3.72 \(\int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=65 \[ \frac {\sec ^4(c+d x)}{4 a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}+\frac {2 \sec (c+d x)}{a^2 d}+\frac {\log (\cos (c+d x))}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 65, 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} \[ \frac {\sec ^4(c+d x)}{4 a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}+\frac {2 \sec (c+d x)}{a^2 d}+\frac {\log (\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

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Mathematica [A]  time = 0.20, size = 83, normalized size = 1.28 \[ \frac {\sec ^4(c+d x) (20 \cos (c+d x)+3 (4 \cos (3 (c+d x))+4 \cos (2 (c+d x)) \log (\cos (c+d x))+\cos (4 (c+d x)) \log (\cos (c+d x))+3 \log (\cos (c+d x))+2))}{24 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^7/(a + a*Sec[c + d*x])^2,x]

[Out]

((20*Cos[c + d*x] + 3*(2 + 4*Cos[3*(c + d*x)] + 3*Log[Cos[c + d*x]] + 4*Cos[2*(c + d*x)]*Log[Cos[c + d*x]] + C
os[4*(c + d*x)]*Log[Cos[c + d*x]]))*Sec[c + d*x]^4)/(24*a^2*d)

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fricas [A]  time = 0.49, size = 55, normalized size = 0.85 \[ \frac {12 \, \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 24 \, \cos \left (d x + c\right )^{3} - 8 \, \cos \left (d x + c\right ) + 3}{12 \, a^{2} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(12*cos(d*x + c)^4*log(-cos(d*x + c)) + 24*cos(d*x + c)^3 - 8*cos(d*x + c) + 3)/(a^2*d*cos(d*x + c)^4)

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giac [B]  time = 8.97, size = 180, normalized size = 2.77 \[ -\frac {\frac {12 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} - \frac {12 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{2}} - \frac {\frac {4 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {54 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {124 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {25 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 7}{a^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/12*(12*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^2 - 12*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1))/a^2 - (4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 54*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 -
124*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 25*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 7)/(a^2*((cos(
d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^4))/d

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maple [A]  time = 0.60, size = 63, normalized size = 0.97 \[ \frac {\sec ^{4}\left (d x +c \right )}{4 a^{2} d}-\frac {2 \left (\sec ^{3}\left (d x +c \right )\right )}{3 a^{2} d}+\frac {2 \sec \left (d x +c \right )}{a^{2} d}-\frac {\ln \left (\sec \left (d x +c \right )\right )}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x)

[Out]

1/4*sec(d*x+c)^4/a^2/d-2/3*sec(d*x+c)^3/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 = 0.44, size = 50, normalized size = 0.77 \[ \frac {\frac {12 \, \log \left (\cos \left (d x + c\right )\right )}{a^{2}} + \frac {24 \, \cos \left (d x + c\right )^{3} - 8 \, \cos \left (d x + c\right ) + 3}{a^{2} \cos \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(12*log(cos(d*x + c))/a^2 + (24*cos(d*x + c)^3 - 8*cos(d*x + c) + 3)/(a^2*cos(d*x + c)^4))/d

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mupad [B]  time = 3.87, size = 135, normalized size = 2.08 \[ \frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {8}{3}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^7/(a + a/cos(c + d*x))^2,x)

[Out]

(8*tan(c/2 + (d*x)/2)^4 - (26*tan(c/2 + (d*x)/2)^2)/3 + 2*tan(c/2 + (d*x)/2)^6 + 8/3)/(d*(6*a^2*tan(c/2 + (d*x
)/2)^4 - 4*a^2*tan(c/2 + (d*x)/2)^2 - 4*a^2*tan(c/2 + (d*x)/2)^6 + a^2*tan(c/2 + (d*x)/2)^8 + a^2)) - (2*atanh
(tan(c/2 + (d*x)/2)^2))/(a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{7}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**7/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**7/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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