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

Optimal. Leaf size=75 \[ \frac{\frac{a}{b^2}+\frac{1}{a}}{d (a \cot (c+d x)+b)}-\frac{2 a \log (\tan (c+d x))}{b^3 d}-\frac{2 a \log (a \cot (c+d x)+b)}{b^3 d}+\frac{\tan (c+d x)}{b^2 d} \]

[Out]

(a^(-1) + a/b^2)/(d*(b + a*Cot[c + d*x])) - (2*a*Log[b + a*Cot[c + d*x]])/(b^3*d) - (2*a*Log[Tan[c + d*x]])/(b
^3*d) + Tan[c + d*x]/(b^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0962912, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {3088, 894} \[ \frac{\frac{a}{b^2}+\frac{1}{a}}{d (a \cot (c+d x)+b)}-\frac{2 a \log (\tan (c+d x))}{b^3 d}-\frac{2 a \log (a \cot (c+d x)+b)}{b^3 d}+\frac{\tan (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(-1) + a/b^2)/(d*(b + a*Cot[c + d*x])) - (2*a*Log[b + a*Cot[c + d*x]])/(b^3*d) - (2*a*Log[Tan[c + d*x]])/(b
^3*d) + Tan[c + d*x]/(b^2*d)

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

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

Mathematica [A]  time = 0.25982, size = 51, normalized size = 0.68 \[ \frac{-\frac{a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))+b \tan (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*Log[a + b*Tan[c + d*x]] + b*Tan[c + d*x] - (a^2 + b^2)/(a + b*Tan[c + d*x]))/(b^3*d)

________________________________________________________________________________________

Maple [A]  time = 0.204, size = 78, normalized size = 1. \begin{align*}{\frac{\tan \left ( dx+c \right ) }{{b}^{2}d}}-2\,{\frac{a\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{3}d}}-{\frac{{a}^{2}}{{b}^{3}d \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{1}{db \left ( a+b\tan \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(d*x+c)/b^2/d-2/d*a/b^3*ln(a+b*tan(d*x+c))-1/d/b^3/(a+b*tan(d*x+c))*a^2-1/d/b/(a+b*tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.19236, size = 81, normalized size = 1.08 \begin{align*} -\frac{\frac{a^{2} + b^{2}}{b^{4} \tan \left (d x + c\right ) + a b^{3}} + \frac{2 \, a \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{3}} - \frac{\tan \left (d x + c\right )}{b^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2 + b^2)/(b^4*tan(d*x + c) + a*b^3) + 2*a*log(b*tan(d*x + c) + a)/b^3 - tan(d*x + c)/b^2)/d

________________________________________________________________________________________

Fricas [B]  time = 0.535613, size = 440, normalized size = 5.87 \begin{align*} -\frac{2 \, b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - b^{2} +{\left (a^{2} \cos \left (d x + c\right )^{2} + a b \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (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 (a^{2} \cos \left (d x + c\right )^{2} + a b \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right )}{a b^{3} d \cos \left (d x + c\right )^{2} + b^{4} d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*b^2*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*sin(d*x + c) - b^2 + (a^2*cos(d*x + c)^2 + a*b*cos(d*x + c)*sin(d*
x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - (a^2*cos(d*x + c)^2 + a*b*co
s(d*x + c)*sin(d*x + c))*log(cos(d*x + c)^2))/(a*b^3*d*cos(d*x + c)^2 + b^4*d*cos(d*x + c)*sin(d*x + c))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.13786, size = 96, normalized size = 1.28 \begin{align*} -\frac{\frac{2 \, a \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{3}} - \frac{\tan \left (d x + c\right )}{b^{2}} - \frac{2 \, a b \tan \left (d x + c\right ) + a^{2} - b^{2}}{{\left (b \tan \left (d x + c\right ) + a\right )} b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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