3.71 \(\int \frac{\tan ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=120 \[ \frac{\sec ^6(c+d x)}{6 a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}-\frac{\sec ^4(c+d x)}{4 a^2 d}+\frac{4 \sec ^3(c+d x)}{3 a^2 d}-\frac{\sec ^2(c+d x)}{2 a^2 d}-\frac{2 \sec (c+d x)}{a^2 d}-\frac{\log (\cos (c+d x))}{a^2 d} \]

[Out]

-(Log[Cos[c + d*x]]/(a^2*d)) - (2*Sec[c + d*x])/(a^2*d) - Sec[c + d*x]^2/(2*a^2*d) + (4*Sec[c + d*x]^3)/(3*a^2
*d) - Sec[c + d*x]^4/(4*a^2*d) - (2*Sec[c + d*x]^5)/(5*a^2*d) + Sec[c + d*x]^6/(6*a^2*d)

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Rubi [A]  time = 0.0736004, antiderivative size = 120, 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, 88} \[ \frac{\sec ^6(c+d x)}{6 a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}-\frac{\sec ^4(c+d x)}{4 a^2 d}+\frac{4 \sec ^3(c+d x)}{3 a^2 d}-\frac{\sec ^2(c+d x)}{2 a^2 d}-\frac{2 \sec (c+d x)}{a^2 d}-\frac{\log (\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^9/(a + a*Sec[c + d*x])^2,x]

[Out]

-(Log[Cos[c + d*x]]/(a^2*d)) - (2*Sec[c + d*x])/(a^2*d) - Sec[c + d*x]^2/(2*a^2*d) + (4*Sec[c + d*x]^3)/(3*a^2
*d) - Sec[c + d*x]^4/(4*a^2*d) - (2*Sec[c + d*x]^5)/(5*a^2*d) + Sec[c + d*x]^6/(6*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\tan ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(a-a x)^4 (a+a x)^2}{x^7} \, dx,x,\cos (c+d x)\right )}{a^8 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{a^6}{x^7}-\frac{2 a^6}{x^6}-\frac{a^6}{x^5}+\frac{4 a^6}{x^4}-\frac{a^6}{x^3}-\frac{2 a^6}{x^2}+\frac{a^6}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^8 d}\\ &=-\frac{\log (\cos (c+d x))}{a^2 d}-\frac{2 \sec (c+d x)}{a^2 d}-\frac{\sec ^2(c+d x)}{2 a^2 d}+\frac{4 \sec ^3(c+d x)}{3 a^2 d}-\frac{\sec ^4(c+d x)}{4 a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{\sec ^6(c+d x)}{6 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.517759, size = 125, normalized size = 1.04 \[ -\frac{\sec ^6(c+d x) (312 \cos (c+d x)+5 (28 \cos (3 (c+d x))+6 \cos (4 (c+d x))+12 \cos (5 (c+d x))+18 \cos (4 (c+d x)) \log (\cos (c+d x))+3 \cos (6 (c+d x)) \log (\cos (c+d x))+30 \log (\cos (c+d x))+9 \cos (2 (c+d x)) (5 \log (\cos (c+d x))+4)+14))}{480 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^9/(a + a*Sec[c + d*x])^2,x]

[Out]

-((312*Cos[c + d*x] + 5*(14 + 28*Cos[3*(c + d*x)] + 6*Cos[4*(c + d*x)] + 12*Cos[5*(c + d*x)] + 30*Log[Cos[c +
d*x]] + 18*Cos[4*(c + d*x)]*Log[Cos[c + d*x]] + 3*Cos[6*(c + d*x)]*Log[Cos[c + d*x]] + 9*Cos[2*(c + d*x)]*(4 +
5*Log[Cos[c + d*x]])))*Sec[c + d*x]^6)/(480*a^2*d)

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Maple [A]  time = 0.084, size = 110, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{6\,d{a}^{2}}}-{\frac{2\, \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{5\,d{a}^{2}}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{4\,d{a}^{2}}}+{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{3\,d{a}^{2}}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{2\,d{a}^{2}}}-2\,{\frac{\sec \left ( dx+c \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)^9/(a+a*sec(d*x+c))^2,x)

[Out]

1/6*sec(d*x+c)^6/d/a^2-2/5*sec(d*x+c)^5/d/a^2-1/4*sec(d*x+c)^4/d/a^2+4/3*sec(d*x+c)^3/d/a^2-1/2*sec(d*x+c)^2/d
/a^2-2*sec(d*x+c)/d/a^2+1/d/a^2*ln(sec(d*x+c))

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Maxima [A]  time = 1.14996, size = 108, normalized size = 0.9 \begin{align*} -\frac{\frac{60 \, \log \left (\cos \left (d x + c\right )\right )}{a^{2}} + \frac{120 \, \cos \left (d x + c\right )^{5} + 30 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )^{2} + 24 \, \cos \left (d x + c\right ) - 10}{a^{2} \cos \left (d x + c\right )^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/60*(60*log(cos(d*x + c))/a^2 + (120*cos(d*x + c)^5 + 30*cos(d*x + c)^4 - 80*cos(d*x + c)^3 + 15*cos(d*x + c
)^2 + 24*cos(d*x + c) - 10)/(a^2*cos(d*x + c)^6))/d

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Fricas [A]  time = 1.22501, size = 234, normalized size = 1.95 \begin{align*} -\frac{60 \, \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) + 120 \, \cos \left (d x + c\right )^{5} + 30 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )^{2} + 24 \, \cos \left (d x + c\right ) - 10}{60 \, a^{2} d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/60*(60*cos(d*x + c)^6*log(-cos(d*x + c)) + 120*cos(d*x + c)^5 + 30*cos(d*x + c)^4 - 80*cos(d*x + c)^3 + 15*
cos(d*x + c)^2 + 24*cos(d*x + c) - 10)/(a^2*d*cos(d*x + c)^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**9/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 11.4248, size = 301, normalized size = 2.51 \begin{align*} \frac{\frac{60 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} - \frac{60 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{2}} + \frac{\frac{234 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{1005 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{2220 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{2925 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{1002 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{147 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 19}{a^{2}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^2 - 60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1))/a^2 + (234*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1005*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^
2 + 2220*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 2925*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 1002*(co
s(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 147*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 19)/(a^2*((cos(d*x +
c) - 1)/(cos(d*x + c) + 1) + 1)^6))/d