3.52 \(\int \frac{\sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=103 \[ \frac{4 \tan ^3(c+d x)}{3 a d}+\frac{4 \tan (c+d x)}{a d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{3 \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{\tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.0954953, antiderivative size = 103, normalized size of antiderivative = 1., 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 = {2768, 2748, 3767, 3768, 3770} \[ \frac{4 \tan ^3(c+d x)}{3 a d}+\frac{4 \tan (c+d x)}{a d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{3 \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{\tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + a*Cos[c + d*x]),x]

[Out]

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

Rule 2768

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

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

Mathematica [B]  time = 4.25736, size = 368, normalized size = 3.57 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (6 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+\frac{1}{8} \sec (c) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-12 \sin (2 c+d x)-6 \sin (c+2 d x)-6 \sin (3 c+2 d x)+20 \sin (2 c+3 d x)+9 \cos (2 c+3 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+9 \cos (4 c+3 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+27 \cos (d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+27 \cos (2 c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-9 \cos (2 c+3 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-9 \cos (4 c+3 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+48 \sin (d x)\right )\right )}{3 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + a*Cos[c + d*x]),x]

[Out]

(Cos[(c + d*x)/2]*(6*Sec[c/2]*Sin[(d*x)/2] + (Cos[(c + d*x)/2]*Sec[c]*Sec[c + d*x]^3*(9*Cos[2*c + 3*d*x]*Log[C
os[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 27*Cos[d*x
]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 27*Cos[2*c + d*x]*(L
og[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 9*Cos[2*c + 3*d*x]*Log[C
os[(c + d*x)/2] + Sin[(c + d*x)/2]] - 9*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 48*Sin[d*x
] - 12*Sin[2*c + d*x] - 6*Sin[c + 2*d*x] - 6*Sin[3*c + 2*d*x] + 20*Sin[2*c + 3*d*x]))/8))/(3*a*d*(1 + Cos[c +
d*x]))

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Maple [A]  time = 0.069, size = 183, normalized size = 1.8 \begin{align*}{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{5}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{1}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{5}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{3}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+cos(d*x+c)*a),x)

[Out]

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

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Maxima [B]  time = 1.17474, size = 277, normalized size = 2.69 \begin{align*} \frac{\frac{2 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac{3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{9 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{9 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{6 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*sin(
d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d*x
+ c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.63692, size = 331, normalized size = 3.21 \begin{align*} -\frac{9 \,{\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \,{\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \, \cos \left (d x + c\right )^{3} + 7 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{12 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(9*(cos(d*x + c)^4 + cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 9*(cos(d*x + c)^4 + cos(d*x + c)^3)*log(-si
n(d*x + c) + 1) - 2*(16*cos(d*x + c)^3 + 7*cos(d*x + c)^2 - cos(d*x + c) + 2)*sin(d*x + c))/(a*d*cos(d*x + c)^
4 + a*d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{4}{\left (c + d x \right )}}{\cos{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+a*cos(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**4/(cos(c + d*x) + 1), x)/a

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Giac [A]  time = 1.38833, size = 154, normalized size = 1.5 \begin{align*} -\frac{\frac{9 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{9 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} + \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, \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 )}^{3} a}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(9*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 9*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 6*tan(1/2*d*x + 1/2*c)
/a + 2*(15*tan(1/2*d*x + 1/2*c)^5 - 16*tan(1/2*d*x + 1/2*c)^3 + 9*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)
^2 - 1)^3*a))/d