3.81 \(\int \frac{\tan ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=34 \[ \frac{\tan (c+d x)}{a^2 d}-\frac{2 \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{x}{a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0653077, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3888, 3773, 3770, 3767, 8} \[ \frac{\tan (c+d x)}{a^2 d}-\frac{2 \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + a*Sec[c + d*x])^2,x]

[Out]

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

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rule 3773

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + (Dist[2*a*b, Int[Csc[c + d*x], x],
x] + Dist[b^2, Int[Csc[c + d*x]^2, x], x]) /; FreeQ[{a, b, c, d}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 8

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

Rubi steps

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

Mathematica [B]  time = 0.48694, size = 177, normalized size = 5.21 \[ \frac{4 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (\frac{\sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+d x\right )}{a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + a*Sec[c + d*x])^2,x]

[Out]

(4*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(d*x + 2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] + 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]))))/(a^2*d*(1 + Sec[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.062, size = 102, normalized size = 3. \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{2}}}-{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+a*sec(d*x+c))^2,x)

[Out]

2/d/a^2*arctan(tan(1/2*d*x+1/2*c))-1/d/a^2/(tan(1/2*d*x+1/2*c)+1)-2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)-1/d/a^2/(ta
n(1/2*d*x+1/2*c)-1)+2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)

________________________________________________________________________________________

Maxima [B]  time = 1.78972, size = 166, normalized size = 4.88 \begin{align*} \frac{2 \,{\left (\frac{\arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac{\sin \left (d x + c\right )}{{\left (a^{2} - \frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

2*(arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + log(sin(d*x +
c)/(cos(d*x + c) + 1) - 1)/a^2 + sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) +
1)))/d

________________________________________________________________________________________

Fricas [A]  time = 1.18004, size = 177, normalized size = 5.21 \begin{align*} \frac{d x \cos \left (d x + c\right ) - \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + \sin \left (d x + c\right )}{a^{2} d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

(d*x*cos(d*x + c) - cos(d*x + c)*log(sin(d*x + c) + 1) + cos(d*x + c)*log(-sin(d*x + c) + 1) + sin(d*x + c))/(
a^2*d*cos(d*x + c))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

________________________________________________________________________________________

Giac [B]  time = 2.48726, size = 107, normalized size = 3.15 \begin{align*} \frac{\frac{d x + c}{a^{2}} - \frac{2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac{2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

((d*x + c)/a^2 - 2*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 + 2*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - 2*tan(1
/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^2))/d