Integrand size = 33, antiderivative size = 211 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {4 a^3 (9 A+5 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (3 A+5 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^3 (21 A+20 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a A \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (9 A+5 B) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \] Output:
-4/5*a^3*(9*A+5*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))* sec(d*x+c)^(1/2)/d+4/3*a^3*(3*A+5*B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2* d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+4/15*a^3*(21*A+20*B)*sec(d*x+c)^(1/2 )*sin(d*x+c)/d+2/5*a*A*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+2/ 15*(9*A+5*B)*sec(d*x+c)^(1/2)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.98 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^3 e^{-i d x} \csc (c) \sec (c) \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (2 (9 A+5 B) e^{-i (c-d x)} \left (-1+e^{4 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\frac {1}{2} \sec ^2(c+d x) \sin (2 c) \left (-18 i (9 A+5 B) \cos (c+d x)-54 i A \cos (3 (c+d x))-30 i B \cos (3 (c+d x))+40 (3 A+5 B) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+66 A \sin (c+d x)+45 B \sin (c+d x)+30 A \sin (2 (c+d x))+10 B \sin (2 (c+d x))+54 A \sin (3 (c+d x))+45 B \sin (3 (c+d x))\right )\right )}{30 d} \] Input:
Integrate[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x ]
Output:
(a^3*Csc[c]*Sec[c]*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*((2*(9*A + 5 *B)*(-1 + E^((4*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2 , 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c - d*x)) + (Sec[c + d*x]^2*Sin[2 *c]*((-18*I)*(9*A + 5*B)*Cos[c + d*x] - (54*I)*A*Cos[3*(c + d*x)] - (30*I) *B*Cos[3*(c + d*x)] + 40*(3*A + 5*B)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x )/2, 2] + 66*A*Sin[c + d*x] + 45*B*Sin[c + d*x] + 30*A*Sin[2*(c + d*x)] + 10*B*Sin[2*(c + d*x)] + 54*A*Sin[3*(c + d*x)] + 45*B*Sin[3*(c + d*x)]))/2) )/(30*d*E^(I*d*x))
Time = 1.32 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3439, 3042, 4506, 27, 3042, 4506, 3042, 4485, 27, 3042, 4274, 3042, 4258, 3042, 3119, 3120}
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.
| \(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^3 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3439 |
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^3 (A \sec (c+d x)+B)}{\sqrt {\sec (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A \csc \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {2}{5} \int -\frac {(\sec (c+d x) a+a)^2 (a (A-5 B)-a (9 A+5 B) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}-\frac {1}{5} \int \frac {(\sec (c+d x) a+a)^2 (a (A-5 B)-a (9 A+5 B) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}-\frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (A-5 B)-a (9 A+5 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left (a^2 (6 A-5 B)-a^2 (21 A+20 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (6 A-5 B)-a^2 (21 A+20 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (2 \int \frac {3 a^3 (9 A+5 B)-5 a^3 (3 A+5 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)}}dx-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (\int \frac {3 a^3 (9 A+5 B)-5 a^3 (3 A+5 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (\int \frac {3 a^3 (9 A+5 B)-5 a^3 (3 A+5 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (3 a^3 (9 A+5 B) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 a^3 (3 A+5 B) \int \sqrt {\sec (c+d x)}dx-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (3 a^3 (9 A+5 B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 a^3 (3 A+5 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (-5 a^3 (3 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^3 (9 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (-5 a^3 (3 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^3 (9 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (-5 a^3 (3 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {6 a^3 (9 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}-\frac {2}{3} \left (-\frac {2 a^3 (21 A+20 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {10 a^3 (3 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^3 (9 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
Input:
Int[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x]
Output:
(2*a*A*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((2 *(9*A + 5*B)*Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(3* d) - (2*((6*a^3*(9*A + 5*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S qrt[Sec[c + d*x]])/d - (10*a^3*(3*A + 5*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (2*a^3*(21*A + 20*B)*Sqrt[Sec[c + d* x]]*Sin[c + d*x])/d))/3)/5
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[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(915\) vs. \(2(190)=380\).
Time = 186.37 (sec) , antiderivative size = 916, normalized size of antiderivative = 4.34
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(916\) |
| parts | \(\text {Expression too large to display}\) | \(1061\) |
Input:
int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x,method=_RETURNV ERBOSE)
Output:
-4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3/(8*sin (1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 2*d*x+1/2*c)^3*(216*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*A*(2*sin( 1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-108*A*(sin(1/2*d*x+1/2*c)^2)^(1/2) *EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*si n(1/2*d*x+1/2*c)^4+180*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-100*B*(si n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2 *d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-60*B*(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)^4-246*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+60*A*( 2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos (1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+108*A*(sin(1/2*d*x+1/2*c)^2) ^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 /2)*sin(1/2*d*x+1/2*c)^2-190*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+100 *B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*s in(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+60*B*(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+72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-1 5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip...
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.15 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (9 \, {\left (6 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 3 \, A a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorith m="fricas")
Output:
-2/15*(5*I*sqrt(2)*(3*A + 5*B)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 5*B)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt( 2)*(9*A + 5*B)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(9*A + 5*B)*a^3*co s(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c ) - I*sin(d*x + c))) - (9*(6*A + 5*B)*a^3*cos(d*x + c)^2 + 5*(3*A + B)*a^3 *cos(d*x + c) + 3*A*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^ 2)
Timed out. \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c)**(7/2),x)
Output:
Timed out
\[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorith m="maxima")
Output:
integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)
\[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorith m="giac")
Output:
integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)
Timed out. \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3 \,d x \] Input:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^3,x)
Output:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^3, x)
\[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=a^{3} \left (3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right ) b +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a \right ) \] Input:
int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x)
Output:
a**3*(3*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*a + int(sqr t(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*b + int(sqrt(sec(c + d*x)) *cos(c + d*x)**4*sec(c + d*x)**3,x)*b + int(sqrt(sec(c + d*x))*cos(c + d*x )**3*sec(c + d*x)**3,x)*a + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**3*sec(c + d*x)**3,x)*b + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3 ,x)*a + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)*b + in t(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*a)