Integrand size = 23, antiderivative size = 185 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {19 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac {13 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {7 \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {\sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\sec (c+d x) \tan (c+d x)}{a d \sqrt {a+a \cos (c+d x)}} \]
19/4*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d-13/4*arc tanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1 /2)-1/2*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-7/4*tan(d*x+c)/a/d/ (a+a*cos(d*x+c))^(1/2)+sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 6.93 (sec) , antiderivative size = 841, normalized size of antiderivative = 4.55 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {19 \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {i e^{i c} x}{-1+e^{i c}}-\frac {\log \left (i-\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )}{d}\right )}{2 \sqrt {2} (a (1+\cos (c+d x)))^{3/2}}+\frac {19 \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {i e^{i c} x}{-1+e^{i c}}+\frac {\log \left (i+\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )}{d}\right )}{2 \sqrt {2} (a (1+\cos (c+d x)))^{3/2}}+\frac {13 \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )}{d (a (1+\cos (c+d x)))^{3/2}}-\frac {13 \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )}{d (a (1+\cos (c+d x)))^{3/2}}-\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )^2}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )^2}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-5 \cos \left (\frac {c}{2}\right )+7 \sin \left (\frac {c}{2}\right )\right )}{2 d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (5 \cos \left (\frac {c}{2}\right )+7 \sin \left (\frac {c}{2}\right )\right )}{2 d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \]
(19*Cos[c/2 + (d*x)/2]^3*((I*E^(I*c)*x)/(-1 + E^(I*c)) - Log[I - Sqrt[2]*E ^((I/2)*(c + d*x)) - I*E^(I*(c + d*x))]/d))/(2*Sqrt[2]*(a*(1 + Cos[c + d*x ]))^(3/2)) + (19*Cos[c/2 + (d*x)/2]^3*(((-I)*E^(I*c)*x)/(-1 + E^(I*c)) + L og[I + Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c + d*x))]/d))/(2*Sqrt[2]*(a* (1 + Cos[c + d*x]))^(3/2)) + (13*Cos[c/2 + (d*x)/2]^3*Log[Cos[c/4 + (d*x)/ 4] - Sin[c/4 + (d*x)/4]])/(d*(a*(1 + Cos[c + d*x]))^(3/2)) - (13*Cos[c/2 + (d*x)/2]^3*Log[Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4]])/(d*(a*(1 + Cos[c + d*x]))^(3/2)) - Cos[c/2 + (d*x)/2]^3/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)* (Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])^2) + Cos[c/2 + (d*x)/2]^3/(2*d*( a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4])^2) + (Cos[c/2 + (d*x)/2]^3*Sin[(d*x)/2])/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[ c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (Cos[c/2 + (d*x)/2]^3*(-5*Cos[c/2] + 7*Sin[c/2]))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)* (Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (Cos[c/ 2 + (d*x)/2]^3*Sin[(d*x)/2])/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2] + S in[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Cos[c/2 + (d*x)/2 ]^3*(5*Cos[c/2] + 7*Sin[c/2]))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))
Time = 1.18 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3042, 3245, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3464, 3042, 3128, 219, 3252, 219}
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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \frac {\int \frac {(8 a-5 a \cos (c+d x)) \sec ^3(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(8 a-5 a \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {8 a-5 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\int -\frac {2 \left (7 a^2-6 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (7 a^2-6 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {7 a^2-6 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {\left (19 a^3-7 a^3 \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (19 a^3-7 a^3 \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {19 a^3-7 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3464 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {19 a^2 \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-26 a^3 \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {19 a^2 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-26 a^3 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {19 a^2 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {52 a^3 \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {19 a^2 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {26 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {38 a^3 \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {26 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {4 a \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {7 a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {38 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {26 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
-1/2*(Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/2)) + ((4*a*Se c[c + d*x]*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]) - (-1/2*((38*a^(5/2) *ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (26*Sqrt[2] *a^(5/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]]) ])/d)/a + (7*a^2*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/a)/(4*a^2)
3.2.38.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*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*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A *b - a*B)/(b*c - a*d) Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c - A*d)/(b*c - a*d) Int[Sqrt[a + b*Sin[e + f*x]]/(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]
Leaf count of result is larger than twice the leaf count of optimal. \(806\) vs. \(2(156)=312\).
Time = 1.81 (sec) , antiderivative size = 807, normalized size of antiderivative = 4.36
-1/2*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(104*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/ 2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^6*a-76*l n(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)* (a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^6*a-76*ln( 4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a* sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^6*a-104*2^(1/ 2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c)) *a*cos(1/2*d*x+1/2*c)^4+28*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)* cos(1/2*d*x+1/2*c)^4+76*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*co s(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos( 1/2*d*x+1/2*c)^4*a+76*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1 /2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2 *d*x+1/2*c)^4*a+26*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+ 2*a)/cos(1/2*d*x+1/2*c))*a*cos(1/2*d*x+1/2*c)^2-22*cos(1/2*d*x+1/2*c)^2*(a *sin(1/2*d*x+1/2*c)^2)^(1/2)*2^(1/2)*a^(1/2)-19*ln(-4/(2*cos(1/2*d*x+1/2*c )-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^ (1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^2*a-19*ln(4/(2*cos(1/2*d*x+1/2*c)+2 ^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/ 2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^2*a+2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2) ^(1/2)*a^(1/2))/a^(5/2)/cos(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)-2^(1/2...
Time = 0.28 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.63 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {26 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 19 \, {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (7 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]
1/16*(26*sqrt(2)*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt (a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*si n(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1) ) + 19*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*log((a *cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)* (cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) - 4*sqrt(a*cos(d*x + c) + a)*(7*cos(d*x + c)^2 + 3*cos(d*x + c) - 2)*sin(d *x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> ECL says: Memory limit reached. Please j ump to an outer pointer, quit program and enlarge thememory limits before executing the program again.
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]