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]

1/2*(a-5*b)*x/(a-b)^3+1/2*(5*a-b)*b^(3/2)*arctan(b^(1/2)*tan(d*x+c)/a^(1/2))/a^(3/2)/(a-b)^3/d+1/2*cos(d*x+c)*
sin(d*x+c)/(a-b)/d/(a+b*tan(d*x+c)^2)+1/2*b*(a+b)*tan(d*x+c)/a/(a-b)^2/d/(a+b*tan(d*x+c)^2)

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Rubi [A]  time = 0.19, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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])

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*}

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Mathematica [A]  time = 1.25, 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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fricas [A]  time = 0.57, size = 614, normalized size = 4.15 \[ \left [\frac {4 \, {\left (a^{3} - 6 \, a^{2} b + 5 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{2} b - 5 \, a b^{2}\right )} d x + {\left (5 \, a b^{2} - b^{3} + {\left (5 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} d\right )}}, \frac {2 \, {\left (a^{3} - 6 \, a^{2} b + 5 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{2} b - 5 \, a b^{2}\right )} d x - {\left (5 \, a b^{2} - b^{3} + {\left (5 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} d\right )}}\right ] \]

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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giac [A]  time = 2.18, size = 211, normalized size = 1.43 \[ \frac {\frac {{\left (d x + c\right )} {\left (a - 5 \, b\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (5 \, a b^{2} - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt {a b}} + \frac {a b \tan \left (d x + c\right )^{3} + b^{2} \tan \left (d x + c\right )^{3} + a^{2} \tan \left (d x + c\right ) + b^{2} \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{4} + a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )}}}{2 \, d} \]

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*((d*x + c)*(a - 5*b)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (5*a*b^2 - b^3)*(pi*floor((d*x + c)/pi + 1/2)*sgn(b
) + arctan(b*tan(d*x + c)/sqrt(a*b)))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*sqrt(a*b)) + (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

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maple [A]  time = 0.77, size = 248, normalized size = 1.68 \[ \frac {b^{2} \tan \left (d x +c \right )}{2 d \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}-\frac {b^{3} \tan \left (d x +c \right )}{2 d \left (a -b \right )^{3} a \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d \left (a -b \right )^{3} \sqrt {a b}}-\frac {b^{3} \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d \left (a -b \right )^{3} a \sqrt {a b}}+\frac {\tan \left (d x +c \right ) a}{2 d \left (a -b \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {\tan \left (d x +c \right ) b}{2 d \left (a -b \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a}{2 d \left (a -b \right )^{3}}-\frac {5 \arctan \left (\tan \left (d x +c \right )\right ) b}{2 d \left (a -b \right )^{3}} \]

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(tan(d*x+c)*b/(a*b)^(1/2))-1/2/d*b^3/(a-b)^3/a/(a*b)^(1/2)*arctan(tan(d*x+c)*b/(a*b)^(
1/2))+1/2/d/(a-b)^3*tan(d*x+c)/(1+tan(d*x+c)^2)*a-1/2/d/(a-b)^3*tan(d*x+c)/(1+tan(d*x+c)^2)*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 [A]  time = 1.05, size = 209, normalized size = 1.41 \[ \frac {\frac {{\left (d x + c\right )} {\left (a - 5 \, b\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (5 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt {a b}} + \frac {{\left (a b + b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (d x + c\right )^{4} + a^{4} - 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3}\right )} \tan \left (d x + c\right )^{2}}}{2 \, d} \]

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]

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

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mupad [B]  time = 16.29, size = 3843, normalized size = 25.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c + d*x)*(a^2 + b^2))/(2*a*(a^2 - 2*a*b + b^2)) + (b*tan(c + d*x)^3*(a + b))/(2*a*(a^2 - 2*a*b + b^2)))/
(d*(a + tan(c + d*x)^2*(a + b) + b*tan(c + d*x)^4)) - (atan(((((((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 172*a^4
*b^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 +
15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) - (tan(c + d*x)*(a - 5*b)*(16*a^2*b^9 - 80*a^3*b^8 + 144*a^4*b^7 - 80*a
^5*b^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^6
- 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(a - 5*b))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - (tan(c
+ d*x)*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*
b^2)))*(a - 5*b)*1i)/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - (((((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 1
72*a^4*b^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3
*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) + (tan(c + d*x)*(a - 5*b)*(16*a^2*b^9 - 80*a^3*b^8 + 144*a^4*b^7
- 80*a^5*b^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)
*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(a - 5*b))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) +
(tan(c + d*x)*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 +
6*a^4*b^2)))*(a - 5*b)*1i)/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)))/((((((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b
^8 - 172*a^4*b^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 -
6*a^3*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) - (tan(c + d*x)*(a - 5*b)*(16*a^2*b^9 - 80*a^3*b^8 + 144*a^
4*b^7 - 80*a^5*b^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(a*b^2*3i - a^2*b*3i + a^3*1i - b
^3*1i)*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(a - 5*b))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1
i)) - (tan(c + d*x)*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*
b^3 + 6*a^4*b^2)))*(a - 5*b))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - ((21*a*b^6)/4 - (5*b^7)/4 + (21*a^
2*b^5)/4 - (5*a^3*b^4)/4)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) + (((((
2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 172*a^4*b^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*a^9
*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) + (tan(c + d*x)*(a - 5*b)*(
16*a^2*b^9 - 80*a^3*b^8 + 144*a^4*b^7 - 80*a^5*b^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(
a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(a - 5*b))/(4*(a*b^
2*3i - a^2*b*3i + a^3*1i - b^3*1i)) + (tan(c + d*x)*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(
a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(a - 5*b))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i))))*(a
- 5*b)*1i)/(2*d*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - (atan((((5*a - b)*(-a^3*b^3)^(1/2)*((tan(c + d*x)*
(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)) -
(((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 172*a^4*b^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*
a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) - (tan(c + d*x)*(5*a - b
)*(-a^3*b^3)^(1/2)*(16*a^2*b^9 - 80*a^3*b^8 + 144*a^4*b^7 - 80*a^5*b^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3
+ 16*a^9*b^2))/(8*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(
5*a - b)*(-a^3*b^3)^(1/2))/(4*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)))*1i)/(4*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*
b^2)) + ((5*a - b)*(-a^3*b^3)^(1/2)*((tan(c + d*x)*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(a
^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)) + (((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 172*a^4*b^7 + 220*a
^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 + 15*a^4*b^4
- 20*a^5*b^3 + 15*a^6*b^2) + (tan(c + d*x)*(5*a - b)*(-a^3*b^3)^(1/2)*(16*a^2*b^9 - 80*a^3*b^8 + 144*a^4*b^7 -
80*a^5*b^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)*(a
^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(5*a - b)*(-a^3*b^3)^(1/2))/(4*(3*a^5*b - a^6 + a^3*b^3 - 3*
a^4*b^2)))*1i)/(4*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)))/(((21*a*b^6)/4 - (5*b^7)/4 + (21*a^2*b^5)/4 - (5*a^3
*b^4)/4)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) + ((5*a - b)*(-a^3*b^3)^
(1/2)*((tan(c + d*x)*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3
*b^3 + 6*a^4*b^2)) - (((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 172*a^4*b^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*
b^4 - 20*a^8*b^3 + 2*a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) - (
tan(c + d*x)*(5*a - b)*(-a^3*b^3)^(1/2)*(16*a^2*b^9 - 80*a^3*b^8 + 144*a^4*b^7 - 80*a^5*b^6 - 80*a^6*b^5 + 144
*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3
*b^3 + 6*a^4*b^2)))*(5*a - b)*(-a^3*b^3)^(1/2))/(4*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2))))/(4*(3*a^5*b - a^6
+ a^3*b^3 - 3*a^4*b^2)) - ((5*a - b)*(-a^3*b^3)^(1/2)*((tan(c + d*x)*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4
+ a^4*b^3))/(2*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)) + (((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 1
72*a^4*b^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3
*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) + (tan(c + d*x)*(5*a - b)*(-a^3*b^3)^(1/2)*(16*a^2*b^9 - 80*a^3*b
^8 + 144*a^4*b^7 - 80*a^5*b^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(3*a^5*b - a^6 + a^3*b
^3 - 3*a^4*b^2)*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(5*a - b)*(-a^3*b^3)^(1/2))/(4*(3*a^5*b -
a^6 + a^3*b^3 - 3*a^4*b^2))))/(4*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2))))*(5*a - b)*(-a^3*b^3)^(1/2)*1i)/(2*d*
(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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