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

Optimal. Leaf size=95 \[ \frac{2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{a C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac{C \tan (c+d x)}{b d} \]

[Out]

-((a*C*ArcTanh[Sin[c + d*x]])/(b^2*d)) + (2*(A*b^2 + a^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]
])/(Sqrt[a - b]*b^2*Sqrt[a + b]*d) + (C*Tan[c + d*x])/(b*d)

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Rubi [A]  time = 0.1929, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4083, 3998, 3770, 3831, 2659, 208} \[ \frac{2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{a C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac{C \tan (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*C*ArcTanh[Sin[c + d*x]])/(b^2*d)) + (2*(A*b^2 + a^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]
])/(Sqrt[a - b]*b^2*Sqrt[a + b]*d) + (C*Tan[c + d*x])/(b*d)

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 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac{C \tan (c+d x)}{b d}+\frac{\int \frac{\sec (c+d x) (A b-a C \sec (c+d x))}{a+b \sec (c+d x)} \, dx}{b}\\ &=\frac{C \tan (c+d x)}{b d}-\frac{(a C) \int \sec (c+d x) \, dx}{b^2}+\left (A+\frac{a^2 C}{b^2}\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx\\ &=-\frac{a C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac{C \tan (c+d x)}{b d}+\frac{\left (A b^2+a^2 C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^3}\\ &=-\frac{a C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac{C \tan (c+d x)}{b d}+\frac{\left (2 \left (A b^2+a^2 C\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^3 d}\\ &=-\frac{a C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac{2 \left (A b^2+a^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^2 \sqrt{a+b} d}+\frac{C \tan (c+d x)}{b d}\\ \end{align*}

Mathematica [C]  time = 2.30183, size = 331, normalized size = 3.48 \[ \frac{2 \cos (c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac{2 i (\cos (c)-i \sin (c)) \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}+a C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-a C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b C \sin \left (\frac{d x}{2}\right )}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{b C \sin \left (\frac{d x}{2}\right )}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}\right )}{b^2 d (a+b \sec (c+d x)) (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]*(b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*(a*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - a*
C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - ((2*I)*(A*b^2 + a^2*C)*ArcTan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-
b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2])]*(Cos[c] - I*Sin[c]))/(Sqrt[a^2 - b
^2]*Sqrt[(Cos[c] - I*Sin[c])^2]) + (b*C*Sin[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)
/2])) + (b*C*Sin[(d*x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(b^2*d*(A + 2*C + A
*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x]))

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Maple [B]  time = 0.072, size = 183, normalized size = 1.9 \begin{align*} 2\,{\frac{A}{d\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{{a}^{2}C}{d{b}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{C}{db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{aC}{d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{aC}{d{b}^{2}}\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+b*sec(d*x+c)),x)

[Out]

2/d/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+2/d/b^2/((a+b)*(a-b))^(1/2)*ar
ctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*a^2*C-1/d/b/(tan(1/2*d*x+1/2*c)+1)*C-1/d*a/b^2*ln(tan(1/2*
d*x+1/2*c)+1)*C-1/d/b/(tan(1/2*d*x+1/2*c)-1)*C+1/d*a/b^2*ln(tan(1/2*d*x+1/2*c)-1)*C

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.98854, size = 967, normalized size = 10.18 \begin{align*} \left [\frac{{\left (C a^{2} + A b^{2}\right )} \sqrt{a^{2} - b^{2}} \cos \left (d x + c\right ) \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) -{\left (C a^{3} - C a b^{2}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (C a^{3} - C a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}, \frac{2 \,{\left (C a^{2} + A b^{2}\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) -{\left (C a^{3} - C a b^{2}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (C a^{3} - C a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} 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+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*((C*a^2 + A*b^2)*sqrt(a^2 - b^2)*cos(d*x + c)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*
sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^
2)) - (C*a^3 - C*a*b^2)*cos(d*x + c)*log(sin(d*x + c) + 1) + (C*a^3 - C*a*b^2)*cos(d*x + c)*log(-sin(d*x + c)
+ 1) + 2*(C*a^2*b - C*b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c)), 1/2*(2*(C*a^2 + A*b^2)*sqrt(-a^2 +
b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c) - (C*a^3 - C*a*b^2
)*cos(d*x + c)*log(sin(d*x + c) + 1) + (C*a^3 - C*a*b^2)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(C*a^2*b - C*
b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c))]

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

[Out]

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

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Giac [A]  time = 1.39042, size = 220, normalized size = 2.32 \begin{align*} -\frac{\frac{C a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac{C a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \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 )} b} + \frac{2 \,{\left (C a^{2} + A b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\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 )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{\sqrt{-a^{2} + b^{2}} b^{2}}}{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+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-(C*a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^2 - C*a*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2 + 2*C*tan(1/2*d*x +
1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*b) + 2*(C*a^2 + A*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) +
arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)*b^2))/d