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

Optimal. Leaf size=93 \[ \frac{\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right )}+\frac{b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{a x \left (a^2+3 b^2\right )}{2 \left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.136052, antiderivative size = 93, 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} \[ \frac{\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right )}+\frac{b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{a x \left (a^2+3 b^2\right )}{2 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.22499, size = 143, normalized size = 1.54 \[ \frac{b \left (a^2+b^2\right ) \cos (2 (c+d x))+a^3 \sin (2 (c+d x))+2 a^3 c+2 a^3 d x+a b^2 \sin (2 (c+d x))+2 b^3 \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+6 a b^2 c+6 a b^2 d x-4 i b^3 \tan ^{-1}(\tan (c+d x))+4 i b^3 c+4 i b^3 d x}{4 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.098, size = 236, normalized size = 2.5 \begin{align*}{\frac{{a}^{3}\tan \left ( dx+c \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{a\tan \left ( dx+c \right ){b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{b{a}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{3}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \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)),x)

[Out]

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

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Maxima [A]  time = 1.64093, size = 190, normalized size = 2.04 \begin{align*} \frac{\frac{2 \, b^{3} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{3} + 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{a \tan \left (d x + c\right ) + b}{{\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2} + b^{2}}}{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)),x, algorithm="maxima")

[Out]

1/2*(2*b^3*log(b*tan(d*x + c) + a)/(a^4 + 2*a^2*b^2 + b^4) - b^3*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^
4) + (a^3 + 3*a*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + (a*tan(d*x + c) + b)/((a^2 + b^2)*tan(d*x + c)^2 + a^
2 + b^2))/d

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Fricas [A]  time = 1.93546, size = 278, normalized size = 2.99 \begin{align*} \frac{b^{3} \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} + 3 \, a b^{2}\right )} d x +{\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} 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)),x, algorithm="fricas")

[Out]

1/2*(b^3*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 + 3*a*b^2)*d*x + (a^2*
b + b^3)*cos(d*x + c)^2 + (a^3 + a*b^2)*cos(d*x + c)*sin(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d)

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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)),x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.60913, size = 246, normalized size = 2.65 \begin{align*} \frac{\frac{2 \, b^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac{b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{3} + 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{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 ) + a^{2} b + 2 \, b^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (d x + c\right )^{2} + 1\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)),x, algorithm="giac")

[Out]

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