\(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx\) [592]

Optimal result

Integrand size = 29, antiderivative size = 183 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}+\frac {2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \] Output:

-2*a^(5/2)*arctan(a^(1/2)*d^(1/2)*cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/(c+d*s 
in(f*x+e))^(1/2))/d^(5/2)/f+2/3*a^2*(c-d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2 
)/d/(c+d)/f/(c+d*sin(f*x+e))^(3/2)+2/3*a^3*(c-d)*(3*c+7*d)*cos(f*x+e)/d^2/ 
(c+d)^2/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 8.23 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.43 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {(a (1+\sin (e+f x)))^{5/2} \left (\frac {2 \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )}{d^{5/2}}+\frac {2 (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 c^2+8 c d+d^2+4 d (c+2 d) \sin (e+f x)\right )}{3 d^2 (c+d)^2 (c+d \sin (e+f x))^{3/2}}\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(5/2)*((2*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 
2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + ArcTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - 
 Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] - Log[Sqrt[2]*Sqrt[d]*Cos[(2*e 
- Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]])/d^(5/2) + (2*(c - d)*(Cos[(e 
 + f*x)/2] - Sin[(e + f*x)/2])*(3*c^2 + 8*c*d + d^2 + 4*d*(c + 2*d)*Sin[e 
+ f*x]))/(3*d^2*(c + d)^2*(c + d*Sin[e + f*x])^(3/2))))/(f*(Cos[(e + f*x)/ 
2] + Sin[(e + f*x)/2])^5)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3241, 27, 3042, 3459, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}-\frac {2 a \int \frac {\sqrt {\sin (e+f x) a+a} (a (c-7 d)-3 a (c+d) \sin (e+f x))}{2 (c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a} (a (c-7 d)-3 a (c+d) \sin (e+f x))}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a} (a (c-7 d)-3 a (c+d) \sin (e+f x))}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}-\frac {a \left (-\frac {3 a (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {2 a^2 (c-d) (3 c+7 d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}-\frac {a \left (-\frac {3 a (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {2 a^2 (c-d) (3 c+7 d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}\)

\(\Big \downarrow \) 3254

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}-\frac {a \left (\frac {6 a^2 (c+d) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{d f}-\frac {2 a^2 (c-d) (3 c+7 d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}-\frac {a \left (\frac {6 a^{3/2} (c+d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{d^{3/2} f}-\frac {2 a^2 (c-d) (3 c+7 d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}\)

Input:

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

Output:

(2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*d*(c + d)*f*(c + 
d*Sin[e + f*x])^(3/2)) - (a*((6*a^(3/2)*(c + d)*ArcTan[(Sqrt[a]*Sqrt[d]*Co 
s[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(d^(3/2) 
*f) - (2*a^2*(c - d)*(3*c + 7*d)*Cos[e + f*x])/(d*(c + d)*f*Sqrt[a + a*Sin 
[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])))/(3*d*(c + 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 3042

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

rule 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b 
*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* 
d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d))   Int[(a + b*Sin[e + f*x])^(m - 
 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* 
c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || 
 (IntegerQ[m] && EqQ[c, 0]))
 

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]
 

rule 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   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, A, B}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1]
 

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

int((a+a*sin(f*x+e))^(5/2)/(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. 919 vs. \(2 (159) = 318\).

Time = 0.49 (sec) , antiderivative size = 2297, normalized size of antiderivative = 12.55 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/12*(3*(a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4 - 
 (a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^3 - (2*a^2*c^3*d + 5*a 
^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^2 + (a^2*c^4 + 2*a^2*c^3* 
d + 2*a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e) + (a^2*c^4 + 4*a^2 
*c^3*d + 6*a^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^ 
3 + a^2*d^4)*cos(f*x + e)^2 + 2*(a^2*c^3*d + 2*a^2*c^2*d^2 + a^2*c*d^3)*co 
s(f*x + e))*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)*c 
os(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 + 289*a*d^4)*cos(f*x + e) + (128*a*d^4*c 
os(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(...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\sqrt {a}\, a^{2} \left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3} d^{3}+3 \sin \left (f x +e \right )^{2} c \,d^{2}+3 \sin \left (f x +e \right ) c^{2} d +c^{3}}d x +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3} d^{3}+3 \sin \left (f x +e \right )^{2} c \,d^{2}+3 \sin \left (f x +e \right ) c^{2} d +c^{3}}d x \right )+\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3} d^{3}+3 \sin \left (f x +e \right )^{2} c \,d^{2}+3 \sin \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \] Input:

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

Output:

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