3.255 \(\int \frac{\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=94 \[ \frac{a^2 \log (a \cos (c+d x)+b)}{b d \left (a^2-b^2\right )}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)}-\frac{\log (\cos (c+d x)+1)}{2 d (a-b)}-\frac{\log (\cos (c+d x))}{b d} \]

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) - Log[Cos[c + d*x]]/(b*d) - Log[1 + Cos[c + d*x]]/(2*(a - b)*d) + (a^2*Log
[b + a*Cos[c + d*x]])/(b*(a^2 - b^2)*d)

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Rubi [A]  time = 0.270216, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4397, 2837, 12, 894} \[ \frac{a^2 \log (a \cos (c+d x)+b)}{b d \left (a^2-b^2\right )}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)}-\frac{\log (\cos (c+d x)+1)}{2 d (a-b)}-\frac{\log (\cos (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) - Log[Cos[c + d*x]]/(b*d) - Log[1 + Cos[c + d*x]]/(2*(a - b)*d) + (a^2*Log
[b + a*Cos[c + d*x]])/(b*(a^2 - b^2)*d)

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.150821, size = 103, normalized size = 1.1 \[ 2 \left (-\frac{a^2 \log (a \cos (c+d x)+b)}{2 b d \left (b^2-a^2\right )}+\frac{\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (a+b)}+\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (b-a)}-\frac{\log (\cos (c+d x))}{2 b d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*(Log[Cos[(c + d*x)/2]]/(2*(-a + b)*d) - Log[Cos[c + d*x]]/(2*b*d) - (a^2*Log[b + a*Cos[c + d*x]])/(2*b*(-a^2
+ b^2)*d) + Log[Sin[(c + d*x)/2]]/(2*(a + b)*d))

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Maple [A]  time = 0.137, size = 95, normalized size = 1. \begin{align*} -{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{d \left ( 2\,a-2\,b \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{d \left ( 2\,a+2\,b \right ) }}-{\frac{\ln \left ( \cos \left ( dx+c \right ) \right ) }{db}}+{\frac{{a}^{2}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{db \left ( a+b \right ) \left ( a-b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

-1/d/(2*a-2*b)*ln(cos(d*x+c)+1)+1/d/(2*a+2*b)*ln(-1+cos(d*x+c))-ln(cos(d*x+c))/d/b+1/d*a^2/b/(a+b)/(a-b)*ln(b+
a*cos(d*x+c))

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Maxima [A]  time = 1.15958, size = 169, normalized size = 1.8 \begin{align*} \frac{\frac{a^{2} \log \left (a + b - \frac{{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} b - b^{3}} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b} + \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

(a^2*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2*b - b^3) - log(sin(d*x + c)/(cos(d*x + c) +
1) + 1)/b - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b + log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b))/d

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Fricas [A]  time = 0.769623, size = 236, normalized size = 2.51 \begin{align*} \frac{2 \, a^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) - 2 \,{\left (a^{2} - b^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) -{\left (a b + b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (a b - b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{2} b - b^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*a^2*log(a*cos(d*x + c) + b) - 2*(a^2 - b^2)*log(-cos(d*x + c)) - (a*b + b^2)*log(1/2*cos(d*x + c) + 1/2
) + (a*b - b^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^2*b - b^3)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**2/(a*sin(c + d*x) + b*tan(c + d*x)), x)

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Giac [A]  time = 1.26786, size = 180, normalized size = 1.91 \begin{align*} \frac{\frac{2 \, a^{2} \log \left ({\left | -a - b - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} b - b^{3}} + \frac{\log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} - \frac{2 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*a^2*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))
)/(a^2*b - b^3) + log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b) - 2*log(abs(-(cos(d*x + c) - 1)/(c
os(d*x + c) + 1) - 1))/b)/d