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

Optimal. Leaf size=152 \[ \frac {\cos ^4(c+d x) (a \tan (c+d x)+b)}{4 d \left (a^2+b^2\right )}+\frac {b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {\cos ^2(c+d x) \left (a \left (3 a^2+7 b^2\right ) \tan (c+d x)+4 b^3\right )}{8 d \left (a^2+b^2\right )^2}+\frac {a x \left (3 a^4+10 a^2 b^2+15 b^4\right )}{8 \left (a^2+b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3506, 741, 823, 801, 635, 203, 260} \[ \frac {\cos ^4(c+d x) (a \tan (c+d x)+b)}{4 d \left (a^2+b^2\right )}+\frac {\cos ^2(c+d x) \left (a \left (3 a^2+7 b^2\right ) \tan (c+d x)+4 b^3\right )}{8 d \left (a^2+b^2\right )^2}+\frac {b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {a x \left (10 a^2 b^2+3 a^4+15 b^4\right )}{8 \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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]

Rubi steps

\begin {align*} \int \frac {\cos ^4(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 )^3} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \frac {-4-\frac {3 a^2}{b^2}-\frac {3 a x}{b^2}}{(a+x) \left (1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{4 \left (a^2+b^2\right ) d}\\ &=\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d}+\frac {\cos ^2(c+d x) \left (4 b^3+a \left (3 a^2+7 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}+\frac {b^5 \operatorname {Subst}\left (\int \frac {\frac {3 a^4+7 a^2 b^2+8 b^4}{b^6}+\frac {a \left (3 a^2+7 b^2\right ) x}{b^6}}{(a+x) \left (1+\frac {x^2}{b^2}\right )} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}\\ &=\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d}+\frac {\cos ^2(c+d x) \left (4 b^3+a \left (3 a^2+7 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}+\frac {b^5 \operatorname {Subst}\left (\int \left (\frac {8}{\left (a^2+b^2\right ) (a+x)}+\frac {3 a^5+10 a^3 b^2+15 a b^4-8 b^4 x}{b^4 \left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}\\ &=\frac {b^5 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d}+\frac {\cos ^2(c+d x) \left (4 b^3+a \left (3 a^2+7 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}+\frac {b \operatorname {Subst}\left (\int \frac {3 a^5+10 a^3 b^2+15 a b^4-8 b^4 x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d}\\ &=\frac {b^5 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d}+\frac {\cos ^2(c+d x) \left (4 b^3+a \left (3 a^2+7 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}-\frac {b^5 \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 (a b \left (3 a^4+10 a^2 b^2+15 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d}\\ &=\frac {a \left (3 a^4+10 a^2 b^2+15 b^4\right ) x}{8 \left (a^2+b^2\right )^3}+\frac {b^5 \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {b^5 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d}+\frac {\cos ^2(c+d x) \left (4 b^3+a \left (3 a^2+7 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 218, normalized size = 1.43 \[ \frac {8 a^5 \sin (2 (c+d x))+a^5 \sin (4 (c+d x))+12 a^5 c+12 a^5 d x+24 a^3 b^2 \sin (2 (c+d x))+2 a^3 b^2 \sin (4 (c+d x))+40 a^3 b^2 c+40 a^3 b^2 d x+b \left (a^2+b^2\right )^2 \cos (4 (c+d x))+4 b \left (a^4+4 a^2 b^2+3 b^4\right ) \cos (2 (c+d x))+32 b^5 \log (a \cos (c+d x)+b \sin (c+d x))+16 a b^4 \sin (2 (c+d x))+a b^4 \sin (4 (c+d x))+60 a b^4 c+60 a b^4 d x}{32 d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^5*c + 40*a^3*b^2*c + 60*a*b^4*c + 12*a^5*d*x + 40*a^3*b^2*d*x + 60*a*b^4*d*x + 4*b*(a^4 + 4*a^2*b^2 + 3*
b^4)*Cos[2*(c + d*x)] + b*(a^2 + b^2)^2*Cos[4*(c + d*x)] + 32*b^5*Log[a*Cos[c + d*x] + b*Sin[c + d*x]] + 8*a^5
*Sin[2*(c + d*x)] + 24*a^3*b^2*Sin[2*(c + d*x)] + 16*a*b^4*Sin[2*(c + d*x)] + a^5*Sin[4*(c + d*x)] + 2*a^3*b^2
*Sin[4*(c + d*x)] + a*b^4*Sin[4*(c + d*x)])/(32*(a^2 + b^2)^3*d)

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fricas [A]  time = 1.00, size = 208, normalized size = 1.37 \[ \frac {4 \, b^{5} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 4 \, {\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*b^5*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) + 2*(a^4*b + 2*a^2*b^3 + b^
5)*cos(d*x + c)^4 + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*d*x + 4*(a^2*b^3 + b^5)*cos(d*x + c)^2 + (2*(a^5 + 2*a^3*b
^2 + a*b^4)*cos(d*x + c)^3 + (3*a^5 + 10*a^3*b^2 + 7*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^6 + 3*a^4*b^2 + 3*
a^2*b^4 + b^6)*d)

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giac [B]  time = 1.63, size = 322, normalized size = 2.12 \[ \frac {\frac {8 \, b^{6} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac {4 \, b^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {6 \, b^{5} \tan \left (d x + c\right )^{4} + 3 \, a^{5} \tan \left (d x + c\right )^{3} + 10 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} + 7 \, a b^{4} \tan \left (d x + c\right )^{3} + 4 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} + 16 \, b^{5} \tan \left (d x + c\right )^{2} + 5 \, a^{5} \tan \left (d x + c\right ) + 14 \, a^{3} b^{2} \tan \left (d x + c\right ) + 9 \, a b^{4} \tan \left (d x + c\right ) + 2 \, a^{4} b + 8 \, a^{2} b^{3} + 12 \, b^{5}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*b^6*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) - 4*b^5*log(tan(d*x + c)^2 + 1)/
(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 +
b^6) + (6*b^5*tan(d*x + c)^4 + 3*a^5*tan(d*x + c)^3 + 10*a^3*b^2*tan(d*x + c)^3 + 7*a*b^4*tan(d*x + c)^3 + 4*
a^2*b^3*tan(d*x + c)^2 + 16*b^5*tan(d*x + c)^2 + 5*a^5*tan(d*x + c) + 14*a^3*b^2*tan(d*x + c) + 9*a*b^4*tan(d*
x + c) + 2*a^4*b + 8*a^2*b^3 + 12*b^5)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(tan(d*x + c)^2 + 1)^2))/d

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maple [B]  time = 0.54, size = 524, normalized size = 3.45 \[ \frac {b^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \left (\tan ^{3}\left (d x +c \right )\right ) a^{5}}{8 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {5 \left (\tan ^{3}\left (d x +c \right )\right ) b^{2} a^{3}}{4 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {7 \left (\tan ^{3}\left (d x +c \right )\right ) a \,b^{4}}{8 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {\left (\tan ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{2 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {\left (\tan ^{2}\left (d x +c \right )\right ) b^{5}}{2 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {7 \tan \left (d x +c \right ) b^{2} a^{3}}{4 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {9 \tan \left (d x +c \right ) a \,b^{4}}{8 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {5 \tan \left (d x +c \right ) a^{5}}{8 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {a^{4} b}{4 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {a^{2} b^{3}}{d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {3 b^{5}}{4 d \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {b^{5} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {15 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{4}}{8 d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \arctan \left (\tan \left (d x +c \right )\right ) a^{5}}{8 d \left (a^{2}+b^{2}\right )^{3}}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right ) b^{2} a^{3}}{4 d \left (a^{2}+b^{2}\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)),x)

[Out]

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

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maxima [A]  time = 0.47, size = 271, normalized size = 1.78 \[ \frac {\frac {8 \, b^{5} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {4 \, b^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {4 \, b^{3} \tan \left (d x + c\right )^{2} + {\left (3 \, a^{3} + 7 \, a b^{2}\right )} \tan \left (d x + c\right )^{3} + 2 \, a^{2} b + 6 \, b^{3} + {\left (5 \, a^{3} + 9 \, a b^{2}\right )} \tan \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{4} + a^{4} + 2 \, a^{2} b^{2} + b^{4} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*b^5*log(b*tan(d*x + c) + a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 4*b^5*log(tan(d*x + c)^2 + 1)/(a^6 +
3*a^4*b^2 + 3*a^2*b^4 + b^6) + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) +
(4*b^3*tan(d*x + c)^2 + (3*a^3 + 7*a*b^2)*tan(d*x + c)^3 + 2*a^2*b + 6*b^3 + (5*a^3 + 9*a*b^2)*tan(d*x + c))/
((a^4 + 2*a^2*b^2 + b^4)*tan(d*x + c)^4 + a^4 + 2*a^2*b^2 + b^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*tan(d*x + c)^2))/d

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mupad [B]  time = 4.20, size = 318, normalized size = 2.09 \[ \frac {\frac {a^2\,b+3\,b^3}{4\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^3+7\,a\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (5\,a^3+9\,a\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-a^2\,3{}\mathrm {i}+9\,a\,b+b^2\,8{}\mathrm {i}\right )}{16\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )}+\frac {b^5\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,{\left (a^2+b^2\right )}^3}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-3\,a^2+a\,b\,9{}\mathrm {i}+8\,b^2\right )}{16\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*b + 3*b^3)/(4*(a^4 + b^4 + 2*a^2*b^2)) + (b^3*tan(c + d*x)^2)/(2*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x
)^3*(7*a*b^2 + 3*a^3))/(8*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x)*(9*a*b^2 + 5*a^3))/(8*(a^4 + b^4 + 2*a^2*b^
2)))/(d*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1)) - (log(tan(c + d*x) - 1i)*(9*a*b - a^2*3i + b^2*8i))/(16*d*(3
*a*b^2 - a^2*b*3i - a^3 + b^3*1i)) + (b^5*log(a + b*tan(c + d*x)))/(d*(a^2 + b^2)^3) - (log(tan(c + d*x) + 1i)
*(a*b*9i - 3*a^2 + 8*b^2))/(16*d*(a*b^2*3i - 3*a^2*b - a^3*1i + b^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{4}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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