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]

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

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Rubi [A]  time = 0.07, antiderivative size = 120, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.53, 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]

-1/480*((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[Co
s[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)/(a^2*d)

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fricas [A]  time = 0.50, size = 85, normalized size = 0.71 \[ -\frac {60 \, \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) + 120 \, \cos \left (d x + c\right )^{5} + 30 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )^{2} + 24 \, \cos \left (d x + c\right ) - 10}{60 \, a^{2} d \cos \left (d x + c\right )^{6}} \]

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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giac [B]  time = 26.20, size = 223, normalized size = 1.86 \[ \frac {\frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} - \frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{2}} + \frac {\frac {234 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {1005 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2220 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2925 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1002 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {147 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 19}{a^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{6}}}{60 \, d} \]

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

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maple [A]  time = 0.63, size = 110, normalized size = 0.92 \[ \frac {\sec ^{6}\left (d x +c \right )}{6 a^{2} d}-\frac {2 \left (\sec ^{5}\left (d x +c \right )\right )}{5 a^{2} d}-\frac {\sec ^{4}\left (d x +c \right )}{4 a^{2} d}+\frac {4 \left (\sec ^{3}\left (d x +c \right )\right )}{3 a^{2} d}-\frac {\sec ^{2}\left (d x +c \right )}{2 a^{2} d}-\frac {2 \sec \left (d x +c \right )}{a^{2} d}+\frac {\ln \left (\sec \left (d x +c \right )\right )}{a^{2} d} \]

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

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maxima [A]  time = 0.33, size = 80, normalized size = 0.67 \[ -\frac {\frac {60 \, \log \left (\cos \left (d x + c\right )\right )}{a^{2}} + \frac {120 \, \cos \left (d x + c\right )^{5} + 30 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )^{2} + 24 \, \cos \left (d x + c\right ) - 10}{a^{2} \cos \left (d x + c\right )^{6}}}{60 \, d} \]

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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mupad [B]  time = 5.18, size = 193, normalized size = 1.61 \[ \frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^2\,d}+\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {54\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-\frac {32}{15}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^9/(a + a/cos(c + d*x))^2,x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2)^2))/(a^2*d) + ((54*tan(c/2 + (d*x)/2)^2)/5 - 20*tan(c/2 + (d*x)/2)^4 + 12*tan(c/2
+ (d*x)/2)^6 + 12*tan(c/2 + (d*x)/2)^8 - 2*tan(c/2 + (d*x)/2)^10 - 32/15)/(d*(15*a^2*tan(c/2 + (d*x)/2)^4 - 6*
a^2*tan(c/2 + (d*x)/2)^2 - 20*a^2*tan(c/2 + (d*x)/2)^6 + 15*a^2*tan(c/2 + (d*x)/2)^8 - 6*a^2*tan(c/2 + (d*x)/2
)^10 + a^2*tan(c/2 + (d*x)/2)^12 + a^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{9}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

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]

Integral(tan(c + d*x)**9/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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