3.86 \(\int \frac {\tan ^{11}(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=137 \[ \frac {\sec ^7(c+d x)}{7 a^3 d}-\frac {\sec ^6(c+d x)}{2 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}+\frac {5 \sec ^4(c+d x)}{4 a^3 d}-\frac {5 \sec ^3(c+d x)}{3 a^3 d}-\frac {\sec ^2(c+d x)}{2 a^3 d}+\frac {3 \sec (c+d x)}{a^3 d}+\frac {\log (\cos (c+d x))}{a^3 d} \]

[Out]

ln(cos(d*x+c))/a^3/d+3*sec(d*x+c)/a^3/d-1/2*sec(d*x+c)^2/a^3/d-5/3*sec(d*x+c)^3/a^3/d+5/4*sec(d*x+c)^4/a^3/d+1
/5*sec(d*x+c)^5/a^3/d-1/2*sec(d*x+c)^6/a^3/d+1/7*sec(d*x+c)^7/a^3/d

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Rubi [A]  time = 0.08, antiderivative size = 137, 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} \[ \frac {\sec ^7(c+d x)}{7 a^3 d}-\frac {\sec ^6(c+d x)}{2 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}+\frac {5 \sec ^4(c+d x)}{4 a^3 d}-\frac {5 \sec ^3(c+d x)}{3 a^3 d}-\frac {\sec ^2(c+d x)}{2 a^3 d}+\frac {3 \sec (c+d x)}{a^3 d}+\frac {\log (\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^11/(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/(2*a^3*d) - (5*Sec[c + d*x]^3)/(3*a^3*d)
+ (5*Sec[c + d*x]^4)/(4*a^3*d) + Sec[c + d*x]^5/(5*a^3*d) - Sec[c + d*x]^6/(2*a^3*d) + Sec[c + d*x]^7/(7*a^3*
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 ^{11}(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(a-a x)^5 (a+a x)^2}{x^8} \, dx,x,\cos (c+d x)\right )}{a^{10} d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {a^7}{x^8}-\frac {3 a^7}{x^7}+\frac {a^7}{x^6}+\frac {5 a^7}{x^5}-\frac {5 a^7}{x^4}-\frac {a^7}{x^3}+\frac {3 a^7}{x^2}-\frac {a^7}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^{10} 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)}{2 a^3 d}-\frac {5 \sec ^3(c+d x)}{3 a^3 d}+\frac {5 \sec ^4(c+d x)}{4 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}-\frac {\sec ^6(c+d x)}{2 a^3 d}+\frac {\sec ^7(c+d x)}{7 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 140, normalized size = 1.02 \[ \frac {\sec ^7(c+d x) (4522 \cos (2 (c+d x))+1050 \cos (3 (c+d x))+2380 \cos (4 (c+d x))-210 \cos (5 (c+d x))+630 \cos (6 (c+d x))+2205 \cos (3 (c+d x)) \log (\cos (c+d x))+735 \cos (5 (c+d x)) \log (\cos (c+d x))+105 \cos (7 (c+d x)) \log (\cos (c+d x))+105 \cos (c+d x) (35 \log (\cos (c+d x))+8)+3732)}{6720 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^11/(a + a*Sec[c + d*x])^3,x]

[Out]

((3732 + 4522*Cos[2*(c + d*x)] + 1050*Cos[3*(c + d*x)] + 2380*Cos[4*(c + d*x)] - 210*Cos[5*(c + d*x)] + 630*Co
s[6*(c + d*x)] + 2205*Cos[3*(c + d*x)]*Log[Cos[c + d*x]] + 735*Cos[5*(c + d*x)]*Log[Cos[c + d*x]] + 105*Cos[7*
(c + d*x)]*Log[Cos[c + d*x]] + 105*Cos[c + d*x]*(8 + 35*Log[Cos[c + d*x]]))*Sec[c + d*x]^7)/(6720*a^3*d)

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fricas [A]  time = 0.66, size = 95, normalized size = 0.69 \[ \frac {420 \, \cos \left (d x + c\right )^{7} \log \left (-\cos \left (d x + c\right )\right ) + 1260 \, \cos \left (d x + c\right )^{6} - 210 \, \cos \left (d x + c\right )^{5} - 700 \, \cos \left (d x + c\right )^{4} + 525 \, \cos \left (d x + c\right )^{3} + 84 \, \cos \left (d x + c\right )^{2} - 210 \, \cos \left (d x + c\right ) + 60}{420 \, a^{3} d \cos \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^11/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/420*(420*cos(d*x + c)^7*log(-cos(d*x + c)) + 1260*cos(d*x + c)^6 - 210*cos(d*x + c)^5 - 700*cos(d*x + c)^4 +
525*cos(d*x + c)^3 + 84*cos(d*x + c)^2 - 210*cos(d*x + c) + 60)/(a^3*d*cos(d*x + c)^7)

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giac [A]  time = 93.14, size = 246, normalized size = 1.80 \[ -\frac {\frac {420 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} - \frac {420 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{3}} - \frac {\frac {1393 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {819 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6755 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {20195 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {28749 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {8463 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {1089 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 319}{a^{3} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{7}}}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^11/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/420*(420*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^3 - 420*log(abs(-(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 1))/a^3 - (1393*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 819*(cos(d*x + c) - 1)^2/(cos(d*x + c) +
1)^2 - 6755*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 20195*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 287
49*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 - 8463*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 - 1089*(cos(d*x
+ c) - 1)^7/(cos(d*x + c) + 1)^7 + 319)/(a^3*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^7))/d

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maple [A]  time = 0.80, size = 127, normalized size = 0.93 \[ \frac {\sec ^{7}\left (d x +c \right )}{7 a^{3} d}-\frac {\sec ^{6}\left (d x +c \right )}{2 a^{3} d}+\frac {\sec ^{5}\left (d x +c \right )}{5 a^{3} d}+\frac {5 \left (\sec ^{4}\left (d x +c \right )\right )}{4 a^{3} d}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{3 a^{3} d}-\frac {\sec ^{2}\left (d x +c \right )}{2 a^{3} d}+\frac {3 \sec \left (d x +c \right )}{a^{3} d}-\frac {\ln \left (\sec \left (d x +c \right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^11/(a+a*sec(d*x+c))^3,x)

[Out]

1/7*sec(d*x+c)^7/a^3/d-1/2*sec(d*x+c)^6/a^3/d+1/5*sec(d*x+c)^5/a^3/d+5/4*sec(d*x+c)^4/a^3/d-5/3*sec(d*x+c)^3/a
^3/d-1/2*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 = 0.42, size = 90, normalized size = 0.66 \[ \frac {\frac {420 \, \log \left (\cos \left (d x + c\right )\right )}{a^{3}} + \frac {1260 \, \cos \left (d x + c\right )^{6} - 210 \, \cos \left (d x + c\right )^{5} - 700 \, \cos \left (d x + c\right )^{4} + 525 \, \cos \left (d x + c\right )^{3} + 84 \, \cos \left (d x + c\right )^{2} - 210 \, \cos \left (d x + c\right ) + 60}{a^{3} \cos \left (d x + c\right )^{7}}}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^11/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/420*(420*log(cos(d*x + c))/a^3 + (1260*cos(d*x + c)^6 - 210*cos(d*x + c)^5 - 700*cos(d*x + c)^4 + 525*cos(d*
x + c)^3 + 84*cos(d*x + c)^2 - 210*cos(d*x + c) + 60)/(a^3*cos(d*x + c)^7))/d

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mupad [B]  time = 5.29, size = 225, normalized size = 1.64 \[ -\frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^3\,d}-\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {128\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}-\frac {224\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {282\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}-\frac {322\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {352}{105}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^11/(a + a/cos(c + d*x))^3,x)

[Out]

- (2*atanh(tan(c/2 + (d*x)/2)^2))/(a^3*d) - ((282*tan(c/2 + (d*x)/2)^4)/5 - (322*tan(c/2 + (d*x)/2)^2)/15 - (2
24*tan(c/2 + (d*x)/2)^6)/3 + (128*tan(c/2 + (d*x)/2)^8)/3 + 14*tan(c/2 + (d*x)/2)^10 - 2*tan(c/2 + (d*x)/2)^12
+ 352/105)/(d*(7*a^3*tan(c/2 + (d*x)/2)^2 - 21*a^3*tan(c/2 + (d*x)/2)^4 + 35*a^3*tan(c/2 + (d*x)/2)^6 - 35*a^
3*tan(c/2 + (d*x)/2)^8 + 21*a^3*tan(c/2 + (d*x)/2)^10 - 7*a^3*tan(c/2 + (d*x)/2)^12 + a^3*tan(c/2 + (d*x)/2)^1
4 - a^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{11}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**11/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**11/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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