Integrand size = 33, antiderivative size = 153 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\frac {3 (A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(5 A-3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {(5 A-3 B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {3 (A-B) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \] Output:
3*(A-B)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+1/3*(5*A-3*B)*InverseJac obiAM(1/2*d*x+1/2*c,2^(1/2))/a/d+1/3*(5*A-3*B)*sin(d*x+c)/a/d/cos(d*x+c)^( 3/2)-3*(A-B)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)-(A-B)*sin(d*x+c)/d/cos(d*x+c) ^(3/2)/(a+a*cos(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.71 (sec) , antiderivative size = 931, normalized size of antiderivative = 6.08 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx =\text {Too large to display} \] Input:
Integrate[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])),x ]
Output:
(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(-(((A - B)*(2 + Cos[c])*Csc[c/2] *Sec[c/2]*Sec[c])/d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] - B* Sin[(d*x)/2]))/d + (4*A*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (4*Sec[c]* Sec[c + d*x]*(A*Sin[c] - 3*A*Sin[d*x] + 3*B*Sin[d*x]))/(3*d)))/(a + a*Cos[ c + d*x]) - (5*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2 }, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]* Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d *x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Co s[c + d*x])*Sqrt[1 + Cot[c]^2]) + (B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Hyperge ometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d* x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Co t[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c ]]]])/(d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (3*A*Cos[c/2 + (d*x)/2 ]^2*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + A rcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + Ar cTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + A rcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTa n[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c ]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTa n[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(a + a*Cos[c + d*x])) + (3*B*Cos[...
Time = 0.65 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 3457, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}
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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {a (5 A-3 B)-3 a (A-B) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (5 A-3 B)-3 a (A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (5 A-3 B)-3 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {a (5 A-3 B) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-3 a (A-B) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (5 A-3 B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-3 a (A-B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {a (5 A-3 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (5 A-3 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a (5 A-3 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {a (5 A-3 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
Input:
Int[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])),x]
Output:
-(((A - B)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x]))) + (a *(5*A - 3*B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d* Cos[c + d*x]^(3/2))) - 3*a*(A - B)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2* Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/(2*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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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])
Leaf count of result is larger than twice the leaf count of optimal. \(465\) vs. \(2(146)=292\).
Time = 4.56 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.05
| method | result | size |
| default | \(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {\left (-2 A +2 B \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {\left (A -B \right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+2 A \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(466\) |
Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c)),x,method=_RETURNVER BOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((-2*A+2*B)/s in(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+si n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin( 1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2 *d*x+1/2*c),2^(1/2)))+(A-B)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1) ^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) -EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x +1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)+2*A*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x +1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^( 1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d* x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2* c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.09 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=-\frac {2 \, {\left (9 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (-5 i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-5 i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (5 i \, A - 3 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (5 i \, A - 3 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c)),x, algorithm= "fricas")
Output:
-1/6*(2*(9*(A - B)*cos(d*x + c)^2 + 2*(2*A - 3*B)*cos(d*x + c) - 2*A)*sqrt (cos(d*x + c))*sin(d*x + c) - (sqrt(2)*(-5*I*A + 3*I*B)*cos(d*x + c)^3 + s qrt(2)*(-5*I*A + 3*I*B)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - (sqrt(2)*(5*I*A - 3*I*B)*cos(d*x + c)^3 + sqrt(2 )*(5*I*A - 3*I*B)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*(sqrt(2)*(-I*A + I*B)*cos(d*x + c)^3 + sqrt(2)*(-I*A + I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*(sqrt(2)*(I*A - I*B)*cos(d*x + c)^3 + sqrt(2)*(I*A - I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c)^3 + a*d*cos (d*x + c)^2)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c)),x)
Output:
Timed out
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c)),x, algorithm= "maxima")
Output:
integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)*cos(d*x + c)^(5/2)), x)
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c)),x, algorithm= "giac")
Output:
integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)*cos(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\left (a+a\,\cos \left (c+d\,x\right )\right )} \,d x \] Input:
int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))),x)
Output:
int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))), x)
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}+\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}}d x \right ) b}{a} \] Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c)),x)
Output:
(int(sqrt(cos(c + d*x))/(cos(c + d*x)**4 + cos(c + d*x)**3),x)*a + int(sqr t(cos(c + d*x))/(cos(c + d*x)**3 + cos(c + d*x)**2),x)*b)/a