\(\int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) [1181]

Optimal result

Integrand size = 29, antiderivative size = 254 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {8 \left (32 a^2-9 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{15 b^5 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {8 a \left (32 a^2-17 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{15 b^5 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}+\frac {4 \cos (c+d x) \left (32 a^2-9 b^2+8 a b \sin (c+d x)\right )}{15 b^4 d \sqrt {a+b \sin (c+d x)}} \] Output:

-8/15*(32*a^2-9*b^2)*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)/b^5/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-8/15*a 
*(32*a^2-17*b^2)*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)/b^5/d/(a+b*sin(d*x+c))^(1/2)+2/15*cos( 
d*x+c)^3*(8*a+3*b*sin(d*x+c))/b^2/d/(a+b*sin(d*x+c))^(3/2)+4/15*cos(d*x+c) 
*(32*a^2-9*b^2+8*a*b*sin(d*x+c))/b^4/d/(a+b*sin(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 6.93 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {-32 (a+b)^2 \left (32 a^2-9 b^2\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}+32 a (a+b) \left (32 a^2-17 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}+2 b \cos (c+d x) \left (256 a^3-24 a b^2-16 a b^2 \cos (2 (c+d x))+b \left (320 a^2-69 b^2\right ) \sin (c+d x)+3 b^3 \sin (3 (c+d x))\right )}{60 b^5 d (a+b \sin (c+d x))^{3/2}} \] Input:

Integrate[(Cos[c + d*x]^4*Sin[c + d*x])/(a + b*Sin[c + d*x])^(5/2),x]
 

Output:

(-32*(a + b)^2*(32*a^2 - 9*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a 
+ b)]*((a + b*Sin[c + d*x])/(a + b))^(3/2) + 32*a*(a + b)*(32*a^2 - 17*b^2 
)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a 
 + b))^(3/2) + 2*b*Cos[c + d*x]*(256*a^3 - 24*a*b^2 - 16*a*b^2*Cos[2*(c + 
d*x)] + b*(320*a^2 - 69*b^2)*Sin[c + d*x] + 3*b^3*Sin[3*(c + d*x)]))/(60*b 
^5*d*(a + b*Sin[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.517, Rules used = {3042, 3342, 27, 3042, 3342, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3342

\(\displaystyle \frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}-\frac {4 \int -\frac {\cos ^2(c+d x) (3 b+8 a \sin (c+d x))}{2 (a+b \sin (c+d x))^{3/2}}dx}{5 b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \int \frac {\cos ^2(c+d x) (3 b+8 a \sin (c+d x))}{(a+b \sin (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {\cos (c+d x)^2 (3 b+8 a \sin (c+d x))}{(a+b \sin (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3342

\(\displaystyle \frac {2 \left (\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {4 \int -\frac {8 a b+\left (32 a^2-9 b^2\right ) \sin (c+d x)}{2 \sqrt {a+b \sin (c+d x)}}dx}{3 b^2}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {2 \int \frac {8 a b+\left (32 a^2-9 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \int \frac {8 a b+\left (32 a^2-9 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {\left (32 a^2-9 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx}{b}-\frac {a \left (32 a^2-17 b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {\left (32 a^2-9 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx}{b}-\frac {a \left (32 a^2-17 b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {\left (32 a^2-9 b^2\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (32 a^2-17 b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {\left (32 a^2-9 b^2\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (32 a^2-17 b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {2 \left (32 a^2-9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (32 a^2-17 b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {2 \left (32 a^2-9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (32 a^2-17 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{b \sqrt {a+b \sin (c+d x)}}\right )}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {2 \left (32 a^2-9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (32 a^2-17 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{b \sqrt {a+b \sin (c+d x)}}\right )}{3 b^2}+\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {2 \left (\frac {2 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (\frac {2 \left (32 a^2-9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 a \left (32 a^2-17 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \sin (c+d x)}}\right )}{3 b^2}\right )}{5 b^2}+\frac {2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}\)

Input:

Int[(Cos[c + d*x]^4*Sin[c + d*x])/(a + b*Sin[c + d*x])^(5/2),x]
 

Output:

(2*Cos[c + d*x]^3*(8*a + 3*b*Sin[c + d*x]))/(15*b^2*d*(a + b*Sin[c + d*x]) 
^(3/2)) + (2*((2*Cos[c + d*x]*(32*a^2 - 9*b^2 + 8*a*b*Sin[c + d*x]))/(3*b^ 
2*d*Sqrt[a + b*Sin[c + d*x]]) + (2*((2*(32*a^2 - 9*b^2)*EllipticE[(c - Pi/ 
2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(b*d*Sqrt[(a + b*Sin[ 
c + d*x])/(a + b)]) - (2*a*(32*a^2 - 17*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 
 (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(b*d*Sqrt[a + b*Sin[c 
+ d*x]])))/(3*b^2)))/(5*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 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 3342

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*C 
os[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d*p 
 + b*d*(m + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Simp[g^2*(( 
p - 1)/(b^2*(m + 1)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin 
[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Sin[e + f*x 
], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[m, -1] && GtQ[p, 1] && NeQ[m + p + 1, 0] && IntegerQ[2*m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1429\) vs. \(2(239)=478\).

Time = 2.34 (sec) , antiderivative size = 1430, normalized size of antiderivative = 5.63

method	result	size
default	\(\text {Expression too large to display}\)	\(1430\)

Input:

int(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE 
)
 

Output:

2/15*(3*b^5*cos(d*x+c)^4*sin(d*x+c)-8*a*b^4*cos(d*x+c)^4+(80*a^2*b^3-18*b^ 
5)*cos(d*x+c)^2*sin(d*x+c)+(64*a^3*b^2-2*a*b^4)*cos(d*x+c)^2-4*(b/(a-b)*si 
n(d*x+c)+a/(a-b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(a+b)*sin( 
d*x+c)+b/(a+b))^(1/2)*b*(32*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),( 
(a-b)/(a+b))^(1/2))*a^4-41*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),(( 
a-b)/(a+b))^(1/2))*a^2*b^2+9*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2), 
((a-b)/(a+b))^(1/2))*b^4-32*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),( 
(a-b)/(a+b))^(1/2))*a^3*b+24*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2), 
((a-b)/(a+b))^(1/2))*a^2*b^2+17*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/ 
2),((a-b)/(a+b))^(1/2))*a*b^3-9*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/ 
2),((a-b)/(a+b))^(1/2))*b^4)*sin(d*x+c)+128*(b/(a-b)*sin(d*x+c)+a/(a-b))^( 
1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/ 
2)*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b 
-96*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2) 
*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b) 
)^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2-68*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2) 
*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*E 
llipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3+3 
6*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*( 
-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 750, normalized size of antiderivative = 2.95 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x, algorithm="fri 
cas")
 

Output:

2/45*(8*(32*a^5 + 11*a^3*b^2 - 21*a*b^4 - (32*a^3*b^2 - 21*a*b^4)*cos(d*x 
+ c)^2 + 2*(32*a^4*b - 21*a^2*b^3)*sin(d*x + c))*sqrt(1/2*I*b)*weierstrass 
PInverse(-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) + 8*(32*a^5 + 11*a^3*b^2 
- 21*a*b^4 - (32*a^3*b^2 - 21*a*b^4)*cos(d*x + c)^2 + 2*(32*a^4*b - 21*a^2 
*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) + 12*(32*I*a^4*b + 23*I*a^2*b^3 - 9*I*b^5 + (-32*I*a 
^2*b^3 + 9*I*b^5)*cos(d*x + c)^2 + 2*(32*I*a^3*b^2 - 9*I*a*b^4)*sin(d*x + 
c))*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)) + 12*(-32*I*a^4*b - 23*I*a^2*b^3 + 9*I*b^5 + (32*I*a^2*b^3 - 9*I*b^5 
)*cos(d*x + c)^2 + 2*(-32*I*a^3*b^2 + 9*I*a*b^4)*sin(d*x + c))*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*(8 
*a*b^4*cos(d*x + c)^3 - 2*(32*a^3*b^2 - a*b^4)*cos(d*x + c) - (3*b^5*cos(d 
*x + c)^3 + 2*(40*a^2*b^3 - 9*b^5)*cos(d*x + c))*sin(d*x + c))*sqrt(b*sin( 
d*x + c) + a))/(b^8*d*cos(d*x + c)^2 - 2*a*b^7*d*sin(d*x + c) - (a^2*b^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**4*sin(d*x+c)/(a+b*sin(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x, algorithm="max 
ima")
 

Output:

integrate(cos(d*x + c)^4*sin(d*x + c)/(b*sin(d*x + c) + a)^(5/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x, algorithm="gia 
c")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int((cos(c + d*x)^4*sin(c + d*x))/(a + b*sin(c + d*x))^(5/2),x)
 

Output:

int((cos(c + d*x)^4*sin(c + d*x))/(a + b*sin(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\sin \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )}{\sin \left (d x +c \right )^{3} b^{3}+3 \sin \left (d x +c \right )^{2} a \,b^{2}+3 \sin \left (d x +c \right ) a^{2} b +a^{3}}d x \] Input:

int(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x)
 

Output:

int((sqrt(sin(c + d*x)*b + a)*cos(c + d*x)**4*sin(c + d*x))/(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)