3.472 \(\int \frac{\cos ^2(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)
Optimal. Leaf size=148 \[ \frac{b^{3/2} (5 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^3}+\frac{b (a+b) \tan (c+d x)}{2 a d (a-b)^2 \left (a+b \tan ^2(c+d x)\right )}+\frac{\sin (c+d x) \cos (c+d x)}{2 d (a-b) \left (a+b \tan ^2(c+d x)\right )}+\frac{x (a-5 b)}{2 (a-b)^3} \]
[Out]
((a - 5*b)*x)/(2*(a - b)^3) + ((5*a - b)*b^(3/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a - b)^3*
d) + (Cos[c + d*x]*Sin[c + d*x])/(2*(a - b)*d*(a + b*Tan[c + d*x]^2)) + (b*(a + b)*Tan[c + d*x])/(2*a*(a - b)^
2*d*(a + b*Tan[c + d*x]^2))
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Rubi [A] time = 0.186098, antiderivative size = 148, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used =
{3675, 414, 527, 522, 203, 205} \[ \frac{b^{3/2} (5 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^3}+\frac{b (a+b) \tan (c+d x)}{2 a d (a-b)^2 \left (a+b \tan ^2(c+d x)\right )}+\frac{\sin (c+d x) \cos (c+d x)}{2 d (a-b) \left (a+b \tan ^2(c+d x)\right )}+\frac{x (a-5 b)}{2 (a-b)^3} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^2/(a + b*Tan[c + d*x]^2)^2,x]
[Out]
((a - 5*b)*x)/(2*(a - b)^3) + ((5*a - b)*b^(3/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a - b)^3*
d) + (Cos[c + d*x]*Sin[c + d*x])/(2*(a - b)*d*(a + b*Tan[c + d*x]^2)) + (b*(a + b)*Tan[c + d*x])/(2*a*(a - b)^
2*d*(a + b*Tan[c + d*x]^2))
Rule 3675
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Rule 414
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]
Rule 527
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
Rule 522
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]
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 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^2 \left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\cos (c+d x) \sin (c+d x)}{2 (a-b) d \left (a+b \tan ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-a+2 b-3 b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{2 (a-b) d}\\ &=\frac{\cos (c+d x) \sin (c+d x)}{2 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{b (a+b) \tan (c+d x)}{2 a (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2 \left (a^2-4 a b+b^2\right )-2 b (a+b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{4 a (a-b)^2 d}\\ &=\frac{\cos (c+d x) \sin (c+d x)}{2 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{b (a+b) \tan (c+d x)}{2 a (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac{(a-5 b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 (a-b)^3 d}+\frac{\left ((5 a-b) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{2 a (a-b)^3 d}\\ &=\frac{(a-5 b) x}{2 (a-b)^3}+\frac{(5 a-b) b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} (a-b)^3 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{b (a+b) \tan (c+d x)}{2 a (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 1.14763, size = 116, normalized size = 0.78 \[ \frac{-\frac{2 b^{3/2} (b-5 a) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{3/2}}+\frac{2 b^2 (a-b) \sin (2 (c+d x))}{a ((a-b) \cos (2 (c+d x))+a+b)}+2 (a-5 b) (c+d x)+(a-b) \sin (2 (c+d x))}{4 d (a-b)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^2/(a + b*Tan[c + d*x]^2)^2,x]
[Out]
(2*(a - 5*b)*(c + d*x) - (2*b^(3/2)*(-5*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(3/2) + (a - b)*Sin[2
*(c + d*x)] + (2*(a - b)*b^2*Sin[2*(c + d*x)])/(a*(a + b + (a - b)*Cos[2*(c + d*x)])))/(4*(a - b)^3*d)
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Maple [A] time = 0.084, size = 248, normalized size = 1.7 \begin{align*}{\frac{{b}^{2}\tan \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{3}\tan \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{3}a \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{5\,{b}^{2}}{2\,d \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{{b}^{3}}{2\,d \left ( a-b \right ) ^{3}a}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{a\tan \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}-{\frac{b\tan \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{2\,d \left ( a-b \right ) ^{3}}}-{\frac{5\,\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{2\,d \left ( a-b \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)^2,x)
[Out]
1/2/d*b^2/(a-b)^3*tan(d*x+c)/(a+b*tan(d*x+c)^2)-1/2/d*b^3/(a-b)^3/a*tan(d*x+c)/(a+b*tan(d*x+c)^2)+5/2/d*b^2/(a
-b)^3/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))-1/2/d*b^3/(a-b)^3/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(
1/2))+1/2/d/(a-b)^3*tan(d*x+c)/(tan(d*x+c)^2+1)*a-1/2/d/(a-b)^3*tan(d*x+c)/(tan(d*x+c)^2+1)*b+1/2/d/(a-b)^3*ar
ctan(tan(d*x+c))*a-5/2/d/(a-b)^3*arctan(tan(d*x+c))*b
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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)^2/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.95393, size = 1370, normalized size = 9.26 \begin{align*} \text{result too large to display} \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)^2,x, algorithm="fricas")
[Out]
[1/8*(4*(a^3 - 6*a^2*b + 5*a*b^2)*d*x*cos(d*x + c)^2 + 4*(a^2*b - 5*a*b^2)*d*x + (5*a*b^2 - b^3 + (5*a^2*b - 6
*a*b^2 + b^3)*cos(d*x + c)^2)*sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b^2)*cos(d*x + c
)^2 - 4*((a^2 + a*b)*cos(d*x + c)^3 - a*b*cos(d*x + c))*sqrt(-b/a)*sin(d*x + c) + b^2)/((a^2 - 2*a*b + b^2)*co
s(d*x + c)^4 + 2*(a*b - b^2)*cos(d*x + c)^2 + b^2)) + 4*((a^3 - 2*a^2*b + a*b^2)*cos(d*x + c)^3 + (a^2*b - b^3
)*cos(d*x + c))*sin(d*x + c))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d*cos(d*x + c)^2 + (a^4*b - 3*a
^3*b^2 + 3*a^2*b^3 - a*b^4)*d), 1/4*(2*(a^3 - 6*a^2*b + 5*a*b^2)*d*x*cos(d*x + c)^2 + 2*(a^2*b - 5*a*b^2)*d*x
- (5*a*b^2 - b^3 + (5*a^2*b - 6*a*b^2 + b^3)*cos(d*x + c)^2)*sqrt(b/a)*arctan(1/2*((a + b)*cos(d*x + c)^2 - b)
*sqrt(b/a)/(b*cos(d*x + c)*sin(d*x + c))) + 2*((a^3 - 2*a^2*b + a*b^2)*cos(d*x + c)^3 + (a^2*b - b^3)*cos(d*x
+ c))*sin(d*x + c))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d*cos(d*x + c)^2 + (a^4*b - 3*a^3*b^2 + 3
*a^2*b^3 - a*b^4)*d)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**2,x)
[Out]
Timed out
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Giac [B] time = 147.814, size = 1083, normalized size = 7.32 \begin{align*} \text{result too large to display} \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)^2,x, algorithm="giac")
[Out]
-1/2*((a^5*b - 12*a^4*b^2 + 22*a^3*b^3 - 12*a^2*b^4 + a*b^5 - a*b*abs(-a^4 + 3*a^3*b - 3*a^2*b^2 + a*b^3) - b^
2*abs(-a^4 + 3*a^3*b - 3*a^2*b^2 + a*b^3))*(pi*floor((d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(d*x + c)/sqr
t((a^4 - a^3*b - a^2*b^2 + a*b^3 + sqrt((a^4 - a^3*b - a^2*b^2 + a*b^3)^2 - 4*(a^4 - 2*a^3*b + a^2*b^2)*(a^3*b
- 2*a^2*b^2 + a*b^3)))/(a^3*b - 2*a^2*b^2 + a*b^3))))/(a^4*abs(-a^4 + 3*a^3*b - 3*a^2*b^2 + a*b^3) - a^3*b*ab
s(-a^4 + 3*a^3*b - 3*a^2*b^2 + a*b^3) - a^2*b^2*abs(-a^4 + 3*a^3*b - 3*a^2*b^2 + a*b^3) + a*b^3*abs(-a^4 + 3*a
^3*b - 3*a^2*b^2 + a*b^3) + (a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)^2) + (sqrt(a*b)*(a + b)*abs(-a^4 + 3*a^3*b - 3
*a^2*b^2 + a*b^3)*abs(b) + (a^5 - 12*a^4*b + 22*a^3*b^2 - 12*a^2*b^3 + a*b^4)*sqrt(a*b)*abs(b))*(pi*floor((d*x
+ c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(d*x + c)/sqrt((a^4 - a^3*b - a^2*b^2 + a*b^3 - sqrt((a^4 - a^3*b - a^
2*b^2 + a*b^3)^2 - 4*(a^4 - 2*a^3*b + a^2*b^2)*(a^3*b - 2*a^2*b^2 + a*b^3)))/(a^3*b - 2*a^2*b^2 + a*b^3))))/((
a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)^2*b - (a^4*b - a^3*b^2 - a^2*b^3 + a*b^4)*abs(-a^4 + 3*a^3*b - 3*a^2*b^2 +
a*b^3)) - (a*b*tan(d*x + c)^3 + b^2*tan(d*x + c)^3 + a^2*tan(d*x + c) + b^2*tan(d*x + c))/((b*tan(d*x + c)^4 +
a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)*(a^3 - 2*a^2*b + a*b^2)))/d