Integrand size = 45, antiderivative size = 253 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\frac {a^{5/2} (304 A+200 B+163 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 d}+\frac {a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {a^2 (16 A+24 B+17 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{32 d \sqrt {\sec (c+d x)}}+\frac {a (8 B+5 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt {\sec (c+d x)}}+\frac {C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}} \] Output:
1/64*a^(5/2)*(304*A+200*B+163*C)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c) )^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/192*a^3*(432*A+392*B+299*C) *sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+1/32*a^2*(16*A+24*B+ 17*C)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+1/24*a*(8*B+5*C )*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+1/4*C*(a+a*cos(d*x+ c))^(5/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)
Time = 1.55 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.66 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (3 \sqrt {2} (304 A+200 B+163 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (528 A+632 B+581 C+(96 A+272 B+362 C) \cos (c+d x)+4 (8 B+23 C) \cos (2 (c+d x))+12 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{384 d} \] Input:
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sqrt[Sec[c + d*x]],x]
Output:
(a^2*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[S ec[c + d*x]]*(3*Sqrt[2]*(304*A + 200*B + 163*C)*ArcSin[Sqrt[2]*Sin[(c + d* x)/2]] + 2*Sqrt[Cos[c + d*x]]*(528*A + 632*B + 581*C + (96*A + 272*B + 362 *C)*Cos[c + d*x] + 4*(8*B + 23*C)*Cos[2*(c + d*x)] + 12*C*Cos[3*(c + d*x)] )*Sin[(c + d*x)/2]))/(384*d)
Time = 1.68 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.356, Rules used = {3042, 4709, 3042, 3524, 27, 3042, 3455, 27, 3042, 3455, 27, 3042, 3460, 3042, 3253, 223}
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.
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{5/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\sqrt {\cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3524 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (a (8 A+C)+a (8 B+5 C) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx}{4 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (a (8 A+C)+a (8 B+5 C) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (8 A+C)+a (8 B+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((48 A+8 B+11 C) a^2+3 (16 A+24 B+17 C) \cos (c+d x) a^2\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((48 A+8 B+11 C) a^2+3 (16 A+24 B+17 C) \cos (c+d x) a^2\right )}{\sqrt {\cos (c+d x)}}dx+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((48 A+8 B+11 C) a^2+3 (16 A+24 B+17 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((240 A+104 B+95 C) a^3+(432 A+392 B+299 C) \cos (c+d x) a^3\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {3 a^3 (16 A+24 B+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((240 A+104 B+95 C) a^3+(432 A+392 B+299 C) \cos (c+d x) a^3\right )}{\sqrt {\cos (c+d x)}}dx+\frac {3 a^3 (16 A+24 B+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((240 A+104 B+95 C) a^3+(432 A+392 B+299 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {3 a^3 (16 A+24 B+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (304 A+200 B+163 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a^4 (432 A+392 B+299 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^3 (16 A+24 B+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (304 A+200 B+163 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^4 (432 A+392 B+299 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^3 (16 A+24 B+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {a^4 (432 A+392 B+299 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 a^3 (304 A+200 B+163 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^3 (16 A+24 B+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (8 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}+\frac {1}{6} \left (\frac {3 a^3 (16 A+24 B+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{7/2} (304 A+200 B+163 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a^4 (432 A+392 B+299 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )\right )}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\right )\) |
Input:
Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sqr t[Sec[c + d*x]],x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(4*d) + ((a^2*(8*B + 5*C)*Sqrt[Cos[c + d*x]]*( a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3*a^3*(16*A + 24*B + 17* C)*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + ((3*a ^(7/2)*(304*A + 200*B + 163*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Co s[c + d*x]]])/d + (a^4*(432*A + 392*B + 299*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/4)/6)/(8*a))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_. ) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n} , x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !Lt Q[m, -2^(-1)] && NeQ[m + n + 2, 0]
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]
Time = 9.54 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.19
| method | result | size |
| default | \(\frac {\sqrt {2}\, a^{2} \left (912 A \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+600 B \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+489 C \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (96 \cos \left (d x +c \right )+528\right ) \sin \left (d x +c \right ) A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (64 \cos \left (d x +c \right )^{2}+272 \cos \left (d x +c \right )+600\right ) \sin \left (d x +c \right ) B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (48 \cos \left (d x +c \right )^{3}+184 \cos \left (d x +c \right )^{2}+326 \cos \left (d x +c \right )+489\right ) \sin \left (d x +c \right ) C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (d x +c \right ) \sqrt {\sec \left (d x +c \right )}}{192 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(302\) |
| parts | \(\frac {A \,a^{2} \left (19 \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\sin \left (2 d x +2 c \right )+11 \sin \left (d x +c \right )\right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right ) \sqrt {\sec \left (d x +c \right )}}{4 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {B \sqrt {2}\, a^{2} \left (75 \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (8 \cos \left (d x +c \right )^{2}+34 \cos \left (d x +c \right )+75\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (d x +c \right ) \sqrt {\sec \left (d x +c \right )}}{24 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {C \sqrt {2}\, a^{2} \left (489 \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (48 \cos \left (d x +c \right )^{3}+184 \cos \left (d x +c \right )^{2}+326 \cos \left (d x +c \right )+489\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (d x +c \right ) \sqrt {\sec \left (d x +c \right )}}{192 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(436\) |
Input:
int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2 ),x,method=_RETURNVERBOSE)
Output:
1/192/d*2^(1/2)*a^2*(912*A*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d* x+c))+600*B*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+489*C*arc tan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+(96*cos(d*x+c)+528)*sin( d*x+c)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+(64*cos(d*x+c)^2+272*cos(d*x+c) +600)*sin(d*x+c)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+(48*cos(d*x+c)^3+184* cos(d*x+c)^2+326*cos(d*x+c)+489)*sin(d*x+c)*C*(cos(d*x+c)/(1+cos(d*x+c)))^ (1/2))*(cos(1/2*d*x+1/2*c)^2*a)^(1/2)*cos(d*x+c)*sec(d*x+c)^(1/2)/(1+cos(d *x+c))/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
Time = 0.36 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.77 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=-\frac {3 \, {\left ({\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (48 \, C a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (8 \, B + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 136 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (176 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(1/2),x, algorithm="fricas")
Output:
-1/192*(3*((304*A + 200*B + 163*C)*a^2*cos(d*x + c) + (304*A + 200*B + 163 *C)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt( a)*sin(d*x + c))) - (48*C*a^2*cos(d*x + c)^4 + 8*(8*B + 23*C)*a^2*cos(d*x + c)^3 + 2*(48*A + 136*B + 163*C)*a^2*cos(d*x + c)^2 + 3*(176*A + 200*B + 163*C)*a^2*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d* x + c)))/(d*cos(d*x + c) + d)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x +c)**(1/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 10517 vs. \(2 (217) = 434\).
Time = 1.00 (sec) , antiderivative size = 10517, normalized size of antiderivative = 41.57 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(1/2),x, algorithm="maxima")
Output:
1/768*(48*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin (2*d*x + 2*c) + a^2*sin(2*d*x + 2*c) - (a^2*cos(2*d*x + 2*c) - 10*a^2)*sin (1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d *x + 2*c), cos(2*d*x + 2*c) + 1)) + (a^2*sin(2*d*x + 2*c)*sin(1/2*arctan2( sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - a^2*cos(2*d*x + 2*c) + 10*a^2 + (a^ 2*cos(2*d*x + 2*c) - 10*a^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 19*(a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*si n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(s in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d* x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2 (sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c) , cos(2*d*x + 2*c)))) + 1) - a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1 )) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*a...
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\int \sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \] Input:
int((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
Output:
int((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)
\[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \, dx=\sqrt {a}\, a^{2} \left (2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}d x \right ) a \right ) \] Input:
int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2 ),x)
Output:
sqrt(a)*a**2*(2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x) ,x)*a + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x),x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**4,x)*c + int(s qrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3,x)*b + 2*int(sqrt (sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3,x)*c + int(sqrt(sec( c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2,x)*a + 2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x) )*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2,x)*c + int(sqrt(sec(c + d*x))*sqr t(cos(c + d*x) + 1),x)*a)