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

Optimal. Leaf size=63 \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d}+\frac {C \tan ^3(c+d x)}{3 d}+\frac {C \tan (c+d x)}{d} \]

[Out]

1/2*B*arctanh(sin(d*x+c))/d+C*tan(d*x+c)/d+1/2*B*sec(d*x+c)*tan(d*x+c)/d+1/3*C*tan(d*x+c)^3/d

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Rubi [A]  time = 0.06, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4047, 3768, 3770, 12, 3767} \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d}+\frac {C \tan ^3(c+d x)}{3 d}+\frac {C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[c + d*x]])/(2*d) + (C*Tan[c + d*x])/d + (B*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (C*Tan[c + d*x]^3
)/(3*d)

Rule 12

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 60, normalized size = 0.95 \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d}+\frac {C \left (\frac {1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (C*(Tan[c + d*x] + Tan[c + d*x]^3/3))/
d

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fricas [A]  time = 0.44, size = 88, normalized size = 1.40 \[ \frac {3 \, B \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*B*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*B*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(4*C*cos(d*x +
c)^2 + 3*B*cos(d*x + c) + 2*C)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [B]  time = 0.49, size = 122, normalized size = 1.94 \[ \frac {3 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*B*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(3*B*tan(1/2*d*x + 1/
2*c)^5 - 6*C*tan(1/2*d*x + 1/2*c)^5 + 4*C*tan(1/2*d*x + 1/2*c)^3 - 3*B*tan(1/2*d*x + 1/2*c) - 6*C*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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maple [A]  time = 1.22, size = 72, normalized size = 1.14 \[ \frac {B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 C \tan \left (d x +c \right )}{3 d}+\frac {C \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 70, normalized size = 1.11 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C - 3 \, B {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C - 3*B*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1)
+ log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 4.19, size = 109, normalized size = 1.73 \[ \frac {B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {\left (B-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {4\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\left (-B-2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/cos(c + d*x)^2,x)

[Out]

(B*atanh(tan(c/2 + (d*x)/2)))/d + ((4*C*tan(c/2 + (d*x)/2)^3)/3 - tan(c/2 + (d*x)/2)*(B + 2*C) + tan(c/2 + (d*
x)/2)^5*(B - 2*C))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)**3, x)

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