\(\int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx\) [530]

Optimal result

Integrand size = 27, antiderivative size = 150 \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx=-\frac {24 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{35 f \sqrt {3+3 \sin (e+f x)}}-\frac {2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{35 f}-\frac {4 (7 c-d) d \cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (3+3 \sin (e+f x))^{5/2}}{21 f} \]

[Out]

-4/35*(7*c-d)*d*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-2/7*d^2*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f-8/105*a^2*(3
5*c^2+42*c*d+19*d^2)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-2/105*a*(35*c^2+42*c*d+19*d^2)*cos(f*x+e)*(a+a*sin(f*
x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2840, 2830, 2726, 2725} \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx=-\frac {8 a^2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{105 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{105 f}-\frac {4 d (7 c-d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f} \]

[In]

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

[Out]

(-8*a^2*(35*c^2 + 42*c*d + 19*d^2)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(35*c^2 + 42*c*d + 19
*d^2)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(105*f) - (4*(7*c - d)*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))
/(35*f) - (2*d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*a*f)

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 2726

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[a*((2*n - 1)/n), Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac {2 \int (a+a \sin (e+f x))^{3/2} \left (\frac {1}{2} a \left (7 c^2+5 d^2\right )+a (7 c-d) d \sin (e+f x)\right ) \, dx}{7 a} \\ & = -\frac {4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac {1}{35} \left (35 c^2+42 c d+19 d^2\right ) \int (a+a \sin (e+f x))^{3/2} \, dx \\ & = -\frac {2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac {1}{105} \left (4 a \left (35 c^2+42 c d+19 d^2\right )\right ) \int \sqrt {a+a \sin (e+f x)} \, dx \\ & = -\frac {8 a^2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92 \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx=-\frac {\sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{3/2} \left (700 c^2+1092 c d+494 d^2-6 d (14 c+13 d) \cos (2 (e+f x))+\left (140 c^2+504 c d+253 d^2\right ) \sin (e+f x)-15 d^2 \sin (3 (e+f x))\right )}{70 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^2,x]

[Out]

-1/70*(Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(3/2)*(700*c^2 + 1092*c*d + 494*d^2 -
6*d*(14*c + 13*d)*Cos[2*(e + f*x)] + (140*c^2 + 504*c*d + 253*d^2)*Sin[e + f*x] - 15*d^2*Sin[3*(e + f*x)]))/(f
*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {2 \left (\sin \left (f x +e \right )+1\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (15 d^{2} \left (\sin ^{3}\left (f x +e \right )\right )+42 c d \left (\sin ^{2}\left (f x +e \right )\right )+39 \left (\sin ^{2}\left (f x +e \right )\right ) d^{2}+35 c^{2} \sin \left (f x +e \right )+126 c d \sin \left (f x +e \right )+52 \sin \left (f x +e \right ) d^{2}+175 c^{2}+252 c d +104 d^{2}\right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(130\)
parts	\(\frac {2 c^{2} \left (\sin \left (f x +e \right )+1\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )+5\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 d^{2} \left (\sin \left (f x +e \right )+1\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (15 \left (\sin ^{3}\left (f x +e \right )\right )+39 \left (\sin ^{2}\left (f x +e \right )\right )+52 \sin \left (f x +e \right )+104\right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {4 c d \left (\sin \left (f x +e \right )+1\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin ^{2}\left (f x +e \right )+3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(198\)

[In]

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

[Out]

2/105*(sin(f*x+e)+1)*a^2*(sin(f*x+e)-1)*(15*d^2*sin(f*x+e)^3+42*c*d*sin(f*x+e)^2+39*sin(f*x+e)^2*d^2+35*c^2*si
n(f*x+e)+126*c*d*sin(f*x+e)+52*sin(f*x+e)*d^2+175*c^2+252*c*d+104*d^2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.53 \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx=\frac {2 \, {\left (15 \, a d^{2} \cos \left (f x + e\right )^{4} + 3 \, {\left (14 \, a c d + 13 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} - 140 \, a c^{2} - 168 \, a c d - 76 \, a d^{2} - {\left (35 \, a c^{2} + 84 \, a c d + 43 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (175 \, a c^{2} + 294 \, a c d + 143 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (15 \, a d^{2} \cos \left (f x + e\right )^{3} + 140 \, a c^{2} + 168 \, a c d + 76 \, a d^{2} - 6 \, {\left (7 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (35 \, a c^{2} + 126 \, a c d + 67 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

[In]

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

[Out]

2/105*(15*a*d^2*cos(f*x + e)^4 + 3*(14*a*c*d + 13*a*d^2)*cos(f*x + e)^3 - 140*a*c^2 - 168*a*c*d - 76*a*d^2 - (
35*a*c^2 + 84*a*c*d + 43*a*d^2)*cos(f*x + e)^2 - (175*a*c^2 + 294*a*c*d + 143*a*d^2)*cos(f*x + e) + (15*a*d^2*
cos(f*x + e)^3 + 140*a*c^2 + 168*a*c*d + 76*a*d^2 - 6*(7*a*c*d + 4*a*d^2)*cos(f*x + e)^2 - (35*a*c^2 + 126*a*c
*d + 67*a*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)

Sympy [F]

\[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sin {\left (e + f x \right )}\right )^{2}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{2} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.58 \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx=\frac {\sqrt {2} {\left (15 \, a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) + 105 \, {\left (12 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 16 \, a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 7 \, a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 35 \, {\left (4 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 12 \, a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) + 21 \, {\left (4 \, a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right )\right )} \sqrt {a}}{420 \, f} \]

[In]

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

[Out]

1/420*sqrt(2)*(15*a*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*(12*a*c^2*sgn
(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 16*a*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 7*a*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*a*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*a*c*d*s
gn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*a*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*a*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*a*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-5/4*p
i + 5/2*f*x + 5/2*e))*sqrt(a)/f

Mupad [F(-1)]

Timed out. \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]

[In]

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

[Out]

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