\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [487]

Optimal result

Integrand size = 33, antiderivative size = 261 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {(49 A-9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(49 A-9 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-1/5*(A-B)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(8*A-3*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+
a*sec(d*x+c))^2-1/6*(13*A-3*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))+1/10*(49*A-9*B)*sin(d*x+c)*s
ec(d*x+c)^(1/2)/a^3/d-1/10*(49*A-9*B)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/d-1/6*(13*A-3*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+
1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/d

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3039, 4104, 3872, 3856, 2720, 3853, 2719} \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {(13 A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{6 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(49 A-9 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(49 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(8 A-3 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2))/(a + a*Cos[c + d*x])^3,x]

[Out]

-1/10*((49*A - 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*d) - ((13*A - 3*B)*S
qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + ((49*A - 9*B)*Sqrt[Sec[c + d*x]]*S
in[c + d*x])/(10*a^3*d) - ((A - B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((8*A - 3*B
)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((13*A - 3*B)*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(6*d*(a^3 + a^3*Sec[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 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4104

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(
a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sec ^{\frac {7}{2}}(c+d x) (B+A \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx \\ & = -\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (-\frac {5}{2} a (A-B)+\frac {1}{2} a (11 A-B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (-\frac {3}{2} a^2 (8 A-3 B)+\frac {1}{2} a^2 (41 A-6 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \sqrt {\sec (c+d x)} \left (-\frac {5}{4} a^3 (13 A-3 B)+\frac {3}{4} a^3 (49 A-9 B) \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(49 A-9 B) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{20 a^3}-\frac {(13 A-3 B) \int \sqrt {\sec (c+d x)} \, dx}{12 a^3} \\ & = \frac {(49 A-9 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(49 A-9 B) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}-\frac {\left ((13 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3} \\ & = -\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(49 A-9 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\left ((49 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3} \\ & = -\frac {(49 A-9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(49 A-9 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.35 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (-i (49 A-9 B) e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right )^5 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+160 (13 A-3 B) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-i \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 i (642 A-102 B+(1082 A-207 B) \cos (c+d x)+6 (87 A-17 B) \cos (2 (c+d x))+106 A \cos (3 (c+d x))-21 B \cos (3 (c+d x))+161 i A \sin (c+d x)-6 i B \sin (c+d x)+148 i A \sin (2 (c+d x))-18 i B \sin (2 (c+d x))+41 i A \sin (3 (c+d x))-6 i B \sin (3 (c+d x)))\right ) \left (\cos \left (\frac {1}{2} (c+3 d x)\right )+i \sin \left (\frac {1}{2} (c+3 d x)\right )\right )}{120 a^3 d (1+\cos (c+d x))^3} \]

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2))/(a + a*Cos[c + d*x])^3,x]

[Out]

-1/120*(Cos[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(((-I)*(49*A - 9*B)*(1 + E^(I*(c + d*x)))^5*Sqrt[1 + E^((2*I)*(c +
d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((2*I)*(c + d*x)) + 160*(13*A - 3*B)*Cos[(c
+ d*x)/2]^5*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[(c + d*x)/2] - I*Sin[(c + d*x)/2]) + (2*I)*(642*
A - 102*B + (1082*A - 207*B)*Cos[c + d*x] + 6*(87*A - 17*B)*Cos[2*(c + d*x)] + 106*A*Cos[3*(c + d*x)] - 21*B*C
os[3*(c + d*x)] + (161*I)*A*Sin[c + d*x] - (6*I)*B*Sin[c + d*x] + (148*I)*A*Sin[2*(c + d*x)] - (18*I)*B*Sin[2*
(c + d*x)] + (41*I)*A*Sin[3*(c + d*x)] - (6*I)*B*Sin[3*(c + d*x)]))*(Cos[(c + 3*d*x)/2] + I*Sin[(c + 3*d*x)/2]
))/(a^3*d*E^(I*d*x)*(1 + Cos[c + d*x])^3)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(684\) vs. \(2(285)=570\).

Time = 6.00 (sec) , antiderivative size = 685, normalized size of antiderivative = 2.62

method	result	size
default	\(\text {Expression too large to display}\)	\(685\)

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

-1/60*(-2*(2*sin(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)*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^4+4*(2*sin(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)*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*EllipticE(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*(2*sin(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)*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*
B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+12*(-2*
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(49*A-9*B)*sin(1/2*d*x+1/2*c)^8-2*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*(817*A-147*B)*sin(1/2*d*x+1/2*c)^6+6*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^
(1/2)*(248*A-43*B)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(439*A-69*B)*sin(
1/2*d*x+1/2*c)^2)/a^3/cos(1/2*d*x+1/2*c)^5/(-2*sin(1/2*d*x+1/2*c)^4+sin(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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 481, normalized size of antiderivative = 1.84 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (-13 i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-13 i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-13 i \, A + 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-13 i \, A + 3 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (13 i \, A - 3 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (13 i \, A - 3 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (13 i \, A - 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (13 i \, A - 3 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (49 i \, A - 9 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (49 i \, A - 9 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (49 i \, A - 9 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (49 i \, A - 9 i \, B\right )}\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 ) + 3 \, {\left (\sqrt {2} {\left (-49 i \, A + 9 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-49 i \, A + 9 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-49 i \, A + 9 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-49 i \, A + 9 i \, B\right )}\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 ) - \frac {2 \, {\left (3 \, {\left (49 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (188 \, A - 33 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (59 \, A - 9 \, B\right )} \cos \left (d x + c\right ) + 60 \, A\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\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 )}} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(5*(sqrt(2)*(-13*I*A + 3*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-13*I*A + 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-
13*I*A + 3*I*B)*cos(d*x + c) + sqrt(2)*(-13*I*A + 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x
+ c)) + 5*(sqrt(2)*(13*I*A - 3*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(13*I*A - 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(13
*I*A - 3*I*B)*cos(d*x + c) + sqrt(2)*(13*I*A - 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c
)) + 3*(sqrt(2)*(49*I*A - 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(49*I*A - 9*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(49*I*
A - 9*I*B)*cos(d*x + c) + sqrt(2)*(49*I*A - 9*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) + I*sin(d*x + c))) + 3*(sqrt(2)*(-49*I*A + 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-49*I*A + 9*I*B)*cos(d*x +
c)^2 + 3*sqrt(2)*(-49*I*A + 9*I*B)*cos(d*x + c) + sqrt(2)*(-49*I*A + 9*I*B))*weierstrassZeta(-4, 0, weierstras
sPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(49*A - 9*B)*cos(d*x + c)^3 + 2*(188*A - 33*B)*cos(d*x
+ c)^2 + 5*(59*A - 9*B)*cos(d*x + c) + 60*A)*sin(d*x + c)/sqrt(cos(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)

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**(3/2)/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

[In]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + a*cos(c + d*x))^3,x)

[Out]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + a*cos(c + d*x))^3, x)