3.2 \(\int \sec ^5(c+d x) (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=98 \[ \frac{(6 A+5 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{(6 A+5 C) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{(6 A+5 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{C \tan (c+d x) \sec ^5(c+d x)}{6 d} \]

[Out]

((6*A + 5*C)*ArcTanh[Sin[c + d*x]])/(16*d) + ((6*A + 5*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((6*A + 5*C)*Sec
[c + d*x]^3*Tan[c + d*x])/(24*d) + (C*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0631755, antiderivative size = 98, 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 = {4046, 3768, 3770} \[ \frac{(6 A+5 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{(6 A+5 C) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{(6 A+5 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{C \tan (c+d x) \sec ^5(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]

[Out]

((6*A + 5*C)*ArcTanh[Sin[c + d*x]])/(16*d) + ((6*A + 5*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((6*A + 5*C)*Sec
[c + d*x]^3*Tan[c + d*x])/(24*d) + (C*Sec[c + d*x]^5*Tan[c + d*x])/(6*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin{align*} \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{6} (6 A+5 C) \int \sec ^5(c+d x) \, dx\\ &=\frac{(6 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{C \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{8} (6 A+5 C) \int \sec ^3(c+d x) \, dx\\ &=\frac{(6 A+5 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{(6 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{C \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{16} (6 A+5 C) \int \sec (c+d x) \, dx\\ &=\frac{(6 A+5 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{(6 A+5 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{(6 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{C \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.337091, size = 75, normalized size = 0.77 \[ \frac{3 (6 A+5 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x) \left (2 (6 A+5 C) \sec ^2(c+d x)+3 (6 A+5 C)+8 C \sec ^4(c+d x)\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]

[Out]

(3*(6*A + 5*C)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(3*(6*A + 5*C) + 2*(6*A + 5*C)*Sec[c + d*x]^2 + 8*C*Sec[c
+ d*x]^4)*Tan[c + d*x])/(48*d)

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 138, normalized size = 1.4 \begin{align*}{\frac{A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{C \left ( \sec \left ( dx+c \right ) \right ) ^{5}\tan \left ( dx+c \right ) }{6\,d}}+{\frac{5\,C \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{24\,d}}+{\frac{5\,C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x)

[Out]

1/4/d*A*tan(d*x+c)*sec(d*x+c)^3+3/8/d*A*tan(d*x+c)*sec(d*x+c)+3/8/d*A*ln(sec(d*x+c)+tan(d*x+c))+1/6*C*sec(d*x+
c)^5*tan(d*x+c)/d+5/24*C*sec(d*x+c)^3*tan(d*x+c)/d+5/16*C*sec(d*x+c)*tan(d*x+c)/d+5/16/d*C*ln(sec(d*x+c)+tan(d
*x+c))

________________________________________________________________________________________

Maxima [A]  time = 0.934614, size = 170, normalized size = 1.73 \begin{align*} \frac{3 \,{\left (6 \, A + 5 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (6 \, A + 5 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )^{5} - 8 \,{\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )^{3} + 3 \,{\left (10 \, A + 11 \, C\right )} \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}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(3*(6*A + 5*C)*log(sin(d*x + c) + 1) - 3*(6*A + 5*C)*log(sin(d*x + c) - 1) - 2*(3*(6*A + 5*C)*sin(d*x + c
)^5 - 8*(6*A + 5*C)*sin(d*x + c)^3 + 3*(10*A + 11*C)*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(
d*x + c)^2 - 1))/d

________________________________________________________________________________________

Fricas [A]  time = 0.524048, size = 293, normalized size = 2.99 \begin{align*} \frac{3 \,{\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \,{\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, C\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/96*(3*(6*A + 5*C)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(6*A + 5*C)*cos(d*x + c)^6*log(-sin(d*x + c) + 1)
+ 2*(3*(6*A + 5*C)*cos(d*x + c)^4 + 2*(6*A + 5*C)*cos(d*x + c)^2 + 8*C)*sin(d*x + c))/(d*cos(d*x + c)^6)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**5, x)

________________________________________________________________________________________

Giac [A]  time = 1.25582, size = 163, normalized size = 1.66 \begin{align*} \frac{3 \,{\left (6 \, A + 5 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \,{\left (6 \, A + 5 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (18 \, A \sin \left (d x + c\right )^{5} + 15 \, C \sin \left (d x + c\right )^{5} - 48 \, A \sin \left (d x + c\right )^{3} - 40 \, C \sin \left (d x + c\right )^{3} + 30 \, A \sin \left (d x + c\right ) + 33 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/96*(3*(6*A + 5*C)*log(abs(sin(d*x + c) + 1)) - 3*(6*A + 5*C)*log(abs(sin(d*x + c) - 1)) - 2*(18*A*sin(d*x +
c)^5 + 15*C*sin(d*x + c)^5 - 48*A*sin(d*x + c)^3 - 40*C*sin(d*x + c)^3 + 30*A*sin(d*x + c) + 33*C*sin(d*x + c)
)/(sin(d*x + c)^2 - 1)^3)/d