\(\int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [80]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac {5 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}} \] Output:

-5/32*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/ 
2)/a^(5/2)/d+1/4*cos(d*x+c)/d/(a+a*sin(d*x+c))^(5/2)-5/16*cos(d*x+c)/a/d/( 
a+a*sin(d*x+c))^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.83 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-8 \sin \left (\frac {1}{2} (c+d x)\right )+4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+10 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2-5 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3+(5+5 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4\right )}{16 d (a (1+\sin (c+d x)))^{5/2}} \] Input:

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

Output:

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-8*Sin[(c + d*x)/2] + 4*(Cos[(c + 
d*x)/2] + Sin[(c + d*x)/2]) + 10*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[ 
(c + d*x)/2])^2 - 5*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (5 + 5*I)*(- 
1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + 
 d*x)/2] + Sin[(c + d*x)/2])^4))/(16*d*(a*(1 + Sin[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3229, 3042, 3129, 3042, 3128, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3229

\(\displaystyle \frac {5 \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {5 \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {5 \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{2 a d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

Input:

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

Output:

Cos[c + d*x]/(4*d*(a + a*Sin[c + d*x])^(5/2)) + (5*(-1/2*ArcTanh[(Sqrt[a]* 
Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])]/(Sqrt[2]*a^(3/2)*d) - Co 
s[c + d*x]/(2*d*(a + a*Sin[c + d*x])^(3/2))))/(8*a)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

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

rule 3129

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

rule 3229

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(192\) vs. \(2(88)=176\).

Time = 0.28 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.80

method	result	size
default	\(-\frac {\left (-5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \cos \left (d x +c \right )^{2}+10 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \sin \left (d x +c \right )-10 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{\frac {3}{2}}+12 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {5}{2}}+10 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {11}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(193\)

Input:

int(sin(d*x+c)/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/32/a^(11/2)*(-5*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1 
/2))*a^3*cos(d*x+c)^2+10*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2 
)/a^(1/2))*a^3*sin(d*x+c)-10*(a-a*sin(d*x+c))^(3/2)*a^(3/2)+12*(a-a*sin(d* 
x+c))^(1/2)*a^(5/2)+10*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/ 
a^(1/2))*a^3)*(-a*(sin(d*x+c)-1))^(1/2)/(1+sin(d*x+c))/cos(d*x+c)/(a+a*sin 
(d*x+c))^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (88) = 176\).

Time = 0.09 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.97 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {5 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (5 \, \cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) - 4\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:

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

Output:

1/64*(5*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2*c 
os(d*x + c) - 4)*sin(d*x + c) - 2*cos(d*x + c) - 4)*sqrt(a)*log(-(a*cos(d* 
x + c)^2 - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin( 
d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2 
*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) 
 + 4*(5*cos(d*x + c)^2 + (5*cos(d*x + c) + 4)*sin(d*x + c) + cos(d*x + c) 
- 4)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c 
)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos 
(d*x + c) - 4*a^3*d)*sin(d*x + c))
 

Sympy [F]

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

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

Output:

Integral(sin(c + d*x)/(a*(sin(c + d*x) + 1))**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.76 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left (5 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{32 \, {\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:

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

Output:

-1/32*sqrt(2)*(5*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 3*sqrt(a)*sin( 
-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^3* 
d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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