\(\int \cos ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) [498]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 299 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{21 d}-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}+\frac {4 \left (5 a^4+102 a^2 b^2+21 b^4\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{315 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{315 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (a \left (5 a^2+27 b^2\right )+3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b d} \] Output:

-8/21*a*b*cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2)/d-2/9*b*cos(d*x+c)^3*(a+b*si 
n(d*x+c))^(3/2)/d-4/315*(5*a^4+102*a^2*b^2+21*b^4)*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^2/d/((a+b*s 
in(d*x+c))/(a+b))^(1/2)-4/315*a*(5*a^4+22*a^2*b^2-27*b^4)*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^2/d/(a+b*sin(d*x+c))^(1/2)+2/315*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)*(a 
*(5*a^2+27*b^2)+3*b*(25*a^2+7*b^2)*sin(d*x+c))/b/d
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {-16 \left (16 b \left (5 a^3 b+3 a b^3\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right )+\left (5 a^4+102 a^2 b^2+21 b^4\right ) \left ((a+b) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right )\right )\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+b (a+b \sin (c+d x)) \left (\left (40 a^3-354 a b^2\right ) \cos (c+d x)+2 b \left (-95 a b \cos (3 (c+d x))+\left (150 a^2+7 b^2-35 b^2 \cos (2 (c+d x))\right ) \sin (2 (c+d x))\right )\right )}{1260 b^2 d \sqrt {a+b \sin (c+d x)}} \] Input:

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

Output:

(-16*(16*b*(5*a^3*b + 3*a*b^3)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + 
 b)] + (5*a^4 + 102*a^2*b^2 + 21*b^4)*((a + b)*EllipticE[(-2*c + Pi - 2*d* 
x)/4, (2*b)/(a + b)] - a*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)])) 
*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + b*(a + b*Sin[c + d*x])*((40*a^3 - 35 
4*a*b^2)*Cos[c + d*x] + 2*b*(-95*a*b*Cos[3*(c + d*x)] + (150*a^2 + 7*b^2 - 
 35*b^2*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])))/(1260*b^2*d*Sqrt[a + b*Sin[c 
 + d*x]])
 

Rubi [A] (verified)

Time = 1.72 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.03, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 3171, 27, 3042, 3341, 27, 3042, 3344, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3171

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+4 b \sin (c+d x) a+b^2\right )dx-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \cos (c+d x)^2 \sqrt {a+b \sin (c+d x)} \left (3 a^2+4 b \sin (c+d x) a+b^2\right )dx-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3341

\(\displaystyle \frac {1}{3} \left (\frac {2}{7} \int \frac {\cos ^2(c+d x) \left (a \left (21 a^2+11 b^2\right )+b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{2 \sqrt {a+b \sin (c+d x)}}dx-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \int \frac {\cos ^2(c+d x) \left (a \left (21 a^2+11 b^2\right )+b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}}dx-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \int \frac {\cos (c+d x)^2 \left (a \left (21 a^2+11 b^2\right )+b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}}dx-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3344

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {4 \int \frac {16 a \left (5 a^2+3 b^2\right ) b^2+\left (5 a^4+102 b^2 a^2+21 b^4\right ) \sin (c+d x) b}{2 \sqrt {a+b \sin (c+d x)}}dx}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \int \frac {16 a \left (5 a^2+3 b^2\right ) b^2+\left (5 a^4+102 b^2 a^2+21 b^4\right ) \sin (c+d x) b}{\sqrt {a+b \sin (c+d x)}}dx}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \int \frac {16 a \left (5 a^2+3 b^2\right ) b^2+\left (5 a^4+102 b^2 a^2+21 b^4\right ) \sin (c+d x) b}{\sqrt {a+b \sin (c+d x)}}dx}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (\left (5 a^4+102 a^2 b^2+21 b^4\right ) \int \sqrt {a+b \sin (c+d x)}dx-a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (\left (5 a^4+102 a^2 b^2+21 b^4\right ) \int \sqrt {a+b \sin (c+d x)}dx-a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (\frac {\left (5 a^4+102 a^2 b^2+21 b^4\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (\frac {\left (5 a^4+102 a^2 b^2+21 b^4\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (\frac {2 \left (5 a^4+102 a^2 b^2+21 b^4\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (\frac {2 \left (5 a^4+102 a^2 b^2+21 b^4\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (\frac {2 \left (5 a^4+102 a^2 b^2+21 b^4\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{15 b d}+\frac {2 \left (\frac {2 \left (5 a^4+102 a^2 b^2+21 b^4\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}\right )}{15 b^2}\right )-\frac {8 a b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\right )-\frac {2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}\)

Input:

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

Output:

(-2*b*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^(3/2))/(9*d) + ((-8*a*b*Cos[c + 
d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(7*d) + ((2*Cos[c + d*x]*Sqrt[a + b*Sin[c 
 + d*x]]*(a*(5*a^2 + 27*b^2) + 3*b*(25*a^2 + 7*b^2)*Sin[c + d*x]))/(15*b*d 
) + (2*((2*(5*a^4 + 102*a^2*b^2 + 21*b^4)*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*(5*a^4 + 22*a^2*b^2 - 27*b^4)*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] 
])))/(15*b^2))/7)/3
 

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 3171
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[1/(m + p)   Int[(g*Cos[e + f*x])^p* 
(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1) 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] 
 && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m])
 

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 3341
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* 
(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S 
imp[1/(m + p + 1)   Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim 
p[a*c*(m + p + 1) + b*d*m + (a*d*m + b*c*(m + p + 1))*Sin[e + f*x], x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] 
 &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S 
implerQ[c + d*x, a + b*x])
 

rule 3344
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* 
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* 
p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( 
(p - 1)/(b^2*(m + p)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si 
n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 
2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 
, 0] && IntegerQ[2*m]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1189\) vs. \(2(280)=560\).

Time = 3.64 (sec) , antiderivative size = 1190, normalized size of antiderivative = 3.98

method result size
default \(\text {Expression too large to display}\) \(1190\)

Input:

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

Output:

2/315*(-35*b^6*sin(d*x+c)^6-130*a*b^5*sin(d*x+c)^5+10*((a+b*sin(d*x+c))/(a 
-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2) 
*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5*b+150*( 
(a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d 
*x+c))/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b)) 
^(1/2))*a^4*b^2+44*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b) 
)^(1/2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b)) 
^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^3-108*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b 
*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2)*EllipticF(((a 
+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^4-54*((a+b*sin(d*x+ 
c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d*x+c))/(a-b)) 
^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^5 
-42*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+ 
sin(d*x+c))/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/( 
a+b))^(1/2))*b^6-10*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b 
))^(1/2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b) 
)^(1/2),((a-b)/(a+b))^(1/2))*a^6-194*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(s 
in(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2)*EllipticE(((a+b* 
sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^2+162*((a+b*sin(d*x+c) 
)/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d*x+c))/(a-b)...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.75 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/945*(4*(5*a^5 - 18*a^3*b^2 - 51*a*b^4)*sqrt(1/2*I*b)*weierstrassPInvers 
e(-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) + 4*(5*a^5 - 18*a^3*b^2 - 51*a*b 
^4)*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) + 6*(5*I*a^4*b + 102*I*a^2*b^3 + 21*I*b^5)*sqrt(1/2*I*b)*weierstrass 
Zeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, weierstras 
sPInverse(-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)) + 6*(-5*I*a^4*b - 102*I 
*a^2*b^3 - 21*I*b^5)*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*(95*a*b^4*cos(d*x + c)^3 - (5*a^3*b^2 + 27*a 
*b^4)*cos(d*x + c) + (35*b^5*cos(d*x + c)^3 - 3*(25*a^2*b^3 + 7*b^5)*cos(d 
*x + c))*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/(b^3*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sin(c + d*x)*b + a)*cos(c + d*x)**2*sin(c + d*x)**2,x)*b**2 + 2*i 
nt(sqrt(sin(c + d*x)*b + a)*cos(c + d*x)**2*sin(c + d*x),x)*a*b + int(sqrt 
(sin(c + d*x)*b + a)*cos(c + d*x)**2,x)*a**2