3.1 \(\int \sec ^6(c+d x) (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=87 \[ \frac{(7 A+6 C) \tan ^5(c+d x)}{35 d}+\frac{2 (7 A+6 C) \tan ^3(c+d x)}{21 d}+\frac{(7 A+6 C) \tan (c+d x)}{7 d}+\frac{C \tan (c+d x) \sec ^6(c+d x)}{7 d} \]

[Out]

((7*A + 6*C)*Tan[c + d*x])/(7*d) + (C*Sec[c + d*x]^6*Tan[c + d*x])/(7*d) + (2*(7*A + 6*C)*Tan[c + d*x]^3)/(21*
d) + ((7*A + 6*C)*Tan[c + d*x]^5)/(35*d)

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Rubi [A]  time = 0.0559017, antiderivative size = 87, 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 = {4046, 3767} \[ \frac{(7 A+6 C) \tan ^5(c+d x)}{35 d}+\frac{2 (7 A+6 C) \tan ^3(c+d x)}{21 d}+\frac{(7 A+6 C) \tan (c+d x)}{7 d}+\frac{C \tan (c+d x) \sec ^6(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(A + C*Sec[c + d*x]^2),x]

[Out]

((7*A + 6*C)*Tan[c + d*x])/(7*d) + (C*Sec[c + d*x]^6*Tan[c + d*x])/(7*d) + (2*(7*A + 6*C)*Tan[c + d*x]^3)/(21*
d) + ((7*A + 6*C)*Tan[c + d*x]^5)/(35*d)

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{1}{7} (7 A+6 C) \int \sec ^6(c+d x) \, dx\\ &=\frac{C \sec ^6(c+d x) \tan (c+d x)}{7 d}-\frac{(7 A+6 C) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{7 d}\\ &=\frac{(7 A+6 C) \tan (c+d x)}{7 d}+\frac{C \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac{2 (7 A+6 C) \tan ^3(c+d x)}{21 d}+\frac{(7 A+6 C) \tan ^5(c+d x)}{35 d}\\ \end{align*}

Mathematica [A]  time = 0.302337, size = 81, normalized size = 0.93 \[ \frac{A \left (\frac{1}{5} \tan ^5(c+d x)+\frac{2}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{C \left (\frac{1}{7} \tan ^7(c+d x)+\frac{3}{5} \tan ^5(c+d x)+\tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(A + C*Sec[c + d*x]^2),x]

[Out]

(A*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/d + (C*(Tan[c + d*x] + Tan[c + d*x]^3 + (3*Tan[c
+ d*x]^5)/5 + Tan[c + d*x]^7/7))/d

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Maple [A]  time = 0.031, size = 78, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -A \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 ) -C \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)^6*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*(-A*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)-C*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/3
5*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 0.936541, size = 81, normalized size = 0.93 \begin{align*} \frac{15 \, C \tan \left (d x + c\right )^{7} + 21 \,{\left (A + 3 \, C\right )} \tan \left (d x + c\right )^{5} + 35 \,{\left (2 \, A + 3 \, C\right )} \tan \left (d x + c\right )^{3} + 105 \,{\left (A + C\right )} \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+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/105*(15*C*tan(d*x + c)^7 + 21*(A + 3*C)*tan(d*x + c)^5 + 35*(2*A + 3*C)*tan(d*x + c)^3 + 105*(A + C)*tan(d*x
+ c))/d

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Fricas [A]  time = 0.908206, size = 188, normalized size = 2.16 \begin{align*} \frac{{\left (8 \,{\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \,{\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \, C\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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/105*(8*(7*A + 6*C)*cos(d*x + c)^6 + 4*(7*A + 6*C)*cos(d*x + c)^4 + 3*(7*A + 6*C)*cos(d*x + c)^2 + 15*C)*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 + C \sec ^{2}{\left (c + d x \right )}\right ) \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+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**6, x)

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Giac [A]  time = 1.17973, size = 107, normalized size = 1.23 \begin{align*} \frac{15 \, C \tan \left (d x + c\right )^{7} + 21 \, A \tan \left (d x + c\right )^{5} + 63 \, C \tan \left (d x + c\right )^{5} + 70 \, A \tan \left (d x + c\right )^{3} + 105 \, C \tan \left (d x + c\right )^{3} + 105 \, A \tan \left (d x + c\right ) + 105 \, C \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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/105*(15*C*tan(d*x + c)^7 + 21*A*tan(d*x + c)^5 + 63*C*tan(d*x + c)^5 + 70*A*tan(d*x + c)^3 + 105*C*tan(d*x +
c)^3 + 105*A*tan(d*x + c) + 105*C*tan(d*x + c))/d