\(\int \frac {1}{(a+b \sin (c+d x))^{5/2}} \, dx\) [82]

Optimal result

Integrand size = 14, antiderivative size = 231 \[ \int \frac {1}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}+\frac {8 a E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}} \] Output:

2/3*b*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^(3/2)+8/3*a*b*cos(d*x+c)/(a^ 
2-b^2)^2/d/(a+b*sin(d*x+c))^(1/2)-8/3*a*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x 
),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)^2/d/((a+b*sin( 
d*x+c))/(a+b))^(1/2)-2/3*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2)*(b/( 
a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/(a^2-b^2)/d/(a+b*sin(d*x+c))^( 
1/2)
 

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {2 \left (-4 a (a+b)^2 E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}+(a-b) (a+b)^2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}+b \cos (c+d x) \left (5 a^2-b^2+4 a b \sin (c+d x)\right )\right )}{3 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))^{3/2}} \] Input:

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

Output:

(2*(-4*a*(a + b)^2*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b 
*Sin[c + d*x])/(a + b))^(3/2) + (a - b)*(a + b)^2*EllipticF[(-2*c + Pi - 2 
*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a + b))^(3/2) + b*Cos[c + d 
*x]*(5*a^2 - b^2 + 4*a*b*Sin[c + d*x])))/(3*(a - b)^2*(a + b)^2*d*(a + b*S 
in[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.071, Rules used = {3042, 3143, 27, 3042, 3233, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

\(\displaystyle \int \frac {1}{(a+b \sin (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3143

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3231

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

\(\displaystyle \frac {2 b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}+\frac {\frac {8 a b \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}+\frac {\frac {8 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}}{3 \left (a^2-b^2\right )}\)

Input:

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

Output:

(2*b*Cos[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^(3/2)) + ((8*a*b* 
Cos[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) + ((8*a*EllipticE[( 
c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(d*Sqrt[(a + b 
*Sin[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2 
*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x 
]]))/(a^2 - b^2))/(3*(a^2 - b^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 3042

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

rule 3132

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

rule 3134

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

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

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

rule 3143

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

rule 3231

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

rule 3233

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 + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(494\) vs. \(2(216)=432\).

Time = 0.24 (sec) , antiderivative size = 495, normalized size of antiderivative = 2.14

method	result	size
default	\(\frac {\sqrt {-\left (-a -b \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}}\, \left (\frac {2 \sqrt {-\left (-a -b \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}}}{3 \left (a^{2}-b^{2}\right ) b \left (\sin \left (d x +c \right )+\frac {a}{b}\right )^{2}}+\frac {8 b \cos \left (d x +c \right )^{2} a}{3 \left (a^{2}-b^{2}\right )^{2} \sqrt {-\left (-a -b \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}}}+\frac {2 \left (3 a^{2}+b^{2}\right ) \left (\frac {a}{b}-1\right ) \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {\frac {b \left (1-\sin \left (d x +c \right )\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )}{\left (3 a^{4}-6 b^{2} a^{2}+3 b^{4}\right ) \sqrt {-\left (-a -b \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}}}+\frac {8 a b \left (\frac {a}{b}-1\right ) \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {\frac {b \left (1-\sin \left (d x +c \right )\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \left (\left (-\frac {a}{b}-1\right ) \operatorname {EllipticE}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )+\operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )\right )}{3 \left (a^{2}-b^{2}\right )^{2} \sqrt {-\left (-a -b \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}}}\right )}{\cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\)	\(495\)

Input:

int(1/(a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*(2/3/(a^2-b^2)/b*(-(-a-b*sin(d*x+c 
))*cos(d*x+c)^2)^(1/2)/(sin(d*x+c)+a/b)^2+8/3*b*cos(d*x+c)^2/(a^2-b^2)^2*a 
/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)+2*(3*a^2+b^2)/(3*a^4-6*a^2*b^2+3* 
b^4)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2) 
*(-b*(1+sin(d*x+c))/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*E 
llipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+8/3*a*b/(a^2- 
b^2)^2*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/ 
2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2) 
*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+E 
llipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))))/cos(d*x+c)/ 
(a+b*sin(d*x+c))^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 687, normalized size of antiderivative = 2.97 \[ \int \frac {1}{(a+b \sin (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

-2/9*((a^4 + 4*a^2*b^2 + 3*b^4 - (a^2*b^2 + 3*b^4)*cos(d*x + c)^2 + 2*(a^3 
*b + 3*a*b^3)*sin(d*x + c))*sqrt(1/2*I*b)*weierstrassPInverse(-4/3*(4*a^2 
- 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I 
*b*sin(d*x + c) - 2*I*a)/b) + (a^4 + 4*a^2*b^2 + 3*b^4 - (a^2*b^2 + 3*b^4) 
*cos(d*x + c)^2 + 2*(a^3*b + 3*a*b^3)*sin(d*x + c))*sqrt(-1/2*I*b)*weierst 
rassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1 
/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b) - 12*(-I*a*b^3*cos(d 
*x + c)^2 + 2*I*a^2*b^2*sin(d*x + c) + I*a^3*b + I*a*b^3)*sqrt(1/2*I*b)*we 
ierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, w 
eierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b 
^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b)) - 12*(I*a*b^3* 
cos(d*x + c)^2 - 2*I*a^2*b^2*sin(d*x + c) - I*a^3*b - I*a*b^3)*sqrt(-1/2*I 
*b)*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2) 
/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I* 
a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b)) + 3*(4 
*a*b^3*cos(d*x + c)*sin(d*x + c) + (5*a^2*b^2 - b^4)*cos(d*x + c))*sqrt(b* 
sin(d*x + c) + a))/((a^4*b^3 - 2*a^2*b^5 + b^7)*d*cos(d*x + c)^2 - 2*(a^5* 
b^2 - 2*a^3*b^4 + a*b^6)*d*sin(d*x + c) - (a^6*b - a^4*b^3 - a^2*b^5 + b^7 
)*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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