Integrand size = 23, antiderivative size = 203 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d}+\frac {\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\frac {\left (a^2+3 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {a \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{d \sqrt {a+b \sin (c+d x)}} \]
sec(d*x+c)*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^(3/2)/d+a*b*cos(d*x+c)*(a+b*s in(d*x+c))^(1/2)/d+(a^2+3*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2 *c+1/4*Pi+1/2*d*x)*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*sin(d*x+c))/(a+b))^(1/2)-a*(a^2-b^2)* (sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(co s(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)
Time = 0.66 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 a^2 b \sec (c+d x)+\left (a^3+a^2 b+3 a b^2+3 b^3\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}-a \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+a^3 \tan (c+d x)+3 a b^2 \tan (c+d x)+a^2 b \sin (c+d x) \tan (c+d x)+b^3 \sin (c+d x) \tan (c+d x)}{d \sqrt {a+b \sin (c+d x)}} \]
(2*a^2*b*Sec[c + d*x] + (a^3 + a^2*b + 3*a*b^2 + 3*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - a*(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^3*Tan[c + d*x] + 3*a*b^2*Tan[c + d*x] + a^2*b*Sin[c + d*x]*Tan[c + d*x] + b^3*Sin[c + d*x]*Tan[c + d*x])/(d*Sqrt[a + b*Sin[c + d *x]])
Time = 1.05 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3170, 27, 3042, 3232, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^{5/2}}{\cos (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3170 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\int \frac {3}{2} \sqrt {a+b \sin (c+d x)} \left (b^2+a \sin (c+d x) b\right )dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \int \sqrt {a+b \sin (c+d x)} \left (b^2+a \sin (c+d x) b\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \int \sqrt {a+b \sin (c+d x)} \left (b^2+a \sin (c+d x) b\right )dx\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {2}{3} \int \frac {4 a b^2+\left (a^2+3 b^2\right ) \sin (c+d x) b}{2 \sqrt {a+b \sin (c+d x)}}dx-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \int \frac {4 a b^2+\left (a^2+3 b^2\right ) \sin (c+d x) b}{\sqrt {a+b \sin (c+d x)}}dx-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \int \frac {4 a b^2+\left (a^2+3 b^2\right ) \sin (c+d x) b}{\sqrt {a+b \sin (c+d x)}}dx-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \left (\left (a^2+3 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx-a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \left (\left (a^2+3 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx-a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \left (\frac {\left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \left (\frac {\left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \left (\frac {2 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \left (\frac {2 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}\right )-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \left (\frac {2 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}\right )-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d}-\frac {3}{2} \left (\frac {1}{3} \left (\frac {2 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}\right )-\frac {2 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\right )\) |
(Sec[c + d*x]*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(3/2))/d - (3*((-2 *a*b*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(3*d) + ((2*(a^2 + 3*b^2)*Elli pticE[(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*(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]]))/3))/2
3.6.2.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
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]
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])
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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[1/(m + 1) Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ [{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]
Leaf count of result is larger than twice the leaf count of optimal. \(1038\) vs. \(2(257)=514\).
Time = 6.27 (sec) , antiderivative size = 1039, normalized size of antiderivative = 5.12
1/b*(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^(1/2)*((b/(a-b)*sin(d*x+c)+ a/(a-b))^(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)+a/(a-b))^(1/2),((a-b)/(a+b))^(1 /2))*a^4+2*(b/(a-b)*sin(d*x+c)+a/(a-b))^(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)+ a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^2-3*(b/(a-b)*sin(d*x+c)+a/(a-b)) ^(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)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^4 -(b/(a-b)*sin(d*x+c)+a/(a-b))^(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)+a/(a-b))^( 1/2),((a-b)/(a+b))^(1/2))*a^3*b-3*(b/(a-b)*sin(d*x+c)+a/(a-b))^(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)*Ellipti cF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^2+(b/(a-b )*sin(d*x+c)+a/(a-b))^(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)+a/(a-b))^(1/2),((a -b)/(a+b))^(1/2))*a*b^3+3*(b/(a-b)*sin(d*x+c)+a/(a-b))^(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)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^4-a^2*b^2*cos(d*x+c)^ 2-b^4*cos(d*x+c)^2+a^3*b*sin(d*x+c)+3*a*b^3*sin(d*x+c)+3*a^2*b^2+b^4)/(-(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...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 474, normalized size of antiderivative = 2.33 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 \, \sqrt {2} {\left (a^{3} - 3 \, a b^{2}\right )} \sqrt {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 ) + 2 \, \sqrt {2} {\left (a^{3} - 3 \, a b^{2}\right )} \sqrt {-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 \, \sqrt {2} {\left (-i \, a^{2} b - 3 i \, b^{3}\right )} \sqrt {i \, 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 \, \sqrt {2} {\left (i \, a^{2} b + 3 i \, b^{3}\right )} \sqrt {-i \, 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 ) + 6 \, {\left (2 \, a b^{2} + {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{6 \, b d \cos \left (d x + c\right )} \]
1/6*(2*sqrt(2)*(a^3 - 3*a*b^2)*sqrt(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*sqrt(2)*(a^3 - 3*a*b^2)*sqrt(- 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) - 3*sqrt(2)*(-I*a^2*b - 3*I*b^3)*sqrt(I*b)*cos(d*x + c)*weierstrassZe ta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, weierstrassP Inverse(-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*sqrt(2)*(I*a^2*b + 3* I*b^3)*sqrt(-I*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)) + 6*(2*a*b^2 + (a^2*b + b^3)*sin(d*x + c))*sqrt(b*sin (d*x + c) + a))/(b*d*cos(d*x + c))
Timed out. \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]