\(\int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx\) [572]

Optimal result

Integrand size = 29, antiderivative size = 203 \[ \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx=-\frac {5 \sqrt {a} (c+d)^3 \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{8 \sqrt {d} f}-\frac {5 a (c+d)^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{8 f \sqrt {a+a \sin (e+f x)}}-\frac {5 a (c+d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{12 f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.00 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.93 \[ \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx=\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) \sqrt {a (1+\sin (e+f x))} \left (\frac {5 i (c+d)^3 e^{\frac {1}{2} i (e+f x)} \sqrt {2 c-i d e^{-i (e+f x)} \left (-1+e^{2 i (e+f x)}\right )} \left ((-1)^{3/4} \sqrt {2} \arctan \left (\frac {\sqrt [4]{-1} \left (d-i c e^{i (e+f x)}\right )}{\sqrt {d} \sqrt {-2 c e^{i (e+f x)}+i d \left (-1+e^{2 i (e+f x)}\right )}}\right )-(1+i) \text {arctanh}\left (\frac {(-1)^{3/4} \left (c-i d e^{i (e+f x)}\right )}{\sqrt {d} \sqrt {-2 c e^{i (e+f x)}+i d \left (-1+e^{2 i (e+f x)}\right )}}\right )\right )}{\sqrt {d} \sqrt {-2 c e^{i (e+f x)}+i d \left (-1+e^{2 i (e+f x)}\right )} f}-\frac {\left (\frac {2}{3}-\frac {2 i}{3}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c+d \sin (e+f x)} \left (33 c^2+40 c d+19 d^2-4 d^2 \cos (2 (e+f x))+2 d (13 c+5 d) \sin (e+f x)\right )}{f}\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )} \] Input:

Integrate[Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2),x]
 

Output:

((1/32 + I/32)*Sqrt[a*(1 + Sin[e + f*x])]*(((5*I)*(c + d)^3*E^((I/2)*(e + 
f*x))*Sqrt[2*c - (I*d*(-1 + E^((2*I)*(e + f*x))))/E^(I*(e + f*x))]*((-1)^( 
3/4)*Sqrt[2]*ArcTan[((-1)^(1/4)*(d - I*c*E^(I*(e + f*x))))/(Sqrt[d]*Sqrt[- 
2*c*E^(I*(e + f*x)) + I*d*(-1 + E^((2*I)*(e + f*x)))])] - (1 + I)*ArcTanh[ 
((-1)^(3/4)*(c - I*d*E^(I*(e + f*x))))/(Sqrt[d]*Sqrt[-2*c*E^(I*(e + f*x)) 
+ I*d*(-1 + E^((2*I)*(e + f*x)))])]))/(Sqrt[d]*Sqrt[-2*c*E^(I*(e + f*x)) + 
 I*d*(-1 + E^((2*I)*(e + f*x)))]*f) - ((2/3 - (2*I)/3)*(Cos[(e + f*x)/2] - 
 Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]*(33*c^2 + 40*c*d + 19*d^2 - 4* 
d^2*Cos[2*(e + f*x)] + 2*d*(13*c + 5*d)*Sin[e + f*x]))/f))/(Cos[(e + f*x)/ 
2] + Sin[(e + f*x)/2])
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3249, 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.

\(\displaystyle \int \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}dx\)

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 218

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

Input:

Int[Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2),x]
 

Output:

-1/3*(a*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(f*Sqrt[a + a*Sin[e + f*x 
]]) + (5*(c + 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]*Sqr 
t[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
 

Defintions of rubi rules used

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 3042

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

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))^(1/2)*(c+d*sin(f*x+e))^(5/2),x)
 

Output:

int((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (171) = 342\).

Time = 0.44 (sec) , antiderivative size = 1257, normalized size of antiderivative = 6.19 \[ \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(5/2),x, algorithm="fric 
as")
 

Output:

[1/192*(15*(c^3 + 3*c^2*d + 3*c*d^2 + d^3 + (c^3 + 3*c^2*d + 3*c*d^2 + d^3 
)*cos(f*x + e) + (c^3 + 3*c^2*d + 3*c*d^2 + 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)*cos(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 - 32*(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 + 2 
89*a*d^4)*cos(f*x + e) + (128*a*d^4*cos(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)*cos(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*d^2*cos(f*x + e)^3 - 2*(13*c*d + d^2)*cos(f*x + 
e)^2 - 33*c^2 - 14*c*d - 13*d^2 - (33*c^2 + 40*c*d + 23*d^2)*cos(f*x + e) 
- (8*d^2*cos(f*x + e)^2 - 33*c^2 - 14*c*d - 13*d^2 + 2*(13*c*d + 5*d^2)*co 
s(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) ...
 

Sympy [F(-1)]

Timed out. \[ \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(1/2)*(c+d*sin(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx=\int { \sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 

Output:

integrate(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2), x)
 

Giac [F]

\[ \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx=\int { \sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(5/2),x, algorithm="giac 
")
 

Output:

integrate(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx=\int \sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \] Input:

int((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^(5/2),x)
 

Output:

int((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) d^{2}+2 \left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) c d +\left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}d x \right ) c^{2}\right ) \] Input:

int((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(5/2),x)
 

Output:

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