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

Optimal. Leaf size=157 \[ \frac{\cos (c+d x) (a \tan (c+d x)+b)}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{b \left (a^2-2 b^2\right ) \sec (c+d x)}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{3 a b^2 \cos (c+d x) \sqrt{\sec ^2(c+d x)} \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

[Out]

(-3*a*b^2*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Cos[c + d*x]*Sqrt[Sec[c + d*x]^
2])/((a^2 + b^2)^(5/2)*d) + (b*(a^2 - 2*b^2)*Sec[c + d*x])/((a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) + (Cos[c + d
*x]*(b + a*Tan[c + d*x]))/((a^2 + b^2)*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.126985, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3512, 741, 807, 725, 206} \[ \frac{\cos (c+d x) (a \tan (c+d x)+b)}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{b \left (a^2-2 b^2\right ) \sec (c+d x)}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{3 a b^2 \cos (c+d x) \sqrt{\sec ^2(c+d x)} \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*b^2*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Cos[c + d*x]*Sqrt[Sec[c + d*x]^
2])/((a^2 + b^2)^(5/2)*d) + (b*(a^2 - 2*b^2)*Sec[c + d*x])/((a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) + (Cos[c + d
*x]*(b + a*Tan[c + d*x]))/((a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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{\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac{\left (\cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x)^2 \left (1+\frac{x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\left (b \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{-2-\frac{a x}{b^2}}{(a+x)^2 \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{b \left (a^2-2 b^2\right ) \sec (c+d x)}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (3 a b \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^2 d}\\ &=\frac{b \left (a^2-2 b^2\right ) \sec (c+d x)}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\left (3 a b \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a^2}{b^2}-x^2} \, dx,x,\frac{1-\frac{a \tan (c+d x)}{b}}{\sqrt{\sec ^2(c+d x)}}\right )}{\left (a^2+b^2\right )^2 d}\\ &=-\frac{3 a b^2 \tanh ^{-1}\left (\frac{b \left (1-\frac{a \tan (c+d x)}{b}\right )}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)}}{\left (a^2+b^2\right )^{5/2} d}+\frac{b \left (a^2-2 b^2\right ) \sec (c+d x)}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.529381, size = 153, normalized size = 0.97 \[ \frac{\sec (c+d x) \left (\left (a^2+b^2\right ) \left (a \left (a^2+b^2\right ) \sin (2 (c+d x))+b \left (a^2+b^2\right ) \cos (2 (c+d x))+3 b \left (a^2-b^2\right )\right )+12 a b^2 \sqrt{a^2+b^2} (a \cos (c+d x)+b \sin (c+d x)) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )\right )}{2 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(12*a*b^2*Sqrt[a^2 + b^2]*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]]*(a*Cos[c + d*x] + b
*Sin[c + d*x]) + (a^2 + b^2)*(3*b*(a^2 - b^2) + b*(a^2 + b^2)*Cos[2*(c + d*x)] + a*(a^2 + b^2)*Sin[2*(c + d*x)
])))/(2*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x]))

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Maple [A]  time = 0.098, size = 172, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{ \left ( -{a}^{2}+{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) -2\,ab}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ({\frac{1}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a} \left ( -{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{a}}-b \right ) }-3\,{\frac{a}{\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 ) } \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*(-2/(a^4+2*a^2*b^2+b^4)*((-a^2+b^2)*tan(1/2*d*x+1/2*c)-2*a*b)/(1+tan(1/2*d*x+1/2*c)^2)-2*b^2/(a^2+b^2)^2*(
(-b^2/a*tan(1/2*d*x+1/2*c)-b)/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)-3*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))))

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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 = 1.94141, size = 697, normalized size = 4.44 \begin{align*} \frac{2 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \,{\left (a^{2} b^{2} \cos \left (d x + c\right ) + a b^{3} \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 )}{2 \,{\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} d \sin \left (d x + c\right )\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*(2*a^4*b - 2*a^2*b^3 - 4*b^5 + 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*co
s(d*x + c)*sin(d*x + c) + 3*(a^2*b^2*cos(d*x + c) + a*b^3*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^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b
^6)*d*cos(d*x + c) + (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*d*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{2}}\, 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 = 1.49833, size = 386, normalized size = 2.46 \begin{align*} -\frac{\frac{3 \, a b^{2} \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 )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a^{3} b + a b^{3}\right )}}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}}}{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]

-(3*a*b^2*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*s
qrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(a^4*tan(1/2*d*x + 1/2*c)^3 - a^2*b^2*tan(1/2*d
*x + 1/2*c)^3 + b^4*tan(1/2*d*x + 1/2*c)^3 + 3*a*b^3*tan(1/2*d*x + 1/2*c)^2 - a^4*tan(1/2*d*x + 1/2*c) - 3*a^2
*b^2*tan(1/2*d*x + 1/2*c) + b^4*tan(1/2*d*x + 1/2*c) - 2*a^3*b + a*b^3)/((a^5 + 2*a^3*b^2 + a*b^4)*(a*tan(1/2*
d*x + 1/2*c)^4 - 2*b*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 1/2*c) - a)))/d