3.122 \(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=264 \[ \frac {(39 A-20 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(31 A-15 B) \tan (c+d x) \sec (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(19 A-11 B) \tan (c+d x) \sec (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

1/4*(39*A-20*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d-1/32*(219*A-115*B)*arctanh(1/2*si
n(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-B)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(
d*x+c))^(5/2)-1/16*(19*A-11*B)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)-7/16*(9*A-5*B)*tan(d*x+c)/a^2/
d/(a+a*cos(d*x+c))^(1/2)+1/16*(31*A-15*B)*sec(d*x+c)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.92, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2978, 2984, 2985, 2649, 206, 2773} \[ -\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(39 A-20 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(31 A-15 B) \tan (c+d x) \sec (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(19 A-11 B) \tan (c+d x) \sec (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

((39*A - 20*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(5/2)*d) - ((219*A - 115*B)*ArcT
anh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (7*(9*A - 5*B)*Tan[c
+ d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]*Tan[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5
/2)) - ((19*A - 11*B)*Sec[c + d*x]*Tan[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + ((31*A - 15*B)*Sec[c +
d*x]*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2773

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

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] + Dist[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])

Rule 2984

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

Rule 2985

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

Rubi steps

\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\left (2 a (3 A-B)-\frac {7}{2} a (A-B) \cos (c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (a^2 (31 A-15 B)-\frac {5}{4} a^2 (19 A-11 B) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-7 a^3 (9 A-5 B)+\frac {3}{2} a^3 (31 A-15 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{16 a^5}\\ &=-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (2 a^4 (39 A-20 B)-\frac {7}{2} a^4 (9 A-5 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{16 a^6}\\ &=-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(219 A-115 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}+\frac {(39 A-20 B) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{8 a^3}\\ &=-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(219 A-115 B) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{16 a^2 d}-\frac {(39 A-20 B) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^2 d}\\ &=\frac {(39 A-20 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 6.20, size = 178, normalized size = 0.67 \[ \frac {-8 (219 A-115 B) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 \sqrt {2} (39 A-20 B) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) ((169 B-269 A) \cos (c+d x)+(110 B-190 A) \cos (2 (c+d x))-63 A \cos (3 (c+d x))-158 A+35 B \cos (3 (c+d x))+110 B)}{64 a d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(-8*(219*A - 115*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + 32*Sqrt[2]*(39*A - 20*B)*ArcTanh[Sqrt[2]*Si
n[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (-158*A + 110*B + (-269*A + 169*B)*Cos[c + d*x] + (-190*A + 110*B)*Cos[2*
(c + d*x)] - 63*A*Cos[3*(c + d*x)] + 35*B*Cos[3*(c + d*x)])*Sec[c + d*x]^2*Tan[(c + d*x)/2])/(64*a*d*(a*(1 + C
os[c + d*x]))^(3/2))

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fricas [A]  time = 0.66, size = 428, normalized size = 1.62 \[ -\frac {\sqrt {2} {\left ({\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (7 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (19 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, A - 4 \, B\right )} \cos \left (d x + c\right ) - 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/64*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^5 + 3*(219*A - 115*B)*cos(d*x + c)^4 + 3*(219*A - 115*B)*cos(d*x
+ c)^3 + (219*A - 115*B)*cos(d*x + c)^2)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*s
qrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((39*A - 20*B)*cos(d*
x + c)^5 + 3*(39*A - 20*B)*cos(d*x + c)^4 + 3*(39*A - 20*B)*cos(d*x + c)^3 + (39*A - 20*B)*cos(d*x + c)^2)*sqr
t(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*
x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(7*(9*A - 5*B)*cos(d*x + c)^3 + 5*(19*A - 11*B)*cos(d*x +
c)^2 + 4*(5*A - 4*B)*cos(d*x + c) - 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3
*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of ab
s or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(cos((d*t_nostep+c)/2
))]Discontinuities at zeroes of cos((d*t_nostep+c)/2) were not checkedEvaluation time: 1.15Unable to divide, p
erhaps due to rounding error%%%{%%{[%%%{%%{[663535861056963827345930584064,0]:[1,0,-2]%%},[16]%%%},0]:[1,0,%%%
{-1,[1]%%%}]%%},[0]%%%} / %%%{%%%{%%{[9903520314283042199192993792,0]:[1,0,-2]%%},[16]%%%},[0]%%%} Error: Bad
Argument Value

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maple [B]  time = 1.99, size = 1610, normalized size = 6.10 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x)

[Out]

-1/8*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(320*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^
2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^8*a-624*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^
(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^8
*a+320*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*
x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^8*a-624*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^8*a+624*A*ln(-4*(a*2^(1/2)*cos(1/
2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1
/2*c)^6*a+624*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*
cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^6*a-320*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d
*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^6*a-320*B*ln(4/(2*cos(1/2*d
*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*
d*x+1/2*c)^6*a-2*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-876*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^
2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a+460*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a-188*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a
^(1/2)*cos(1/2*d*x+1/2*c)^4-156*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*
a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^4*a-156*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^
(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^4*a+80*B*
ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*
c)+2*a))*cos(1/2*d*x+1/2*c)^4*a+80*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2
)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^4*a+19*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*
c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2-11*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2+100*
B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4+2*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/
2)*a^(1/2)+219*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1
/2*c)^4*a-115*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/
2*c)^4*a+252*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6-140*B*2^(1/2)*(a*sin(1/2*d*
x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6+876*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1
/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^8*a-460*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/
2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^8*a)/a^(7/2)/cos(1/2*d*x+1/2*c)^3/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^2/(2
*cos(1/2*d*x+1/2*c)+2^(1/2))^2/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(5/2)),x)

[Out]

int((A + B*cos(c + d*x))/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(5/2)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**3/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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