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

Optimal. Leaf size=169 \[ -\frac{a^2}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^3}+\frac{a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}+\frac{b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac{4 a b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}-\frac{x \left (-6 a^2 b^2+a^4+b^4\right )}{\left (a^2+b^2\right )^4} \]

[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) - a^2/(3*b*(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]))

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Rubi [A]  time = 0.276632, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3542, 3529, 3531, 3530} \[ -\frac{a^2}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^3}+\frac{a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}+\frac{b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac{4 a b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}-\frac{x \left (-6 a^2 b^2+a^4+b^4\right )}{\left (a^2+b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2/(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) - a^2/(3*b*(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 3542

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

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 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]

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]

Rubi steps

\begin{align*} \int \frac{\tan ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx &=-\frac{a^2}{3 b \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{a^2}{3 b \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{a^2}{3 b \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{a^2}{3 b \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{a^2}{3 b \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*}

Mathematica [C]  time = 6.19019, size = 324, normalized size = 1.92 \[ \frac{b^2 \tan ^3(c+d x)}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^3}+\frac{\frac{3 a \tan (c+d x)}{d (a+b \tan (c+d x))^2}-\frac{3 a \left (\frac{2 b}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{4 a b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2}-2 a \left (\frac{4 a b}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{2 b \left (3 a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3}+\frac{\log (-\tan (c+d x)+i)}{(-b+i a)^3}-\frac{\log (\tan (c+d x)+i)}{(b+i a)^3}\right )+\frac{i \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac{i \log (\tan (c+d x)+i)}{(a-i b)^2}\right )}{2 d}}{3 a \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*Tan[c + d*x]^3)/(3*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^3) + ((3*a*Tan[c + d*x])/(d*(a + b*Tan[c + d*x])^
2) - (3*a*((I*Log[I - Tan[c + d*x]])/(a + I*b)^2 - (I*Log[I + Tan[c + d*x]])/(a - I*b)^2 - (4*a*b*Log[a + b*Ta
n[c + d*x]])/(a^2 + b^2)^2 + (2*b)/((a^2 + b^2)*(a + b*Tan[c + d*x])) - 2*a*(Log[I - Tan[c + d*x]]/(I*a - b)^3
- Log[I + Tan[c + d*x]]/(I*a + b)^3 - (2*b*(3*a^2 - b^2)*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^3 + b/((a^2 + b
^2)*(a + b*Tan[c + d*x])^2) + (4*a*b)/((a^2 + b^2)^2*(a + b*Tan[c + d*x])))))/(2*d))/(3*a*(a^2 + b^2))

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Maple [A]  time = 0.036, size = 311, normalized size = 1.8 \begin{align*} 2\,{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{3}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-2\,{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) a{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{4}}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+6\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{4}}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-{\frac{{a}^{2}}{3\,b \left ({a}^{2}+{b}^{2} \right ) d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}}+{\frac{ab}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{b{a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-4\,{\frac{b{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+4\,{\frac{{b}^{3}a\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.61005, size = 525, normalized size = 3.11 \begin{align*} -\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{a^{6} - 10 \, a^{4} b^{2} + a^{2} b^{4} - 3 \,{\left (3 \, a^{2} b^{4} - b^{6}\right )} \tan \left (d x + c\right )^{2} - 3 \,{\left (7 \, a^{3} b^{3} - a b^{5}\right )} \tan \left (d x + c\right )}{a^{9} b + 3 \, a^{7} b^{3} + 3 \, a^{5} b^{5} + a^{3} b^{7} +{\left (a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}\right )} \tan \left (d x + c\right )^{3} + 3 \,{\left (a^{7} b^{3} + 3 \, a^{5} b^{5} + 3 \, a^{3} b^{7} + a b^{9}\right )} \tan \left (d x + c\right )^{2} + 3 \,{\left (a^{8} b^{2} + 3 \, a^{6} b^{4} + 3 \, a^{4} b^{6} + a^{2} b^{8}\right )} \tan \left (d x + c\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [B]  time = 2.00467, size = 1130, normalized size = 6.69 \begin{align*} -\frac{6 \, a^{6} b - 15 \, a^{4} b^{3} + a^{2} b^{5} -{\left (a^{5} b^{2} - 15 \, a^{3} b^{4} + 6 \, 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 (a^{6} b - 12 \, a^{4} b^{3} + 8 \, 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 (a^{7} - 8 \, a^{5} b^{2} + 12 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*a^6*b - 15*a^4*b^3 + a^2*b^5 - (a^5*b^2 - 15*a^3*b^4 + 6*a*b^6 - 3*(a^4*b^3 - 6*a^2*b^5 + b^7)*d*x)*ta
n(d*x + c)^3 + 3*(a^7 - 6*a^5*b^2 + a^3*b^4)*d*x - 3*(a^6*b - 12*a^4*b^3 + 8*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*(a^7 - 8*a^5*b^2 + 12*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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.57182, size = 508, normalized size = 3.01 \begin{align*} -\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^{5} \tan \left (d x + c\right )^{3} - 22 \, a b^{7} \tan \left (d x + c\right )^{3} + 75 \, a^{4} b^{4} \tan \left (d x + c\right )^{2} - 60 \, a^{2} b^{6} \tan \left (d x + c\right )^{2} - 3 \, b^{8} \tan \left (d x + c\right )^{2} + 87 \, a^{5} b^{3} \tan \left (d x + c\right ) - 48 \, a^{3} b^{5} \tan \left (d x + c\right ) - 3 \, a b^{7} \tan \left (d x + c\right ) - a^{8} + 31 \, a^{6} b^{2} - 13 \, a^{4} b^{4} - a^{2} b^{6}}{{\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(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)*
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) + 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^5*tan(d*x + c)^3 - 22*a*b^7*tan(d*
x + c)^3 + 75*a^4*b^4*tan(d*x + c)^2 - 60*a^2*b^6*tan(d*x + c)^2 - 3*b^8*tan(d*x + c)^2 + 87*a^5*b^3*tan(d*x +
c) - 48*a^3*b^5*tan(d*x + c) - 3*a*b^7*tan(d*x + c) - a^8 + 31*a^6*b^2 - 13*a^4*b^4 - a^2*b^6)/((a^8*b + 4*a^
6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9)*(b*tan(d*x + c) + a)^3))/d