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

Optimal. Leaf size=119 \[ -\frac {16 \tan (c+d x)}{3 a^2 d}+\frac {7 \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {7 \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {8 \tan (c+d x) \sec (c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {\tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

7/2*arctanh(sin(d*x+c))/a^2/d-16/3*tan(d*x+c)/a^2/d+7/2*sec(d*x+c)*tan(d*x+c)/a^2/d-8/3*sec(d*x+c)*tan(d*x+c)/
a^2/d/(1+cos(d*x+c))-1/3*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^2

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2766, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac {16 \tan (c+d x)}{3 a^2 d}+\frac {7 \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {7 \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {8 \tan (c+d x) \sec (c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {\tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*ArcTanh[Sin[c + d*x]])/(2*a^2*d) - (16*Tan[c + d*x])/(3*a^2*d) + (7*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) -
(8*Sec[c + d*x]*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - (Sec[c + d*x]*Tan[c + d*x])/(3*d*(a + a*Cos[c + d
*x])^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac {\sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {(5 a-3 a \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {8 \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \left (21 a^2-16 a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{3 a^4}\\ &=-\frac {8 \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {16 \int \sec ^2(c+d x) \, dx}{3 a^2}+\frac {7 \int \sec ^3(c+d x) \, dx}{a^2}\\ &=\frac {7 \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {8 \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {7 \int \sec (c+d x) \, dx}{2 a^2}+\frac {16 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=\frac {7 \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac {16 \tan (c+d x)}{3 a^2 d}+\frac {7 \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {8 \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.81, size = 292, normalized size = 2.45 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (-2 \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )-2 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+3 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {8 \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {1}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-14 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+14 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-40 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(-2*Sec[c/2]*Sin[(d*x)/2] - 40*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 3*Cos[(c + d*x)/2]
^3*(-14*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 14*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (Cos[(c + d*x
)/2] - Sin[(c + d*x)/2])^(-2) - (Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^(-2) - (8*Sin[d*x])/((Cos[c/2] - Sin[c/2
])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) - 2*Cos
[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)

________________________________________________________________________________________

fricas [A]  time = 1.10, size = 162, normalized size = 1.36 \[ \frac {21 \, {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \, {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (32 \, \cos \left (d x + c\right )^{3} + 43 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) - 3\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(21*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 21*(cos(d*x + c)^4 + 2*c
os(d*x + c)^3 + cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(32*cos(d*x + c)^3 + 43*cos(d*x + c)^2 + 6*cos(d*x
+ c) - 3)*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)

________________________________________________________________________________________

giac [A]  time = 0.60, size = 122, normalized size = 1.03 \[ \frac {\frac {21 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {21 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(21*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 21*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 6*(5*tan(1/2*d*x
+ 1/2*c)^3 - 3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2) - (a^4*tan(1/2*d*x + 1/2*c)^3 + 21*a
^4*tan(1/2*d*x + 1/2*c))/a^6)/d

________________________________________________________________________________________

maple [A]  time = 0.14, size = 162, normalized size = 1.36 \[ -\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d \,a^{2}}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {1}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {5}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}-\frac {1}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3-7/2/d/a^2*tan(1/2*d*x+1/2*c)+1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2+5/2/d/a^2/(tan
(1/2*d*x+1/2*c)-1)-7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)-1/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2+5/2/d/a^2/(tan(1/2*d*
x+1/2*c)+1)+7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)

________________________________________________________________________________________

maxima [A]  time = 0.95, size = 190, normalized size = 1.60 \[ -\frac {\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + sin(
d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c)/(
cos(d*x + c) + 1) - 1)/a^2)/d

________________________________________________________________________________________

mupad [B]  time = 0.43, size = 122, normalized size = 1.03 \[ \frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,a^2\,d}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*atanh(tan(c/2 + (d*x)/2)))/(a^2*d) - tan(c/2 + (d*x)/2)^3/(6*a^2*d) - (3*tan(c/2 + (d*x)/2) - 5*tan(c/2 + (
d*x)/2)^3)/(d*(a^2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c/2 + (d*x)/2)^2 + a^2)) - (7*tan(c/2 + (d*x)/2))/(2*a^2*d
)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________