3.461 \(\int \frac {\cos ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx\)

Optimal. Leaf size=129 \[ \frac {x \left (3 a^2-10 a b+15 b^2\right )}{8 (a-b)^3}-\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} d (a-b)^3}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d (a-b)}+\frac {(3 a-7 b) \sin (c+d x) \cos (c+d x)}{8 d (a-b)^2} \]

[Out]

1/8*(3*a^2-10*a*b+15*b^2)*x/(a-b)^3+1/8*(3*a-7*b)*cos(d*x+c)*sin(d*x+c)/(a-b)^2/d+1/4*cos(d*x+c)^3*sin(d*x+c)/
(a-b)/d-b^(5/2)*arctan(b^(1/2)*tan(d*x+c)/a^(1/2))/(a-b)^3/d/a^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 129, 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} \[ \frac {x \left (3 a^2-10 a b+15 b^2\right )}{8 (a-b)^3}-\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} d (a-b)^3}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d (a-b)}+\frac {(3 a-7 b) \sin (c+d x) \cos (c+d x)}{8 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + b*Tan[c + d*x]^2),x]

[Out]

((3*a^2 - 10*a*b + 15*b^2)*x)/(8*(a - b)^3) - (b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b
)^3*d) + ((3*a - 7*b)*Cos[c + d*x]*Sin[c + d*x])/(8*(a - b)^2*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*(a - b)*d)

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 ^4(c+d x)}{a+b \tan ^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d}-\frac {\operatorname {Subst}\left (\int \frac {-3 a+4 b-3 b x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{4 (a-b) d}\\ &=\frac {(3 a-7 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d}+\frac {\operatorname {Subst}\left (\int \frac {3 a^2-7 a b+8 b^2+(3 a-7 b) b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{8 (a-b)^2 d}\\ &=\frac {(3 a-7 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{(a-b)^3 d}+\frac {\left (3 a^2-10 a b+15 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 (a-b)^3 d}\\ &=\frac {\left (3 a^2-10 a b+15 b^2\right ) x}{8 (a-b)^3}-\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^3 d}+\frac {(3 a-7 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.42, size = 113, normalized size = 0.88 \[ \frac {\sqrt {a} \left (4 \left (3 a^2-10 a b+15 b^2\right ) (c+d x)+8 \left (a^2-3 a b+2 b^2\right ) \sin (2 (c+d x))+(a-b)^2 \sin (4 (c+d x))\right )-32 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{32 \sqrt {a} d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + b*Tan[c + d*x]^2),x]

[Out]

(-32*b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a]*(4*(3*a^2 - 10*a*b + 15*b^2)*(c + d*x) + 8*(a^2
- 3*a*b + 2*b^2)*Sin[2*(c + d*x)] + (a - b)^2*Sin[4*(c + d*x)]))/(32*Sqrt[a]*(a - b)^3*d)

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 401, normalized size = 3.11 \[ \left [-\frac {2 \, b^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - {\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} d x - {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d}, \frac {4 \, b^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + {\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} d x + {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/8*(2*b^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)*cos(d*x + c)^4 + 2*
(a*b - b^2)*cos(d*x + c)^2 + b^2)) - (3*a^2 - 10*a*b + 15*b^2)*d*x - (2*(a^2 - 2*a*b + b^2)*cos(d*x + c)^3 + (
3*a^2 - 10*a*b + 7*b^2)*cos(d*x + c))*sin(d*x + c))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d), 1/8*(4*b^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))) + (3*a^2 - 10*a*b + 15*b^2)*d
*x + (2*(a^2 - 2*a*b + b^2)*cos(d*x + c)^3 + (3*a^2 - 10*a*b + 7*b^2)*cos(d*x + c))*sin(d*x + c))/((a^3 - 3*a^
2*b + 3*a*b^2 - b^3)*d)]

________________________________________________________________________________________

giac [A]  time = 1.46, size = 183, normalized size = 1.42 \[ -\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} b^{3}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {3 \, a \tan \left (d x + c\right )^{3} - 7 \, b \tan \left (d x + c\right )^{3} + 5 \, a \tan \left (d x + c\right ) - 9 \, b \tan \left (d x + c\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

-1/8*(8*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))*b^3/((a^3 - 3*a^2*b + 3*a*b^2
- b^3)*sqrt(a*b)) - (3*a^2 - 10*a*b + 15*b^2)*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*a*tan(d*x + c)^3
- 7*b*tan(d*x + c)^3 + 5*a*tan(d*x + c) - 9*b*tan(d*x + c))/((a^2 - 2*a*b + b^2)*(tan(d*x + c)^2 + 1)^2))/d

________________________________________________________________________________________

maple [B]  time = 0.67, size = 303, normalized size = 2.35 \[ -\frac {b^{3} \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{d \left (a -b \right )^{3} \sqrt {a b}}+\frac {3 \left (\tan ^{3}\left (d x +c \right )\right ) a^{2}}{8 d \left (a -b \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {5 \left (\tan ^{3}\left (d x +c \right )\right ) a b}{4 d \left (a -b \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {7 \left (\tan ^{3}\left (d x +c \right )\right ) b^{2}}{8 d \left (a -b \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {7 \tan \left (d x +c \right ) a b}{4 d \left (a -b \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {9 \tan \left (d x +c \right ) b^{2}}{8 d \left (a -b \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {5 \tan \left (d x +c \right ) a^{2}}{8 d \left (a -b \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {15 \arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{8 d \left (a -b \right )^{3}}+\frac {3 \arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{8 d \left (a -b \right )^{3}}-\frac {5 \arctan \left (\tan \left (d x +c \right )\right ) a b}{4 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+b*tan(d*x+c)^2),x)

[Out]

-1/d*b^3/(a-b)^3/(a*b)^(1/2)*arctan(tan(d*x+c)*b/(a*b)^(1/2))+3/8/d/(a-b)^3/(1+tan(d*x+c)^2)^2*tan(d*x+c)^3*a^
2-5/4/d/(a-b)^3/(1+tan(d*x+c)^2)^2*tan(d*x+c)^3*a*b+7/8/d/(a-b)^3/(1+tan(d*x+c)^2)^2*tan(d*x+c)^3*b^2-7/4/d/(a
-b)^3/(1+tan(d*x+c)^2)^2*tan(d*x+c)*a*b+9/8/d/(a-b)^3/(1+tan(d*x+c)^2)^2*tan(d*x+c)*b^2+5/8/d/(a-b)^3/(1+tan(d
*x+c)^2)^2*tan(d*x+c)*a^2+15/8/d/(a-b)^3*arctan(tan(d*x+c))*b^2+3/8/d/(a-b)^3*arctan(tan(d*x+c))*a^2-5/4/d/(a-
b)^3*arctan(tan(d*x+c))*a*b

________________________________________________________________________________________

maxima [A]  time = 0.65, size = 185, normalized size = 1.43 \[ -\frac {\frac {8 \, b^{3} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (3 \, a - 7 \, b\right )} \tan \left (d x + c\right )^{3} + {\left (5 \, a - 9 \, b\right )} \tan \left (d x + c\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2} - 2 \, a b + b^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/8*(8*b^3*arctan(b*tan(d*x + c)/sqrt(a*b))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)) - (3*a^2 - 10*a*b + 1
5*b^2)*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - ((3*a - 7*b)*tan(d*x + c)^3 + (5*a - 9*b)*tan(d*x + c))/((a
^2 - 2*a*b + b^2)*tan(d*x + c)^4 + 2*(a^2 - 2*a*b + b^2)*tan(d*x + c)^2 + a^2 - 2*a*b + b^2))/d

________________________________________________________________________________________

mupad [B]  time = 15.87, size = 3681, normalized size = 28.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a + b*tan(c + d*x)^2),x)

[Out]

((tan(c + d*x)*(5*a - 9*b))/(8*(a^2 - 2*a*b + b^2)) + (tan(c + d*x)^3*(3*a - 7*b))/(8*(a^2 - 2*a*b + b^2)))/(d
*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1)) - (atan(((((tan(c + d*x)*(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a^3
*b^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (((256*b^10 - 1760*a*b^9 + 5280*a^2*b^8
- 9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5*b^5 + 3040*a^6*b^4 - 800*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6
*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - (tan(c + d*x)*(3*a^2 - 10*a*b + 15*b^2)*(1280*a*b^8 -
256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2
*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(3*a^2 - 10*a*b + 15*b^2))/(16
*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)))*(3*a^2 - 10*a*b + 15*b^2)*1i)/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b
^3*1i)) + (((tan(c + d*x)*(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a^3*b^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4
*a*b^3 + b^4 + 6*a^2*b^2)) - (((256*b^10 - 1760*a*b^9 + 5280*a^2*b^8 - 9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5*
b^5 + 3040*a^6*b^4 - 800*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 +
15*a^4*b^2)) + (tan(c + d*x)*(3*a^2 - 10*a*b + 15*b^2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1
280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 -
4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(3*a^2 - 10*a*b + 15*b^2))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i))
)*(3*a^2 - 10*a*b + 15*b^2)*1i)/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)))/((115*a*b^7 - 105*b^8 - 51*a^2*b
^6 + 9*a^3*b^5)/(32*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - (((tan(c + d*x)*
(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a^3*b^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))
+ (((256*b^10 - 1760*a*b^9 + 5280*a^2*b^8 - 9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5*b^5 + 3040*a^6*b^4 - 800*a
^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - (tan(c + d*
x)*(3*a^2 - 10*a*b + 15*b^2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4
+ 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 +
6*a^2*b^2)))*(3*a^2 - 10*a*b + 15*b^2))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)))*(3*a^2 - 10*a*b + 15*b^2
))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) + (((tan(c + d*x)*(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a^3*
b^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - (((256*b^10 - 1760*a*b^9 + 5280*a^2*b^8 -
9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5*b^5 + 3040*a^6*b^4 - 800*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*
a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(c + d*x)*(3*a^2 - 10*a*b + 15*b^2)*(1280*a*b^8 - 2
56*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2*
3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(3*a^2 - 10*a*b + 15*b^2))/(16*
(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)))*(3*a^2 - 10*a*b + 15*b^2))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1
i))))*(3*a^2 - 10*a*b + 15*b^2)*1i)/(8*d*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - (atan((((-a*b^5)^(1/2)*((t
an(c + d*x)*(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a^3*b^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 +
6*a^2*b^2)) + ((-a*b^5)^(1/2)*((256*b^10 - 1760*a*b^9 + 5280*a^2*b^8 - 9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5
*b^5 + 3040*a^6*b^4 - 800*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 +
15*a^4*b^2)) - (tan(c + d*x)*(-a*b^5)^(1/2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^
5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(64*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)*(a^4 - 4*a^3*b - 4*a*b
^3 + b^4 + 6*a^2*b^2))))/(2*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)))*1i)/(2*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2))
+ ((-a*b^5)^(1/2)*((tan(c + d*x)*(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a^3*b^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^
3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - ((-a*b^5)^(1/2)*((256*b^10 - 1760*a*b^9 + 5280*a^2*b^8 - 9056*a^3*b^7 + 97
60*a^4*b^6 - 6816*a^5*b^5 + 3040*a^6*b^4 - 800*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a
^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(c + d*x)*(-a*b^5)^(1/2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*
a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(64*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)*(
a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(2*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)))*1i)/(2*(a*b^3 + 3*a^3*b
- a^4 - 3*a^2*b^2)))/((115*a*b^7 - 105*b^8 - 51*a^2*b^6 + 9*a^3*b^5)/(32*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*
a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - ((-a*b^5)^(1/2)*((tan(c + d*x)*(289*b^7 - 300*a*b^6 + 190*a^2*b^5 - 60*a
^3*b^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + ((-a*b^5)^(1/2)*((256*b^10 - 1760*a*b^
9 + 5280*a^2*b^8 - 9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5*b^5 + 3040*a^6*b^4 - 800*a^7*b^3 + 96*a^8*b^2)/(64*(
a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - (tan(c + d*x)*(-a*b^5)^(1/2)*(1280*a*
b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(64*(
a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(2*(a*b^3 + 3*a^3*b - a^4 -
3*a^2*b^2))))/(2*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)) + ((-a*b^5)^(1/2)*((tan(c + d*x)*(289*b^7 - 300*a*b^6 +
190*a^2*b^5 - 60*a^3*b^4 + 9*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - ((-a*b^5)^(1/2)*((25
6*b^10 - 1760*a*b^9 + 5280*a^2*b^8 - 9056*a^3*b^7 + 9760*a^4*b^6 - 6816*a^5*b^5 + 3040*a^6*b^4 - 800*a^7*b^3 +
96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(c + d*x)*(-a*b
^5)^(1/2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 2
56*a^7*b^2))/(64*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))))/(2*(a*b^3
+ 3*a^3*b - a^4 - 3*a^2*b^2))))/(2*(a*b^3 + 3*a^3*b - a^4 - 3*a^2*b^2))))*(-a*b^5)^(1/2)*1i)/(d*(a*b^3 + 3*a^3
*b - a^4 - 3*a^2*b^2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________