Integrand size = 44, antiderivative size = 353 \[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (9 a^2 B+3 b^2 B+20 a b C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^2 d}-\frac {2 (a-b) \sqrt {a+b} (9 a B-3 b B-5 a C+15 b C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a d}+\frac {2 a B \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 b B+5 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:
2/15*(a-b)*(a+b)^(1/2)*(9*B*a^2+3*B*b^2+20*C*a*b)*cot(d*x+c)*EllipticE((a+ b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a* (1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d-2/15*(a-b )*(a+b)^(1/2)*(9*B*a-3*B*b-5*C*a+15*C*b)*cot(d*x+c)*EllipticF((a+b*cos(d*x +c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d* x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/d+2/5*a*B*(a+b*cos(d*x +c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/15*(6*B*b+5*C*a)*(a+b*cos(d*x+c ))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)
Result contains complex when optimal does not.
Time = 6.91 (sec) , antiderivative size = 1314, normalized size of antiderivative = 3.72 \[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[((a + b*Cos[c + d*x])^(3/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Cos[c + d*x]^(9/2),x]
Output:
-1/15*((-4*a*(-3*a^2*b*B + 3*b^3*B - 5*a^3*C + 5*a*b^2*C)*Sqrt[((a + b)*Co t[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^ 2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*Elli pticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], ( -2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b *Cos[c + d*x]]) - 4*a*(9*a^3*B + 3*a*b^2*B + 20*a^2*b*C)*((Sqrt[((a + b)*C ot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2] ^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*Ell ipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[ (c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos [c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x) /2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(9*a^2*b*B + 3 *b^3*B + 20*a*b^2*C)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*Ellipti cE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x] )*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + ...
Time = 1.67 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.295, Rules used = {3042, 3508, 3042, 3468, 27, 3042, 3534, 27, 3042, 3477, 3042, 3295, 3473}
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.
\(\displaystyle \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {(a+b \cos (c+d x))^{3/2} (B+C \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3468 |
\(\displaystyle \frac {2}{5} \int \frac {b (2 a B+5 b C) \cos ^2(c+d x)+\left (3 B a^2+10 b C a+5 b^2 B\right ) \cos (c+d x)+a (6 b B+5 a C)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {b (2 a B+5 b C) \cos ^2(c+d x)+\left (3 B a^2+10 b C a+5 b^2 B\right ) \cos (c+d x)+a (6 b B+5 a C)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {b (2 a B+5 b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 B a^2+10 b C a+5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a (6 b B+5 a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {1}{5} \left (\frac {2 \int \frac {a \left (9 B a^2+20 b C a+3 b^2 B\right )+a \left (5 C a^2+12 b B a+15 b^2 C\right ) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 (5 a C+6 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {\int \frac {a \left (9 B a^2+20 b C a+3 b^2 B\right )+a \left (5 C a^2+12 b B a+15 b^2 C\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 (5 a C+6 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {\int \frac {a \left (9 B a^2+20 b C a+3 b^2 B\right )+a \left (5 C a^2+12 b B a+15 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 (5 a C+6 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \frac {1}{5} \left (\frac {a \left (9 a^2 B+20 a b C+3 b^2 B\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a (a-b) (9 a B-5 a C-3 b B+15 b C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 (5 a C+6 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {a \left (9 a^2 B+20 a b C+3 b^2 B\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a (a-b) (9 a B-5 a C-3 b B+15 b C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 (5 a C+6 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \frac {1}{5} \left (\frac {a \left (9 a^2 B+20 a b C+3 b^2 B\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (9 a B-5 a C-3 b B+15 b C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{3 a}+\frac {2 (5 a C+6 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \frac {1}{5} \left (\frac {\frac {2 (a-b) \sqrt {a+b} \left (9 a^2 B+20 a b C+3 b^2 B\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} (9 a B-5 a C-3 b B+15 b C) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{3 a}+\frac {2 (5 a C+6 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[((a + b*Cos[c + d*x])^(3/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]
Output:
(2*a*B*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ( ((2*(a - b)*Sqrt[a + b]*(9*a^2*B + 3*b^2*B + 20*a*b*C)*Cot[c + d*x]*Ellipt icE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -(( a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d) - (2*(a - b)*Sqrt[a + b]*(9*a*B - 3*b*B - 5*a*C + 15*b*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b ]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d)/(3*a) + (2*(6*b*B + 5*a*C) *Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 , 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*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] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Leaf count of result is larger than twice the leaf count of optimal. \(1618\) vs. \(2(315)=630\).
Time = 19.59 (sec) , antiderivative size = 1619, normalized size of antiderivative = 4.59
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1619\) |
default | \(\text {Expression too large to display}\) | \(1621\) |
Input:
int((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2), x,method=_RETURNVERBOSE)
Output:
2/5*B/d/a*((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+ cos(d*x+c)))^(1/2)*a^3*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/ 2))*(3*cos(d*x+c)^4+6*cos(d*x+c)^3+3*cos(d*x+c)^2)+(cos(d*x+c)/(1+cos(d*x+ c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b*EllipticE (-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(3*cos(d*x+c)^4+6*cos(d*x+c) ^3+3*cos(d*x+c)^2)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x +c))/(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/ (a+b))^(1/2))*(cos(d*x+c)^4+2*cos(d*x+c)^3+cos(d*x+c)^2)+(cos(d*x+c)/(1+co s(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^3*Ellip ticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)^4+2*cos(d*x+ c)^3+cos(d*x+c)^2)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x +c))/(1+cos(d*x+c)))^(1/2)*a^3*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a +b))^(1/2))*(-3*cos(d*x+c)^4-6*cos(d*x+c)^3-3*cos(d*x+c)^2)+(cos(d*x+c)/(1 +cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b* EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-4*cos(d*x+c)^4-8* cos(d*x+c)^3-4*cos(d*x+c)^2)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a +b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c ),(-(a-b)/(a+b))^(1/2))*(-cos(d*x+c)^4-2*cos(d*x+c)^3-cos(d*x+c)^2)+(3*cos (d*x+c)^2+cos(d*x+c)+1)*sin(d*x+c)*a^3+sin(d*x+c)*cos(d*x+c)*(3*cos(d*x+c) ^2+3*cos(d*x+c)+3)*a^2*b+sin(d*x+c)*cos(d*x+c)^2*(2*cos(d*x+c)+3)*a*b^2...
\[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (9/2),x, algorithm="fricas")
Output:
integral((C*b*cos(d*x + c)^2 + B*a + (C*a + B*b)*cos(d*x + c))*sqrt(b*cos( d*x + c) + a)/cos(d*x + c)^(7/2), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c )**(9/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (9/2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^(3/2)/c os(d*x + c)^(9/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (9/2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^(3/2)/c os(d*x + c)^(9/2), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \] Input:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(3/2))/cos(c + d*x)^(9/2),x)
Output:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(3/2))/cos(c + d*x)^(9/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) b^{2}+\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) b c \] Input:
int((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2), x)
Output:
int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**4,x)*a*b + int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**3,x)*a*c + int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**3,x)*b** 2 + int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**2,x)*b *c