Integrand size = 25, antiderivative size = 197 \[ \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx=\frac {13 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 \sqrt {2} a^{7/2} d}+\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}+\frac {\sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {5 \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \] Output:
13/128*arctan(1/2*a^(1/2)*sin(d*x+c)*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)*2^(1/2)/a^(7/2)/d+1/6*sin(d* x+c)/d/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2)+1/16*sin(d*x+c)/a/d/(a+a*co s(d*x+c))^(5/2)/sec(d*x+c)^(1/2)-5/192*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^( 3/2)/sec(d*x+c)^(1/2)
Time = 1.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx=\frac {-312 \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cot \left (\frac {1}{2} (c+d x)\right ) \sqrt {2-2 \sec (c+d x)}+(73+4 \cos (c+d x)-5 \cos (2 (c+d x))) \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{3072 a^3 d \sqrt {a (1+\cos (c+d x))} \sqrt {\sec (c+d x)}} \] Input:
Integrate[1/((a + a*Cos[c + d*x])^(7/2)*Sqrt[Sec[c + d*x]]),x]
Output:
(-312*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]]*Cot[(c + d*x)/2]*S qrt[2 - 2*Sec[c + d*x]] + (73 + 4*Cos[c + d*x] - 5*Cos[2*(c + d*x)])*Sec[( c + d*x)/2]^6*Sin[c + d*x])/(3072*a^3*d*Sqrt[a*(1 + Cos[c + d*x])]*Sqrt[Se c[c + d*x]])
Time = 1.05 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4710, 3042, 3243, 27, 3042, 3457, 27, 3042, 3457, 27, 3042, 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.
\(\displaystyle \int \frac {1}{\sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{7/2}}dx\) |
\(\Big \downarrow \) 4710 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)}}{(\cos (c+d x) a+a)^{7/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3243 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {4 \cos (c+d x) a+a}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx}{6 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {4 \cos (c+d x) a+a}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {4 \sin \left (c+d x+\frac {\pi }{2}\right ) a+a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {6 \cos (c+d x) a^2+11 a^2}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}+\frac {3 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {6 \cos (c+d x) a^2+11 a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}+\frac {3 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {6 \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+11 a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}+\frac {3 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {39 a^3}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {5 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {3 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {39}{4} a \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-\frac {5 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {3 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {39}{4} a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {5 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {3 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {-\frac {39 a^2 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}-\frac {5 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {3 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {39 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} d}-\frac {5 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {3 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
Input:
Int[1/((a + a*Cos[c + d*x])^(7/2)*Sqrt[Sec[c + d*x]]),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((Sqrt[Cos[c + d*x]]*Sin[c + d*x])/( 6*d*(a + a*Cos[c + d*x])^(7/2)) + ((3*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/( 4*d*(a + a*Cos[c + d*x])^(5/2)) + ((39*Sqrt[a]*ArcTan[(Sqrt[a]*Sin[c + d*x ])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*d) - (5*a^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2))) /(8*a^2))/(12*a^2))
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[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*d*n - b*c *(m + 1) - b*d*(m + n + 1)*Sin[e + f*x], x], 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] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c , 0]))
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[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*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] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 4.61 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {\left (\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-39\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (39 \cos \left (d x +c \right )^{3}+117 \cos \left (d x +c \right )^{2}+117 \cos \left (d x +c \right )+39\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{384 d \left (\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{4}}\) | \(188\) |
Input:
int(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/384/d*(sin(d*x+c)*(5*cos(d*x+c)^2-2*cos(d*x+c)-39)*2^(1/2)*(cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)+(39*cos(d*x+c)^3+117*cos(d*x+c)^2+117*cos(d*x+c)+39) *arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)/(cos(d*x+ c)^4+4*cos(d*x+c)^3+6*cos(d*x+c)^2+4*cos(d*x+c)+1)/sec(d*x+c)^(1/2)/(cos(d *x+c)/(cos(d*x+c)+1))^(1/2)/a^4
Time = 0.11 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {39 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} - 39 \, \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:
integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")
Output:
-1/384*(39*sqrt(2)*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt( cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*(5*cos(d*x + c)^3 - 2*cos(d*x + c)^2 - 39*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)
Timed out. \[ \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+a*cos(d*x+c))**(7/2)/sec(d*x+c)**(1/2),x)
Output:
Timed out
\[ \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")
Output:
integrate(1/((a*cos(d*x + c) + a)^(7/2)*sqrt(sec(d*x + c))), x)
Timed out. \[ \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:
int(1/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(7/2)),x)
Output:
int(1/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(7/2)), x)
\[ \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+4 \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )+6 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )+4 \cos \left (d x +c \right ) \sec \left (d x +c \right )+\sec \left (d x +c \right )}d x \right )}{a^{4}} \] Input:
int(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2),x)
Output:
(sqrt(a)*int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1))/(cos(c + d*x)**4* sec(c + d*x) + 4*cos(c + d*x)**3*sec(c + d*x) + 6*cos(c + d*x)**2*sec(c + d*x) + 4*cos(c + d*x)*sec(c + d*x) + sec(c + d*x)),x))/a**4