3.279 \(\int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx\)

Optimal. Leaf size=157 \[ \frac{a^2 \tan ^5(c+d x)}{5 d}-\frac{a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}-a^2 x-\frac{5 a b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a b \tan ^5(c+d x) \sec (c+d x)}{3 d}-\frac{5 a b \tan ^3(c+d x) \sec (c+d x)}{12 d}+\frac{5 a b \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b^2 \tan ^7(c+d x)}{7 d} \]

[Out]

-(a^2*x) - (5*a*b*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*Tan[c + d*x])/d + (5*a*b*Sec[c + d*x]*Tan[c + d*x])/(8*d
) - (a^2*Tan[c + d*x]^3)/(3*d) - (5*a*b*Sec[c + d*x]*Tan[c + d*x]^3)/(12*d) + (a^2*Tan[c + d*x]^5)/(5*d) + (a*
b*Sec[c + d*x]*Tan[c + d*x]^5)/(3*d) + (b^2*Tan[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.197791, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3886, 3473, 8, 2611, 3770, 2607, 30} \[ \frac{a^2 \tan ^5(c+d x)}{5 d}-\frac{a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}-a^2 x-\frac{5 a b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a b \tan ^5(c+d x) \sec (c+d x)}{3 d}-\frac{5 a b \tan ^3(c+d x) \sec (c+d x)}{12 d}+\frac{5 a b \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b^2 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^2*Tan[c + d*x]^6,x]

[Out]

-(a^2*x) - (5*a*b*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*Tan[c + d*x])/d + (5*a*b*Sec[c + d*x]*Tan[c + d*x])/(8*d
) - (a^2*Tan[c + d*x]^3)/(3*d) - (5*a*b*Sec[c + d*x]*Tan[c + d*x]^3)/(12*d) + (a^2*Tan[c + d*x]^5)/(5*d) + (a*
b*Sec[c + d*x]*Tan[c + d*x]^5)/(3*d) + (b^2*Tan[c + d*x]^7)/(7*d)

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx &=\int \left (a^2 \tan ^6(c+d x)+2 a b \sec (c+d x) \tan ^6(c+d x)+b^2 \sec ^2(c+d x) \tan ^6(c+d x)\right ) \, dx\\ &=a^2 \int \tan ^6(c+d x) \, dx+(2 a b) \int \sec (c+d x) \tan ^6(c+d x) \, dx+b^2 \int \sec ^2(c+d x) \tan ^6(c+d x) \, dx\\ &=\frac{a^2 \tan ^5(c+d x)}{5 d}+\frac{a b \sec (c+d x) \tan ^5(c+d x)}{3 d}-a^2 \int \tan ^4(c+d x) \, dx-\frac{1}{3} (5 a b) \int \sec (c+d x) \tan ^4(c+d x) \, dx+\frac{b^2 \operatorname{Subst}\left (\int x^6 \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{a^2 \tan ^3(c+d x)}{3 d}-\frac{5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac{a^2 \tan ^5(c+d x)}{5 d}+\frac{a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac{b^2 \tan ^7(c+d x)}{7 d}+a^2 \int \tan ^2(c+d x) \, dx+\frac{1}{4} (5 a b) \int \sec (c+d x) \tan ^2(c+d x) \, dx\\ &=\frac{a^2 \tan (c+d x)}{d}+\frac{5 a b \sec (c+d x) \tan (c+d x)}{8 d}-\frac{a^2 \tan ^3(c+d x)}{3 d}-\frac{5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac{a^2 \tan ^5(c+d x)}{5 d}+\frac{a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac{b^2 \tan ^7(c+d x)}{7 d}-a^2 \int 1 \, dx-\frac{1}{8} (5 a b) \int \sec (c+d x) \, dx\\ &=-a^2 x-\frac{5 a b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{5 a b \sec (c+d x) \tan (c+d x)}{8 d}-\frac{a^2 \tan ^3(c+d x)}{3 d}-\frac{5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac{a^2 \tan ^5(c+d x)}{5 d}+\frac{a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac{b^2 \tan ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 1.35533, size = 293, normalized size = 1.87 \[ \frac{-2100 \sec ^6(c+d x) \left (7 a^2 (c+d x)-\left (a^2+b^2\right ) \tan (c+d x)\right )+\sec ^7(c+d x) \left (-\left (-3444 a^2 \sin (3 (c+d x))-1988 a^2 \sin (5 (c+d x))-644 a^2 \sin (7 (c+d x))+8820 a^2 (c+d x) \cos (3 (c+d x))+2940 a^2 (c+d x) \cos (5 (c+d x))+420 a^2 c \cos (7 (c+d x))+420 a^2 d x \cos (7 (c+d x))-980 a b \sin (4 (c+d x))-1155 a b \sin (6 (c+d x))+1260 b^2 \sin (3 (c+d x))-420 b^2 \sin (5 (c+d x))+60 b^2 \sin (7 (c+d x))\right )\right )+5950 a b \tan (c+d x) \sec ^5(c+d x)+16800 a b \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{26880 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^2*Tan[c + d*x]^6,x]

[Out]

(16800*a*b*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c + d*x
]^7*(8820*a^2*(c + d*x)*Cos[3*(c + d*x)] + 2940*a^2*(c + d*x)*Cos[5*(c + d*x)] + 420*a^2*c*Cos[7*(c + d*x)] +
420*a^2*d*x*Cos[7*(c + d*x)] - 3444*a^2*Sin[3*(c + d*x)] + 1260*b^2*Sin[3*(c + d*x)] - 980*a*b*Sin[4*(c + d*x)
] - 1988*a^2*Sin[5*(c + d*x)] - 420*b^2*Sin[5*(c + d*x)] - 1155*a*b*Sin[6*(c + d*x)] - 644*a^2*Sin[7*(c + d*x)
] + 60*b^2*Sin[7*(c + d*x)]) + 5950*a*b*Sec[c + d*x]^5*Tan[c + d*x] - 2100*Sec[c + d*x]^6*(7*a^2*(c + d*x) - (
a^2 + b^2)*Tan[c + d*x]))/(26880*d)

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Maple [A]  time = 0.051, size = 219, normalized size = 1.4 \begin{align*}{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d}}-{a}^{2}x-{\frac{{a}^{2}c}{d}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{12\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d}}+{\frac{5\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,ab\sin \left ( dx+c \right ) }{8\,d}}-{\frac{5\,ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{7\,d \left ( \cos \left ( dx+c \right ) \right ) ^{7}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^2*tan(d*x+c)^6,x)

[Out]

1/5*a^2*tan(d*x+c)^5/d-1/3*a^2*tan(d*x+c)^3/d+a^2*tan(d*x+c)/d-a^2*x-1/d*a^2*c+1/3/d*a*b*sin(d*x+c)^7/cos(d*x+
c)^6-1/12/d*a*b*sin(d*x+c)^7/cos(d*x+c)^4+1/8/d*a*b*sin(d*x+c)^7/cos(d*x+c)^2+1/8/d*a*b*sin(d*x+c)^5+5/24/d*a*
b*sin(d*x+c)^3+5/8/d*a*b*sin(d*x+c)-5/8/d*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/7/d*b^2*sin(d*x+c)^7/cos(d*x+c)^7

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Maxima [A]  time = 1.46323, size = 203, normalized size = 1.29 \begin{align*} \frac{240 \, b^{2} \tan \left (d x + c\right )^{7} + 112 \,{\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a^{2} - 35 \, a b{\left (\frac{2 \,{\left (33 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{1680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*tan(d*x+c)^6,x, algorithm="maxima")

[Out]

1/1680*(240*b^2*tan(d*x + c)^7 + 112*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a
^2 - 35*a*b*(2*(33*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 15*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 +
3*sin(d*x + c)^2 - 1) + 15*log(sin(d*x + c) + 1) - 15*log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 0.834162, size = 494, normalized size = 3.15 \begin{align*} -\frac{1680 \, a^{2} d x \cos \left (d x + c\right )^{7} + 525 \, a b \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 525 \, a b \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (1155 \, a b \cos \left (d x + c\right )^{5} + 8 \,{\left (161 \, a^{2} - 15 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 910 \, a b \cos \left (d x + c\right )^{3} - 8 \,{\left (77 \, a^{2} - 45 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 280 \, a b \cos \left (d x + c\right ) + 24 \,{\left (7 \, a^{2} - 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 120 \, b^{2}\right )} \sin \left (d x + c\right )}{1680 \, d \cos \left (d x + c\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*tan(d*x+c)^6,x, algorithm="fricas")

[Out]

-1/1680*(1680*a^2*d*x*cos(d*x + c)^7 + 525*a*b*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 525*a*b*cos(d*x + c)^7*l
og(-sin(d*x + c) + 1) - 2*(1155*a*b*cos(d*x + c)^5 + 8*(161*a^2 - 15*b^2)*cos(d*x + c)^6 - 910*a*b*cos(d*x + c
)^3 - 8*(77*a^2 - 45*b^2)*cos(d*x + c)^4 + 280*a*b*cos(d*x + c) + 24*(7*a^2 - 15*b^2)*cos(d*x + c)^2 + 120*b^2
)*sin(d*x + c))/(d*cos(d*x + c)^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{2} \tan ^{6}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**2*tan(d*x+c)**6,x)

[Out]

Integral((a + b*sec(c + d*x))**2*tan(c + d*x)**6, x)

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Giac [A]  time = 5.02981, size = 381, normalized size = 2.43 \begin{align*} -\frac{840 \,{\left (d x + c\right )} a^{2} + 525 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 525 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (840 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} - 525 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} - 6160 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 3500 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 19768 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 9905 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 28896 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 7680 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 19768 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9905 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6160 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3500 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 840 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 525 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{7}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*tan(d*x+c)^6,x, algorithm="giac")

[Out]

-1/840*(840*(d*x + c)*a^2 + 525*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 525*a*b*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(840*a^2*tan(1/2*d*x + 1/2*c)^13 - 525*a*b*tan(1/2*d*x + 1/2*c)^13 - 6160*a^2*tan(1/2*d*x + 1/2*c)^1
1 + 3500*a*b*tan(1/2*d*x + 1/2*c)^11 + 19768*a^2*tan(1/2*d*x + 1/2*c)^9 - 9905*a*b*tan(1/2*d*x + 1/2*c)^9 - 28
896*a^2*tan(1/2*d*x + 1/2*c)^7 + 7680*b^2*tan(1/2*d*x + 1/2*c)^7 + 19768*a^2*tan(1/2*d*x + 1/2*c)^5 + 9905*a*b
*tan(1/2*d*x + 1/2*c)^5 - 6160*a^2*tan(1/2*d*x + 1/2*c)^3 - 3500*a*b*tan(1/2*d*x + 1/2*c)^3 + 840*a^2*tan(1/2*
d*x + 1/2*c) + 525*a*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d