3.517 \(\int \sec ^8(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=119 \[ \frac{\left (a^2+3 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac{3 \left (a^2+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{\left (3 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{a b \sec ^8(c+d x)}{4 d}+\frac{b^2 \tan ^9(c+d x)}{9 d} \]

[Out]

(a*b*Sec[c + d*x]^8)/(4*d) + (a^2*Tan[c + d*x])/d + ((3*a^2 + b^2)*Tan[c + d*x]^3)/(3*d) + (3*(a^2 + b^2)*Tan[
c + d*x]^5)/(5*d) + ((a^2 + 3*b^2)*Tan[c + d*x]^7)/(7*d) + (b^2*Tan[c + d*x]^9)/(9*d)

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Rubi [A]  time = 0.106558, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3506, 696, 1810} \[ \frac{\left (a^2+3 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac{3 \left (a^2+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{\left (3 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{a b \sec ^8(c+d x)}{4 d}+\frac{b^2 \tan ^9(c+d x)}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + b*Tan[c + d*x])^2,x]

[Out]

(a*b*Sec[c + d*x]^8)/(4*d) + (a^2*Tan[c + d*x])/d + ((3*a^2 + b^2)*Tan[c + d*x]^3)/(3*d) + (3*(a^2 + b^2)*Tan[
c + d*x]^5)/(5*d) + ((a^2 + 3*b^2)*Tan[c + d*x]^7)/(7*d) + (b^2*Tan[c + d*x]^9)/(9*d)

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 696

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*m*d^(m - 1)*(a + c*x^2)^(p + 1))
/(2*c*(p + 1)), x] + Int[((d + e*x)^m - e*m*d^(m - 1)*x)*(a + c*x^2)^p, x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*
d^2 + a*e^2, 0] && IGtQ[p, 1] && IGtQ[m, 0] && LeQ[m, p]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

Mathematica [A]  time = 0.482637, size = 133, normalized size = 1.12 \[ \frac{\tan (c+d x) \left (180 \left (a^2+3 b^2\right ) \tan ^6(c+d x)+756 \left (a^2+b^2\right ) \tan ^4(c+d x)+420 \left (3 a^2+b^2\right ) \tan ^2(c+d x)+1260 a^2+315 a b \tan ^7(c+d x)+1260 a b \tan ^5(c+d x)+1890 a b \tan ^3(c+d x)+1260 a b \tan (c+d x)+140 b^2 \tan ^8(c+d x)\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + b*Tan[c + d*x])^2,x]

[Out]

(Tan[c + d*x]*(1260*a^2 + 1260*a*b*Tan[c + d*x] + 420*(3*a^2 + b^2)*Tan[c + d*x]^2 + 1890*a*b*Tan[c + d*x]^3 +
756*(a^2 + b^2)*Tan[c + d*x]^4 + 1260*a*b*Tan[c + d*x]^5 + 180*(a^2 + 3*b^2)*Tan[c + d*x]^6 + 315*a*b*Tan[c +
d*x]^7 + 140*b^2*Tan[c + d*x]^8))/(1260*d)

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Maple [A]  time = 0.054, size = 138, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{9\, \left ( \cos \left ( dx+c \right ) \right ) ^{9}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{21\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{16\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{315\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{ab}{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}-{a}^{2} \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) \tan \left ( dx+c \right ) \right ) } \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))^2,x)

[Out]

1/d*(b^2*(1/9*sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x+c)^3/cos(d*x+c)^5+16/315*
sin(d*x+c)^3/cos(d*x+c)^3)+1/4*a*b/cos(d*x+c)^8-a^2*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)
^2)*tan(d*x+c))

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Maxima [A]  time = 1.59889, size = 180, normalized size = 1.51 \begin{align*} \frac{140 \, b^{2} \tan \left (d x + c\right )^{9} + 315 \, a b \tan \left (d x + c\right )^{8} + 1260 \, a b \tan \left (d x + c\right )^{6} + 180 \,{\left (a^{2} + 3 \, b^{2}\right )} \tan \left (d x + c\right )^{7} + 1890 \, a b \tan \left (d x + c\right )^{4} + 756 \,{\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{5} + 1260 \, a b \tan \left (d x + c\right )^{2} + 420 \,{\left (3 \, a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} \tan \left (d x + c\right )}{1260 \, 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))^2,x, algorithm="maxima")

[Out]

1/1260*(140*b^2*tan(d*x + c)^9 + 315*a*b*tan(d*x + c)^8 + 1260*a*b*tan(d*x + c)^6 + 180*(a^2 + 3*b^2)*tan(d*x
+ c)^7 + 1890*a*b*tan(d*x + c)^4 + 756*(a^2 + b^2)*tan(d*x + c)^5 + 1260*a*b*tan(d*x + c)^2 + 420*(3*a^2 + b^2
)*tan(d*x + c)^3 + 1260*a^2*tan(d*x + c))/d

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Fricas [A]  time = 2.1421, size = 282, normalized size = 2.37 \begin{align*} \frac{315 \, a b \cos \left (d x + c\right ) + 4 \,{\left (16 \,{\left (9 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{8} + 8 \,{\left (9 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 6 \,{\left (9 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 5 \,{\left (9 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 35 \, b^{2}\right )} \sin \left (d x + c\right )}{1260 \, d \cos \left (d x + c\right )^{9}} \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))^2,x, algorithm="fricas")

[Out]

1/1260*(315*a*b*cos(d*x + c) + 4*(16*(9*a^2 - b^2)*cos(d*x + c)^8 + 8*(9*a^2 - b^2)*cos(d*x + c)^6 + 6*(9*a^2
- b^2)*cos(d*x + c)^4 + 5*(9*a^2 - b^2)*cos(d*x + c)^2 + 35*b^2)*sin(d*x + c))/(d*cos(d*x + c)^9)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{2} \sec ^{8}{\left (c + d x \right )}\, dx \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))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sec(c + d*x)**8, x)

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Giac [A]  time = 1.33762, size = 211, normalized size = 1.77 \begin{align*} \frac{140 \, b^{2} \tan \left (d x + c\right )^{9} + 315 \, a b \tan \left (d x + c\right )^{8} + 180 \, a^{2} \tan \left (d x + c\right )^{7} + 540 \, b^{2} \tan \left (d x + c\right )^{7} + 1260 \, a b \tan \left (d x + c\right )^{6} + 756 \, a^{2} \tan \left (d x + c\right )^{5} + 756 \, b^{2} \tan \left (d x + c\right )^{5} + 1890 \, a b \tan \left (d x + c\right )^{4} + 1260 \, a^{2} \tan \left (d x + c\right )^{3} + 420 \, b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a b \tan \left (d x + c\right )^{2} + 1260 \, a^{2} \tan \left (d x + c\right )}{1260 \, 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))^2,x, algorithm="giac")

[Out]

1/1260*(140*b^2*tan(d*x + c)^9 + 315*a*b*tan(d*x + c)^8 + 180*a^2*tan(d*x + c)^7 + 540*b^2*tan(d*x + c)^7 + 12
60*a*b*tan(d*x + c)^6 + 756*a^2*tan(d*x + c)^5 + 756*b^2*tan(d*x + c)^5 + 1890*a*b*tan(d*x + c)^4 + 1260*a^2*t
an(d*x + c)^3 + 420*b^2*tan(d*x + c)^3 + 1260*a*b*tan(d*x + c)^2 + 1260*a^2*tan(d*x + c))/d