Integrand size = 25, antiderivative size = 237 \[ \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac {6 \left (a^4-4 a^2 b^2-4 b^4\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {6 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^5}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}-\frac {6 (a+b \sin (c+d x))^2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}} \]
2/5*a*b*(3*a^2-10*b^2)*(e*cos(d*x+c))^(3/2)/d/e^5+6/5*b*(a^2-2*b^2)*(e*cos (d*x+c))^(3/2)*(a+b*sin(d*x+c))/d/e^5+2/5*(b+a*sin(d*x+c))*(a+b*sin(d*x+c) )^3/d/e/(e*cos(d*x+c))^(5/2)-6/5*(a+b*sin(d*x+c))^2*(a*b-(a^2-2*b^2)*sin(d *x+c))/d/e^3/(e*cos(d*x+c))^(1/2)-6/5*(a^4-4*a^2*b^2-4*b^4)*(cos(1/2*d*x+1 /2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e *cos(d*x+c))^(1/2)/d/e^4/cos(d*x+c)^(1/2)
Time = 1.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.64 \[ \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 \left (-20 a b^3-3 \left (a^4-4 a^2 b^2-4 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+4 a b \left (a^2+b^2\right ) \sec ^2(c+d x)+3 a^4 \sin (c+d x)-12 a^2 b^2 \sin (c+d x)-7 b^4 \sin (c+d x)+\left (a^4+6 a^2 b^2+b^4\right ) \sec (c+d x) \tan (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}} \]
(2*(-20*a*b^3 - 3*(a^4 - 4*a^2*b^2 - 4*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[( c + d*x)/2, 2] + 4*a*b*(a^2 + b^2)*Sec[c + d*x]^2 + 3*a^4*Sin[c + d*x] - 1 2*a^2*b^2*Sin[c + d*x] - 7*b^4*Sin[c + d*x] + (a^4 + 6*a^2*b^2 + b^4)*Sec[ c + d*x]*Tan[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c + d*x]])
Time = 1.20 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3170, 27, 3042, 3340, 27, 3042, 3341, 27, 3042, 3148, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3170 |
\(\displaystyle \frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}-\frac {2 \int -\frac {3 (a+b \sin (c+d x))^2 \left (a^2-b \sin (c+d x) a-2 b^2\right )}{2 (e \cos (c+d x))^{3/2}}dx}{5 e^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \int \frac {(a+b \sin (c+d x))^2 \left (a^2-b \sin (c+d x) a-2 b^2\right )}{(e \cos (c+d x))^{3/2}}dx}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {(a+b \sin (c+d x))^2 \left (a^2-b \sin (c+d x) a-2 b^2\right )}{(e \cos (c+d x))^{3/2}}dx}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3340 |
\(\displaystyle \frac {3 \left (-\frac {2 \int \frac {1}{2} \sqrt {e \cos (c+d x)} (a+b \sin (c+d x)) \left (a \left (a^2-6 b^2\right )+5 b \left (a^2-2 b^2\right ) \sin (c+d x)\right )dx}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \left (-\frac {\int \sqrt {e \cos (c+d x)} (a+b \sin (c+d x)) \left (a \left (a^2-6 b^2\right )+5 b \left (a^2-2 b^2\right ) \sin (c+d x)\right )dx}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (-\frac {\int \sqrt {e \cos (c+d x)} (a+b \sin (c+d x)) \left (a \left (a^2-6 b^2\right )+5 b \left (a^2-2 b^2\right ) \sin (c+d x)\right )dx}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3341 |
\(\displaystyle \frac {3 \left (-\frac {\frac {2}{5} \int \frac {5}{2} \sqrt {e \cos (c+d x)} \left (a^4-4 b^2 a^2+b \left (3 a^2-10 b^2\right ) \sin (c+d x) a-4 b^4\right )dx-\frac {2 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \left (-\frac {\int \sqrt {e \cos (c+d x)} \left (a^4-4 b^2 a^2+b \left (3 a^2-10 b^2\right ) \sin (c+d x) a-4 b^4\right )dx-\frac {2 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (-\frac {\int \sqrt {e \cos (c+d x)} \left (a^4-4 b^2 a^2+b \left (3 a^2-10 b^2\right ) \sin (c+d x) a-4 b^4\right )dx-\frac {2 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle \frac {3 \left (-\frac {\left (a^4-4 a^2 b^2-4 b^4\right ) \int \sqrt {e \cos (c+d x)}dx-\frac {2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e}-\frac {2 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (-\frac {\left (a^4-4 a^2 b^2-4 b^4\right ) \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e}-\frac {2 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {3 \left (-\frac {\frac {\left (a^4-4 a^2 b^2-4 b^4\right ) \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)}}-\frac {2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e}-\frac {2 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (-\frac {\frac {\left (a^4-4 a^2 b^2-4 b^4\right ) \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)}}-\frac {2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e}-\frac {2 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {3 \left (-\frac {2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {-\frac {2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e}-\frac {2 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}+\frac {2 \left (a^4-4 a^2 b^2-4 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}}{e^2}\right )}{5 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}\) |
(2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^3)/(5*d*e*(e*Cos[c + d*x])^(5 /2)) + (3*((-2*(a + b*Sin[c + d*x])^2*(a*b - (a^2 - 2*b^2)*Sin[c + d*x]))/ (d*e*Sqrt[e*Cos[c + d*x]]) - ((-2*a*b*(3*a^2 - 10*b^2)*(e*Cos[c + d*x])^(3 /2))/(3*d*e) + (2*(a^4 - 4*a^2*b^2 - 4*b^4)*Sqrt[e*Cos[c + d*x]]*EllipticE [(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]) - (2*b*(a^2 - 2*b^2)*(e*Cos[c + d *x])^(3/2)*(a + b*Sin[c + d*x]))/(d*e))/e^2))/(5*e^2)
3.6.71.3.1 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[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(g* Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^m*((d + c*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*Si n[e + f*x])^(m - 1)*Simp[a*c*(p + 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ [m, 0] && LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[1/(m + p + 1) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim p[a*c*(m + p + 1) + b*d*m + (a*d*m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && !LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S implerQ[c + d*x, a + b*x])
Leaf count of result is larger than twice the leaf count of optimal. \(873\) vs. \(2(241)=482\).
Time = 11.41 (sec) , antiderivative size = 874, normalized size of antiderivative = 3.69
method | result | size |
default | \(\text {Expression too large to display}\) | \(874\) |
parts | \(\text {Expression too large to display}\) | \(1170\) |
2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)/( -2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^3*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+ 1/2*c)^6*a^4-96*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a^2*b^2-56*cos(1/2 *d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*b^4-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*( sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2* d*x+1/2*c)^4*a^4+48*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2 )^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4*a^2*b^2 +48*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic E(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4*b^4-24*cos(1/2*d*x+1/2* c)*sin(1/2*d*x+1/2*c)^4*a^4+96*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2 *b^2+56*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^4+12*(2*sin(1/2*d*x+1/2* c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^ (1/2))*sin(1/2*d*x+1/2*c)^2*a^4-48*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1 /2*c)^2*a^2*b^2-48*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2) ^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2*b^4-80*s in(1/2*d*x+1/2*c)^5*a*b^3+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^4-12 *cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^2*b^2-12*cos(1/2*d*x+1/2*c)*sin (1/2*d*x+1/2*c)^2*b^4-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) ^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^4+12*(sin(1/2*d*x+1...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (i \, a^{4} - 4 i \, a^{2} b^{2} - 4 i \, b^{4}\right )} \sqrt {e} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, a^{4} + 4 i \, a^{2} b^{2} + 4 i \, b^{4}\right )} \sqrt {e} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (20 \, a b^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} b - 4 \, a b^{3} - {\left (a^{4} + 6 \, a^{2} b^{2} + b^{4} + {\left (3 \, a^{4} - 12 \, a^{2} b^{2} - 7 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{5 \, d e^{4} \cos \left (d x + c\right )^{3}} \]
-1/5*(3*sqrt(2)*(I*a^4 - 4*I*a^2*b^2 - 4*I*b^4)*sqrt(e)*cos(d*x + c)^3*wei erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-I*a^4 + 4*I*a^2*b^2 + 4*I*b^4)*sqrt(e)*cos(d*x + c)^3*w eierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(20*a*b^3*cos(d*x + c)^2 - 4*a^3*b - 4*a*b^3 - (a^4 + 6*a^2*b^2 + b^4 + (3*a^4 - 12*a^2*b^2 - 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(e *cos(d*x + c)))/(d*e^4*cos(d*x + c)^3)
Timed out. \[ \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]