\(\int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) [521]

Optimal result

Integrand size = 23, antiderivative size = 160 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 \cos (c+d x)}{b d \sqrt {a+b \sin (c+d x)}}-\frac {4 E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {4 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{b^2 d \sqrt {a+b \sin (c+d x)}} \] Output:

-2*cos(d*x+c)/b/d/(a+b*sin(d*x+c))^(1/2)+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*sin(d*x+c 
))/(a+b))^(1/2)+4*a*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)
 

Mathematica [A] (verified)

Time = 3.01 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {4 (a+b) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}-2 \left (b \cos (c+d x)+2 a \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}\right )}{b^2 d \sqrt {a+b \sin (c+d x)}} \] Input:

Integrate[Cos[c + d*x]^2/(a + b*Sin[c + d*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3172, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3172

\(\displaystyle -\frac {2 \int \frac {\sin (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {2 \cos (c+d x)}{b d \sqrt {a+b \sin (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 \int \frac {\sin (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {2 \cos (c+d x)}{b d \sqrt {a+b \sin (c+d x)}}\)

\(\Big \downarrow \) 3231

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

Input:

Int[Cos[c + d*x]^2/(a + b*Sin[c + d*x])^(3/2),x]
 

Output:

(-2*Cos[c + d*x])/(b*d*Sqrt[a + b*Sin[c + d*x]]) - (2*((2*EllipticE[(c - P 
i/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(b*d*Sqrt[(a + b*Si 
n[c + d*x])/(a + b)]) - (2*a*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]])))/b
 

Defintions of rubi rules used

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 3172

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(433\) vs. \(2(155)=310\).

Time = 0.86 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.71

method	result	size
default	\(\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticE}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2}-4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticE}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2}-4 a \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b +4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2}+2 b^{2} \sin \left (d x +c \right )^{2}-2 b^{2}}{b^{3} \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\)	\(434\)

Input:

int(cos(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2*(2*((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^2-2*((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))*b^2-2*a*((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)*EllipticF(((a+b* 
sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b+2*((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)*Ell 
ipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^2+b^2*sin(d*x 
+c)^2-b^2)/b^3/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 = 0.11 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.78 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {b \sin \left (d x + c\right ) + a} b^{2} \cos \left (d x + c\right ) - 4 \, {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {\frac {1}{2} i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right ) - 4 \, {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {-\frac {1}{2} i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right ) + 6 \, {\left (-i \, b^{2} \sin \left (d x + c\right ) - i \, a b\right )} \sqrt {\frac {1}{2} i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right )\right ) + 6 \, {\left (i \, b^{2} \sin \left (d x + c\right ) + i \, a b\right )} \sqrt {-\frac {1}{2} i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right )\right )}}{3 \, {\left (b^{4} d \sin \left (d x + c\right ) + a b^{3} d\right )}} \] Input:

integrate(cos(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-2/3*(3*sqrt(b*sin(d*x + c) + a)*b^2*cos(d*x + c) - 4*(a*b*sin(d*x + c) + 
a^2)*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) - 4*(a*b*sin(d*x + c) + a^2)*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*(-I*b^2*sin(d*x + c) - I*a*b)*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)) + 
6*(I*b^2*sin(d*x + c) + I*a*b)*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)))/(b^4*d*sin(d*x + c) + a*b^3*d)
 

Sympy [F]

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

integrate(cos(d*x+c)**2/(a+b*sin(d*x+c))**(3/2),x)
 

Output:

Integral(cos(c + d*x)**2/(a + b*sin(c + d*x))**(3/2), x)
 

Maxima [F]

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

integrate(cos(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(cos(d*x + c)^2/(b*sin(d*x + c) + a)^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sin(c + d*x)*b + a)*cos(c + d*x)**2)/(sin(c + d*x)**2*b**2 + 2*s 
in(c + d*x)*a*b + a**2),x)