\(\int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2} \, dx\) [580]

Optimal result

Integrand size = 29, antiderivative size = 228 \[ \int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2} \, dx=\frac {a^{3/2} (c-11 d) (c+d)^2 \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{8 d^{3/2} f}+\frac {a^2 (c-11 d) (c+d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{8 d f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-11 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{12 d f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a+a \sin (e+f x)}} \] Output:

1/8*a^(3/2)*(c-11*d)*(c+d)^2*arctan(a^(1/2)*d^(1/2)*cos(f*x+e)/(a+a*sin(f* 
x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))/d^(3/2)/f+1/8*a^2*(c-11*d)*(c+d)*cos(f 
*x+e)*(c+d*sin(f*x+e))^(1/2)/d/f/(a+a*sin(f*x+e))^(1/2)+1/12*a^2*(c-11*d)* 
cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d/f/(a+a*sin(f*x+e))^(1/2)-1/3*a^2*cos(f 
*x+e)*(c+d*sin(f*x+e))^(5/2)/d/f/(a+a*sin(f*x+e))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.54 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.23 \[ \int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2} \, dx=\frac {(a (1+\sin (e+f x)))^{3/2} \left (\frac {(c-11 d) (c+d)^2 \left (-2 \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+\log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )\right )}{d^{3/2}}-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c+d \sin (e+f x)} \left (3 c^2+52 c d+37 d^2-4 d^2 \cos (2 (e+f x))+2 d (7 c+11 d) \sin (e+f x)\right )}{3 d}\right )}{16 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(3/2)*(((c - 11*d)*(c + d)^2*(-2*ArcTan[(Sqrt[2]*S 
qrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] - ArcTanh[(Sqr 
t[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + Log[Sq 
rt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]))/d^(3 
/2) - (2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]*(3 
*c^2 + 52*c*d + 37*d^2 - 4*d^2*Cos[2*(e + f*x)] + 2*d*(7*c + 11*d)*Sin[e + 
 f*x]))/(3*d)))/(16*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {3042, 3242, 27, 2011, 3042, 3249, 3042, 3249, 3042, 3254, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

\(\displaystyle \frac {\int -\frac {\left ((c-11 d) a^2+(c-11 d) \sin (e+f x) a^2\right ) (c+d \sin (e+f x))^{3/2}}{2 \sqrt {\sin (e+f x) a+a}}dx}{3 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\left ((c-11 d) a^2+(c-11 d) \sin (e+f x) a^2\right ) (c+d \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx}{6 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 2011

\(\displaystyle -\frac {a (c-11 d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}dx}{6 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (c-11 d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}dx}{6 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3249

\(\displaystyle -\frac {a (c-11 d) \left (\frac {3}{4} (c+d) \int \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}dx-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (c-11 d) \left (\frac {3}{4} (c+d) \int \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}dx-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3249

\(\displaystyle -\frac {a (c-11 d) \left (\frac {3}{4} (c+d) \left (\frac {1}{2} (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (c-11 d) \left (\frac {3}{4} (c+d) \left (\frac {1}{2} (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3254

\(\displaystyle -\frac {a (c-11 d) \left (\frac {3}{4} (c+d) \left (-\frac {a (c+d) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt {a \sin (e+f x)+a}}-\frac {a (c-11 d) \left (\frac {3}{4} (c+d) \left (-\frac {\sqrt {a} (c+d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {d} f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{6 d}\)

Input:

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

Output:

-1/3*(a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(d*f*Sqrt[a + a*Sin[e + 
 f*x]]) - (a*(c - 11*d)*(-1/2*(a*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/ 
(f*Sqrt[a + a*Sin[e + f*x]]) + (3*(c + d)*(-((Sqrt[a]*(c + d)*ArcTan[(Sqrt 
[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x 
]])])/(Sqrt[d]*f)) - (a*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a + 
 a*Sin[e + f*x]])))/4))/(6*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 0]))
 

rule 3249

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

rule 3254

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b + d*x^2), x], 
x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))], x 
] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] 
 && NeQ[c^2 - d^2, 0]
 

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (196) = 392\).

Time = 0.48 (sec) , antiderivative size = 1337, normalized size of antiderivative = 5.86 \[ \int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/192*(3*(a*c^3 - 9*a*c^2*d - 21*a*c*d^2 - 11*a*d^3 + (a*c^3 - 9*a*c^2*d 
 - 21*a*c*d^2 - 11*a*d^3)*cos(f*x + e) + (a*c^3 - 9*a*c^2*d - 21*a*c*d^2 - 
 11*a*d^3)*sin(f*x + e))*sqrt(-a/d)*log((128*a*d^4*cos(f*x + e)^5 + a*c^4 
+ 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 + 128*(2*a*c*d^3 - a*d^4)*co 
s(f*x + e)^4 - 32*(5*a*c^2*d^2 - 14*a*c*d^3 + 13*a*d^4)*cos(f*x + e)^3 - 3 
2*(a*c^3*d - 2*a*c^2*d^2 + 9*a*c*d^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16*d^4 
*cos(f*x + e)^4 - c^3*d + 17*c^2*d^2 - 59*c*d^3 + 51*d^4 + 24*(c*d^3 - d^4 
)*cos(f*x + e)^3 - 2*(5*c^2*d^2 - 26*c*d^3 + 33*d^4)*cos(f*x + e)^2 - (c^3 
*d - 7*c^2*d^2 + 31*c*d^3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)^3 
+ c^3*d - 17*c^2*d^2 + 59*c*d^3 - 51*d^4 - 8*(3*c*d^3 - 5*d^4)*cos(f*x + e 
)^2 - 2*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a 
*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-a/d) + (a*c^4 - 28*a*c^3 
*d + 230*a*c^2*d^2 - 476*a*c*d^3 + 289*a*d^4)*cos(f*x + e) + (128*a*d^4*co 
s(f*x + e)^4 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - 256*( 
a*c*d^3 - a*d^4)*cos(f*x + e)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 + 5*a*d^4)*c 
os(f*x + e)^2 + 32*(a*c^3*d - 7*a*c^2*d^2 + 15*a*c*d^3 - 9*a*d^4)*cos(f*x 
+ e))*sin(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)) - 8*(8*a*d^2*cos(f* 
x + e)^3 - 3*a*c^2 - 38*a*c*d - 19*a*d^2 - 14*(a*c*d + a*d^2)*cos(f*x + e) 
^2 - (3*a*c^2 + 52*a*c*d + 41*a*d^2)*cos(f*x + e) - (8*a*d^2*cos(f*x + e)^ 
2 - 3*a*c^2 - 38*a*c*d - 19*a*d^2 + 2*(7*a*c*d + 11*a*d^2)*cos(f*x + e)...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int(sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1)*sin(e + f*x 
)**2,x)*d + int(sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1)*sin(e + f* 
x),x)*c + int(sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1)*sin(e + f*x) 
,x)*d + int(sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1),x)*c)