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

Optimal. Leaf size=56 \[ \frac{a (B+C) \tan (c+d x)}{d}+\frac{a (2 B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a C \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0586209, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4048, 3770, 3767, 8} \[ \frac{a (B+C) \tan (c+d x)}{d}+\frac{a (2 B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4048

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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]

Rubi steps

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

Mathematica [A]  time = 0.0375233, size = 75, normalized size = 1.34 \[ \frac{a B \tan (c+d x)}{d}+\frac{a B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a C \tan (c+d x)}{d}+\frac{a C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 86, normalized size = 1.5 \begin{align*}{\frac{Ba\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{aC\tan \left ( dx+c \right ) }{d}}+{\frac{Ba\tan \left ( dx+c \right ) }{d}}+{\frac{aC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*B*a*ln(sec(d*x+c)+tan(d*x+c))+a*C*tan(d*x+c)/d+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))

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Maxima [A]  time = 0.930831, size = 119, normalized size = 2.12 \begin{align*} -\frac{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 )} - 4 \, B a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, B a \tan \left (d x + c\right ) - 4 \, C a \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(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)) - 4*B*a*log(se
c(d*x + c) + tan(d*x + c)) - 4*B*a*tan(d*x + c) - 4*C*a*tan(d*x + c))/d

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Fricas [A]  time = 0.500579, size = 239, normalized size = 4.27 \begin{align*} \frac{{\left (2 \, B + C\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, B + C\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (B + C\right )} a \cos \left (d x + c\right ) + C a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(B*sec(c + d*x), x) + Integral(B*sec(c + d*x)**2, x) + Integral(C*sec(c + d*x)**2, x) + Integral(C*
sec(c + d*x)**3, x))

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Giac [B]  time = 1.15362, size = 167, normalized size = 2.98 \begin{align*} \frac{{\left (2 \, B a + C a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, B a + C a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (2 \, 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} - 2 \, 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 )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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