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

Optimal. Leaf size=152 \[ \frac{b \left (a^2-3 b^2\right )}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{4 a b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (6 a^2 b^2+a^4-3 b^4\right )}{2 \left (a^2+b^2\right )^3} \]

[Out]

((a^4 + 6*a^2*b^2 - 3*b^4)*x)/(2*(a^2 + b^2)^3) + (4*a*b^3*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^
3*d) + (b*(a^2 - 3*b^2))/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) + (Cos[c + d*x]^2*(b + a*Tan[c + d*x]))/(2*(
a^2 + b^2)*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.162816, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3506, 741, 801, 635, 203, 260} \[ \frac{b \left (a^2-3 b^2\right )}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{4 a b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (6 a^2 b^2+a^4-3 b^4\right )}{2 \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^4 + 6*a^2*b^2 - 3*b^4)*x)/(2*(a^2 + b^2)^3) + (4*a*b^3*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^
3*d) + (b*(a^2 - 3*b^2))/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) + (Cos[c + d*x]^2*(b + a*Tan[c + d*x]))/(2*(
a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, 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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

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

Rule 203

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{(a+x)^2 \left (1+\frac{x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{b \operatorname{Subst}\left (\int \frac{-3-\frac{a^2}{b^2}-\frac{2 a x}{b^2}}{(a+x)^2 \left (1+\frac{x^2}{b^2}\right )} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac{\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{b \operatorname{Subst}\left (\int \left (\frac{a^2-3 b^2}{\left (a^2+b^2\right ) (a+x)^2}-\frac{8 a b^2}{\left (a^2+b^2\right )^2 (a+x)}+\frac{-a^4-6 a^2 b^2+3 b^4+8 a b^2 x}{\left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac{4 a b^3 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{b \left (a^2-3 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{b \operatorname{Subst}\left (\int \frac{-a^4-6 a^2 b^2+3 b^4+8 a b^2 x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=\frac{4 a b^3 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{b \left (a^2-3 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int \frac{x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^3 d}+\frac{\left (b \left (a^4+6 a^2 b^2-3 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=\frac{\left (a^4+6 a^2 b^2-3 b^4\right ) x}{2 \left (a^2+b^2\right )^3}+\frac{4 a b^3 \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{4 a b^3 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{b \left (a^2-3 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.75131, size = 304, normalized size = 2. \[ \frac{-\frac{a b \left (\left (\sqrt{-b^2}-a\right ) \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )-2 \sqrt{-b^2} \log (a+b \tan (c+d x))+\left (a+\sqrt{-b^2}\right ) \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )\right )}{\sqrt{-b^2} \left (a^2+b^2\right )}+\frac{b \left (a^2-3 b^2\right ) \left (\frac{2 \left (a^2+b^2\right )}{a+b \tan (c+d x)}+\left (\frac{b^2-a^2}{\sqrt{-b^2}}+2 a\right ) \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+\left (\frac{a^2-b^2}{\sqrt{-b^2}}+2 a\right ) \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )-4 a \log (a+b \tan (c+d x))\right )}{2 \left (a^2+b^2\right )^2}+\frac{\cos ^2(c+d x) (a \tan (c+d x)+b)}{a+b \tan (c+d x)}}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*b*((-a + Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 2*Sqrt[-b^2]*Log[a + b*Tan[c + d*x]] + (a + Sqrt
[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]]))/(Sqrt[-b^2]*(a^2 + b^2))) + (Cos[c + d*x]^2*(b + a*Tan[c + d*x]))/(
a + b*Tan[c + d*x]) + (b*(a^2 - 3*b^2)*((2*a + (-a^2 + b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 4*a
*Log[a + b*Tan[c + d*x]] + (2*a + (a^2 - b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + (2*(a^2 + b^2))/(
a + b*Tan[c + d*x])))/(2*(a^2 + b^2)^2))/(2*(a^2 + b^2)*d)

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Maple [A]  time = 0.102, size = 292, normalized size = 1.9 \begin{align*}{\frac{\tan \left ( dx+c \right ){a}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{\tan \left ( dx+c \right ){b}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{b{a}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{3}a}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-2\,{\frac{{b}^{3}a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+4\,{\frac{{b}^{3}a\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/(a^2+b^2)^3/(1+tan(d*x+c)^2)*tan(d*x+c)*a^4-1/2/d/(a^2+b^2)^3/(1+tan(d*x+c)^2)*tan(d*x+c)*b^4+1/d/(a^2+b
^2)^3/(1+tan(d*x+c)^2)*b*a^3+1/d/(a^2+b^2)^3/(1+tan(d*x+c)^2)*b^3*a-2/d/(a^2+b^2)^3*b^3*a*ln(1+tan(d*x+c)^2)+3
/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a^2*b^2-3/2/d/(a^2+b^2)^3*arctan(tan(d*x+c))*b^4+1/2/d/(a^2+b^2)^3*arctan(ta
n(d*x+c))*a^4-1/d*b^3/(a^2+b^2)^2/(a+b*tan(d*x+c))+4/d*b^3/(a^2+b^2)^3*a*ln(a+b*tan(d*x+c))

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Maxima [A]  time = 1.66835, size = 381, normalized size = 2.51 \begin{align*} \frac{\frac{8 \, a b^{3} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{4 \, a b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{4} + 6 \, a^{2} b^{2} - 3 \, b^{4}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \, a^{2} b - 2 \, b^{3} +{\left (a^{2} b - 3 \, b^{3}\right )} \tan \left (d x + c\right )^{2} +{\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )^{3} +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \tan \left (d x + c\right )^{2} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(8*a*b^3*log(b*tan(d*x + c) + a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 4*a*b^3*log(tan(d*x + c)^2 + 1)/(a^
6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^4 + 6*a^2*b^2 - 3*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (
2*a^2*b - 2*b^3 + (a^2*b - 3*b^3)*tan(d*x + c)^2 + (a^3 + a*b^2)*tan(d*x + c))/(a^5 + 2*a^3*b^2 + a*b^4 + (a^4
*b + 2*a^2*b^3 + b^5)*tan(d*x + c)^3 + (a^5 + 2*a^3*b^2 + a*b^4)*tan(d*x + c)^2 + (a^4*b + 2*a^2*b^3 + b^5)*ta
n(d*x + c)))/d

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Fricas [A]  time = 2.12177, size = 614, normalized size = 4.04 \begin{align*} \frac{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} b^{3} + 3 \, b^{5} -{\left (a^{5} + 6 \, a^{3} b^{2} - 3 \, a b^{4}\right )} d x\right )} \cos \left (d x + c\right ) + 4 \,{\left (a^{2} b^{3} \cos \left (d x + c\right ) + a b^{4} \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^{3} b^{2} - a b^{4} -{\left (a^{4} b + 6 \, a^{2} b^{3} - 3 \, b^{5}\right )} d x -{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\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)^2/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*((a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^3 - (a^2*b^3 + 3*b^5 - (a^5 + 6*a^3*b^2 - 3*a*b^4)*d*x)*cos(d*x +
c) + 4*(a^2*b^3*cos(d*x + c) + a*b^4*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^3*b^2 - a*b^4 - (a^4*b + 6*a^2*b^3 - 3*b^5)*d*x - (a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c)^2)*
sin(d*x + c))/((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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.43003, size = 338, normalized size = 2.22 \begin{align*} \frac{\frac{8 \, a b^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{4 \, a b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{4} + 6 \, a^{2} b^{2} - 3 \, b^{4}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{a^{2} b \tan \left (d x + c\right )^{2} - 3 \, b^{3} \tan \left (d x + c\right )^{2} + a^{3} \tan \left (d x + c\right ) + a b^{2} \tan \left (d x + c\right ) + 2 \, a^{2} b - 2 \, b^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \tan \left (d x + c\right )^{3} + a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(8*a*b^4*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) - 4*a*b^3*log(tan(d*x + c)^2 +
1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^4 + 6*a^2*b^2 - 3*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6) + (a^2*b*tan(d*x + c)^2 - 3*b^3*tan(d*x + c)^2 + a^3*tan(d*x + c) + a*b^2*tan(d*x + c) + 2*a^2*b - 2*b^3)/
((a^4 + 2*a^2*b^2 + b^4)*(b*tan(d*x + c)^3 + a*tan(d*x + c)^2 + b*tan(d*x + c) + a)))/d