3.492 \(\int \frac {1}{(a+b \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=165 \[ -\frac {b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac {a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}-\frac {b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^3}+\frac {4 a b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac {x \left (a^4-6 a^2 b^2+b^4\right )}{\left (a^2+b^2\right )^4} \]

[Out]

(a^4-6*a^2*b^2+b^4)*x/(a^2+b^2)^4+4*a*b*(a^2-b^2)*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^4/d-1/3*b/(a^2+b^2)/
d/(a+b*tan(d*x+c))^3-a*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^2-b*(3*a^2-b^2)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.23, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3483, 3529, 3531, 3530} \[ -\frac {b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac {a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}-\frac {b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^3}+\frac {4 a b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac {x \left (-6 a^2 b^2+a^4+b^4\right )}{\left (a^2+b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-4),x]

[Out]

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

Rule 3483

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rubi steps

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

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Mathematica [C]  time = 3.02, size = 176, normalized size = 1.07 \[ \frac {\frac {2 b \left (12 a \left (a^2-b^2\right ) \log (a+b \tan (c+d x))-\frac {\left (a^2+b^2\right ) \left (13 a^4+3 a b \left (7 a^2-b^2\right ) \tan (c+d x)+2 a^2 b^2+\left (9 a^2 b^2-3 b^4\right ) \tan ^2(c+d x)+b^4\right )}{(a+b \tan (c+d x))^3}\right )}{\left (a^2+b^2\right )^4}-\frac {3 i \log (-\tan (c+d x)+i)}{(a+i b)^4}+\frac {3 i \log (\tan (c+d x)+i)}{(a-i b)^4}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)*Log[I - Tan[c + d*x]])/(a + I*b)^4 + ((3*I)*Log[I + Tan[c + d*x]])/(a - I*b)^4 + (2*b*(12*a*(a^2 - b^
2)*Log[a + b*Tan[c + d*x]] - ((a^2 + b^2)*(13*a^4 + 2*a^2*b^2 + b^4 + 3*a*b*(7*a^2 - b^2)*Tan[c + d*x] + (9*a^
2*b^2 - 3*b^4)*Tan[c + d*x]^2))/(a + b*Tan[c + d*x])^3))/(a^2 + b^2)^4)/(6*d)

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fricas [B]  time = 0.49, size = 511, normalized size = 3.10 \[ -\frac {24 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7} - {\left (13 \, a^{3} b^{4} - 9 \, a b^{6} + 3 \, {\left (a^{4} b^{3} - 6 \, a^{2} b^{5} + b^{7}\right )} d x\right )} \tan \left (d x + c\right )^{3} - 3 \, {\left (a^{7} - 6 \, a^{5} b^{2} + a^{3} b^{4}\right )} d x - 3 \, {\left (10 \, a^{4} b^{3} - 11 \, a^{2} b^{5} + b^{7} + 3 \, {\left (a^{5} b^{2} - 6 \, a^{3} b^{4} + a b^{6}\right )} d x\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (a^{6} b - a^{4} b^{3} + {\left (a^{3} b^{4} - a b^{6}\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (a^{4} b^{3} - a^{2} b^{5}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{5} b^{2} - a^{3} b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \, {\left (6 \, a^{5} b^{2} - 15 \, a^{3} b^{4} + a b^{6} + 3 \, {\left (a^{6} b - 6 \, a^{4} b^{3} + a^{2} b^{5}\right )} d x\right )} \tan \left (d x + c\right )}{3 \, {\left ({\left (a^{8} b^{3} + 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} + 4 \, a^{2} b^{9} + b^{11}\right )} d \tan \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} + 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} + 4 \, a^{3} b^{8} + a b^{10}\right )} d \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b + 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} + 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \tan \left (d x + c\right ) + {\left (a^{11} + 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} + 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/3*(24*a^4*b^3 + 3*a^2*b^5 + b^7 - (13*a^3*b^4 - 9*a*b^6 + 3*(a^4*b^3 - 6*a^2*b^5 + b^7)*d*x)*tan(d*x + c)^3
- 3*(a^7 - 6*a^5*b^2 + a^3*b^4)*d*x - 3*(10*a^4*b^3 - 11*a^2*b^5 + b^7 + 3*(a^5*b^2 - 6*a^3*b^4 + a*b^6)*d*x)
*tan(d*x + c)^2 - 6*(a^6*b - a^4*b^3 + (a^3*b^4 - a*b^6)*tan(d*x + c)^3 + 3*(a^4*b^3 - a^2*b^5)*tan(d*x + c)^2
+ 3*(a^5*b^2 - a^3*b^4)*tan(d*x + c))*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1
)) - 3*(6*a^5*b^2 - 15*a^3*b^4 + a*b^6 + 3*(a^6*b - 6*a^4*b^3 + a^2*b^5)*d*x)*tan(d*x + c))/((a^8*b^3 + 4*a^6*
b^5 + 6*a^4*b^7 + 4*a^2*b^9 + b^11)*d*tan(d*x + c)^3 + 3*(a^9*b^2 + 4*a^7*b^4 + 6*a^5*b^6 + 4*a^3*b^8 + a*b^10
)*d*tan(d*x + c)^2 + 3*(a^10*b + 4*a^8*b^3 + 6*a^6*b^5 + 4*a^4*b^7 + a^2*b^9)*d*tan(d*x + c) + (a^11 + 4*a^9*b
^2 + 6*a^7*b^4 + 4*a^5*b^6 + a^3*b^8)*d)

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giac [B]  time = 1.01, size = 370, normalized size = 2.24 \[ \frac {\frac {3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {12 \, {\left (a^{3} b^{2} - a b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}} - \frac {22 \, a^{3} b^{4} \tan \left (d x + c\right )^{3} - 22 \, a b^{6} \tan \left (d x + c\right )^{3} + 75 \, a^{4} b^{3} \tan \left (d x + c\right )^{2} - 60 \, a^{2} b^{5} \tan \left (d x + c\right )^{2} - 3 \, b^{7} \tan \left (d x + c\right )^{2} + 87 \, a^{5} b^{2} \tan \left (d x + c\right ) - 48 \, a^{3} b^{4} \tan \left (d x + c\right ) - 3 \, a b^{6} \tan \left (d x + c\right ) + 35 \, a^{6} b - 7 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(a^4 - 6*a^2*b^2 + b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 6*(a^3*b - a*b^3)*l
og(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 12*(a^3*b^2 - a*b^4)*log(abs(b*tan(d*
x + c) + a))/(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9) - (22*a^3*b^4*tan(d*x + c)^3 - 22*a*b^6*tan(d*x
+ c)^3 + 75*a^4*b^3*tan(d*x + c)^2 - 60*a^2*b^5*tan(d*x + c)^2 - 3*b^7*tan(d*x + c)^2 + 87*a^5*b^2*tan(d*x +
c) - 48*a^3*b^4*tan(d*x + c) - 3*a*b^6*tan(d*x + c) + 35*a^6*b - 7*a^4*b^3 + 3*a^2*b^5 + b^7)/((a^8 + 4*a^6*b^
2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(b*tan(d*x + c) + a)^3))/d

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maple [A]  time = 0.18, size = 304, normalized size = 1.84 \[ -\frac {b}{3 \left (a^{2}+b^{2}\right ) d \left (a +b \tan \left (d x +c \right )\right )^{3}}-\frac {3 a^{2} b}{d \left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {b^{3}}{d \left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}-\frac {a b}{\left (a^{2}+b^{2}\right )^{2} d \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {4 b \,a^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{4}}-\frac {4 a \,b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{4}}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{4}}{d \left (a^{2}+b^{2}\right )^{4}}-\frac {6 \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b^{2}}{d \left (a^{2}+b^{2}\right )^{4}}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) b^{4}}{d \left (a^{2}+b^{2}\right )^{4}}-\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} b}{d \left (a^{2}+b^{2}\right )^{4}}+\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a \,b^{3}}{d \left (a^{2}+b^{2}\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c))^4,x)

[Out]

-1/3*b/(a^2+b^2)/d/(a+b*tan(d*x+c))^3-3/d*a^2/(a^2+b^2)^3*b/(a+b*tan(d*x+c))+1/d*b^3/(a^2+b^2)^3/(a+b*tan(d*x+
c))-a*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^2+4/d*b*a^3/(a^2+b^2)^4*ln(a+b*tan(d*x+c))-4/d*a*b^3/(a^2+b^2)^4*ln(a+b
*tan(d*x+c))+1/d/(a^2+b^2)^4*arctan(tan(d*x+c))*a^4-6/d/(a^2+b^2)^4*arctan(tan(d*x+c))*a^2*b^2+1/d/(a^2+b^2)^4
*arctan(tan(d*x+c))*b^4-2/d/(a^2+b^2)^4*ln(1+tan(d*x+c)^2)*a^3*b+2/d/(a^2+b^2)^4*ln(1+tan(d*x+c)^2)*a*b^3

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maxima [B]  time = 0.56, size = 385, normalized size = 2.33 \[ \frac {\frac {3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {12 \, {\left (a^{3} b - a b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {13 \, a^{4} b + 2 \, a^{2} b^{3} + b^{5} + 3 \, {\left (3 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (7 \, a^{3} b^{2} - a b^{4}\right )} \tan \left (d x + c\right )}{a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6} + {\left (a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (a^{7} b^{2} + 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{8} b + 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} + a^{2} b^{7}\right )} \tan \left (d x + c\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/3*(3*(a^4 - 6*a^2*b^2 + b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 12*(a^3*b - a*b^3)*
log(b*tan(d*x + c) + a)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 6*(a^3*b - a*b^3)*log(tan(d*x + c)^2
+ 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - (13*a^4*b + 2*a^2*b^3 + b^5 + 3*(3*a^2*b^3 - b^5)*tan(
d*x + c)^2 + 3*(7*a^3*b^2 - a*b^4)*tan(d*x + c))/(a^9 + 3*a^7*b^2 + 3*a^5*b^4 + a^3*b^6 + (a^6*b^3 + 3*a^4*b^5
+ 3*a^2*b^7 + b^9)*tan(d*x + c)^3 + 3*(a^7*b^2 + 3*a^5*b^4 + 3*a^3*b^6 + a*b^8)*tan(d*x + c)^2 + 3*(a^8*b + 3
*a^6*b^3 + 3*a^4*b^5 + a^2*b^7)*tan(d*x + c)))/d

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mupad [B]  time = 4.28, size = 333, normalized size = 2.02 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {4\,a\,b}{{\left (a^2+b^2\right )}^3}-\frac {8\,a\,b^3}{{\left (a^2+b^2\right )}^4}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{2\,d\,\left (a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}\right )}+\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (b^5-3\,a^2\,b^3\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}-\frac {13\,a^4\,b+2\,a^2\,b^3+b^5}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a\,b^4-7\,a^3\,b^2\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}}{d\,\left (a^3+3\,a^2\,b\,\mathrm {tan}\left (c+d\,x\right )+3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2+b^3\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a^4-a^3\,b\,4{}\mathrm {i}-6\,a^2\,b^2+a\,b^3\,4{}\mathrm {i}+b^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(c + d*x))^4,x)

[Out]

(log(a + b*tan(c + d*x))*((4*a*b)/(a^2 + b^2)^3 - (8*a*b^3)/(a^2 + b^2)^4))/d + log(tan(c + d*x) - 1i)/(2*d*(4
*a*b^3 - 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i)) + (log(tan(c + d*x) + 1i)*1i)/(2*d*(a*b^3*4i - a^3*b*4i + a^
4 + b^4 - 6*a^2*b^2)) + ((tan(c + d*x)^2*(b^5 - 3*a^2*b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (13*a^4*b +
b^5 + 2*a^2*b^3)/(3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (tan(c + d*x)*(a*b^4 - 7*a^3*b^2))/(a^6 + b^6 + 3*a
^2*b^4 + 3*a^4*b^2))/(d*(a^3 + b^3*tan(c + d*x)^3 + 3*a*b^2*tan(c + d*x)^2 + 3*a^2*b*tan(c + d*x)))

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: AttributeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))**4,x)

[Out]

Exception raised: AttributeError

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