\(\int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [79]

Optimal result

Integrand size = 23, antiderivative size = 107 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {19 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}} \] Output:

-19/32*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)+13/16*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 = 0.51 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.83 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (8 \sin \left (\frac {1}{2} (c+d x)\right )-4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-26 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+13 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3+(19+19 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4\right )}{16 d (a (1+\sin (c+d x)))^{5/2}} \] Input:

Integrate[Sin[c + d*x]^2/(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]) - 26*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[( 
c + d*x)/2])^2 + 13*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (19 + 19*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 = 0.48 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3237, 27, 3042, 3229, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3237

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3229

\(\displaystyle -\frac {-\frac {19}{4} \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx-\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {19}{4} \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx-\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {\frac {19 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{2 d}-\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {19 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*Cos[c + d*x]/(d*(a + a*Sin[c + d*x])^(5/2)) - ((19*ArcTanh[(Sqrt[a]*C 
os[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(2*Sqrt[2]*Sqrt[a]*d) - 
(13*a*Cos[c + d*x])/(2*d*(a + a*Sin[c + d*x])^(3/2)))/(8*a^2)
 

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 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 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)]
 

rule 3237

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), 
x_Symbol] :> Simp[b*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), 
x] - Simp[1/(a^2*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2* 
m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && 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 = 0.32 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.80

method	result	size
default	\(-\frac {\left (-19 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \cos \left (d x +c \right )^{2}+38 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )+38 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+26 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}-44 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(193\)

Input:

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

Output:

-1/32/a^(9/2)*(-19*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1 
/2))*a^2*cos(d*x+c)^2+38*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2 
)/a^(1/2))*a^2*sin(d*x+c)+38*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^ 
(1/2)/a^(1/2))*a^2+26*(a-a*sin(d*x+c))^(3/2)*a^(1/2)-44*(a-a*sin(d*x+c))^( 
1/2)*a^(3/2))*(-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. 320 vs. \(2 (88) = 176\).

Time = 0.10 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.99 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {19 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (13 \, \cos \left (d x + c\right )^{2} + {\left (13 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) + 9 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:

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

Output:

1/64*(19*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2* 
cos(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*(13*cos(d*x + c)^2 + (13*cos(d*x + c) + 4)*sin(d*x + c) + 9*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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\frac {19 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {19 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} {\left (13 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \] Input:

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

Output:

1/64*(19*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos( 
-1/4*pi + 1/2*d*x + 1/2*c))) - 19*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2 
*c) + 1)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 2*sqrt(2)*(13*sqr 
t(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 11*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*sgn(cos(-1/4*pi + 1 
/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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