Integrand size = 33, antiderivative size = 172 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 a (3 A+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 A b \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:
-2/5*a*(3*A+5*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*se c(d*x+c)^(1/2)/d+2/3*b*(A+3*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/ 2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+2/5*a*(3*A+5*C)*sec(d*x+c)^(1/2)*sin(d*x+c )/d+2/3*A*b*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*A*sec(d*x+c)^(5/2)*sin(d*x +c)/d
Time = 3.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.71 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {\sec ^{\frac {5}{2}}(c+d x) \left (-12 a (3 A+5 C) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 b (A+3 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (10 A b \cos (c+d x)+3 a (5 (A+C)+(3 A+5 C) \cos (2 (c+d x)))) \sin (c+d x)\right )}{30 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x ]
Output:
(Sec[c + d*x]^(5/2)*(-12*a*(3*A + 5*C)*Cos[c + d*x]^(5/2)*EllipticE[(c + d *x)/2, 2] + 20*b*(A + 3*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*(10*A*b*Cos[c + d*x] + 3*a*(5*(A + C) + (3*A + 5*C)*Cos[2*(c + d*x)]))*S in[c + d*x]))/(30*d)
Time = 0.89 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.88, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4709, 3042, 3511, 27, 3042, 3500, 27, 3042, 3227, 3042, 3116, 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.
| \(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{7/2} (a+b \cos (c+d x)) \left (A+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(a+b \cos (c+d x)) \left (C \cos ^2(c+d x)+A\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3511 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{5} \int \frac {5 b C \cos ^2(c+d x)+a (3 A+5 C) \cos (c+d x)+5 A b}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {5 b C \cos ^2(c+d x)+a (3 A+5 C) \cos (c+d x)+5 A b}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {5 b C \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 A b}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {3 a (3 A+5 C)+5 b (A+3 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {3 a (3 A+5 C)+5 b (A+3 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {3 a (3 A+5 C)+5 b (A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (3 a (3 A+5 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+5 b (A+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (3 a (3 A+5 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+5 b (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (3 a (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )+5 b (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (3 a (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+5 b (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 b (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (3 a (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {10 b (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {10 A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
Input:
Int[(a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*Sin[c + d*x])/(5*d*Cos[c + d *x]^(5/2)) + ((10*A*b*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((10*b*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/d + 3*a*(3*A + 5*C)*((-2*EllipticE[(c + d *x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ (-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/( b^2*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d )) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] + b *C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e , f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(704\) vs. \(2(151)=302\).
Time = 395.03 (sec) , antiderivative size = 705, normalized size of antiderivative = 4.10
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(705\) |
| parts | \(\text {Expression too large to display}\) | \(892\) |
Input:
int((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x,method=_RETURNV ERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*b*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2 *c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2* A*b*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2 )^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2* cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*C*a/sin(1/2*d*x+1/2*c)^2 /(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2) ^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(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 ))+2/5*a*A/(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)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c )-12*(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)^4-24*cos(1/2*d*x+1/2*c)* sin(1/2*d*x+1/2*c)^4+12*(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+8*s in(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli pticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))*(-2*si n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2))/sin(1/2*d*x+1/2*c)/(2*cos( 1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.28 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {-5 i \, \sqrt {2} {\left (A + 3 \, C\right )} b \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (A + 3 \, C\right )} b \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, {\left (3 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{2} + 5 \, A b \cos \left (d x + c\right ) + 3 \, A a\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorith m="fricas")
Output:
1/15*(-5*I*sqrt(2)*(A + 3*C)*b*cos(d*x + c)^2*weierstrassPInverse(-4, 0, c os(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(A + 3*C)*b*cos(d*x + c)^2*wei erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(3*A + 5*C)*a*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(3*A + 5*C)*a*cos(d*x + c)^2 *weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c))) + 2*(3*(3*A + 5*C)*a*cos(d*x + c)^2 + 5*A*b*cos(d*x + c) + 3*A*a) *sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)
Output:
Timed out
\[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorith m="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)
\[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorith m="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)
Timed out. \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \] Input:
int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x)),x)
Output:
int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x)), x)
\[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) a b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right ) b c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a^{2} \] Input:
int((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)
Output:
int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*a*b + int(sqrt(sec( c + d*x))*cos(c + d*x)**3*sec(c + d*x)**3,x)*b*c + int(sqrt(sec(c + d*x))* cos(c + d*x)**2*sec(c + d*x)**3,x)*a*c + int(sqrt(sec(c + d*x))*sec(c + d* x)**3,x)*a**2