\(\int \cos ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\) [487]

Optimal result

Integrand size = 23, antiderivative size = 247 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}+\frac {4 a \left (3 a^2+29 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{105 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 \left (3 a^4+2 a^2 b^2-5 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{105 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+5 b^2+24 a b \sin (c+d x)\right )}{105 b d} \] Output:

-2/7*b*cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2)/d-4/105*a*(3*a^2+29*b^2)*Ellipt 
icE(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)-4/105*(3*a^4+2*a^2*b^2-5*b^4)*Inv 
erseJacobiAM(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)+2/105*cos(d*x+c)*(a+b*sin(d*x 
+c))^(1/2)*(3*a^2+5*b^2+24*a*b*sin(d*x+c))/b/d
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\frac {-8 a \left (3 a^3+3 a^2 b+29 a b^2+29 b^3\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+8 \left (3 a^4+2 a^2 b^2-5 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+b \cos (c+d x) \left (12 a^3+38 a b^2-78 a b^2 \cos (2 (c+d x))+b \left (108 a^2+5 b^2\right ) \sin (c+d x)-15 b^3 \sin (3 (c+d x))\right )}{210 b^2 d \sqrt {a+b \sin (c+d x)}} \] Input:

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

Output:

(-8*a*(3*a^3 + 3*a^2*b + 29*a*b^2 + 29*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/ 
4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + 8*(3*a^4 + 2*a^2*b^ 
2 - 5*b^4)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin 
[c + d*x])/(a + b)] + b*Cos[c + d*x]*(12*a^3 + 38*a*b^2 - 78*a*b^2*Cos[2*( 
c + d*x)] + b*(108*a^2 + 5*b^2)*Sin[c + d*x] - 15*b^3*Sin[3*(c + d*x)]))/( 
210*b^2*d*Sqrt[a + b*Sin[c + d*x]])
 

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.01, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3171, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3344

\(\displaystyle \frac {1}{7} \left (\frac {4 \int \frac {\left (27 a^2+5 b^2\right ) b^2+a \left (3 a^2+29 b^2\right ) \sin (c+d x) b}{2 \sqrt {a+b \sin (c+d x)}}dx}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {2 \int \frac {\left (27 a^2+5 b^2\right ) b^2+a \left (3 a^2+29 b^2\right ) \sin (c+d x) b}{\sqrt {a+b \sin (c+d x)}}dx}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {2 \int \frac {\left (27 a^2+5 b^2\right ) b^2+a \left (3 a^2+29 b^2\right ) \sin (c+d x) b}{\sqrt {a+b \sin (c+d x)}}dx}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {1}{7} \left (\frac {2 \left (a \left (3 a^2+29 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx-\left (3 a^4+2 a^2 b^2-5 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {2 \left (a \left (3 a^2+29 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx-\left (3 a^4+2 a^2 b^2-5 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{7} \left (\frac {2 \left (\frac {a \left (3 a^2+29 b^2\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\left (3 a^4+2 a^2 b^2-5 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {2 \left (\frac {a \left (3 a^2+29 b^2\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\left (3 a^4+2 a^2 b^2-5 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{7} \left (\frac {2 \left (\frac {2 a \left (3 a^2+29 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\left (3 a^4+2 a^2 b^2-5 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{7} \left (\frac {2 \left (\frac {2 a \left (3 a^2+29 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (3 a^4+2 a^2 b^2-5 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {2 \left (\frac {2 a \left (3 a^2+29 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (3 a^4+2 a^2 b^2-5 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}\right )}{15 b^2}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{7} \left (\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{15 b d}+\frac {2 \left (\frac {2 a \left (3 a^2+29 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \left (3 a^4+2 a^2 b^2-5 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}\right )}{15 b^2}\right )-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}\)

Input:

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

Output:

(-2*b*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(7*d) + ((2*Cos[c + d*x]*Sq 
rt[a + b*Sin[c + d*x]]*(3*a^2 + 5*b^2 + 24*a*b*Sin[c + d*x]))/(15*b*d) + ( 
2*((2*a*(3*a^2 + 29*b^2)*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*(3*a^4 + 
 2*a^2*b^2 - 5*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
 

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 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. \(942\) vs. \(2(232)=464\).

Time = 1.45 (sec) , antiderivative size = 943, normalized size of antiderivative = 3.82

method	result	size
default	\(\text {Expression too large to display}\)	\(943\)

Input:

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

Output:

2/105*(-15*b^5*sin(d*x+c)^5+6*((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+48*((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)*Ellip 
ticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2+4*((a+b*s 
in(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^3-48*((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^4-10*((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))*b^5-6*((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)*Ellipti 
cE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5-52*((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^3 
*b^2+58*((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*b^4-39*a*b^4*sin(d*x+c)^4-27*a^2*b^3*sin(d*x+c)^3+25*b^ 
5*sin(d*x+c)^3-3*a^3*b^2*sin(d*x+c)^2+49*a*b^4*sin(d*x+c)^2+27*a^2*b^3*...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 481, normalized size of antiderivative = 1.95 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/315*(2*(6*a^4 - 23*a^2*b^2 - 15*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) + 2*(6*a^4 - 23*a^2*b^2 - 15*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*(3*I*a^3*b + 29*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)) + 6*(-3*I*a^3*b - 29*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*(1 
5*b^4*cos(d*x + c)^3 - 24*a*b^3*cos(d*x + c)*sin(d*x + c) - (3*a^2*b^2 + 5 
*b^4)*cos(d*x + c))*sqrt(b*sin(d*x + c) + a))/(b^3*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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