3.58 \(\int \frac{\tan ^5(c+d x)}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=66 \[ \frac{\sec ^3(c+d x)}{3 a d}-\frac{\sec ^2(c+d x)}{2 a d}-\frac{\sec (c+d x)}{a d}-\frac{\log (\cos (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.0572111, antiderivative size = 66, 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, 75} \[ \frac{\sec ^3(c+d x)}{3 a d}-\frac{\sec ^2(c+d x)}{2 a d}-\frac{\sec (c+d x)}{a d}-\frac{\log (\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + a*Sec[c + d*x]),x]

[Out]

-(Log[Cos[c + d*x]]/(a*d)) - Sec[c + d*x]/(a*d) - Sec[c + d*x]^2/(2*a*d) + Sec[c + d*x]^3/(3*a*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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.198023, size = 65, normalized size = 0.98 \[ -\frac{\sec ^3(c+d x) (6 \cos (2 (c+d x))+3 \cos (3 (c+d x)) \log (\cos (c+d x))+\cos (c+d x) (9 \log (\cos (c+d x))+6)+2)}{12 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + a*Sec[c + d*x]),x]

[Out]

-((2 + 6*Cos[2*(c + d*x)] + 3*Cos[3*(c + d*x)]*Log[Cos[c + d*x]] + Cos[c + d*x]*(6 + 9*Log[Cos[c + d*x]]))*Sec
[c + d*x]^3)/(12*a*d)

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Maple [A]  time = 0.065, size = 62, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{3\,da}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{2\,da}}-{\frac{\sec \left ( dx+c \right ) }{da}}+{\frac{\ln \left ( \sec \left ( dx+c \right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+a*sec(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.18203, size = 68, normalized size = 1.03 \begin{align*} -\frac{\frac{6 \, \log \left (\cos \left (d x + c\right )\right )}{a} + \frac{6 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 2}{a \cos \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.175, size = 142, normalized size = 2.15 \begin{align*} -\frac{6 \, \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) + 6 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 2}{6 \, a d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(6*cos(d*x + c)^3*log(-cos(d*x + c)) + 6*cos(d*x + c)^2 + 3*cos(d*x + c) - 2)/(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{\tan ^{5}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+a*sec(d*x+c)),x)

[Out]

Integral(tan(c + d*x)**5/(sec(c + d*x) + 1), x)/a

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Giac [B]  time = 3.77887, size = 212, normalized size = 3.21 \begin{align*} \frac{\frac{6 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a} - \frac{6 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a} + \frac{\frac{21 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{45 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{11 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 3}{a{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a - 6*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) +
1) - 1))/a + (21*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 45*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 11*(co
s(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 3)/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^3))/d