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

Optimal. Leaf size=185 \[ \frac {19 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {13 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {7 \tan (c+d x)}{4 a d \sqrt {a \cos (c+d x)+a}}+\frac {\tan (c+d x) \sec (c+d x)}{a d \sqrt {a \cos (c+d x)+a}}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

19/4*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d-13/4*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/
(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-7/4*tan(d*x+c)/a/
d/(a+a*cos(d*x+c))^(1/2)+sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.50, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2766, 2984, 2985, 2649, 206, 2773} \[ \frac {19 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {13 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {7 \tan (c+d x)}{4 a d \sqrt {a \cos (c+d x)+a}}+\frac {\tan (c+d x) \sec (c+d x)}{a d \sqrt {a \cos (c+d x)+a}}-\frac {\tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(19*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(3/2)*d) - (13*ArcTanh[(Sqrt[a]*Sin[c + d*x
])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - (7*Tan[c + d*x])/(4*a*d*Sqrt[a + a*Cos[c + d*x
]]) - (Sec[c + d*x]*Tan[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + (Sec[c + d*x]*Tan[c + d*x])/(a*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 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*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] + Dis
t[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*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 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 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 {\sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac {\sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (4 a-\frac {5}{2} a \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {\sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\sec (c+d x) \tan (c+d x)}{a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-7 a^2+6 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^3}\\ &=-\frac {7 \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {\sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\sec (c+d x) \tan (c+d x)}{a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (\frac {19 a^3}{2}-\frac {7}{2} a^3 \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^4}\\ &=-\frac {7 \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {\sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\sec (c+d x) \tan (c+d x)}{a d \sqrt {a+a \cos (c+d x)}}+\frac {19 \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{8 a^2}-\frac {13 \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac {7 \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {\sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\sec (c+d x) \tan (c+d x)}{a d \sqrt {a+a \cos (c+d x)}}-\frac {19 \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 d}+\frac {13 \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 )}{2 a d}\\ &=\frac {19 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac {13 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {7 \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {\sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\sec (c+d x) \tan (c+d x)}{a d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 28.43, size = 1941, normalized size = 10.49 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-19/8 + (19*I)/8)*(1 + E^(I*c))*(Sqrt[2] - (1 - I)*E^((I/2)*c) + (16 - 16*I)*E^(((3*I)/2)*c + I*d*x) + (20 +
20*I)*Sqrt[2]*E^((2*I)*c + ((3*I)/2)*d*x) - (34 - 34*I)*E^(((5*I)/2)*c + (2*I)*d*x) - (20 + 20*I)*Sqrt[2]*E^(
(3*I)*c + ((5*I)/2)*d*x) + (16 - 16*I)*E^(((7*I)/2)*c + (3*I)*d*x) + (4 + 4*I)*Sqrt[2]*E^((4*I)*c + ((7*I)/2)*
d*x) - (1 - I)*E^(((9*I)/2)*c + (4*I)*d*x) + (8*I)*E^((I/2)*(c + d*x)) - 16*Sqrt[2]*E^(I*(c + d*x)) - (40*I)*E
^(((3*I)/2)*(c + d*x)) + 34*Sqrt[2]*E^((2*I)*(c + d*x)) + (40*I)*E^(((5*I)/2)*(c + d*x)) - 16*Sqrt[2]*E^((3*I)
*(c + d*x)) - (8*I)*E^(((7*I)/2)*(c + d*x)) + Sqrt[2]*E^((4*I)*(c + d*x)) - (4 + 4*I)*Sqrt[2]*E^((I/2)*(2*c +
d*x)))*x*Cos[c/2 + (d*x)/2]^3)/(((-1 - I) + Sqrt[2]*E^((I/2)*c))*(-1 + E^(I*c))*(I - 2*Sqrt[2]*E^((I/2)*(c + d
*x)) - (4*I)*E^(I*(c + d*x)) + 2*Sqrt[2]*E^(((3*I)/2)*(c + d*x)) + I*E^((2*I)*(c + d*x)))^2*(a*(1 + Cos[c + d*
x]))^(3/2)) - (((19*I)/2)*ArcTan[(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4] - Sqrt[2]*Sin[c/4 + (d*x)/4])/(-Cos[
c/4 + (d*x)/4] + Sqrt[2]*Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])]*Cos[c/2 + (d*x)/2]^3)/(Sqrt[2]*d*(a*(1 + Co
s[c + d*x]))^(3/2)) - (((19*I)/2)*ArcTan[(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4] - Sqrt[2]*Sin[c/4 + (d*x)/4]
)/(Cos[c/4 + (d*x)/4] + Sqrt[2]*Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])]*Cos[c/2 + (d*x)/2]^3)/(Sqrt[2]*d*(a*
(1 + Cos[c + d*x]))^(3/2)) + (13*Cos[c/2 + (d*x)/2]^3*Log[Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4]])/(d*(a*(1 +
Cos[c + d*x]))^(3/2)) - (13*Cos[c/2 + (d*x)/2]^3*Log[Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4]])/(d*(a*(1 + Cos
[c + d*x]))^(3/2)) - (19*Cos[c/2 + (d*x)/2]^3*Log[2 - Sqrt[2]*Cos[c/2 + (d*x)/2] - Sqrt[2]*Sin[c/2 + (d*x)/2]]
)/(4*Sqrt[2]*d*(a*(1 + Cos[c + d*x]))^(3/2)) - (19*Cos[c/2 + (d*x)/2]^3*Log[2 + Sqrt[2]*Cos[c/2 + (d*x)/2] - S
qrt[2]*Sin[c/2 + (d*x)/2]])/(4*Sqrt[2]*d*(a*(1 + Cos[c + d*x]))^(3/2)) - ((19*I)*ArcTan[((2*I)*Cos[c/2] - I*(-
Sqrt[2] + 2*Sin[c/2])*Tan[(d*x)/4])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]]*Cos[c/2 + (d*x)/2]^3*Cot[c/2])/(d*
(a*(1 + Cos[c + d*x]))^(3/2)*Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]) + (19*Cos[c/2 + (d*x)/2]^3*Csc[c/2]*(-(d*
x*Cos[c/2]) + 2*Log[Sqrt[2] + 2*Cos[(d*x)/2]*Sin[c/2] + 2*Cos[c/2]*Sin[(d*x)/2]]*Sin[c/2] + ((4*I)*Sqrt[2]*Arc
Tan[((2*I)*Cos[c/2] - I*(-Sqrt[2] + 2*Sin[c/2])*Tan[(d*x)/4])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]]*Cos[c/2]
)/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]))/(Sqrt[2]*d*(a*(1 + Cos[c + d*x]))^(3/2)*(4*Cos[c/2]^2 + 4*Sin[c/2]^
2)) - Cos[c/2 + (d*x)/2]^3/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])^2) + Co
s[c/2 + (d*x)/2]^3/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4])^2) + (Cos[c/2 +
(d*x)/2]^3*Sin[(d*x)/2])/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2
+ (d*x)/2])^2) + (Cos[c/2 + (d*x)/2]^3*(-5*Cos[c/2] + 7*Sin[c/2]))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2]
- Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (Cos[c/2 + (d*x)/2]^3*Sin[(d*x)/2])/(d*(a*(1 + Cos[c
+ d*x]))^(3/2)*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Cos[c/2 + (d*x)/2]^3*(5*
Cos[c/2] + 7*Sin[c/2]))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2
+ (d*x)/2]))

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fricas [A]  time = 1.63, size = 302, normalized size = 1.63 \[ \frac {26 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 19 \, {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (7 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(26*sqrt(2)*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(
2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) +
1)) + 19*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(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*sqrt(a*cos(d*x + c) + a)*(7*cos(d*x + c)^2 + 3*cos(d*x + c) - 2)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 +
2*a^2*d*cos(d*x + c)^3 + a^2*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(sec(d*x+c)^3/(a+a*cos(d*x+c))^(3/2),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)Warning, integration of abs or sign assumes cons
tant 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 checkedUnable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{
[350488137400481480704,0]:[1,0,-2]%%},[16]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0]%%%} / %%%{%%%{%%{[18446744073709
551616,0]:[1,0,-2]%%},[16]%%%},[0]%%%} Error: Bad Argument Value

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maple [B]  time = 1.05, size = 807, normalized size = 4.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(104*2^(1/2)*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))*cos(1/2*d*x+1/2*c)^6*a-76*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-76*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-104*2
^(1/2)*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))*a*cos(1/2*d*x+1/2*c)^4+28*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+76*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+76*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+26*2^(1/2)*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))*a*c
os(1/2*d*x+1/2*c)^2-22*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)^2-19*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)^2*a-19*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)^2*a+2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(5/2)/
cos(1/2*d*x+1/2*c)/(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(sec(d*x+c)^3/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**3/(a*(cos(c + d*x) + 1))**(3/2), x)

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