3.453 \(\int \frac{\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx\)

Optimal. Leaf size=60 \[ \frac{\sin (c+d x)}{d (a-b)}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{3/2}} \]

[Out]

-((b*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(3/2)*d)) + Sin[c + d*x]/((a - b)*d)

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Rubi [A]  time = 0.0797293, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3676, 388, 208} \[ \frac{\sin (c+d x)}{d (a-b)}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(3/2)*d)) + Sin[c + d*x]/((a - b)*d)

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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{\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{a-(a-b) x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\sin (c+d x)}{(a-b) d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{a+(-a+b) x^2} \, dx,x,\sin (c+d x)\right )}{(a-b) d}\\ &=-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} (a-b)^{3/2} d}+\frac{\sin (c+d x)}{(a-b) d}\\ \end{align*}

Mathematica [A]  time = 0.0937933, size = 60, normalized size = 1. \[ \frac{\sin (c+d x)}{d (a-b)}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(3/2)*d)) + Sin[c + d*x]/((a - b)*d)

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Maple [A]  time = 0.063, size = 61, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}{\it Artanh} \left ({ \left ( a-b \right ) \sin \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53467, size = 414, normalized size = 6.9 \begin{align*} \left [-\frac{\sqrt{a^{2} - a b} b \log \left (-\frac{{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) - 2 \,{\left (a^{2} - a b\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} d}, \frac{\sqrt{-a^{2} + a b} b \arctan \left (\frac{\sqrt{-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) +{\left (a^{2} - a b\right )} \sin \left (d x + c\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.70302, size = 99, normalized size = 1.65 \begin{align*} -\frac{\frac{b \arctan \left (-\frac{a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt{-a^{2} + a b}}\right )}{\sqrt{-a^{2} + a b}{\left (a - b\right )}} - \frac{\sin \left (d x + c\right )}{a - b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*(a - b)) - sin(d*x + c)/(a -
b))/d