3.124 \(\int \frac{\sec (c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=57 \[ \frac{(A+C) \tan (c+d x)}{a d (\sec (c+d x)+1)}+\frac{C \tan (c+d x)}{a d}-\frac{C \tanh ^{-1}(\sin (c+d x))}{a d} \]

[Out]

-((C*ArcTanh[Sin[c + d*x]])/(a*d)) + (C*Tan[c + d*x])/(a*d) + ((A + C)*Tan[c + d*x])/(a*d*(1 + Sec[c + d*x]))

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Rubi [A]  time = 0.151981, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {4083, 3998, 3770, 3794} \[ \frac{(A+C) \tan (c+d x)}{a d (\sec (c+d x)+1)}+\frac{C \tan (c+d x)}{a d}-\frac{C \tanh ^{-1}(\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((C*ArcTanh[Sin[c + d*x]])/(a*d)) + (C*Tan[c + d*x])/(a*d) + ((A + C)*Tan[c + d*x])/(a*d*(1 + Sec[c + d*x]))

Rule 4083

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [B]  time = 1.79207, size = 227, normalized size = 3.98 \[ \frac{4 \cos \left (\frac{1}{2} (c+d x)\right ) \cos (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left ((A+C) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+C \cos \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{a d (\sec (c+d x)+1) (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*((A + C)*Sec[c/2]*Sin[(d*x)/2] + C*Cos[(c + d*x)/2]*(L
og[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin
[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))))/(
a*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x]))

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Maple [B]  time = 0.051, size = 121, normalized size = 2.1 \begin{align*}{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{C}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{C}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/d*A*tan(1/2*d*x+1/2*c)+1/a/d*C*tan(1/2*d*x+1/2*c)-1/a/d/(tan(1/2*d*x+1/2*c)+1)*C-1/a/d*ln(tan(1/2*d*x+1/2*
c)+1)*C-1/a/d/(tan(1/2*d*x+1/2*c)-1)*C+1/a/d*ln(tan(1/2*d*x+1/2*c)-1)*C

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Maxima [B]  time = 0.940495, size = 194, normalized size = 3.4 \begin{align*} -\frac{C{\left (\frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (d x + c\right )}{{\left (a - \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - \frac{A \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(C*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - 2*sin(d*x + c)/
((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))) - A*si
n(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 0.496646, size = 288, normalized size = 5.05 \begin{align*} -\frac{{\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left ({\left (A + 2 \, C\right )} \cos \left (d x + c\right ) + C\right )} \sin \left (d x + c\right )}{2 \,{\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{3}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**3/(sec(c + d*x) + 1), x))/a

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Giac [A]  time = 1.20918, size = 136, normalized size = 2.39 \begin{align*} -\frac{\frac{C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} + \frac{2 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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