3.2.43 \(\int (a+b \sin ^2(c+d x))^{5/2} \, dx\) [143]

3.2.43.1 Optimal result

Integrand size = 16, antiderivative size = 210 \[ \int \left (a+b \sin ^2(c+d x)\right )^{5/2} \, dx=-\frac {4 b (2 a+b) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{15 d}-\frac {b \cos (c+d x) \sin (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}+\frac {\left (23 a^2+23 a b+8 b^2\right ) E\left (c+d x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(c+d x)}}{15 d \sqrt {1+\frac {b \sin ^2(c+d x)}{a}}}-\frac {4 a (a+b) (2 a+b) \operatorname {EllipticF}\left (c+d x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(c+d x)}{a}}}{15 d \sqrt {a+b \sin ^2(c+d x)}} \]

output

-1/5*b*cos(d*x+c)*sin(d*x+c)*(a+b*sin(d*x+c)^2)^(3/2)/d-4/15*b*(2*a+b)*cos 
(d*x+c)*sin(d*x+c)*(a+b*sin(d*x+c)^2)^(1/2)/d+1/15*(23*a^2+23*a*b+8*b^2)*( 
cos(d*x+c)^2)^(1/2)/cos(d*x+c)*EllipticE(sin(d*x+c),(-b/a)^(1/2))*(a+b*sin 
(d*x+c)^2)^(1/2)/d/(1+b*sin(d*x+c)^2/a)^(1/2)-4/15*a*(a+b)*(2*a+b)*(cos(d* 
x+c)^2)^(1/2)/cos(d*x+c)*EllipticF(sin(d*x+c),(-b/a)^(1/2))*(1+b*sin(d*x+c 
)^2/a)^(1/2)/d/(a+b*sin(d*x+c)^2)^(1/2)
 

3.2.43.2 Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.92 \[ \int \left (a+b \sin ^2(c+d x)\right )^{5/2} \, dx=\frac {16 a \left (23 a^2+23 a b+8 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (c+d x))}{a}} E\left (c+d x\left |-\frac {b}{a}\right .\right )-64 a \left (2 a^2+3 a b+b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (c+d x))}{a}} \operatorname {EllipticF}\left (c+d x,-\frac {b}{a}\right )-\sqrt {2} b \left (88 a^2+88 a b+25 b^2-28 b (2 a+b) \cos (2 (c+d x))+3 b^2 \cos (4 (c+d x))\right ) \sin (2 (c+d x))}{240 d \sqrt {2 a+b-b \cos (2 (c+d x))}} \]

input

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

output

(16*a*(23*a^2 + 23*a*b + 8*b^2)*Sqrt[(2*a + b - b*Cos[2*(c + d*x)])/a]*Ell 
ipticE[c + d*x, -(b/a)] - 64*a*(2*a^2 + 3*a*b + b^2)*Sqrt[(2*a + b - b*Cos 
[2*(c + d*x)])/a]*EllipticF[c + d*x, -(b/a)] - Sqrt[2]*b*(88*a^2 + 88*a*b 
+ 25*b^2 - 28*b*(2*a + b)*Cos[2*(c + d*x)] + 3*b^2*Cos[4*(c + d*x)])*Sin[2 
*(c + d*x)])/(240*d*Sqrt[2*a + b - b*Cos[2*(c + d*x)]])
 

3.2.43.3 Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {3042, 3659, 3042, 3649, 3042, 3651, 3042, 3657, 3042, 3656, 3662, 3042, 3661}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3659

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3649

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {b \left (23 a^2+23 b a+8 b^2\right ) \sin ^2(c+d x)+a \left (15 a^2+11 b a+4 b^2\right )}{\sqrt {b \sin ^2(c+d x)+a}}dx-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {b \left (23 a^2+23 b a+8 b^2\right ) \sin (c+d x)^2+a \left (15 a^2+11 b a+4 b^2\right )}{\sqrt {b \sin (c+d x)^2+a}}dx-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3651

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (23 a^2+23 a b+8 b^2\right ) \int \sqrt {b \sin ^2(c+d x)+a}dx-4 a (a+b) (2 a+b) \int \frac {1}{\sqrt {b \sin ^2(c+d x)+a}}dx\right )-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (23 a^2+23 a b+8 b^2\right ) \int \sqrt {b \sin (c+d x)^2+a}dx-4 a (a+b) (2 a+b) \int \frac {1}{\sqrt {b \sin (c+d x)^2+a}}dx\right )-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3657

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (23 a^2+23 a b+8 b^2\right ) \sqrt {a+b \sin ^2(c+d x)} \int \sqrt {\frac {b \sin ^2(c+d x)}{a}+1}dx}{\sqrt {\frac {b \sin ^2(c+d x)}{a}+1}}-4 a (a+b) (2 a+b) \int \frac {1}{\sqrt {b \sin (c+d x)^2+a}}dx\right )-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (23 a^2+23 a b+8 b^2\right ) \sqrt {a+b \sin ^2(c+d x)} \int \sqrt {\frac {b \sin (c+d x)^2}{a}+1}dx}{\sqrt {\frac {b \sin ^2(c+d x)}{a}+1}}-4 a (a+b) (2 a+b) \int \frac {1}{\sqrt {b \sin (c+d x)^2+a}}dx\right )-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3656

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (23 a^2+23 a b+8 b^2\right ) \sqrt {a+b \sin ^2(c+d x)} E\left (c+d x\left |-\frac {b}{a}\right .\right )}{d \sqrt {\frac {b \sin ^2(c+d x)}{a}+1}}-4 a (a+b) (2 a+b) \int \frac {1}{\sqrt {b \sin (c+d x)^2+a}}dx\right )-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3662

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (23 a^2+23 a b+8 b^2\right ) \sqrt {a+b \sin ^2(c+d x)} E\left (c+d x\left |-\frac {b}{a}\right .\right )}{d \sqrt {\frac {b \sin ^2(c+d x)}{a}+1}}-\frac {4 a (a+b) (2 a+b) \sqrt {\frac {b \sin ^2(c+d x)}{a}+1} \int \frac {1}{\sqrt {\frac {b \sin ^2(c+d x)}{a}+1}}dx}{\sqrt {a+b \sin ^2(c+d x)}}\right )-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (23 a^2+23 a b+8 b^2\right ) \sqrt {a+b \sin ^2(c+d x)} E\left (c+d x\left |-\frac {b}{a}\right .\right )}{d \sqrt {\frac {b \sin ^2(c+d x)}{a}+1}}-\frac {4 a (a+b) (2 a+b) \sqrt {\frac {b \sin ^2(c+d x)}{a}+1} \int \frac {1}{\sqrt {\frac {b \sin (c+d x)^2}{a}+1}}dx}{\sqrt {a+b \sin ^2(c+d x)}}\right )-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

\(\Big \downarrow \) 3661

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (23 a^2+23 a b+8 b^2\right ) \sqrt {a+b \sin ^2(c+d x)} E\left (c+d x\left |-\frac {b}{a}\right .\right )}{d \sqrt {\frac {b \sin ^2(c+d x)}{a}+1}}-\frac {4 a (a+b) (2 a+b) \sqrt {\frac {b \sin ^2(c+d x)}{a}+1} \operatorname {EllipticF}\left (c+d x,-\frac {b}{a}\right )}{d \sqrt {a+b \sin ^2(c+d x)}}\right )-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{3 d}\right )-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}\)

input

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

output

-1/5*(b*Cos[c + d*x]*Sin[c + d*x]*(a + b*Sin[c + d*x]^2)^(3/2))/d + ((-4*b 
*(2*a + b)*Cos[c + d*x]*Sin[c + d*x]*Sqrt[a + b*Sin[c + d*x]^2])/(3*d) + ( 
((23*a^2 + 23*a*b + 8*b^2)*EllipticE[c + d*x, -(b/a)]*Sqrt[a + b*Sin[c + d 
*x]^2])/(d*Sqrt[1 + (b*Sin[c + d*x]^2)/a]) - (4*a*(a + b)*(2*a + b)*Ellipt 
icF[c + d*x, -(b/a)]*Sqrt[1 + (b*Sin[c + d*x]^2)/a])/(d*Sqrt[a + b*Sin[c + 
 d*x]^2]))/3)/5
 

3.2.43.3.1 Defintions of rubi rules used

rule 3042

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

rule 3649

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]^2), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*Sin[e + f*x]*((a + b* 
Sin[e + f*x]^2)^p/(2*f*(p + 1))), x] + Simp[1/(2*(p + 1))   Int[(a + b*Sin[ 
e + f*x]^2)^(p - 1)*Simp[a*B + 2*a*A*(p + 1) + (2*A*b*(p + 1) + B*(b + 2*a* 
p + 2*b*p))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && G 
tQ[p, 0]
 

rule 3651

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + 
 (f_.)*(x_)]^2], x_Symbol] :> Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]^2], x] 
, x] + Simp[(A*b - a*B)/b   Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /; Fre 
eQ[{a, b, e, f, A, B}, x]
 

rule 3656

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a 
]/f)*EllipticE[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
 

rule 3657

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

rule 3659

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C 
os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p - 1)/(2*f*p)), x] + Sim 
p[1/(2*p)   Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(2*a 
+ b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[ 
a + b, 0] && GtQ[p, 1]
 

rule 3661

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(S 
qrt[a]*f))*EllipticF[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 
 0]
 

rule 3662

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[Sqrt[ 
1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a + b*Sin[e + f*x]^2]   Int[1/Sqrt[1 + (b*Si 
n[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]
 

3.2.43.4 Maple [A] (verified)

Time = 4.22 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.08

method	result	size
default	\(\frac {-\frac {b^{3} \left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{5}+\frac {\left (14 a \,b^{2}+10 b^{3}\right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}+\frac {\left (-11 a^{2} b -18 a \,b^{2}-7 b^{3}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}-\frac {8 a^{3} \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right )}{15}-\frac {4 a^{2} \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) b}{5}-\frac {4 a \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) b^{2}}{15}+\frac {23 \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, E\left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}}{15}+\frac {23 \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, E\left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b}{15}+\frac {8 \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, E\left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}}{15}}{\cos \left (d x +c \right ) \sqrt {a +\left (\sin ^{2}\left (d x +c \right )\right ) b}\, d}\)	\(437\)

input

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

output

(-1/5*b^3*cos(d*x+c)^6*sin(d*x+c)+1/15*(14*a*b^2+10*b^3)*cos(d*x+c)^4*sin( 
d*x+c)+1/15*(-11*a^2*b-18*a*b^2-7*b^3)*cos(d*x+c)^2*sin(d*x+c)-8/15*a^3*(c 
os(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticF(sin(d*x+c), 
(-1/a*b)^(1/2))-4/5*a^2*(cos(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^( 
1/2)*EllipticF(sin(d*x+c),(-1/a*b)^(1/2))*b-4/15*a*(cos(d*x+c)^2)^(1/2)*(- 
b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticF(sin(d*x+c),(-1/a*b)^(1/2))*b^2+2 
3/15*(cos(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticE(sin( 
d*x+c),(-1/a*b)^(1/2))*a^3+23/15*(cos(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+( 
a+b)/a)^(1/2)*EllipticE(sin(d*x+c),(-1/a*b)^(1/2))*a^2*b+8/15*(cos(d*x+c)^ 
2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticE(sin(d*x+c),(-1/a*b)^( 
1/2))*a*b^2)/cos(d*x+c)/(a+b*sin(d*x+c)^2)^(1/2)/d
 

3.2.43.5 Fricas [F]

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

input

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

output

integral((b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b 
+ b^2)*sqrt(-b*cos(d*x + c)^2 + a + b), x)
 

3.2.43.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.2.43.7 Maxima [F]

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

input

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

output

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

3.2.43.8 Giac [F]

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

input

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

output

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

3.2.43.9 Mupad [F(-1)]

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

input

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

output

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