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

Optimal. Leaf size=97 \[ \frac {\sec ^5(c+d x)}{5 a d}-\frac {\sec ^4(c+d x)}{4 a d}-\frac {2 \sec ^3(c+d x)}{3 a d}+\frac {\sec ^2(c+d x)}{a d}+\frac {\sec (c+d x)}{a d}+\frac {\log (\cos (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 97, 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} \[ \frac {\sec ^5(c+d x)}{5 a d}-\frac {\sec ^4(c+d x)}{4 a d}-\frac {2 \sec ^3(c+d x)}{3 a d}+\frac {\sec ^2(c+d x)}{a d}+\frac {\sec (c+d x)}{a d}+\frac {\log (\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.27, size = 103, normalized size = 1.06 \[ \frac {\sec ^5(c+d x) (40 \cos (2 (c+d x))+60 \cos (3 (c+d x))+30 \cos (4 (c+d x))+75 \cos (3 (c+d x)) \log (\cos (c+d x))+15 \cos (5 (c+d x)) \log (\cos (c+d x))+30 \cos (c+d x) (5 \log (\cos (c+d x))+4)+58)}{240 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((58 + 40*Cos[2*(c + d*x)] + 60*Cos[3*(c + d*x)] + 30*Cos[4*(c + d*x)] + 75*Cos[3*(c + d*x)]*Log[Cos[c + d*x]]
+ 15*Cos[5*(c + d*x)]*Log[Cos[c + d*x]] + 30*Cos[c + d*x]*(4 + 5*Log[Cos[c + d*x]]))*Sec[c + d*x]^5)/(240*a*d
)

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fricas [A]  time = 0.50, size = 75, normalized size = 0.77 \[ \frac {60 \, \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) + 60 \, \cos \left (d x + c\right )^{4} + 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} - 15 \, \cos \left (d x + c\right ) + 12}{60 \, a d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*cos(d*x + c)^5*log(-cos(d*x + c)) + 60*cos(d*x + c)^4 + 60*cos(d*x + c)^3 - 40*cos(d*x + c)^2 - 15*co
s(d*x + c) + 12)/(a*d*cos(d*x + c)^5)

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giac [B]  time = 6.79, size = 201, normalized size = 2.07 \[ -\frac {\frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a} - \frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a} + \frac {\frac {485 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {1330 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1970 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {805 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {137 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 73}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a - 60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c
) + 1) - 1))/a + (485*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1330*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 +
1970*(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 + 73)/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^5))/d

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maple [A]  time = 0.56, size = 93, normalized size = 0.96 \[ \frac {\sec ^{5}\left (d x +c \right )}{5 d a}-\frac {\sec ^{4}\left (d x +c \right )}{4 d a}-\frac {2 \left (\sec ^{3}\left (d x +c \right )\right )}{3 d a}+\frac {\sec ^{2}\left (d x +c \right )}{d a}+\frac {\sec \left (d x +c \right )}{d a}-\frac {\ln \left (\sec \left (d x +c \right )\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 70, normalized size = 0.72 \[ \frac {\frac {60 \, \log \left (\cos \left (d x + c\right )\right )}{a} + \frac {60 \, \cos \left (d x + c\right )^{4} + 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} - 15 \, \cos \left (d x + c\right ) + 12}{a \cos \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*log(cos(d*x + c))/a + (60*cos(d*x + c)^4 + 60*cos(d*x + c)^3 - 40*cos(d*x + c)^2 - 15*cos(d*x + c) +
12)/(a*cos(d*x + c)^5))/d

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mupad [B]  time = 6.02, size = 153, normalized size = 1.58 \[ \frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {16}{15}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^4)/3 - (10*tan(c/2 + (d*x)/2)^2)/3 + 10*tan(c/2 + (d*x)/2)^6 - 2*tan(c/2 + (d*x)/2)^8 +
16/15)/(d*(a - 5*a*tan(c/2 + (d*x)/2)^2 + 10*a*tan(c/2 + (d*x)/2)^4 - 10*a*tan(c/2 + (d*x)/2)^6 + 5*a*tan(c/2
+ (d*x)/2)^8 - a*tan(c/2 + (d*x)/2)^10)) - (2*atanh(tan(c/2 + (d*x)/2)^2))/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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