\(\int \sec ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\) [488]

Optimal result

Integrand size = 23, antiderivative size = 168 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\frac {\sec (c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{d}-\frac {a E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{d \sqrt {a+b \sin (c+d x)}} \] Output:

sec(d*x+c)*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/d+a*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)/d/((a+b*s 
in(d*x+c))/(a+b))^(1/2)+(a^2-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)/d/(a+b*sin(d*x+c))^(1 
/2)
 

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.97 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\frac {a b \sec (c+d x)+a (a+b) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}-\left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+a^2 \tan (c+d x)+b^2 \tan (c+d x)+a b \sin (c+d x) \tan (c+d x)}{d \sqrt {a+b \sin (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3170, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3231

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

Input:

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

Output:

(Sec[c + d*x]*(b + a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/d + ((-2*a*El 
lipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(d*Sq 
rt[(a + b*Sin[c + d*x])/(a + b)]) + (2*(a^2 - b^2)*EllipticF[(c - Pi/2 + d 
*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Si 
n[c + d*x]]))/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 3170

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(632\) vs. \(2(162)=324\).

Time = 0.84 (sec) , antiderivative size = 633, normalized size of antiderivative = 3.77

method	result	size
default	\(\frac {\sqrt {\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) b +\cos \left (d x +c \right )^{2} a}\, \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \operatorname {EllipticE}\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}-\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \operatorname {EllipticE}\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}-\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b +\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}-a \,b^{2} \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right ) a^{2} b +b^{3} \sin \left (d x +c \right )+2 a \,b^{2}\right )}{b \sqrt {-\left (a +b \sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\)	\(633\)

Input:

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

Output:

1/b*(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*((b/(a-b)*sin(d*x+c)+ 
1/(a-b)*a)^(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)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b) 
)^(1/2))*a^3-(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(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)*EllipticE((b/(a-b)*sin(d*x 
+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a*b^2-(b/(a-b)*sin(d*x+c)+1/(a-b 
)*a)^(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)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2 
))*a^2*b+(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(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)+ 
1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*b^3-a*b^2*cos(d*x+c)^2+sin(d*x+c)*a^ 
2*b+b^3*sin(d*x+c)+2*a*b^2)/(-(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(1+sin(d*x+c 
)))^(1/2)/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 = 0.15 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.57 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\frac {3 i \, a \sqrt {\frac {1}{2} i \, b} b \cos \left (d x + c\right ) {\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 ) - 3 i \, a \sqrt {-\frac {1}{2} i \, b} b \cos \left (d x + c\right ) {\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 ) + {\left (2 \, a^{2} - 3 \, b^{2}\right )} \sqrt {\frac {1}{2} i \, b} \cos \left (d x + c\right ) {\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 ) + {\left (2 \, a^{2} - 3 \, b^{2}\right )} \sqrt {-\frac {1}{2} i \, b} \cos \left (d x + c\right ) {\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 ) + 3 \, {\left (a b \sin \left (d x + c\right ) + b^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{3 \, b d \cos \left (d x + c\right )} \] Input:

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

Output:

1/3*(3*I*a*sqrt(1/2*I*b)*b*cos(d*x + c)*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*I*a*sqrt(-1/2*I*b)*b*cos(d*x + c)*weierstr 
assZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, weiers 
trassPInverse(-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*a^2 - 3*b^2)* 
sqrt(1/2*I*b)*cos(d*x + c)*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*a^2 - 3*b^2)*sqrt(-1/2*I*b)*cos(d*x + c)*weierstrassPInv 
erse(-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*(a*b*sin(d*x + c) + b^2) 
*sqrt(b*sin(d*x + c) + a))/(b*d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(sec(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)*sec(d*x + c)^2, x)
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sin(c + d*x)*b + a)*sec(c + d*x)**2*sin(c + d*x),x)*b + int(sqrt( 
sin(c + d*x)*b + a)*sec(c + d*x)**2,x)*a