Integrand size = 33, antiderivative size = 152 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {3 (A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {(3 A-5 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {(3 A-5 B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {3 (A-B) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}+\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \] Output:
-3*(A-B)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d-1/3*(3*A-5*B)*InverseJa cobiAM(1/2*d*x+1/2*c,2^(1/2))/a/d-1/3*(3*A-5*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 = 7.55 (sec) , antiderivative size = 1041, normalized size of antiderivative = 6.85 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx =\text {Too large to display} \] Input:
Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x ]
Output:
(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(A + B*Sec[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*B*Sec[c]*Sec[c + d*x]^2*Sin [d*x])/(3*d) + (4*Sec[c]*Sec[c + d*x]*(B*Sin[c] + 3*A*Sin[d*x] - 3*B*Sin[d *x]))/(3*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) + (A*Cos[c/2 + ( d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[C ot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x])*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 - Ar cTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A*Cos[c + d*x ])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) - (5*B*Cos[c/2 + (d*x)/2]^2*Cs c[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*S ec[c/2]*(A + B*Sec[c + d*x])*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*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) + (3*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec [c/2]*(A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d* x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d* x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcT...
Time = 0.76 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.94, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3433, 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.
\(\displaystyle \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
\(\Big \downarrow \) 3433 |
\(\displaystyle \int \frac {A \cos (c+d x)+B}{\cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\int -\frac {a (3 A-5 B)-3 a (A-B) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}+\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {\int \frac {a (3 A-5 B)-3 a (A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {\int \frac {a (3 A-5 B)-3 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-3 a (A-B) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-3 a (A-B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}\) |
Input:
Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Sec[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*(3 *A - 5*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_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[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[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. \(2(145)=290\).
Time = 3.40 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.07
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 (-A +B \right ) \left (\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}\, \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}}}+\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 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}\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 )}+2 B \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*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(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*((-A+B)*(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)*(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-2*B)/sin(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+sin(1/2*d*x+1/2 *c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-EllipticE(cos(1/2* d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1 )^(1/2))+2*B*(-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 = 0.09 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.09 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {2 \, {\left (9 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (3 i \, A - 5 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (3 i \, A - 5 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-3 i \, A + 5 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-3 i \, A + 5 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= "fricas")
Output:
1/6*(2*(9*(A - B)*cos(d*x + c)^2 + 2*(3*A - 2*B)*cos(d*x + c) + 2*B)*sqrt( cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(3*I*A - 5*I*B)*cos(d*x + c)^3 + sqr t(2)*(3*I*A - 5*I*B)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(-3*I*A + 5*I*B)*cos(d*x + c)^3 + sqrt(2)* (-3*I*A + 5*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 + sq rt(2)*(-I*A + I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInve rse(-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. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\text {Timed out} \] Input:
integrate((A+B*sec(d*x+c))/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c)),x)
Output:
Timed out
\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= "maxima")
Output:
integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*cos(d*x + c)^(5/2)), x)
\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= "giac")
Output:
integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*cos(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{5/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )} \,d x \] 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)
\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{3}}d x \right ) b}{a} \] Input:
int((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x)
Output:
(int(sqrt(cos(c + d*x))/(cos(c + d*x)**3*sec(c + d*x) + cos(c + d*x)**3),x )*a + int((sqrt(cos(c + d*x))*sec(c + d*x))/(cos(c + d*x)**3*sec(c + d*x) + cos(c + d*x)**3),x)*b)/a