3.54 \(\int \sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=122 \[ \frac{(5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac{(5 A+4 C) \tan (c+d x)}{5 d}+\frac{3 B \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 B \tan (c+d x) \sec (c+d x)}{8 d}+\frac{C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0985237, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {4047, 3768, 3770, 4046, 3767} \[ \frac{(5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac{(5 A+4 C) \tan (c+d x)}{5 d}+\frac{3 B \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 B \tan (c+d x) \sec (c+d x)}{8 d}+\frac{C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

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]

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 ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=B \int \sec ^5(c+d x) \, dx+\int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} (3 B) \int \sec ^3(c+d x) \, dx+\frac{1}{5} (5 A+4 C) \int \sec ^4(c+d x) \, dx\\ &=\frac{3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac{B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{8} (3 B) \int \sec (c+d x) \, dx-\frac{(5 A+4 C) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{3 B \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(5 A+4 C) \tan (c+d x)}{5 d}+\frac{3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac{B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{(5 A+4 C) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.5791, size = 80, normalized size = 0.66 \[ \frac{\tan (c+d x) \left (8 \left (5 (A+2 C) \tan ^2(c+d x)+15 (A+C)+3 C \tan ^4(c+d x)\right )+30 B \sec ^3(c+d x)+45 B \sec (c+d x)\right )+45 B \tanh ^{-1}(\sin (c+d x))}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(45*B*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(45*B*Sec[c + d*x] + 30*B*Sec[c + d*x]^3 + 8*(15*(A + C) + 5*(A + 2
*C)*Tan[c + d*x]^2 + 3*C*Tan[c + d*x]^4)))/(120*d)

________________________________________________________________________________________

Maple [A]  time = 0.032, size = 144, normalized size = 1.2 \begin{align*}{\frac{2\,A\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{B \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,C\tan \left ( dx+c \right ) }{15\,d}}+{\frac{C \left ( \sec \left ( dx+c \right ) \right ) ^{4}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{4\,C \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

2/3/d*A*tan(d*x+c)+1/3/d*A*tan(d*x+c)*sec(d*x+c)^2+1/4*B*sec(d*x+c)^3*tan(d*x+c)/d+3/8*B*sec(d*x+c)*tan(d*x+c)
/d+3/8/d*B*ln(sec(d*x+c)+tan(d*x+c))+8/15*C*tan(d*x+c)/d+1/5*C*sec(d*x+c)^4*tan(d*x+c)/d+4/15*C*sec(d*x+c)^2*t
an(d*x+c)/d

________________________________________________________________________________________

Maxima [A]  time = 0.935019, size = 171, normalized size = 1.4 \begin{align*} \frac{80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A + 16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C - 15 \, B{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C
- 15*B*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1
) + 3*log(sin(d*x + c) - 1)))/d

________________________________________________________________________________________

Fricas [A]  time = 0.516688, size = 329, normalized size = 2.7 \begin{align*} \frac{45 \, B \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, B \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} + 45 \, B \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 30 \, B \cos \left (d x + c\right ) + 24 \, C\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(45*B*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*B*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(5*A + 4
*C)*cos(d*x + c)^4 + 45*B*cos(d*x + c)^3 + 8*(5*A + 4*C)*cos(d*x + c)^2 + 30*B*cos(d*x + c) + 24*C)*sin(d*x +
c))/(d*cos(d*x + c)^5)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4, x)

________________________________________________________________________________________

Giac [B]  time = 1.21692, size = 332, normalized size = 2.72 \begin{align*} \frac{45 \, B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (120 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 75 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 120 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 320 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 30 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 160 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 400 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 464 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 320 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 30 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 160 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 75 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 120 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(45*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*B*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*tan(1/2*d*
x + 1/2*c)^9 - 75*B*tan(1/2*d*x + 1/2*c)^9 + 120*C*tan(1/2*d*x + 1/2*c)^9 - 320*A*tan(1/2*d*x + 1/2*c)^7 + 30*
B*tan(1/2*d*x + 1/2*c)^7 - 160*C*tan(1/2*d*x + 1/2*c)^7 + 400*A*tan(1/2*d*x + 1/2*c)^5 + 464*C*tan(1/2*d*x + 1
/2*c)^5 - 320*A*tan(1/2*d*x + 1/2*c)^3 - 30*B*tan(1/2*d*x + 1/2*c)^3 - 160*C*tan(1/2*d*x + 1/2*c)^3 + 120*A*ta
n(1/2*d*x + 1/2*c) + 75*B*tan(1/2*d*x + 1/2*c) + 120*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d