Integrand size = 45, antiderivative size = 241 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\frac {2 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{5/2} d}+\frac {(5 A+3 B-43 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(5 A+3 B-11 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \]
-1/4*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(3/2)+1/16*(5* A+3*B-11*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+2*C*arc sin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c) ^(1/2)/a^(5/2)/d+1/32*(5*A+3*B-43*C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2) /cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2 )/a^(5/2)/d*2^(1/2)
Time = 5.13 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\left ((5 A+3 B-43 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+32 \sqrt {2} C \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {1+\sec (c+d x)}}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {1}{4} (5 A+3 B-11 C+(A+7 B-15 C) \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{4 d (a (1+\cos (c+d x)))^{5/2}} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^(5 /2)*Sqrt[Sec[c + d*x]]),x]
(Cos[(c + d*x)/2]^5*((((5*A + 3*B - 43*C)*ArcSin[Tan[(c + d*x)/2]] + 32*Sq rt[2]*C*ArcTan[Tan[(c + d*x)/2]/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]])*Sq rt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[1 + Sec[c + d*x] ])/Sqrt[Sec[(c + d*x)/2]^2] + ((5*A + 3*B - 11*C + (A + 7*B - 15*C)*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(-Sin[(c + d*x)/2] + Sin[(3* (c + d*x))/2]))/4))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2))
Time = 1.43 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4709, 3042, 3520, 27, 3042, 3456, 27, 3042, 3461, 3042, 3253, 223, 3261, 218}
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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos (c+d x)^2}{\sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{(\cos (c+d x) a+a)^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (a (5 A+3 B-3 C)+8 a C \cos (c+d x))}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (a (5 A+3 B-3 C)+8 a C \cos (c+d x))}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a (5 A+3 B-3 C)+8 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3456 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {(5 A+3 B-11 C) a^2+32 C \cos (c+d x) a^2}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {(5 A+3 B-11 C) a^2+32 C \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {(5 A+3 B-11 C) a^2+32 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3461 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (5 A+3 B-43 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx+32 a C \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx}{4 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (5 A+3 B-43 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+32 a C \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3253 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (5 A+3 B-43 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {64 a C \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{4 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (5 A+3 B-43 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {64 a^{3/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {64 a^{3/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^3 (5 A+3 B-43 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}}{4 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sqrt {2} a^{3/2} (5 A+3 B-43 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {64 a^{3/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {a (5 A+3 B-11 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^(5/2)*Sq rt[Sec[c + d*x]]),x]
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A - B + C)*Cos[c + d*x]^(3/2 )*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) + (((64*a^(3/2)*C*ArcSin[(S qrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (Sqrt[2]*a^(3/2)*(5*A + 3*B - 43*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sq rt[a + a*Cos[c + d*x]])])/d)/(4*a^2) + (a*(5*A + 3*B - 11*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)))/(8*a^2))
3.14.57.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & & NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim p[(A*b - a*B)/b Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) , x], x] + Simp[B/b Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] , x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-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. \(578\) vs. \(2(202)=404\).
Time = 1.89 (sec) , antiderivative size = 579, normalized size of antiderivative = 2.40
method | result | size |
default | \(-\frac {\sqrt {2}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-1\right ) \sqrt {\frac {a}{\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}}\, \left (2 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \left (\csc ^{3}\left (d x +c \right )\right )-2 B \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \left (\csc ^{3}\left (d x +c \right )\right )+2 C \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \left (\csc ^{3}\left (d x +c \right )\right )+3 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+5 B \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-32 C \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-1}\right )-13 C \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-5 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-3 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+43 C \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{32 a^{3} d \sqrt {-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-1}}\, \left (-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1\right )^{\frac {3}{2}}}\) | \(579\) |
parts | \(\frac {A \sqrt {2}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+5 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-10 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}}{32 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {B \sqrt {2}\, \left (7 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-6 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}}{32 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}-\frac {C \sqrt {2}\, \left (15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-32 \sqrt {2}\, \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 (\cos ^{2}\left (d x +c \right )\right )+11 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-43 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-64 \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \cos \left (d x +c \right )-86 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-32 \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )-43 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}}{32 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(699\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2 ),x,method=_RETURNVERBOSE)
-1/32/a^3/d*2^(1/2)/(-((1-cos(d*x+c))^2*csc(d*x+c)^2+1)/((1-cos(d*x+c))^2* csc(d*x+c)^2-1))^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(3/2)*((1-cos(d* x+c))^2*csc(d*x+c)^2+1)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(a/((1-cos(d*x+c ))^2*csc(d*x+c)^2+1))^(1/2)*(2*A*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)* (1-cos(d*x+c))^3*csc(d*x+c)^3-2*B*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2) *(1-cos(d*x+c))^3*csc(d*x+c)^3+2*C*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2 )*(1-cos(d*x+c))^3*csc(d*x+c)^3+3*A*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/ 2)*(-cot(d*x+c)+csc(d*x+c))+5*B*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*( -cot(d*x+c)+csc(d*x+c))-32*C*2^(1/2)*arctan(2^(1/2)*(-(1-cos(d*x+c))^2*csc (d*x+c)^2+1)^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(-cot(d*x+c)+csc(d*x+ c)))-13*C*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-cot(d*x+c)+csc(d*x+c) )-5*A*arcsin(cot(d*x+c)-csc(d*x+c))-3*B*arcsin(cot(d*x+c)-csc(d*x+c))+43*C *arcsin(cot(d*x+c)-csc(d*x+c)))
Time = 36.82 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.22 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {\sqrt {2} {\left ({\left (5 \, A + 3 \, B - 43 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 3 \, B - 43 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 3 \, B - 43 \, C\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B - 43 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 64 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left ({\left (A + 7 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 3 \, B - 11 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c )^(1/2),x, algorithm="fricas")
-1/32*(sqrt(2)*((5*A + 3*B - 43*C)*cos(d*x + c)^3 + 3*(5*A + 3*B - 43*C)*c os(d*x + c)^2 + 3*(5*A + 3*B - 43*C)*cos(d*x + c) + 5*A + 3*B - 43*C)*sqrt (a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*si n(d*x + c))) + 64*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c ) + C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a) *sin(d*x + c))) - 2*((A + 7*B - 15*C)*cos(d*x + c)^2 + (5*A + 3*B - 11*C)* cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a ^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3* d)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c )^(1/2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(5 /2)*sqrt(sec(d*x + c))), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c )^(1/2),x, algorithm="giac")
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(1/2)*(a + a *cos(c + d*x))^(5/2)),x)