3.88 \(\int \frac{\tan ^7(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0568776, antiderivative size = 65, normalized size of antiderivative = 1., 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, 43} \[ \frac{\sec ^3(c+d x)}{3 a^3 d}-\frac{3 \sec ^2(c+d x)}{2 a^3 d}+\frac{3 \sec (c+d x)}{a^3 d}+\frac{\log (\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^7/(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) - (3*Sec[c + d*x]^2)/(2*a^3*d) + Sec[c + d*x]^3/(3*a^3*d)

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]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^7(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(a-a x)^3}{x^4} \, dx,x,\cos (c+d x)\right )}{a^6 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{a^3}{x^4}-\frac{3 a^3}{x^3}+\frac{3 a^3}{x^2}-\frac{a^3}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^6 d}\\ &=\frac{\log (\cos (c+d x))}{a^3 d}+\frac{3 \sec (c+d x)}{a^3 d}-\frac{3 \sec ^2(c+d x)}{2 a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.183869, size = 64, normalized size = 0.98 \[ \frac{\sec ^3(c+d x) (18 \cos (2 (c+d x))+9 \cos (c+d x) (\log (\cos (c+d x))-2)+3 \cos (3 (c+d x)) \log (\cos (c+d x))+22)}{12 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^7/(a + a*Sec[c + d*x])^3,x]

[Out]

((22 + 18*Cos[2*(c + d*x)] + 9*Cos[c + d*x]*(-2 + Log[Cos[c + d*x]]) + 3*Cos[3*(c + d*x)]*Log[Cos[c + d*x]])*S
ec[c + d*x]^3)/(12*a^3*d)

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Maple [A]  time = 0.075, size = 63, normalized size = 1. \begin{align*}{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{3\,d{a}^{3}}}-{\frac{3\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{2\,d{a}^{3}}}+3\,{\frac{\sec \left ( dx+c \right ) }{d{a}^{3}}}-{\frac{\ln \left ( \sec \left ( dx+c \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^7/(a+a*sec(d*x+c))^3,x)

[Out]

1/3*sec(d*x+c)^3/a^3/d-3/2*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 = 1.07671, size = 68, normalized size = 1.05 \begin{align*} \frac{\frac{6 \, \log \left (\cos \left (d x + c\right )\right )}{a^{3}} + \frac{18 \, \cos \left (d x + c\right )^{2} - 9 \, \cos \left (d x + c\right ) + 2}{a^{3} \cos \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/6*(6*log(cos(d*x + c))/a^3 + (18*cos(d*x + c)^2 - 9*cos(d*x + c) + 2)/(a^3*cos(d*x + c)^3))/d

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Fricas [A]  time = 1.18473, size = 144, normalized size = 2.22 \begin{align*} \frac{6 \, \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) + 18 \, \cos \left (d x + c\right )^{2} - 9 \, \cos \left (d x + c\right ) + 2}{6 \, a^{3} d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/6*(6*cos(d*x + c)^3*log(-cos(d*x + c)) + 18*cos(d*x + c)^2 - 9*cos(d*x + c) + 2)/(a^3*d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{7}{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**7/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**7/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [B]  time = 7.96975, size = 213, normalized size = 3.28 \begin{align*} -\frac{\frac{6 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} - \frac{6 \, \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{75 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{51 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{11 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 29}{a^{3}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(6*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^3 - 6*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) - 1))/a^3 - (75*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 51*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1
1*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 29)/(a^3*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^3))/d