\(\int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx\) [453]

Optimal result

Integrand size = 21, antiderivative size = 60 \[ \int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {b \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{3/2} d}+\frac {\sin (c+d x)}{(a-b) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3757, 396, 214} \[ \int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\sin (c+d x)}{d (a-b)}-\frac {b \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} d (a-b)^{3/2}} \]

[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 214

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

Rule 396

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 3757

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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 \text {Subst}\left (\int \frac {1}{a+(-a+b) x^2} \, dx,x,\sin (c+d x)\right )}{(a-b) d} \\ & = -\frac {b \text {arctanh}\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] (verified)

Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {b \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{3/2} d}+\frac {\sin (c+d x)}{(a-b) d} \]

[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)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (d x +c \right )}{a -b}-\frac {b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{\left (a -b \right ) \sqrt {a \left (a -b \right )}}}{d}\)	\(61\)
default	\(\frac {\frac {\sin \left (d x +c \right )}{a -b}-\frac {b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{\left (a -b \right ) \sqrt {a \left (a -b \right )}}}{d}\)	\(61\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 \left (a -b \right ) d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 \left (a -b \right ) d}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\sqrt {a^{2}-a b}}-1\right )}{2 \sqrt {a^{2}-a b}\, \left (a -b \right ) d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\sqrt {a^{2}-a b}}-1\right )}{2 \sqrt {a^{2}-a b}\, \left (a -b \right ) d}\)	\(162\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 3.03 \[ \int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [-\frac {\sqrt {a^{2} - a b} b \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) - 2 \, {\left (a^{2} - a b\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} d}, \frac {\sqrt {-a^{2} + a b} b \arctan \left (\frac {\sqrt {-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) + {\left (a^{2} - a b\right )} \sin \left (d x + c\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} d}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {\cos {\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {\frac {b \arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{\sqrt {-a^{2} + a b} {\left (a - b\right )}} - \frac {\sin \left (d x + c\right )}{a - b}}{d} \]

[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

Mupad [B] (verification not implemented)

Time = 12.54 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )}{d\,\left (a-b\right )}+\frac {b\,\mathrm {atanh}\left (\frac {\sin \left (c+d\,x\right )\,{\left (a-b\right )}^{3/2}}{\sqrt {a}\,b-a^{3/2}}\right )}{\sqrt {a}\,d\,{\left (a-b\right )}^{3/2}} \]

[In]

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

[Out]

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