3.566 \(\int \frac{\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=185 \[ \frac{3 \left (2 a^2+b^2\right ) \tan ^2(c+d x)}{2 b^5 d}-\frac{a \left (10 a^2+9 b^2\right ) \tan (c+d x)}{b^6 d}+\frac{6 a \left (a^2+b^2\right )^2}{b^7 d (a+b \tan (c+d x))}-\frac{\left (a^2+b^2\right )^3}{2 b^7 d (a+b \tan (c+d x))^2}+\frac{3 \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^7 d}-\frac{a \tan ^3(c+d x)}{b^4 d}+\frac{\tan ^4(c+d x)}{4 b^3 d} \]

[Out]

(3*(a^2 + b^2)*(5*a^2 + b^2)*Log[a + b*Tan[c + d*x]])/(b^7*d) - (a*(10*a^2 + 9*b^2)*Tan[c + d*x])/(b^6*d) + (3
*(2*a^2 + b^2)*Tan[c + d*x]^2)/(2*b^5*d) - (a*Tan[c + d*x]^3)/(b^4*d) + Tan[c + d*x]^4/(4*b^3*d) - (a^2 + b^2)
^3/(2*b^7*d*(a + b*Tan[c + d*x])^2) + (6*a*(a^2 + b^2)^2)/(b^7*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.155936, antiderivative size = 185, 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 = {3506, 697} \[ \frac{3 \left (2 a^2+b^2\right ) \tan ^2(c+d x)}{2 b^5 d}-\frac{a \left (10 a^2+9 b^2\right ) \tan (c+d x)}{b^6 d}+\frac{6 a \left (a^2+b^2\right )^2}{b^7 d (a+b \tan (c+d x))}-\frac{\left (a^2+b^2\right )^3}{2 b^7 d (a+b \tan (c+d x))^2}+\frac{3 \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^7 d}-\frac{a \tan ^3(c+d x)}{b^4 d}+\frac{\tan ^4(c+d x)}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + b*Tan[c + d*x])^3,x]

[Out]

(3*(a^2 + b^2)*(5*a^2 + b^2)*Log[a + b*Tan[c + d*x]])/(b^7*d) - (a*(10*a^2 + 9*b^2)*Tan[c + d*x])/(b^6*d) + (3
*(2*a^2 + b^2)*Tan[c + d*x]^2)/(2*b^5*d) - (a*Tan[c + d*x]^3)/(b^4*d) + Tan[c + d*x]^4/(4*b^3*d) - (a^2 + b^2)
^3/(2*b^7*d*(a + b*Tan[c + d*x])^2) + (6*a*(a^2 + b^2)^2)/(b^7*d*(a + b*Tan[c + d*x]))

Rule 3506

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

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\sec ^8(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{b^2}\right )^3}{(a+x)^3} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{-10 a^3-9 a b^2}{b^6}+\frac{3 \left (2 a^2+b^2\right ) x}{b^6}-\frac{3 a x^2}{b^6}+\frac{x^3}{b^6}+\frac{\left (a^2+b^2\right )^3}{b^6 (a+x)^3}-\frac{6 a \left (a^2+b^2\right )^2}{b^6 (a+x)^2}+\frac{3 \left (5 a^4+6 a^2 b^2+b^4\right )}{b^6 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{3 \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^7 d}-\frac{a \left (10 a^2+9 b^2\right ) \tan (c+d x)}{b^6 d}+\frac{3 \left (2 a^2+b^2\right ) \tan ^2(c+d x)}{2 b^5 d}-\frac{a \tan ^3(c+d x)}{b^4 d}+\frac{\tan ^4(c+d x)}{4 b^3 d}-\frac{\left (a^2+b^2\right )^3}{2 b^7 d (a+b \tan (c+d x))^2}+\frac{6 a \left (a^2+b^2\right )^2}{b^7 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.33094, size = 272, normalized size = 1.47 \[ \frac{4 a^2 b^4 \tan ^4(c+d x)-20 a b^3 \left (a^2+b^2\right ) \tan ^3(c+d x)+4 b^2 \tan ^2(c+d x) \left (3 \left (6 a^2 b^2+5 a^4+b^4\right ) \log (a+b \tan (c+d x))-10 a^2 b^2-13 a^4\right )+2 \left (a^2+b^2\right ) \left (6 a^2 \left (5 a^2+b^2\right ) \log (a+b \tan (c+d x))+16 a^2 b^2+19 a^4-3 b^4\right )+4 a b \tan (c+d x) \left (6 \left (6 a^2 b^2+5 a^4+b^4\right ) \log (a+b \tan (c+d x))+17 a^2 b^2+4 a^4+11 b^4\right )+b^4 \sec ^4(c+d x) \left (a^2-2 a b \tan (c+d x)+3 b^2\right )+b^6 \sec ^6(c+d x)}{4 b^7 d (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + b*Tan[c + d*x])^3,x]

[Out]

(2*(a^2 + b^2)*(19*a^4 + 16*a^2*b^2 - 3*b^4 + 6*a^2*(5*a^2 + b^2)*Log[a + b*Tan[c + d*x]]) + b^6*Sec[c + d*x]^
6 + 4*a*b*(4*a^4 + 17*a^2*b^2 + 11*b^4 + 6*(5*a^4 + 6*a^2*b^2 + b^4)*Log[a + b*Tan[c + d*x]])*Tan[c + d*x] + 4
*b^2*(-13*a^4 - 10*a^2*b^2 + 3*(5*a^4 + 6*a^2*b^2 + b^4)*Log[a + b*Tan[c + d*x]])*Tan[c + d*x]^2 - 20*a*b^3*(a
^2 + b^2)*Tan[c + d*x]^3 + 4*a^2*b^4*Tan[c + d*x]^4 + b^4*Sec[c + d*x]^4*(a^2 + 3*b^2 - 2*a*b*Tan[c + d*x]))/(
4*b^7*d*(a + b*Tan[c + d*x])^2)

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Maple [A]  time = 0.128, size = 321, normalized size = 1.7 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,{b}^{3}d}}-{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{{b}^{4}d}}+3\,{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}{a}^{2}}{d{b}^{5}}}+{\frac{3\, \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,{b}^{3}d}}-10\,{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d{b}^{6}}}-9\,{\frac{a\tan \left ( dx+c \right ) }{{b}^{4}d}}-{\frac{{a}^{6}}{2\,d{b}^{7} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{4}}{2\,d{b}^{5} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{2}}{2\,{b}^{3}d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{2\,bd \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+6\,{\frac{{a}^{5}}{d{b}^{7} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+12\,{\frac{{a}^{3}}{d{b}^{5} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+6\,{\frac{a}{{b}^{3}d \left ( a+b\tan \left ( dx+c \right ) \right ) }}+15\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ){a}^{4}}{d{b}^{7}}}+18\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ){a}^{2}}{d{b}^{5}}}+3\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8/(a+b*tan(d*x+c))^3,x)

[Out]

1/4*tan(d*x+c)^4/b^3/d-a*tan(d*x+c)^3/b^4/d+3/d/b^5*tan(d*x+c)^2*a^2+3/2*tan(d*x+c)^2/b^3/d-10/d/b^6*a^3*tan(d
*x+c)-9*a*tan(d*x+c)/b^4/d-1/2/d/b^7/(a+b*tan(d*x+c))^2*a^6-3/2/d/b^5/(a+b*tan(d*x+c))^2*a^4-3/2/d/b^3/(a+b*ta
n(d*x+c))^2*a^2-1/2/b/d/(a+b*tan(d*x+c))^2+6/d*a^5/b^7/(a+b*tan(d*x+c))+12/d*a^3/b^5/(a+b*tan(d*x+c))+6*a/b^3/
d/(a+b*tan(d*x+c))+15/d/b^7*ln(a+b*tan(d*x+c))*a^4+18/d/b^5*ln(a+b*tan(d*x+c))*a^2+3*ln(a+b*tan(d*x+c))/b^3/d

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Maxima [A]  time = 1.03506, size = 270, normalized size = 1.46 \begin{align*} \frac{\frac{2 \,{\left (11 \, a^{6} + 21 \, a^{4} b^{2} + 9 \, a^{2} b^{4} - b^{6} + 12 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )\right )}}{b^{9} \tan \left (d x + c\right )^{2} + 2 \, a b^{8} \tan \left (d x + c\right ) + a^{2} b^{7}} + \frac{b^{3} \tan \left (d x + c\right )^{4} - 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \,{\left (2 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{2} - 4 \,{\left (10 \, a^{3} + 9 \, a b^{2}\right )} \tan \left (d x + c\right )}{b^{6}} + \frac{12 \,{\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{7}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*(2*(11*a^6 + 21*a^4*b^2 + 9*a^2*b^4 - b^6 + 12*(a^5*b + 2*a^3*b^3 + a*b^5)*tan(d*x + c))/(b^9*tan(d*x + c)
^2 + 2*a*b^8*tan(d*x + c) + a^2*b^7) + (b^3*tan(d*x + c)^4 - 4*a*b^2*tan(d*x + c)^3 + 6*(2*a^2*b + b^3)*tan(d*
x + c)^2 - 4*(10*a^3 + 9*a*b^2)*tan(d*x + c))/b^6 + 12*(5*a^4 + 6*a^2*b^2 + b^4)*log(b*tan(d*x + c) + a)/b^7)/
d

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Fricas [B]  time = 2.5378, size = 1076, normalized size = 5.82 \begin{align*} \frac{8 \,{\left (15 \, a^{4} b^{2} + 13 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{6} + b^{6} - 2 \,{\left (45 \, a^{4} b^{2} + 44 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{4} +{\left (5 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 6 \,{\left ({\left (5 \, a^{6} + a^{4} b^{2} - 5 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{6} + 2 \,{\left (5 \, a^{5} b + 6 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) +{\left (5 \, a^{4} b^{2} + 6 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 6 \,{\left ({\left (5 \, a^{6} + a^{4} b^{2} - 5 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{6} + 2 \,{\left (5 \, a^{5} b + 6 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) +{\left (5 \, a^{4} b^{2} + 6 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) - 2 \,{\left (a b^{5} \cos \left (d x + c\right ) + 2 \,{\left (15 \, a^{5} b - 2 \, a^{3} b^{3} - 13 \, a b^{5}\right )} \cos \left (d x + c\right )^{5} + 10 \,{\left (a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \,{\left (2 \, a b^{8} d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + b^{9} d \cos \left (d x + c\right )^{4} +{\left (a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right )^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(8*(15*a^4*b^2 + 13*a^2*b^4)*cos(d*x + c)^6 + b^6 - 2*(45*a^4*b^2 + 44*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + (
5*a^2*b^4 + 3*b^6)*cos(d*x + c)^2 + 6*((5*a^6 + a^4*b^2 - 5*a^2*b^4 - b^6)*cos(d*x + c)^6 + 2*(5*a^5*b + 6*a^3
*b^3 + a*b^5)*cos(d*x + c)^5*sin(d*x + c) + (5*a^4*b^2 + 6*a^2*b^4 + b^6)*cos(d*x + c)^4)*log(2*a*b*cos(d*x +
c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 6*((5*a^6 + a^4*b^2 - 5*a^2*b^4 - b^6)*cos(d*x + c)^6 +
2*(5*a^5*b + 6*a^3*b^3 + a*b^5)*cos(d*x + c)^5*sin(d*x + c) + (5*a^4*b^2 + 6*a^2*b^4 + b^6)*cos(d*x + c)^4)*lo
g(cos(d*x + c)^2) - 2*(a*b^5*cos(d*x + c) + 2*(15*a^5*b - 2*a^3*b^3 - 13*a*b^5)*cos(d*x + c)^5 + 10*(a^3*b^3 +
a*b^5)*cos(d*x + c)^3)*sin(d*x + c))/(2*a*b^8*d*cos(d*x + c)^5*sin(d*x + c) + b^9*d*cos(d*x + c)^4 + (a^2*b^7
- b^9)*d*cos(d*x + c)^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8/(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.43227, size = 328, normalized size = 1.77 \begin{align*} \frac{\frac{12 \,{\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{7}} - \frac{2 \,{\left (45 \, a^{4} b^{2} \tan \left (d x + c\right )^{2} + 54 \, a^{2} b^{4} \tan \left (d x + c\right )^{2} + 9 \, b^{6} \tan \left (d x + c\right )^{2} + 78 \, a^{5} b \tan \left (d x + c\right ) + 84 \, a^{3} b^{3} \tan \left (d x + c\right ) + 6 \, a b^{5} \tan \left (d x + c\right ) + 34 \, a^{6} + 33 \, a^{4} b^{2} + b^{6}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b^{7}} + \frac{b^{9} \tan \left (d x + c\right )^{4} - 4 \, a b^{8} \tan \left (d x + c\right )^{3} + 12 \, a^{2} b^{7} \tan \left (d x + c\right )^{2} + 6 \, b^{9} \tan \left (d x + c\right )^{2} - 40 \, a^{3} b^{6} \tan \left (d x + c\right ) - 36 \, a b^{8} \tan \left (d x + c\right )}{b^{12}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/4*(12*(5*a^4 + 6*a^2*b^2 + b^4)*log(abs(b*tan(d*x + c) + a))/b^7 - 2*(45*a^4*b^2*tan(d*x + c)^2 + 54*a^2*b^4
*tan(d*x + c)^2 + 9*b^6*tan(d*x + c)^2 + 78*a^5*b*tan(d*x + c) + 84*a^3*b^3*tan(d*x + c) + 6*a*b^5*tan(d*x + c
) + 34*a^6 + 33*a^4*b^2 + b^6)/((b*tan(d*x + c) + a)^2*b^7) + (b^9*tan(d*x + c)^4 - 4*a*b^8*tan(d*x + c)^3 + 1
2*a^2*b^7*tan(d*x + c)^2 + 6*b^9*tan(d*x + c)^2 - 40*a^3*b^6*tan(d*x + c) - 36*a*b^8*tan(d*x + c))/b^12)/d