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

Optimal. Leaf size=44 \[ \frac{(B-C) \tan (c+d x)}{a d (\sec (c+d x)+1)}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.0728113, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4050, 3770, 12, 3794} \[ \frac{(B-C) \tan (c+d x)}{a d (\sec (c+d x)+1)}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4050

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> Dist[C/b, Int[Csc[e + f*x], x], x] + Dist[1/b, Int[(A*b + (b*B - a*C)*Csc[e + f*x])/(a +
b*Csc[e + f*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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{B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx &=\frac{\int \frac{(a B-a C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{a}+\frac{C \int \sec (c+d x) \, dx}{a}\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{a d}+(B-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{(B-C) \tan (c+d x)}{d (a+a \sec (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.184871, size = 106, normalized size = 2.41 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left ((B-C) \sin \left (\frac{1}{2} (c+d x)\right )+C \cos \left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*(C*Cos[(c + d*x)/2]*(-Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]]) + (B - C)*Sin[(c + d*x)/2]))/(a*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 0.044, size = 78, normalized size = 1.8 \begin{align*}{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.932191, size = 134, normalized size = 3.05 \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{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac{B \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((B*sec(d*x+c)+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 - sin(d*x + c)/(a*
(cos(d*x + c) + 1))) + B*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 0.495314, size = 197, normalized size = 4.48 \begin{align*} \frac{{\left (C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (B - C\right )} \sin \left (d x + c\right )}{2 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{B \sec{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{2}{\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((B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

(Integral(B*sec(c + d*x)/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x) + 1), x))/a

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Giac [A]  time = 1.14828, size = 95, normalized size = 2.16 \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{B \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}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+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 + (B*tan(1/2*d*x + 1/2*c) - C
*tan(1/2*d*x + 1/2*c))/a)/d