3.3.87 \(\int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx\) [287]

Optimal. Leaf size=192 \[ \frac {2 a (B c-7 A d-6 B d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{105 d f \sqrt {a+a \sin (e+f x)}}+\frac {4 (5 c-d) (B c-7 A d-6 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}+\frac {2 d (B c-7 A d-6 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a+a \sin (e+f x)}} \]

[Out]

2/35*d*(-7*A*d+B*c-6*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/a/f+2/105*a*(-7*A*d+B*c-6*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)-2/7*a*B*cos(f*x+e)*(c+d*sin(f*x+e))^3/d/f/(a+a*sin(f*x+e))^(1/2)+4/10
5*(5*c-d)*(-7*A*d+B*c-6*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.23, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {3060, 2840, 2830, 2725} \begin {gather*} \frac {2 a \left (15 c^2+10 c d+7 d^2\right ) (-7 A d+B c-6 B d) \cos (e+f x)}{105 d f \sqrt {a \sin (e+f x)+a}}+\frac {2 d (-7 A d+B c-6 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 a f}+\frac {4 (5 c-d) (-7 A d+B c-6 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{105 f}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}} \end {gather*}

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])^2,x]

[Out]

(2*a*(B*c - 7*A*d - 6*B*d)*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(105*d*f*Sqrt[a + a*Sin[e + f*x]]) + (4*(5*
c - d)*(B*c - 7*A*d - 6*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(105*f) + (2*d*(B*c - 7*A*d - 6*B*d)*Cos[e
+ f*x]*(a + a*Sin[e + f*x])^(3/2))/(35*a*f) - (2*a*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(7*d*f*Sqrt[a + a*S
in[e + f*x]])

Rule 2725

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 2830

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*S
in[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 2840

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 3060

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)*Sqrt
[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))^2 \, dx &=-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a+a \sin (e+f x)}}+\frac {(7 a A d-B (a c-6 a d)) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx}{7 a d}\\ &=\frac {2 d (B c-7 A d-6 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a+a \sin (e+f x)}}+\frac {(2 (7 a A d-B (a c-6 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}{35 a^2 d}\\ &=\frac {4 (5 c-d) (B c-7 A d-6 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}+\frac {2 d (B c-7 A d-6 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a+a \sin (e+f x)}}+\frac {\left (\left (15 c^2+10 c d+7 d^2\right ) (7 a A d-B (a c-6 a d))\right ) \int \sqrt {a+a \sin (e+f x)} \, dx}{105 a d}\\ &=\frac {2 a (B c-7 A d-6 B d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{105 d f \sqrt {a+a \sin (e+f x)}}+\frac {4 (5 c-d) (B c-7 A d-6 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}+\frac {2 d (B c-7 A d-6 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.50, size = 176, normalized size = 0.92 \begin {gather*} -\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (420 A c^2+280 B c^2+560 A c d+532 B c d+266 A d^2+228 B d^2-6 d (14 B c+7 A d+6 B d) \cos (2 (e+f x))+\left (56 A d (5 c+2 d)+B \left (140 c^2+224 c d+141 d^2\right )\right ) \sin (e+f x)-15 B d^2 \sin (3 (e+f x))\right )}{210 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

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])^2,x]

[Out]

-1/210*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(420*A*c^2 + 280*B*c^2 + 560*A*c*d +
532*B*c*d + 266*A*d^2 + 228*B*d^2 - 6*d*(14*B*c + 7*A*d + 6*B*d)*Cos[2*(e + f*x)] + (56*A*d*(5*c + 2*d) + B*(1
40*c^2 + 224*c*d + 141*d^2))*Sin[e + f*x] - 15*B*d^2*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2
]))

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Maple [A]
time = 6.51, size = 161, normalized size = 0.84

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (-15 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) d^{2}+\left (70 A c d +28 A \,d^{2}+35 B \,c^{2}+56 B c d +39 B \,d^{2}\right ) \sin \left (f x +e \right )+\left (-21 A \,d^{2}-42 B c d -18 B \,d^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+105 A \,c^{2}+140 A c d +77 A \,d^{2}+70 B \,c^{2}+154 B c d +66 B \,d^{2}\right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(1+sin(f*x+e))*a*(sin(f*x+e)-1)*(-15*B*cos(f*x+e)^2*sin(f*x+e)*d^2+(70*A*c*d+28*A*d^2+35*B*c^2+56*B*c*d+
39*B*d^2)*sin(f*x+e)+(-21*A*d^2-42*B*c*d-18*B*d^2)*cos(f*x+e)^2+105*A*c^2+140*A*c*d+77*A*d^2+70*B*c^2+154*B*c*
d+66*B*d^2)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 0.38, size = 317, normalized size = 1.65 \begin {gather*} \frac {2 \, {\left (15 \, B d^{2} \cos \left (f x + e\right )^{4} + 3 \, {\left (14 \, B c d + {\left (7 \, A + 6 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{3} - 35 \, {\left (3 \, A + B\right )} c^{2} - 14 \, {\left (5 \, A + 7 \, B\right )} c d - {\left (49 \, A + 27 \, B\right )} d^{2} - {\left (35 \, B c^{2} + 14 \, {\left (5 \, A + B\right )} c d + {\left (7 \, A + 36 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (35 \, {\left (3 \, A + 2 \, B\right )} c^{2} + 14 \, {\left (10 \, A + 11 \, B\right )} c d + 11 \, {\left (7 \, A + 6 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) + {\left (15 \, B d^{2} \cos \left (f x + e\right )^{3} + 35 \, {\left (3 \, A + B\right )} c^{2} + 14 \, {\left (5 \, A + 7 \, B\right )} c d + {\left (49 \, A + 27 \, B\right )} d^{2} - 3 \, {\left (14 \, B c d + {\left (7 \, A + B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (35 \, B c^{2} + 14 \, {\left (5 \, A + 4 \, B\right )} c d + {\left (28 \, A + 39 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*d^2*cos(f*x + e)^4 + 3*(14*B*c*d + (7*A + 6*B)*d^2)*cos(f*x + e)^3 - 35*(3*A + B)*c^2 - 14*(5*A +
7*B)*c*d - (49*A + 27*B)*d^2 - (35*B*c^2 + 14*(5*A + B)*c*d + (7*A + 36*B)*d^2)*cos(f*x + e)^2 - (35*(3*A + 2*
B)*c^2 + 14*(10*A + 11*B)*c*d + 11*(7*A + 6*B)*d^2)*cos(f*x + e) + (15*B*d^2*cos(f*x + e)^3 + 35*(3*A + B)*c^2
+ 14*(5*A + 7*B)*c*d + (49*A + 27*B)*d^2 - 3*(14*B*c*d + (7*A + B)*d^2)*cos(f*x + e)^2 - (35*B*c^2 + 14*(5*A
+ 4*B)*c*d + (28*A + 39*B)*d^2)*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [A]
time = 0.58, size = 367, normalized size = 1.91 \begin {gather*} \frac {\sqrt {2} {\left (15 \, B d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) + 105 \, {\left (8 \, A c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 4 \, B c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 8 \, A c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 8 \, B c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 4 \, A d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 35 \, {\left (4 \, B c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 8 \, A c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 4 \, B c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, A d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) + 21 \, {\left (4 \, B c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, A d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right )\right )} \sqrt {a}}{420 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*sqrt(2)*(15*B*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-7/4*pi + 7/2*f*x + 7/2*e) + 105*(8*A*c^2*sgn(
cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*B*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 8*A*c*d*sgn(cos(-1/4*pi + 1/2*
f*x + 1/2*e)) + 8*B*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*A*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*
B*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 35*(4*B*c^2*sgn(cos(-1/4*pi + 1/2*
f*x + 1/2*e)) + 8*A*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*B*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*
A*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-3/4*pi + 3/2*f*x
+ 3/2*e) + 21*(4*B*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*A*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + B*
d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-5/4*pi + 5/2*f*x + 5/2*e))*sqrt(a)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 )}^2 \,d x \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))^(1/2)*(c + d*sin(e + f*x))^2,x)

[Out]

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

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