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

Optimal. Leaf size=92 \[ \frac{a \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^2 d \sqrt{a^2+b^2}}-\frac{1}{b d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0794156, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3094, 3770, 3074, 206} \[ \frac{a \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^2 d \sqrt{a^2+b^2}}-\frac{1}{b d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3094

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_)/cos[(c_.) + (d_.)*(x_)], x_Symbol] :>
Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)), x] + (Dist[1/b^2, Int[(a*Cos[c + d*x] + b*Sin[c
+ d*x])^(n + 2)/Cos[c + d*x], x], x] - Dist[a/b^2, Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1), x], x]) /;
FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3770

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

Rule 3074

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.73498, size = 120, normalized size = 1.3 \[ -\frac{\frac{2 a \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}}+\frac{b \sec (c+d x)}{a+b \tan (c+d x)}+\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.22, size = 174, normalized size = 1.9 \begin{align*} 2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) a}}+2\,{\frac{1}{db \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) }}-2\,{\frac{a}{{b}^{2}d\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+{\frac{1}{{b}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{{b}^{2}d}\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*cos(d*x+c)+b*sin(d*x+c))^2,x)

[Out]

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

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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*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.598545, size = 699, normalized size = 7.6 \begin{align*} -\frac{2 \, a^{2} b + 2 \, b^{3} -{\left (a^{2} \cos \left (d x + c\right ) + a b \sin \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) -{\left ({\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right ) +{\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right ) +{\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left ({\left (a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{2} b^{3} + b^{5}\right )} d \sin \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*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(2*a^2*b + 2*b^3 - (a^2*cos(d*x + c) + a*b*sin(d*x + c))*sqrt(a^2 + b^2)*log((2*a*b*cos(d*x + c)*sin(d*x
+ c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*
cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) - ((a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*s
in(d*x + c))*log(sin(d*x + c) + 1) + ((a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))*log(-sin(d*x +
c) + 1))/((a^3*b^2 + a*b^4)*d*cos(d*x + c) + (a^2*b^3 + b^5)*d*sin(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28809, size = 224, normalized size = 2.43 \begin{align*} \frac{\frac{a \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} b^{2}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )} a b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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