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

Optimal. Leaf size=63 \[ \frac{(A-B+C) \tan (c+d x)}{a d (\sec (c+d x)+1)}+\frac{(B-C) \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C \tan (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.168264, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {4082, 3998, 3770, 3794} \[ \frac{(A-B+C) \tan (c+d x)}{a d (\sec (c+d x)+1)}+\frac{(B-C) \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C \tan (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, 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+B \sec (c+d x)+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 (B-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{a}\\ &=\frac{C \tan (c+d x)}{a d}+\frac{(B-C) \int \sec (c+d x) \, dx}{a}+(A-B+C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx\\ &=\frac{(B-C) \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C \tan (c+d x)}{a d}+\frac{(A-B+C) \tan (c+d x)}{d (a+a \sec (c+d x))}\\ \end{align*}

Mathematica [B]  time = 1.39599, size = 255, normalized size = 4.05 \[ \frac{4 \cos \left (\frac{1}{2} (c+d x)\right ) \cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\sec \left (\frac{c}{2}\right ) (A-B+C) \sin \left (\frac{d x}{2}\right )+\cos \left (\frac{1}{2} (c+d x)\right ) \left (\frac{C \sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-(B-C) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )\right )}{a d (\sec (c+d x)+1) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x] + 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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((A - B + C)*Sec[c/2]*Sin[(d*x)/2] +
Cos[(c + d*x)/2]*(-((B - C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]])) + (C*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 + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x]))

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Maple [B]  time = 0.059, size = 180, normalized size = 2.9 \begin{align*}{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{B}{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{B}{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} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c)+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*B*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)*B-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(ta
n(1/2*d*x+1/2*c)-1)*B+1/a/d*ln(tan(1/2*d*x+1/2*c)-1)*C

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Maxima [B]  time = 0.942921, size = 294, normalized size = 4.67 \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 )} - B{\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{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+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 - 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))) - B*(l
og(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))) - A*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [B]  time = 0.512863, size = 324, normalized size = 5.14 \begin{align*} \frac{{\left ({\left (B - C\right )} \cos \left (d x + c\right )^{2} +{\left (B - C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (B - C\right )} \cos \left (d x + c\right )^{2} +{\left (B - C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left ({\left (A - B + 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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(((B - C)*cos(d*x + c)^2 + (B - C)*cos(d*x + c))*log(sin(d*x + c) + 1) - ((B - C)*cos(d*x + c)^2 + (B - C)
*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*((A - B + 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{B \sec ^{2}{\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+B*sec(d*x+c)+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(B*sec(c + d*x)**2/(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 = 1.28263, size = 161, normalized size = 2.56 \begin{align*} \frac{\frac{{\left (B - C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{{\left (B - C\right )} \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 ) - 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} - \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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

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