2*d^2)*f*x - 48*(a^2*c^2 + 2*a^
2*c*d + a^2*d^2)*cos(f*x + e) + 3*(2*a^2*d^2*cos(f*x + e)^3 - (4*a^2*c^2 + 16*a^2*c*d + 9*a^2*d^2)*cos(f*x + e
))*sin(f*x + e))/f

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giac [A]  time = 0.35, size = 208, normalized size = 1.33 \[ -\frac {2 \, a^{2} c d \cos \left (f x + e\right )}{f} + \frac {a^{2} d^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} - \frac {a^{2} d^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac {1}{8} \, {\left (4 \, a^{2} c^{2} + 16 \, a^{2} c d + 3 \, a^{2} d^{2}\right )} x + \frac {1}{2} \, {\left (2 \, a^{2} c^{2} + a^{2} d^{2}\right )} x + \frac {{\left (a^{2} c d + a^{2} d^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{6 \, f} - \frac {{\left (4 \, a^{2} c^{2} + 3 \, a^{2} c d + 3 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )}{2 \, f} - \frac {{\left (a^{2} c^{2} + 4 \, a^{2} c d + a^{2} d^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^2*c*d*cos(f*x + e)/f + 1/32*a^2*d^2*sin(4*f*x + 4*e)/f - 1/4*a^2*d^2*sin(2*f*x + 2*e)/f + 1/8*(4*a^2*c^2
+ 16*a^2*c*d + 3*a^2*d^2)*x + 1/2*(2*a^2*c^2 + a^2*d^2)*x + 1/6*(a^2*c*d + a^2*d^2)*cos(3*f*x + 3*e)/f - 1/2*(
4*a^2*c^2 + 3*a^2*c*d + 3*a^2*d^2)*cos(f*x + e)/f - 1/4*(a^2*c^2 + 4*a^2*c*d + a^2*d^2)*sin(2*f*x + 2*e)/f

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maple [A]  time = 0.26, size = 219, normalized size = 1.40 \[ \frac {a^{2} c^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {2 a^{2} c d \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+a^{2} d^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-2 a^{2} c^{2} \cos \left (f x +e \right )+4 a^{2} c d \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {2 a^{2} d^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+a^{2} c^{2} \left (f x +e \right )-2 a^{2} c d \cos \left (f x +e \right )+a^{2} d^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 211, normalized size = 1.35 \[ \frac {24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c^{2} + 96 \, {\left (f x + e\right )} a^{2} c^{2} + 64 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} c d + 96 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c d + 64 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} d^{2} + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} d^{2} + 24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} d^{2} - 192 \, a^{2} c^{2} \cos \left (f x + e\right ) - 192 \, a^{2} c d \cos \left (f x + e\right )}{96 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.34, size = 440, normalized size = 2.82 \[ \frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (12\,c^2+16\,c\,d+7\,d^2\right )}{4\,\left (3\,a^2\,c^2+4\,a^2\,c\,d+\frac {7\,a^2\,d^2}{4}\right )}\right )\,\left (12\,c^2+16\,c\,d+7\,d^2\right )}{4\,f}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^2\,c^2+4\,a^2\,c\,d+\frac {7\,a^2\,d^2}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (a^2\,c^2+4\,a^2\,c\,d+\frac {7\,a^2\,d^2}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (a^2\,c^2+4\,a^2\,c\,d+\frac {15\,a^2\,d^2}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (a^2\,c^2+4\,a^2\,c\,d+\frac {15\,a^2\,d^2}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (12\,a^2\,c^2+20\,a^2\,c\,d+8\,a^2\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (12\,a^2\,c^2+\frac {68\,a^2\,c\,d}{3}+\frac {32\,a^2\,d^2}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (4\,a^2\,c^2+4\,d\,a^2\,c\right )+4\,a^2\,c^2+\frac {8\,a^2\,d^2}{3}+\frac {20\,a^2\,c\,d}{3}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {a^2\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )\,\left (12\,c^2+16\,c\,d+7\,d^2\right )}{4\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*atan((a^2*tan(e/2 + (f*x)/2)*(16*c*d + 12*c^2 + 7*d^2))/(4*(3*a^2*c^2 + (7*a^2*d^2)/4 + 4*a^2*c*d)))*(16*
c*d + 12*c^2 + 7*d^2))/(4*f) - (tan(e/2 + (f*x)/2)*(a^2*c^2 + (7*a^2*d^2)/4 + 4*a^2*c*d) - tan(e/2 + (f*x)/2)^
7*(a^2*c^2 + (7*a^2*d^2)/4 + 4*a^2*c*d) + tan(e/2 + (f*x)/2)^3*(a^2*c^2 + (15*a^2*d^2)/4 + 4*a^2*c*d) - tan(e/
2 + (f*x)/2)^5*(a^2*c^2 + (15*a^2*d^2)/4 + 4*a^2*c*d) + tan(e/2 + (f*x)/2)^4*(12*a^2*c^2 + 8*a^2*d^2 + 20*a^2*
c*d) + tan(e/2 + (f*x)/2)^2*(12*a^2*c^2 + (32*a^2*d^2)/3 + (68*a^2*c*d)/3) + tan(e/2 + (f*x)/2)^6*(4*a^2*c^2 +
4*a^2*c*d) + 4*a^2*c^2 + (8*a^2*d^2)/3 + (20*a^2*c*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)) - (a^2*(atan(tan(e/2 + (f*x)/2)) - (f*x)/2)*(16*c*d + 12
*c^2 + 7*d^2))/(4*f)

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sympy [A]  time = 2.36, size = 459, normalized size = 2.94 \[ \begin {cases} \frac {a^{2} c^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a^{2} c^{2} x \cos ^{2}{\left (e + f x \right )}}{2} + a^{2} c^{2} x - \frac {a^{2} c^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 a^{2} c^{2} \cos {\left (e + f x \right )}}{f} + 2 a^{2} c d x \sin ^{2}{\left (e + f x \right )} + 2 a^{2} c d x \cos ^{2}{\left (e + f x \right )} - \frac {2 a^{2} c d \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 a^{2} c d \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 a^{2} c d \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {2 a^{2} c d \cos {\left (e + f x \right )}}{f} + \frac {3 a^{2} d^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 a^{2} d^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {a^{2} d^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 a^{2} d^{2} x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {a^{2} d^{2} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {5 a^{2} d^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {2 a^{2} d^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 a^{2} d^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {a^{2} d^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {4 a^{2} d^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (c + d \sin {\relax (e )}\right )^{2} \left (a \sin {\relax (e )} + a\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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