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

Optimal. Leaf size=33 \[ \frac{2 \log (\cos (c+d x)+1)}{a^2 d}-\frac{\log (\cos (c+d x))}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0448387, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3879, 72} \[ \frac{2 \log (\cos (c+d x)+1)}{a^2 d}-\frac{\log (\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.060438, size = 30, normalized size = 0.91 \[ \frac{4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log (\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.072, size = 34, normalized size = 1. \begin{align*} 2\,{\frac{\ln \left ( 1+\sec \left ( dx+c \right ) \right ) }{d{a}^{2}}}-{\frac{\ln \left ( \sec \left ( dx+c \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^2*ln(1+sec(d*x+c))-1/d/a^2*ln(sec(d*x+c))

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Maxima [A]  time = 1.12223, size = 42, normalized size = 1.27 \begin{align*} \frac{\frac{2 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} - \frac{\log \left (\cos \left (d x + c\right )\right )}{a^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.19847, size = 85, normalized size = 2.58 \begin{align*} -\frac{\log \left (-\cos \left (d x + c\right )\right ) - 2 \, \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{3}{\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)**3/(a+a*sec(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.91302, size = 45, normalized size = 1.36 \begin{align*} -\frac{\log \left ({\left | \frac{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1 \right |}\right )}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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