Integrand size = 45, antiderivative size = 254 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(2 B-5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(3 A-43 B+115 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]
(2*B-5*C)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d-1/4* (A-B+C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+1/16*(A+7*B-1 5*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/32*(3*A-43*B +115*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))/a^(5/2)/d*2^(1/2)+1/16*(3*A-11*B+35*C)*sin(d*x+c)*cos(d*x+c)^ (1/2)/a^2/d/(a+a*cos(d*x+c))^(1/2)
Time = 5.07 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\left (-8 (11 B-35 C) \arcsin \left (\sqrt {1-\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )-344 B \arcsin \left (\sqrt {\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+920 C \arcsin \left (\sqrt {\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )-4 \sqrt {2} (3 A-43 B+115 C) \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+14 A \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)-30 B \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+110 C \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+32 C \sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)+6 A \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}-22 B \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}+70 C \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}\right ) \sin (c+d x)}{32 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2}} \]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(5/2),x]
((-8*(11*B - 35*C)*ArcSin[Sqrt[1 - Cos[c + d*x]]]*Cos[(c + d*x)/2]^4 - 344 *B*ArcSin[Sqrt[Cos[c + d*x]]]*Cos[(c + d*x)/2]^4 + 920*C*ArcSin[Sqrt[Cos[c + d*x]]]*Cos[(c + d*x)/2]^4 - 4*Sqrt[2]*(3*A - 43*B + 115*C)*ArcTan[Sqrt[ Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]]*Cos[(c + d*x)/2]^4 + 14*A*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2) - 30*B*Sqrt[1 - Cos[c + d*x]]*Cos[c + d* x]^(3/2) + 110*C*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2) + 32*C*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2) + 6*A*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])] - 22*B*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])] + 70*C*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])])*Sin[c + d*x])/(32*d*Sqrt[1 - Cos[c + d*x] ]*(a*(1 + Cos[c + d*x]))^(5/2))
Time = 1.66 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3520, 27, 3042, 3456, 27, 3042, 3462, 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.
\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (a (3 A+5 B-5 C)+2 a (A-B+5 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (a (3 A+5 B-5 C)+2 a (A-B+5 C) \cos (c+d x))}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a (3 A+5 B-5 C)+2 a (A-B+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3456 |
\(\displaystyle \frac {\frac {\int \frac {\sqrt {\cos (c+d x)} \left (3 (A+7 B-15 C) a^2+2 (3 A-11 B+35 C) \cos (c+d x) a^2\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\sqrt {\cos (c+d x)} \left (3 (A+7 B-15 C) a^2+2 (3 A-11 B+35 C) \cos (c+d x) a^2\right )}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 (A+7 B-15 C) a^2+2 (3 A-11 B+35 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3462 |
\(\displaystyle \frac {\frac {\frac {\int \frac {(3 A-11 B+35 C) a^3+16 (2 B-5 C) \cos (c+d x) a^3}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {(3 A-11 B+35 C) a^3+16 (2 B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3461 |
\(\displaystyle \frac {\frac {\frac {a^3 (3 A-43 B+115 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx+16 a^2 (2 B-5 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {a^3 (3 A-43 B+115 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+16 a^2 (2 B-5 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3253 |
\(\displaystyle \frac {\frac {\frac {a^3 (3 A-43 B+115 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {32 a^2 (2 B-5 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {\frac {a^3 (3 A-43 B+115 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {32 a^{5/2} (2 B-5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle \frac {\frac {\frac {\frac {32 a^{5/2} (2 B-5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^4 (3 A-43 B+115 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\frac {\frac {\sqrt {2} a^{5/2} (3 A-43 B+115 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {32 a^{5/2} (2 B-5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
Int[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Co s[c + d*x])^(5/2),x]
-1/4*((A - B + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) ^(5/2)) + ((a*(A + 7*B - 15*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + (((32*a^(5/2)*(2*B - 5*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (Sqrt[2]*a^(5/2)*(3*A - 43*B + 115*C )*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^2*(3*A - 11*B + 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2))/(8*a^2)
3.6.17.3.1 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_)*((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])
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)]
Leaf count of result is larger than twice the leaf count of optimal. \(728\) vs. \(2(217)=434\).
Time = 30.77 (sec) , antiderivative size = 729, normalized size of antiderivative = 2.87
method | result | size |
default | \(\frac {\left (-3 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+43 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+32 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-115 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+14 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-6 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-30 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+64 B \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+86 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+110 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-160 C \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-230 C \cos \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+6 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-3 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-22 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+128 B \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+43 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+70 C \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-320 C \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-115 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+64 B \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-160 C \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right )}{32 a^{3} d \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(729\) |
parts | \(\text {Expression too large to display}\) | \(872\) |
int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+cos(d*x+c)*a)^(5/2 ),x,method=_RETURNVERBOSE)
1/32/a^3/d*(-3*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2+43*B*2 ^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2+32*C*cos(d*x+c)^2*sin(d* x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-115*C*2^(1/2)*arcsin(cot(d*x+c)-csc (d*x+c))*cos(d*x+c)^2+14*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c) ))^(1/2)-6*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)-30*B*cos(d*x +c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+64*B*cos(d*x+c)^2*arctan( tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+86*B*2^(1/2)*arcsin(cot(d*x+ c)-csc(d*x+c))*cos(d*x+c)+110*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d *x+c)))^(1/2)-160*C*cos(d*x+c)^2*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+ c)))^(1/2))-230*C*cos(d*x+c)*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+6*A*(co s(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-3*A*2^(1/2)*arcsin(cot(d*x+c)-cs c(d*x+c))-22*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+128*B*cos(d*x+ c)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+43*B*2^(1/2)*arcsi n(cot(d*x+c)-csc(d*x+c))+70*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2) -320*C*cos(d*x+c)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-115 *C*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+64*B*arctan(tan(d*x+c)*(cos(d*x+c )/(1+cos(d*x+c)))^(1/2))-160*C*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c) ))^(1/2)))*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^(1/2)/(1+cos(d*x+c))^3/(cos (d*x+c)/(1+cos(d*x+c)))^(1/2)
Time = 47.25 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, B + 115 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left (16 \, C \cos \left (d x + c\right )^{2} + {\left (7 \, A - 15 \, B + 55 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 11 \, B + 35 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 32 \, {\left ({\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 2 \, B - 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="fricas")
-1/32*(sqrt(2)*((3*A - 43*B + 115*C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 115* C)*cos(d*x + c)^2 + 3*(3*A - 43*B + 115*C)*cos(d*x + c) + 3*A - 43*B + 115 *C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sq rt(a)*sin(d*x + c))) - 2*(16*C*cos(d*x + c)^2 + (7*A - 15*B + 55*C)*cos(d* x + c) + 3*A - 11*B + 35*C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*si n(d*x + c) + 32*((2*B - 5*C)*cos(d*x + c)^3 + 3*(2*B - 5*C)*cos(d*x + c)^2 + 3*(2*B - 5*C)*cos(d*x + c) + 2*B - 5*C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(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. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*co s(d*x + c) + a)^(5/2), x)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="giac")
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
int((cos(c + d*x)^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*co s(c + d*x))^(5/2),x)