\(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{3/2}} \, dx\) [591]

Optimal result

Integrand size = 29, antiderivative size = 180 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\frac {a^{5/2} (3 c-5 d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {a^3 (3 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 3.92 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.46 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\frac {(a (1+\sin (e+f x)))^{5/2} \left (\frac {(-3 c+5 d) \left (2 \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )\right )}{d^{5/2}}-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 c^2-3 c d+2 d^2+d (c+d) \sin (e+f x)\right )}{d^2 (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(3/2),x]
 

Output:

((a*(1 + Sin[e + f*x]))^(5/2)*(((-3*c + 5*d)*(2*ArcTan[(Sqrt[2]*Sqrt[d]*Si 
n[(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]*Sqr 
t[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]))/d^(5/2) - (2* 
(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(3*c^2 - 3*c*d + 2*d^2 + d*(c + d)*S 
in[e + f*x]))/(d^2*(c + d)*Sqrt[c + d*Sin[e + f*x]])))/(2*f*(Cos[(e + f*x) 
/2] + Sin[(e + f*x)/2])^5)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.06, 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, 3460, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 218

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

Input:

Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(3/2),x]
 

Output:

(2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(d*(c + d)*f*Sqrt[c 
+ d*Sin[e + f*x]]) - (a*(-((a^(3/2)*(3*c - 5*d)*(c + d)*ArcTan[(Sqrt[a]*Sq 
rt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/ 
(d^(3/2)*f)) + (a^2*(3*c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(d*f* 
Sqrt[a + a*Sin[e + f*x]])))/(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 3460

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

Output:

int((a+a*sin(f*x+e))^(5/2)/(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. 603 vs. \(2 (160) = 320\).

Time = 0.44 (sec) , antiderivative size = 1665, normalized size of antiderivative = 9.25 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="fric 
as")
 

Output:

[1/8*((3*a^2*c^3 + a^2*c^2*d - 7*a^2*c*d^2 - 5*a^2*d^3 - (3*a^2*c^2*d - 2* 
a^2*c*d^2 - 5*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 - 2*a^2*c^2*d - 5*a^2*c 
*d^2)*cos(f*x + e) + (3*a^2*c^3 + a^2*c^2*d - 7*a^2*c*d^2 - 5*a^2*d^3 + (3 
*a^2*c^2*d - 2*a^2*c*d^2 - 5*a^2*d^3)*cos(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)*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 + 1 
3*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*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*(3*a^2*c^2 - 4*a^2*c*d + a^2*d^2 + (a^2*c*d + a^ 
2*d^2)*cos(f*x + e)^2 + (3*a^2*c^2 - 3*a^2*c*d + 2*a^2*d^2)*cos(f*x + e...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(5/2)/(c+d*sin(f*x+e))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxi 
ma")
 

Output:

integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c)^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="giac 
")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int((a + a*sin(e + f*x))^(5/2)/(c + d*sin(e + f*x))^(3/2),x)
 

Output:

int((a + a*sin(e + f*x))^(5/2)/(c + d*sin(e + f*x))^(3/2), x)
 

Reduce [F]

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

int((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(3/2),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)**2*d**2 + 2*sin(e + f*x)*c*d + c**2),x) + 2*int((s 
qrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) 
**2*d**2 + 2*sin(e + f*x)*c*d + c**2),x) + int((sqrt(sin(e + f*x)*d + c)*s 
qrt(sin(e + f*x) + 1))/(sin(e + f*x)**2*d**2 + 2*sin(e + f*x)*c*d + c**2), 
x))