\(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) [586]

Optimal result

Integrand size = 33, antiderivative size = 346 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=-\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^3 (a+b)^2 d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \]

[Out]

-(3*A*a^2*b-2*A*b^3-5*B*a^3+4*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1
/2*c),2^(1/2))/b^3/(a^2-b^2)/d-1/3*(3*A*a*b-5*B*a^2+2*B*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E
llipticF(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/(a^2-b^2)/d-a*(3*A*a^2*b-5*A*b^3-5*B*a^3+7*B*a*b^2)*(cos(1/2*d*x+1/2*
c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/(a-b)/b^3/(a+b)^2/d-1/3*(3*A*a
*b-5*B*a^2+2*B*b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/cos(d*x+c)^(3/2)+a*(A*b-B*a)*sin(d*x+c)/b/(a^2-b^2)/d/cos(d*x+c
)^(3/2)/(b+a*cos(d*x+c))+(3*A*a^2*b-2*A*b^3-5*B*a^3+4*B*a*b^2)*sin(d*x+c)/b^3/(a^2-b^2)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3033, 3079, 3134, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x)}{b d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-5 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}-\frac {a \left (-5 a^3 B+3 a^2 A b+7 a b^2 B-5 A b^3\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 d (a-b) (a+b)^2}+\frac {\left (-5 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}} \]

[In]

Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

-(((3*a^2*A*b - 2*A*b^3 - 5*a^3*B + 4*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(b^3*(a^2 - b^2)*d)) - ((3*a*A*b - 5
*a^2*B + 2*b^2*B)*EllipticF[(c + d*x)/2, 2])/(3*b^2*(a^2 - b^2)*d) - (a*(3*a^2*A*b - 5*A*b^3 - 5*a^3*B + 7*a*b
^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/((a - b)*b^3*(a + b)^2*d) - ((3*a*A*b - 5*a^2*B + 2*b^2*B)*Si
n[c + d*x])/(3*b^2*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)) + ((3*a^2*A*b - 2*A*b^3 - 5*a^3*B + 4*a*b^2*B)*Sin[c + d*
x])/(b^3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]) + (a*(A*b - a*B)*Sin[c + d*x])/(b*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*
(b + a*Cos[c + d*x]))

Rule 2719

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]

Rule 2720

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]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[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] && I
ntegerQ[m] && IntegerQ[n]

Rule 3079

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c +
d*Sin[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*
(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n
, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(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] + D
ist[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])))

Rule 3138

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {B+A \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))^2} \, dx \\ & = \frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{2} \left (3 a A b-5 a^2 B+2 b^2 B\right )+b (A b-a B) \cos (c+d x)-\frac {3}{2} a (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {2 \int \frac {-\frac {3}{4} \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )-\frac {1}{2} b \left (3 a A b-2 a^2 B-b^2 B\right ) \cos (c+d x)+\frac {1}{4} a \left (3 a A b-5 a^2 B+2 b^2 B\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {4 \int \frac {\frac {1}{8} \left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right )+\frac {1}{4} b \left (6 a^2 A b-3 A b^3-10 a^3 B+7 a b^2 B\right ) \cos (c+d x)+\frac {3}{8} a \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}+\frac {4 \int \frac {-\frac {1}{8} a \left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right )-\frac {1}{8} a^2 b \left (3 a A b-5 a^2 B+2 b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )}-\frac {\left (a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^3 (a+b)^2 d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 7.32 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.23 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 \left (-27 a^3 A b+30 a A b^3+45 a^4 B-44 a^2 b^2 B-4 b^4 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {\left (-24 a^2 A b^2+12 A b^4+40 a^3 b B-28 a b^3 B\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {2 \left (-9 a^3 A b+6 a A b^3+15 a^4 B-12 a^2 b^2 B\right ) \cos (2 (c+d x)) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {1-\cos ^2(c+d x)} \left (-1+2 \cos ^2(c+d x)\right )}}{12 (a-b) b^3 (a+b) d}+\frac {\sqrt {\cos (c+d x)} \left (\frac {2 \sec (c+d x) (A b \sin (c+d x)-2 a B \sin (c+d x))}{b^3}+\frac {-a^3 A b \sin (c+d x)+a^4 B \sin (c+d x)}{b^3 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}+\frac {2 B \sec (c+d x) \tan (c+d x)}{3 b^2}\right )}{d} \]

[In]

Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

((2*(-27*a^3*A*b + 30*a*A*b^3 + 45*a^4*B - 44*a^2*b^2*B - 4*b^4*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/
(a + b) + ((-24*a^2*A*b^2 + 12*A*b^4 + 40*a^3*b*B - 28*a*b^3*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi
[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)))/a + (2*(-9*a^3*A*b + 6*a*A*b^3 + 15*a^4*B - 12*a^2*b^2*B)*Cos[2*(c
+ d*x)]*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]],
-1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[1 - Cos[c +
d*x]^2]*(-1 + 2*Cos[c + d*x]^2)))/(12*(a - b)*b^3*(a + b)*d) + (Sqrt[Cos[c + d*x]]*((2*Sec[c + d*x]*(A*b*Sin[c
+ d*x] - 2*a*B*Sin[c + d*x]))/b^3 + (-(a^3*A*b*Sin[c + d*x]) + a^4*B*Sin[c + d*x])/(b^3*(-a^2 + b^2)*(b + a*C
os[c + d*x])) + (2*B*Sec[c + d*x]*Tan[c + d*x])/(3*b^2)))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(996\) vs. \(2(414)=828\).

Time = 31.83 (sec) , antiderivative size = 997, normalized size of antiderivative = 2.88

method	result	size
default	\(\text {Expression too large to display}\)	\(997\)

[In]

int((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/b^2*(-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
)))+2*(A*b-2*B*a)/b^3/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-(sin(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)^2-1)^(1/2))-2*(A*b-B*a)*a/b^2*(a^2/b/(a^2-b^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*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/b*(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))+1/2*a/b/(a^2-b^2)*(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))-1/2*a/b/(a^2-b^2)*(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)*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b)
,2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+2*a^2*(A*b-2*B*a
)/b^3/(a^2-a*b)*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),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

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)**(7/2)/(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2*cos(d*x + c)^(7/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))^2),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))^2), x)