Integrand size = 35, antiderivative size = 237 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(2 A-3 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{3/2} d}-\frac {(5 A-9 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}-\frac {(A-3 B) \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]
1/2*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(3/2)-1/2*(A-3*B) *sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+(2*A-3*B)*arcsin(s in(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^(3/2)/d-1/4*(5*A-9*B)*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^(3/2)/d *2^(1/2)
Result contains complex when optimal does not.
Time = 7.44 (sec) , antiderivative size = 836, normalized size of antiderivative = 3.53 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {i A e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2}}+\frac {3 i B e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2}}+\frac {2 i \sqrt {2} A e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-\text {arcsinh}\left (e^{i (c+d x)}\right )+\sqrt {2} \text {arctanh}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2}}-\frac {3 i \sqrt {2} B e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-\text {arcsinh}\left (e^{i (c+d x)}\right )+\sqrt {2} \text {arctanh}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2}}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\sec (c+d x)} \left (-\frac {2 A \cos \left (\frac {d x}{2}\right ) \sin \left (\frac {c}{2}\right )}{d}+\frac {\sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {c}{2}\right )-B \sin \left (\frac {c}{2}\right )\right )}{d}+\frac {2 B \cos \left (\frac {3 d x}{2}\right ) \sin \left (\frac {3 c}{2}\right )}{d}-\frac {2 A \cos \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}+\frac {\sec \left (\frac {c}{2}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )\right )}{d}+\frac {2 B \cos \left (\frac {3 c}{2}\right ) \sin \left (\frac {3 d x}{2}\right )}{d}\right )}{(a (1+\cos (c+d x)))^{3/2}} \]
((-I)*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)* (c + d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])]*Cos[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a*(1 + Cos[c + d*x] ))^(3/2)) + ((3*I)*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[ 1 + E^((2*I)*(c + d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E ^((2*I)*(c + d*x))])]*Cos[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a*(1 + Cos[c + d*x]))^(3/2)) + ((2*I)*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2* I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d* x))])] + ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Cos[c/2 + (d*x)/2]^3)/(d* E^((I/2)*(c + d*x))*(a*(1 + Cos[c + d*x]))^(3/2)) - ((3*I)*Sqrt[2]*B*Sqrt[ E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*( -ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2 ]*Sqrt[1 + E^((2*I)*(c + d*x))])] + ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] )*Cos[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a*(1 + Cos[c + d*x]))^(3/2 )) + (Cos[c/2 + (d*x)/2]^3*Sqrt[Sec[c + d*x]]*((-2*A*Cos[(d*x)/2]*Sin[c/2] )/d + (Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[c/2] - B*Sin[c/2]))/d + (2*B*Cos [(3*d*x)/2]*Sin[(3*c)/2])/d - (2*A*Cos[c/2]*Sin[(d*x)/2])/d + (Sec[c/2]*Se c[c/2 + (d*x)/2]^2*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/d + (2*B*Cos[(3*c)/2 ]*Sin[(3*d*x)/2])/d))/(a*(1 + Cos[c + d*x]))^(3/2)
Time = 1.42 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.95, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3440, 3042, 3456, 27, 3042, 3462, 25, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3440 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(\cos (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a (A-B)-2 a (A-3 B) \cos (c+d x))}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a (A-B)-2 a (A-3 B) \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a (A-B)-2 a (A-3 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int -\frac {a^2 (A-3 B)-2 a^2 (2 A-3 B) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {\int \frac {a^2 (A-3 B)-2 a^2 (2 A-3 B) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {\int \frac {a^2 (A-3 B)-2 a^2 (2 A-3 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3461 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {a^2 (5 A-9 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-2 a (2 A-3 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {a^2 (5 A-9 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-2 a (2 A-3 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {a^2 (5 A-9 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {4 a (2 A-3 B) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {a^2 (5 A-9 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^{3/2} (2 A-3 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {-\frac {2 a^3 (5 A-9 B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {4 a^{3/2} (2 A-3 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {\frac {\sqrt {2} a^{3/2} (5 A-9 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {4 a^{3/2} (2 A-3 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((A - B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + (-(((-4*a^(3/2)*(2*A - 3*B)*ArcS in[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (Sqrt[2]*a^(3/2)* (5*A - 9*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt [a + a*Cos[c + d*x]])])/d)/a) - (2*a*(A - 3*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2))
3.6.35.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p Int[(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g , m, n, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && !(IntegerQ[m] && I ntegerQ[n])
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_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Sin[e + f*x])^m*(c + d*S in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Time = 20.86 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.63
| method | result | size |
| default | \(\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (2 B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+4 A \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-\tan \left (d x +c \right ) A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-6 B \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+3 \tan \left (d x +c \right ) B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+4 \sec \left (d x +c \right ) A \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+5 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-6 \sec \left (d x +c \right ) B \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-9 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+5 \sec \left (d x +c \right ) A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-9 \sec \left (d x +c \right ) B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{4 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(387\) |
| parts | \(\frac {A \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (4 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-\tan \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+4 \sec \left (d x +c \right ) \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+5 \sec \left (d x +c \right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{2}}+\frac {B \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (2 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-6 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+3 \tan \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-6 \sec \left (d x +c \right ) \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-9 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-9 \sec \left (d x +c \right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{2}}\) | \(440\) |
1/4/a^2/d*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))^2/sec(d*x+c)^(3/2)/(cos( d*x+c)/(1+cos(d*x+c)))^(1/2)*(2*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 )*sin(d*x+c)+4*A*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 2))-tan(d*x+c)*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-6*B*2^(1/2)*arc tan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+3*tan(d*x+c)*B*2^(1/2)*( cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+4*sec(d*x+c)*A*2^(1/2)*arctan(tan(d*x+c)* (cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+5*A*arcsin(cot(d*x+c)-csc(d*x+c))-6*sec (d*x+c)*B*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-9*B *arcsin(cot(d*x+c)-csc(d*x+c))+5*sec(d*x+c)*A*arcsin(cot(d*x+c)-csc(d*x+c) )-9*sec(d*x+c)*B*arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)
Time = 3.75 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.04 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left ({\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 9 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 4 \, {\left ({\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 3 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (2 \, B \cos \left (d x + c\right )^{2} - {\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/4*(sqrt(2)*((5*A - 9*B)*cos(d*x + c)^2 + 2*(5*A - 9*B)*cos(d*x + c) + 5* A - 9*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c) )/(sqrt(a)*sin(d*x + c))) - 4*((2*A - 3*B)*cos(d*x + c)^2 + 2*(2*A - 3*B)* cos(d*x + c) + 2*A - 3*B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos (d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*(2*B*cos(d*x + c)^2 - (A - 3*B)*cos (d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2* d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]