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

Optimal. Leaf size=97 \[ \frac{\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{a b \sec ^6(c+d x)}{3 d}+\frac{b^2 \tan ^7(c+d x)}{7 d} \]

[Out]

(a*b*Sec[c + d*x]^6)/(3*d) + (a^2*Tan[c + d*x])/d + ((2*a^2 + b^2)*Tan[c + d*x]^3)/(3*d) + ((a^2 + 2*b^2)*Tan[
c + d*x]^5)/(5*d) + (b^2*Tan[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.0825734, antiderivative size = 97, 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} \[ \frac{\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{a b \sec ^6(c+d x)}{3 d}+\frac{b^2 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.638971, size = 104, normalized size = 1.07 \[ \frac{\tan (c+d x) \left (21 \left (a^2+2 b^2\right ) \tan ^4(c+d x)+35 \left (2 a^2+b^2\right ) \tan ^2(c+d x)+105 a^2+35 a b \tan ^5(c+d x)+105 a b \tan ^3(c+d x)+105 a b \tan (c+d x)+15 b^2 \tan ^6(c+d x)\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[c + d*x]*(105*a^2 + 105*a*b*Tan[c + d*x] + 35*(2*a^2 + b^2)*Tan[c + d*x]^2 + 105*a*b*Tan[c + d*x]^3 + 21*
(a^2 + 2*b^2)*Tan[c + d*x]^4 + 35*a*b*Tan[c + d*x]^5 + 15*b^2*Tan[c + d*x]^6))/(105*d)

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Maple [A]  time = 0.053, size = 110, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{ab}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-{a}^{2} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \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)^6*(a+b*tan(d*x+c))^2,x)

[Out]

1/d*(b^2*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)+1/3*a*
b/cos(d*x+c)^6-a^2*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.35239, size = 140, normalized size = 1.44 \begin{align*} \frac{15 \, b^{2} \tan \left (d x + c\right )^{7} + 35 \, a b \tan \left (d x + c\right )^{6} + 105 \, a b \tan \left (d x + c\right )^{4} + 21 \,{\left (a^{2} + 2 \, b^{2}\right )} \tan \left (d x + c\right )^{5} + 105 \, a b \tan \left (d x + c\right )^{2} + 35 \,{\left (2 \, a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*b^2*tan(d*x + c)^7 + 35*a*b*tan(d*x + c)^6 + 105*a*b*tan(d*x + c)^4 + 21*(a^2 + 2*b^2)*tan(d*x + c)^
5 + 105*a*b*tan(d*x + c)^2 + 35*(2*a^2 + b^2)*tan(d*x + c)^3 + 105*a^2*tan(d*x + c))/d

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Fricas [A]  time = 1.95376, size = 231, normalized size = 2.38 \begin{align*} \frac{35 \, a b \cos \left (d x + c\right ) +{\left (8 \,{\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 4 \,{\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(35*a*b*cos(d*x + c) + (8*(7*a^2 - b^2)*cos(d*x + c)^6 + 4*(7*a^2 - b^2)*cos(d*x + c)^4 + 3*(7*a^2 - b^2
)*cos(d*x + c)^2 + 15*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 \tan{\left (c + d x \right )}\right )^{2} \sec ^{6}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.39152, size = 159, normalized size = 1.64 \begin{align*} \frac{15 \, b^{2} \tan \left (d x + c\right )^{7} + 35 \, a b \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} + 42 \, b^{2} \tan \left (d x + c\right )^{5} + 105 \, a b \tan \left (d x + c\right )^{4} + 70 \, a^{2} \tan \left (d x + c\right )^{3} + 35 \, b^{2} \tan \left (d x + c\right )^{3} + 105 \, a b \tan \left (d x + c\right )^{2} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*b^2*tan(d*x + c)^7 + 35*a*b*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 + 42*b^2*tan(d*x + c)^5 + 105*a*b
*tan(d*x + c)^4 + 70*a^2*tan(d*x + c)^3 + 35*b^2*tan(d*x + c)^3 + 105*a*b*tan(d*x + c)^2 + 105*a^2*tan(d*x + c
))/d