3.37 \(\int \sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=85 \[ \frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d}+\frac{3 C \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 C \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.0716354, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4047, 3767, 12, 3768, 3770} \[ \frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d}+\frac{3 C \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 C \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

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

Mathematica [A]  time = 0.229007, size = 76, normalized size = 0.89 \[ \frac{B \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 C \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

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Maple [A]  time = 0.03, size = 92, normalized size = 1.1 \begin{align*}{\frac{2\,B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{C \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

2/3*B*tan(d*x+c)/d+1/3/d*B*tan(d*x+c)*sec(d*x+c)^2+1/4*C*sec(d*x+c)^3*tan(d*x+c)/d+3/8*C*sec(d*x+c)*tan(d*x+c)
/d+3/8/d*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.934751, size = 128, normalized size = 1.51 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B - 3 \, C{\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 )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B - 3*C*(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

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Fricas [A]  time = 0.505312, size = 266, normalized size = 3.13 \begin{align*} \frac{9 \, C \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, C \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \, B \cos \left (d x + c\right )^{3} + 9 \, C \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, C\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(9*C*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*C*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*B*cos(d*x +
c)^3 + 9*C*cos(d*x + c)^2 + 8*B*cos(d*x + c) + 6*C)*sin(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (B + C \sec{\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)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

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

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Giac [B]  time = 1.22148, size = 221, normalized size = 2.6 \begin{align*} \frac{9 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, 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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(9*C*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*C*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(24*B*tan(1/2*d*x +
1/2*c)^7 - 15*C*tan(1/2*d*x + 1/2*c)^7 - 40*B*tan(1/2*d*x + 1/2*c)^5 - 9*C*tan(1/2*d*x + 1/2*c)^5 + 40*B*tan(1
/2*d*x + 1/2*c)^3 - 9*C*tan(1/2*d*x + 1/2*c)^3 - 24*B*tan(1/2*d*x + 1/2*c) - 15*C*tan(1/2*d*x + 1/2*c))/(tan(1
/2*d*x + 1/2*c)^2 - 1)^4)/d