3.87 \(\int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=99 \[ \frac {\sec ^5(c+d x)}{5 a^3 d}-\frac {3 \sec ^4(c+d x)}{4 a^3 d}+\frac {2 \sec ^3(c+d x)}{3 a^3 d}+\frac {\sec ^2(c+d x)}{a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}-\frac {\log (\cos (c+d x))}{a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 99, 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, 75} \[ \frac {\sec ^5(c+d x)}{5 a^3 d}-\frac {3 \sec ^4(c+d x)}{4 a^3 d}+\frac {2 \sec ^3(c+d x)}{3 a^3 d}+\frac {\sec ^2(c+d x)}{a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}-\frac {\log (\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^9/(a + a*Sec[c + d*x])^3,x]

[Out]

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

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])

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

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Mathematica [A]  time = 0.37, size = 93, normalized size = 0.94 \[ -\frac {\sec ^5(c+d x) (280 \cos (2 (c+d x))+90 \cos (4 (c+d x))+150 \cos (c+d x) \log (\cos (c+d x))+15 \cos (5 (c+d x)) \log (\cos (c+d x))+15 \cos (3 (c+d x)) (5 \log (\cos (c+d x))-4)+142)}{240 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^9/(a + a*Sec[c + d*x])^3,x]

[Out]

-1/240*((142 + 280*Cos[2*(c + d*x)] + 90*Cos[4*(c + d*x)] + 150*Cos[c + d*x]*Log[Cos[c + d*x]] + 15*Cos[5*(c +
d*x)]*Log[Cos[c + d*x]] + 15*Cos[3*(c + d*x)]*(-4 + 5*Log[Cos[c + d*x]]))*Sec[c + d*x]^5)/(a^3*d)

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fricas [A]  time = 0.71, size = 75, normalized size = 0.76 \[ -\frac {60 \, \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) + 180 \, \cos \left (d x + c\right )^{4} - 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 12}{60 \, a^{3} d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(60*cos(d*x + c)^5*log(-cos(d*x + c)) + 180*cos(d*x + c)^4 - 60*cos(d*x + c)^3 - 40*cos(d*x + c)^2 + 45*
cos(d*x + c) - 12)/(a^3*d*cos(d*x + c)^5)

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giac [B]  time = 20.75, size = 202, normalized size = 2.04 \[ \frac {\frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{3}} - \frac {\frac {475 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {590 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {50 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {805 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {137 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 119}{a^{3} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^3 - 60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1))/a^3 - (475*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 590*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2
- 50*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 805*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 137*(cos(d*x
+ c) - 1)^5/(cos(d*x + c) + 1)^5 + 119)/(a^3*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^5))/d

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maple [A]  time = 0.65, size = 93, normalized size = 0.94 \[ \frac {\sec ^{5}\left (d x +c \right )}{5 a^{3} d}-\frac {3 \left (\sec ^{4}\left (d x +c \right )\right )}{4 a^{3} d}+\frac {2 \left (\sec ^{3}\left (d x +c \right )\right )}{3 a^{3} d}+\frac {\sec ^{2}\left (d x +c \right )}{a^{3} d}-\frac {3 \sec \left (d x +c \right )}{a^{3} d}+\frac {\ln \left (\sec \left (d x +c \right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.33, size = 70, normalized size = 0.71 \[ -\frac {\frac {60 \, \log \left (\cos \left (d x + c\right )\right )}{a^{3}} + \frac {180 \, \cos \left (d x + c\right )^{4} - 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 12}{a^{3} \cos \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(60*log(cos(d*x + c))/a^3 + (180*cos(d*x + c)^4 - 60*cos(d*x + c)^3 - 40*cos(d*x + c)^2 + 45*cos(d*x + c
) - 12)/(a^3*cos(d*x + c)^5))/d

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mupad [B]  time = 5.92, size = 167, normalized size = 1.69 \[ \frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^3\,d}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {98\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {58\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {64}{15}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^9/(a + a/cos(c + d*x))^3,x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2)^2))/(a^3*d) - ((58*tan(c/2 + (d*x)/2)^2)/3 - (98*tan(c/2 + (d*x)/2)^4)/3 + 22*tan(
c/2 + (d*x)/2)^6 + 2*tan(c/2 + (d*x)/2)^8 - 64/15)/(d*(5*a^3*tan(c/2 + (d*x)/2)^2 - 10*a^3*tan(c/2 + (d*x)/2)^
4 + 10*a^3*tan(c/2 + (d*x)/2)^6 - 5*a^3*tan(c/2 + (d*x)/2)^8 + a^3*tan(c/2 + (d*x)/2)^10 - a^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{9}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**9/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**9/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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