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

Optimal. Leaf size=78 \[ \frac{(3 A+2 C) \tan (c+d x)}{3 d}+\frac{B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{B \tan (c+d x) \sec (c+d x)}{2 d}+\frac{C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0805421, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4047, 3768, 3770, 4046, 3767, 8} \[ \frac{(3 A+2 C) \tan (c+d x)}{3 d}+\frac{B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{B \tan (c+d x) \sec (c+d x)}{2 d}+\frac{C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.199872, size = 51, normalized size = 0.65 \[ \frac{\tan (c+d x) \left (6 (A+C)+3 B \sec (c+d x)+2 C \tan ^2(c+d x)\right )+3 B \tanh ^{-1}(\sin (c+d x))}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.028, size = 83, normalized size = 1.1 \begin{align*}{\frac{A\tan \left ( dx+c \right ) }{d}}+{\frac{B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{C \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*A*tan(d*x+c)+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 = 0.931582, size = 107, normalized size = 1.37 \begin{align*} \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 \, A \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+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)) + 12*A*tan(d*x + c))/d

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Fricas [A]  time = 0.507411, size = 250, normalized size = 3.21 \begin{align*} \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 (2 \,{\left (3 \, A + 2 \, C\right )} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+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*(2*(3*A + 2*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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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 ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2, x)

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Giac [B]  time = 1.2104, size = 219, normalized size = 2.81 \begin{align*} \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 (6 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 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} - 12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+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*(6*A*tan(1/2*d*x + 1/
2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5 - 12*A*tan(1/2*d*x + 1/2*c)^3 - 4*C*tan(1/2*d
*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) + 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