3.2.36 \(\int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [136]

Optimal. Leaf size=114 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}} \]

[Out]

2*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d-1/2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-5/4*arc
tanh(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)

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Rubi [A]
time = 0.15, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2845, 3064, 2728, 212, 2852} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{3/2} d}-\frac {5 \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 {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(3/2)*d) - (5*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(
Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - Sin[c + d*x]/(2*d*(a + a*Cos[c + d*x])^(3/2))

Rule 212

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

Rule 2728

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 2845

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 2852

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 3064

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 (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac {\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (2 a-\frac {1}{2} a \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{a^2}-\frac {5 \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac {\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {2 \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a d}+\frac {5 \text {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 {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}-\frac {5 \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 {\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 23.50, size = 1787, normalized size = 15.68 \begin {gather*} \text {Too large to display} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1 + I)*(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)*Sqr
t[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*C
os[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)
) - ((2*I)*Sqrt[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)/(d*(a*(1 + Cos[c + d*x]))^(3
/2)) + (5*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))
- (5*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)) - (Sq
rt[2]*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]])/(d*(a*(1 + Cos[c
+ d*x]))^(3/2)) + ((1 - I)*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*((1 + I)*Cos[c/4] + Sq
rt[2]*Cos[c/4] - (1 - I)*Sin[c/4] - I*Sqrt[2]*Sin[c/4])*((-1 - I)*Cos[c/4] + Sqrt[2]*Cos[c/4] + (1 - I)*Sin[c/
4] - I*Sqrt[2]*Sin[c/4]))/(Sqrt[2]*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2] + Sin[c/2])) - ((1/2 + I/2)*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]]*((1 + I)*Cos[c/4] + Sqrt[2]*Co
s[c/4] - (1 - I)*Sin[c/4] - I*Sqrt[2]*Sin[c/4])*((-1 - I)*Cos[c/4] + Sqrt[2]*Cos[c/4] + (1 - I)*Sin[c/4] - I*S
qrt[2]*Sin[c/4]))/(Sqrt[2]*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2] + Sin[c/2])) - ((8*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]) + (4*Sqrt[2]*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]*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])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]))/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(4*Cos[c/2]^2 + 4*S
in[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) + 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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(289\) vs. \(2(93)=186\).
time = 0.17, size = 290, normalized size = 2.54

method	result	size
default	\(-\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (5 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -4 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -4 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\right )}{4 a^{\frac {5}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(290\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^(5/2)/cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(5*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^
2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^2*a-4*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*co
s(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^2*a-4*ln(-4/(2*cos(1/
2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1
/2*d*x+1/2*c)^2*a+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2))/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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (93) = 186\).
time = 0.42, size = 254, normalized size = 2.23 \begin {gather*} \frac {5 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\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 (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\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} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(5*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*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)) + 4*(cos(d*x
+ c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*s
qrt(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)*
sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.58, size = 137, normalized size = 1.20 \begin {gather*} -\frac {\sqrt {2} {\left (\frac {4 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a} + \frac {5 \, \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a} - \frac {5 \, \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a} - \frac {2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a}\right )}}{8 \, \sqrt {a} d \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*(4*sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c)))/
a + 5*log(sin(1/2*d*x + 1/2*c) + 1)/a - 5*log(-sin(1/2*d*x + 1/2*c) + 1)/a - 2*sin(1/2*d*x + 1/2*c)/((sin(1/2*
d*x + 1/2*c)^2 - 1)*a))/(sqrt(a)*d*sgn(cos(1/2*d*x + 1/2*c)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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