\(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\) [296]

Optimal result

Integrand size = 25, antiderivative size = 101 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=-\frac {8 a^2 (5 A+3 B) \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a (5 A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f} \] Output:

-8/15*a^2*(5*A+3*B)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-2/15*a*(5*A+3*B)*c 
os(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f-2/5*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2) 
/f
 

Mathematica [A] (verified)

Time = 1.99 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (50 A+39 B-3 B \cos (2 (e+f x))+2 (5 A+9 B) \sin (e+f x))}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

Integrate[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]),x]
 

Output:

-1/15*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]* 
(50*A + 39*B - 3*B*Cos[2*(e + f*x)] + 2*(5*A + 9*B)*Sin[e + f*x]))/(f*(Cos 
[(e + f*x)/2] + Sin[(e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3230, 3042, 3126, 3042, 3125}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {1}{5} (5 A+3 B) \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} (5 A+3 B) \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\)

\(\Big \downarrow \) 3126

\(\displaystyle \frac {1}{5} (5 A+3 B) \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} (5 A+3 B) \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\)

\(\Big \downarrow \) 3125

\(\displaystyle \frac {1}{5} (5 A+3 B) \left (-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\)

Input:

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

Output:

(-2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f) + ((5*A + 3*B)*((-8*a 
^2*Cos[e + f*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*Cos[e + f*x]*Sqrt[a 
 + a*Sin[e + f*x]])/(3*f)))/5
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3125

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] && Eq 
Q[a^2 - b^2, 0]
 

rule 3126

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] + Simp[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 3230

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] + Simp[(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)]
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (-3 B \cos \left (f x +e \right )^{2}+\sin \left (f x +e \right ) \left (5 A +9 B \right )+25 A +21 B \right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(77\)
parts	\(\frac {2 A \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )+5\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 B \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(118\)

Input:

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

Output:

2/15*(1+sin(f*x+e))*a^2*(sin(f*x+e)-1)*(-3*B*cos(f*x+e)^2+sin(f*x+e)*(5*A+ 
9*B)+25*A+21*B)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.36 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\frac {2 \, {\left (3 \, B a \cos \left (f x + e\right )^{3} - {\left (5 \, A + 6 \, B\right )} a \cos \left (f x + e\right )^{2} - {\left (25 \, A + 21 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (5 \, A + 3 \, B\right )} a - {\left (3 \, B a \cos \left (f x + e\right )^{2} + {\left (5 \, A + 9 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (5 \, A + 3 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \] Input:

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

Output:

2/15*(3*B*a*cos(f*x + e)^3 - (5*A + 6*B)*a*cos(f*x + e)^2 - (25*A + 21*B)* 
a*cos(f*x + e) - 4*(5*A + 3*B)*a - (3*B*a*cos(f*x + e)^2 + (5*A + 9*B)*a*c 
os(f*x + e) - 4*(5*A + 3*B)*a)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f*c 
os(f*x + e) + f*sin(f*x + e) + f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.38 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\frac {\sqrt {2} {\left (3 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 30 \, {\left (3 \, A a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right )\right )} \sqrt {a}}{30 \, f} \] Input:

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

Output:

1/30*sqrt(2)*(3*B*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-5/4*pi + 5/2* 
f*x + 5/2*e) + 30*(3*A*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*B*a*sgn(c 
os(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 5*(2*A*a* 
sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/ 
2*e)))*sin(-3/4*pi + 3/2*f*x + 3/2*e))*sqrt(a)/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b \right ) \] Input:

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

Output:

sqrt(a)*a*(int(sqrt(sin(e + f*x) + 1),x)*a + int(sqrt(sin(e + f*x) + 1)*si 
n(e + f*x)**2,x)*b + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x),x)*a + int(sq 
rt(sin(e + f*x) + 1)*sin(e + f*x),x)*b)