3.767 \(\int \sec (c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=93 \[ \frac{(3 a B+2 b C) \tan (c+d x)}{3 d}+\frac{(a C+b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(a C+b B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.153376, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4072, 3997, 3787, 3767, 8, 3768, 3770} \[ \frac{(3 a B+2 b C) \tan (c+d x)}{3 d}+\frac{(a C+b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(a C+b B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 (c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^2(c+d x) (a+b \sec (c+d x)) (B+C \sec (c+d x)) \, dx\\ &=\frac{b C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int \sec ^2(c+d x) (3 a B+2 b C+3 (b B+a C) \sec (c+d x)) \, dx\\ &=\frac{b C \sec ^2(c+d x) \tan (c+d x)}{3 d}+(b B+a C) \int \sec ^3(c+d x) \, dx+\frac{1}{3} (3 a B+2 b C) \int \sec ^2(c+d x) \, dx\\ &=\frac{(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{b C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} (b B+a C) \int \sec (c+d x) \, dx-\frac{(3 a B+2 b C) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{(b B+a C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(3 a B+2 b C) \tan (c+d x)}{3 d}+\frac{(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{b C \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.290459, size = 67, normalized size = 0.72 \[ \frac{3 (a C+b B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (a C+b B) \sec (c+d x)+6 a B+2 b C \tan ^2(c+d x)+6 b C\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 128, normalized size = 1.4 \begin{align*}{\frac{Ba\tan \left ( dx+c \right ) }{d}}+{\frac{C\sec \left ( dx+c \right ) a\tan \left ( dx+c \right ) }{2\,d}}+{\frac{aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{Bb\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{Bb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,Cb\tan \left ( dx+c \right ) }{3\,d}}+{\frac{C \left ( \sec \left ( dx+c \right ) \right ) ^{2}b\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)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.986974, size = 171, normalized size = 1.84 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b - 3 \, C a{\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 )} - 3 \, B 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 \, B 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)*(a+b*sec(d*x+c))*(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*b - 3*C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) - 3*B*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*B*a*tan(d*x + c))/d

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Fricas [A]  time = 0.683446, size = 298, normalized size = 3.2 \begin{align*} \frac{3 \,{\left (C a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (C a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (3 \, B a + 2 \, C b\right )} \cos \left (d x + c\right )^{2} + 2 \, C b + 3 \,{\left (C a + B b\right )} \cos \left (d x + c\right )\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)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(3*(C*a + B*b)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(C*a + B*b)*cos(d*x + c)^3*log(-sin(d*x + c) + 1)
+ 2*(2*(3*B*a + 2*C*b)*cos(d*x + c)^2 + 2*C*b + 3*(C*a + B*b)*cos(d*x + 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 (B + C \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\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)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((B + C*sec(c + d*x))*(a + b*sec(c + d*x))*sec(c + d*x)**2, x)

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Giac [B]  time = 1.18757, size = 284, normalized size = 3.05 \begin{align*} \frac{3 \,{\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b \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)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(
6*B*a*tan(1/2*d*x + 1/2*c)^5 - 3*C*a*tan(1/2*d*x + 1/2*c)^5 - 3*B*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*b*tan(1/2*d*x
+ 1/2*c)^5 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 4*C*b*tan(1/2*d*x + 1/2*c)^3 + 6*B*a*tan(1/2*d*x + 1/2*c) + 3*C*
a*tan(1/2*d*x + 1/2*c) + 3*B*b*tan(1/2*d*x + 1/2*c) + 6*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)
^3)/d