3.513 \(\int \frac{a+b \tan (d+e x)}{(b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x))^2} \, dx\)

Optimal. Leaf size=197 \[ -\frac{a^2-b^2}{3 e \left (a^2+b^2\right ) (a \tan (d+e x)+b)^3}+\frac{-6 a^2 b^2+a^4+b^4}{e \left (a^2+b^2\right )^3 (a \tan (d+e x)+b)}-\frac{b \left (3 a^2-b^2\right )}{2 e \left (a^2+b^2\right )^2 (a \tan (d+e x)+b)^2}-\frac{b \left (-10 a^2 b^2+5 a^4+b^4\right ) \log (a \sin (d+e x)+b \cos (d+e x))}{e \left (a^2+b^2\right )^4}+\frac{a x \left (-10 a^2 b^2+a^4+5 b^4\right )}{\left (a^2+b^2\right )^4} \]

[Out]

(a*(a^4 - 10*a^2*b^2 + 5*b^4)*x)/(a^2 + b^2)^4 - (b*(5*a^4 - 10*a^2*b^2 + b^4)*Log[b*Cos[d + e*x] + a*Sin[d +
e*x]])/((a^2 + b^2)^4*e) - (a^2 - b^2)/(3*(a^2 + b^2)*e*(b + a*Tan[d + e*x])^3) - (b*(3*a^2 - b^2))/(2*(a^2 +
b^2)^2*e*(b + a*Tan[d + e*x])^2) + (a^4 - 6*a^2*b^2 + b^4)/((a^2 + b^2)^3*e*(b + a*Tan[d + e*x]))

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Rubi [A]  time = 0.535338, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3708, 3529, 3531, 3530} \[ -\frac{a^2-b^2}{3 e \left (a^2+b^2\right ) (a \tan (d+e x)+b)^3}+\frac{-6 a^2 b^2+a^4+b^4}{e \left (a^2+b^2\right )^3 (a \tan (d+e x)+b)}-\frac{b \left (3 a^2-b^2\right )}{2 e \left (a^2+b^2\right )^2 (a \tan (d+e x)+b)^2}-\frac{b \left (-10 a^2 b^2+5 a^4+b^4\right ) \log (a \sin (d+e x)+b \cos (d+e x))}{e \left (a^2+b^2\right )^4}+\frac{a x \left (-10 a^2 b^2+a^4+5 b^4\right )}{\left (a^2+b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^4 - 10*a^2*b^2 + 5*b^4)*x)/(a^2 + b^2)^4 - (b*(5*a^4 - 10*a^2*b^2 + b^4)*Log[b*Cos[d + e*x] + a*Sin[d +
e*x]])/((a^2 + b^2)^4*e) - (a^2 - b^2)/(3*(a^2 + b^2)*e*(b + a*Tan[d + e*x])^3) - (b*(3*a^2 - b^2))/(2*(a^2 +
b^2)^2*e*(b + a*Tan[d + e*x])^2) + (a^4 - 6*a^2*b^2 + b^4)/((a^2 + b^2)^3*e*(b + a*Tan[d + e*x]))

Rule 3708

Int[((A_) + (B_.)*tan[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*tan[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Tan[d + e*x])*(b + 2*c*Tan[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

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{a+b \tan (d+e x)}{\left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^2} \, dx &=\left (16 a^4\right ) \int \frac{a+b \tan (d+e x)}{\left (2 a b+2 a^2 \tan (d+e x)\right )^4} \, dx\\ &=-\frac{a^2-b^2}{3 \left (a^2+b^2\right ) e (b+a \tan (d+e x))^3}+\frac{\left (4 a^2\right ) \int \frac{4 a^2 b-2 a \left (a^2-b^2\right ) \tan (d+e x)}{\left (2 a b+2 a^2 \tan (d+e x)\right )^3} \, dx}{a^2+b^2}\\ &=-\frac{a^2-b^2}{3 \left (a^2+b^2\right ) e (b+a \tan (d+e x))^3}-\frac{b \left (3 a^2-b^2\right )}{2 \left (a^2+b^2\right )^2 e (b+a \tan (d+e x))^2}+\frac{\int \frac{-4 a^3 \left (a^2-3 b^2\right )-4 a^2 b \left (3 a^2-b^2\right ) \tan (d+e x)}{\left (2 a b+2 a^2 \tan (d+e x)\right )^2} \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac{a^2-b^2}{3 \left (a^2+b^2\right ) e (b+a \tan (d+e x))^3}-\frac{b \left (3 a^2-b^2\right )}{2 \left (a^2+b^2\right )^2 e (b+a \tan (d+e x))^2}+\frac{a^4-6 a^2 b^2+b^4}{\left (a^2+b^2\right )^3 e (b+a \tan (d+e x))}+\frac{\int \frac{-32 a^4 b \left (a^2-b^2\right )+8 a^3 \left (a^4-6 a^2 b^2+b^4\right ) \tan (d+e x)}{2 a b+2 a^2 \tan (d+e x)} \, dx}{4 a^2 \left (a^2+b^2\right )^3}\\ &=\frac{a \left (a^4-10 a^2 b^2+5 b^4\right ) x}{\left (a^2+b^2\right )^4}-\frac{a^2-b^2}{3 \left (a^2+b^2\right ) e (b+a \tan (d+e x))^3}-\frac{b \left (3 a^2-b^2\right )}{2 \left (a^2+b^2\right )^2 e (b+a \tan (d+e x))^2}+\frac{a^4-6 a^2 b^2+b^4}{\left (a^2+b^2\right )^3 e (b+a \tan (d+e x))}-\frac{\left (b \left (5 a^4-10 a^2 b^2+b^4\right )\right ) \int \frac{2 a^2-2 a b \tan (d+e x)}{2 a b+2 a^2 \tan (d+e x)} \, dx}{\left (a^2+b^2\right )^4}\\ &=\frac{a \left (a^4-10 a^2 b^2+5 b^4\right ) x}{\left (a^2+b^2\right )^4}-\frac{b \left (5 a^4-10 a^2 b^2+b^4\right ) \log (b \cos (d+e x)+a \sin (d+e x))}{\left (a^2+b^2\right )^4 e}-\frac{a^2-b^2}{3 \left (a^2+b^2\right ) e (b+a \tan (d+e x))^3}-\frac{b \left (3 a^2-b^2\right )}{2 \left (a^2+b^2\right )^2 e (b+a \tan (d+e x))^2}+\frac{a^4-6 a^2 b^2+b^4}{\left (a^2+b^2\right )^3 e (b+a \tan (d+e x))}\\ \end{align*}

Mathematica [C]  time = 4.89778, size = 308, normalized size = 1.56 \[ \frac{3 b \left (\frac{a \left (-\frac{\left (a^2+b^2\right ) \left (a^2+4 a b \tan (d+e x)+5 b^2\right )}{(a \tan (d+e x)+b)^2}-2 \left (a^2-3 b^2\right ) \log (a \tan (d+e x)+b)\right )}{\left (a^2+b^2\right )^3}+\frac{\log (-\tan (d+e x)+i)}{(a-i b)^3}+\frac{\log (\tan (d+e x)+i)}{(a+i b)^3}\right )-(a-b) (a+b) \left (-\frac{6 a \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3 (a \tan (d+e x)+b)}+\frac{6 a b}{\left (a^2+b^2\right )^2 (a \tan (d+e x)+b)^2}+\frac{2 a}{\left (a^2+b^2\right ) (a \tan (d+e x)+b)^3}+\frac{24 a b (a-b) (a+b) \log (a \tan (d+e x)+b)}{\left (a^2+b^2\right )^4}+\frac{3 i \log (-\tan (d+e x)+i)}{(a-i b)^4}-\frac{3 i \log (\tan (d+e x)+i)}{(a+i b)^4}\right )}{6 a e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a - b)*(a + b)*(((3*I)*Log[I - Tan[d + e*x]])/(a - I*b)^4 - ((3*I)*Log[I + Tan[d + e*x]])/(a + I*b)^4 + (2
4*a*(a - b)*b*(a + b)*Log[b + a*Tan[d + e*x]])/(a^2 + b^2)^4 + (2*a)/((a^2 + b^2)*(b + a*Tan[d + e*x])^3) + (6
*a*b)/((a^2 + b^2)^2*(b + a*Tan[d + e*x])^2) - (6*a*(a^2 - 3*b^2))/((a^2 + b^2)^3*(b + a*Tan[d + e*x])))) + 3*
b*(Log[I - Tan[d + e*x]]/(a - I*b)^3 + Log[I + Tan[d + e*x]]/(a + I*b)^3 + (a*(-2*(a^2 - 3*b^2)*Log[b + a*Tan[
d + e*x]] - ((a^2 + b^2)*(a^2 + 5*b^2 + 4*a*b*Tan[d + e*x]))/(b + a*Tan[d + e*x])^2))/(a^2 + b^2)^3))/(6*a*e)

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Maple [B]  time = 0.059, size = 458, normalized size = 2.3 \begin{align*} -{\frac{{a}^{2}}{3\,e \left ({a}^{2}+{b}^{2} \right ) \left ( b+a\tan \left ( ex+d \right ) \right ) ^{3}}}+{\frac{{b}^{2}}{3\,e \left ({a}^{2}+{b}^{2} \right ) \left ( b+a\tan \left ( ex+d \right ) \right ) ^{3}}}-{\frac{3\,{a}^{2}b}{2\,e \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( b+a\tan \left ( ex+d \right ) \right ) ^{2}}}+{\frac{{b}^{3}}{2\,e \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( b+a\tan \left ( ex+d \right ) \right ) ^{2}}}+{\frac{{a}^{4}}{e \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( b+a\tan \left ( ex+d \right ) \right ) }}-6\,{\frac{{a}^{2}{b}^{2}}{e \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( b+a\tan \left ( ex+d \right ) \right ) }}+{\frac{{b}^{4}}{e \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( b+a\tan \left ( ex+d \right ) \right ) }}-5\,{\frac{b\ln \left ( b+a\tan \left ( ex+d \right ) \right ){a}^{4}}{e \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+10\,{\frac{{b}^{3}\ln \left ( b+a\tan \left ( ex+d \right ) \right ){a}^{2}}{e \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-{\frac{{b}^{5}\ln \left ( b+a\tan \left ( ex+d \right ) \right ) }{e \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+{\frac{5\,\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){a}^{4}b}{2\,e \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-5\,{\frac{\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){a}^{2}{b}^{3}}{e \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){b}^{5}}{2\,e \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ){a}^{5}}{e \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-10\,{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ){a}^{3}{b}^{2}}{e \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+5\,{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ) a{b}^{4}}{e \left ({a}^{2}+{b}^{2} \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/e/(a^2+b^2)/(b+a*tan(e*x+d))^3*a^2+1/3/e/(a^2+b^2)/(b+a*tan(e*x+d))^3*b^2-3/2/e*b/(a^2+b^2)^2/(b+a*tan(e*
x+d))^2*a^2+1/2/e*b^3/(a^2+b^2)^2/(b+a*tan(e*x+d))^2+1/e/(a^2+b^2)^3/(b+a*tan(e*x+d))*a^4-6/e/(a^2+b^2)^3/(b+a
*tan(e*x+d))*a^2*b^2+1/e/(a^2+b^2)^3/(b+a*tan(e*x+d))*b^4-5/e*b/(a^2+b^2)^4*ln(b+a*tan(e*x+d))*a^4+10/e*b^3/(a
^2+b^2)^4*ln(b+a*tan(e*x+d))*a^2-1/e*b^5/(a^2+b^2)^4*ln(b+a*tan(e*x+d))+5/2/e/(a^2+b^2)^4*ln(1+tan(e*x+d)^2)*a
^4*b-5/e/(a^2+b^2)^4*ln(1+tan(e*x+d)^2)*a^2*b^3+1/2/e/(a^2+b^2)^4*ln(1+tan(e*x+d)^2)*b^5+1/e/(a^2+b^2)^4*arcta
n(tan(e*x+d))*a^5-10/e/(a^2+b^2)^4*arctan(tan(e*x+d))*a^3*b^2+5/e/(a^2+b^2)^4*arctan(tan(e*x+d))*a*b^4

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Maxima [B]  time = 1.59177, size = 566, normalized size = 2.87 \begin{align*} \frac{\frac{6 \,{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )}{\left (e x + d\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{6 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \tan \left (e x + d\right ) + b\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac{3 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{2 \, a^{6} + 5 \, a^{4} b^{2} + 40 \, a^{2} b^{4} - 11 \, b^{6} - 6 \,{\left (a^{6} - 6 \, a^{4} b^{2} + a^{2} b^{4}\right )} \tan \left (e x + d\right )^{2} - 3 \,{\left (a^{5} b - 26 \, a^{3} b^{3} + 5 \, a b^{5}\right )} \tan \left (e x + d\right )}{a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9} +{\left (a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6}\right )} \tan \left (e x + d\right )^{3} + 3 \,{\left (a^{8} b + 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} + a^{2} b^{7}\right )} \tan \left (e x + d\right )^{2} + 3 \,{\left (a^{7} b^{2} + 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (e x + d\right )}}{6 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(a^5 - 10*a^3*b^2 + 5*a*b^4)*(e*x + d)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 6*(5*a^4*b - 1
0*a^2*b^3 + b^5)*log(a*tan(e*x + d) + b)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 3*(5*a^4*b - 10*a^2
*b^3 + b^5)*log(tan(e*x + d)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - (2*a^6 + 5*a^4*b^2 + 40*
a^2*b^4 - 11*b^6 - 6*(a^6 - 6*a^4*b^2 + a^2*b^4)*tan(e*x + d)^2 - 3*(a^5*b - 26*a^3*b^3 + 5*a*b^5)*tan(e*x + d
))/(a^6*b^3 + 3*a^4*b^5 + 3*a^2*b^7 + b^9 + (a^9 + 3*a^7*b^2 + 3*a^5*b^4 + a^3*b^6)*tan(e*x + d)^3 + 3*(a^8*b
+ 3*a^6*b^3 + 3*a^4*b^5 + a^2*b^7)*tan(e*x + d)^2 + 3*(a^7*b^2 + 3*a^5*b^4 + 3*a^3*b^6 + a*b^8)*tan(e*x + d)))
/e

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Fricas [B]  time = 2.07852, size = 1266, normalized size = 6.43 \begin{align*} -\frac{2 \, a^{8} + 7 \, a^{6} b^{2} + 66 \, a^{4} b^{4} - 27 \, a^{2} b^{6} +{\left (21 \, a^{7} b - 56 \, a^{5} b^{3} + 11 \, a^{3} b^{5} - 6 \,{\left (a^{8} - 10 \, a^{6} b^{2} + 5 \, a^{4} b^{4}\right )} e x\right )} \tan \left (e x + d\right )^{3} - 6 \,{\left (a^{5} b^{3} - 10 \, a^{3} b^{5} + 5 \, a b^{7}\right )} e x - 3 \,{\left (2 \, a^{8} - 31 \, a^{6} b^{2} + 46 \, a^{4} b^{4} - 9 \, a^{2} b^{6} + 6 \,{\left (a^{7} b - 10 \, a^{5} b^{3} + 5 \, a^{3} b^{5}\right )} e x\right )} \tan \left (e x + d\right )^{2} + 3 \,{\left (5 \, a^{4} b^{4} - 10 \, a^{2} b^{6} + b^{8} +{\left (5 \, a^{7} b - 10 \, a^{5} b^{3} + a^{3} b^{5}\right )} \tan \left (e x + d\right )^{3} + 3 \,{\left (5 \, a^{6} b^{2} - 10 \, a^{4} b^{4} + a^{2} b^{6}\right )} \tan \left (e x + d\right )^{2} + 3 \,{\left (5 \, a^{5} b^{3} - 10 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (e x + d\right )\right )} \log \left (\frac{a^{2} \tan \left (e x + d\right )^{2} + 2 \, a b \tan \left (e x + d\right ) + b^{2}}{\tan \left (e x + d\right )^{2} + 1}\right ) - 3 \,{\left (a^{7} b - 46 \, a^{5} b^{3} + 35 \, a^{3} b^{5} - 6 \, a b^{7} + 6 \,{\left (a^{6} b^{2} - 10 \, a^{4} b^{4} + 5 \, a^{2} b^{6}\right )} e x\right )} \tan \left (e x + d\right )}{6 \,{\left ({\left (a^{11} + 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} + 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} e \tan \left (e x + d\right )^{3} + 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 )} e \tan \left (e x + d\right )^{2} + 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 )} e \tan \left (e x + d\right ) +{\left (a^{8} b^{3} + 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} + 4 \, a^{2} b^{9} + b^{11}\right )} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*a^8 + 7*a^6*b^2 + 66*a^4*b^4 - 27*a^2*b^6 + (21*a^7*b - 56*a^5*b^3 + 11*a^3*b^5 - 6*(a^8 - 10*a^6*b^2
+ 5*a^4*b^4)*e*x)*tan(e*x + d)^3 - 6*(a^5*b^3 - 10*a^3*b^5 + 5*a*b^7)*e*x - 3*(2*a^8 - 31*a^6*b^2 + 46*a^4*b^4
- 9*a^2*b^6 + 6*(a^7*b - 10*a^5*b^3 + 5*a^3*b^5)*e*x)*tan(e*x + d)^2 + 3*(5*a^4*b^4 - 10*a^2*b^6 + b^8 + (5*a
^7*b - 10*a^5*b^3 + a^3*b^5)*tan(e*x + d)^3 + 3*(5*a^6*b^2 - 10*a^4*b^4 + a^2*b^6)*tan(e*x + d)^2 + 3*(5*a^5*b
^3 - 10*a^3*b^5 + a*b^7)*tan(e*x + d))*log((a^2*tan(e*x + d)^2 + 2*a*b*tan(e*x + d) + b^2)/(tan(e*x + d)^2 + 1
)) - 3*(a^7*b - 46*a^5*b^3 + 35*a^3*b^5 - 6*a*b^7 + 6*(a^6*b^2 - 10*a^4*b^4 + 5*a^2*b^6)*e*x)*tan(e*x + d))/((
a^11 + 4*a^9*b^2 + 6*a^7*b^4 + 4*a^5*b^6 + a^3*b^8)*e*tan(e*x + d)^3 + 3*(a^10*b + 4*a^8*b^3 + 6*a^6*b^5 + 4*a
^4*b^7 + a^2*b^9)*e*tan(e*x + d)^2 + 3*(a^9*b^2 + 4*a^7*b^4 + 6*a^5*b^6 + 4*a^3*b^8 + a*b^10)*e*tan(e*x + d) +
(a^8*b^3 + 4*a^6*b^5 + 6*a^4*b^7 + 4*a^2*b^9 + b^11)*e)

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.62634, size = 609, normalized size = 3.09 \begin{align*} \frac{1}{6} \,{\left (\frac{6 \,{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )}{\left (x e + d\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac{3 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (x e + d\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{6 \,{\left (5 \, a^{5} b - 10 \, a^{3} b^{3} + a b^{5}\right )} \log \left ({\left | a \tan \left (x e + d\right ) + b \right |}\right )}{a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}} + \frac{55 \, a^{7} b \tan \left (x e + d\right )^{3} - 110 \, a^{5} b^{3} \tan \left (x e + d\right )^{3} + 11 \, a^{3} b^{5} \tan \left (x e + d\right )^{3} + 6 \, a^{8} \tan \left (x e + d\right )^{2} + 135 \, a^{6} b^{2} \tan \left (x e + d\right )^{2} - 360 \, a^{4} b^{4} \tan \left (x e + d\right )^{2} + 39 \, a^{2} b^{6} \tan \left (x e + d\right )^{2} + 3 \, a^{7} b \tan \left (x e + d\right ) + 90 \, a^{5} b^{3} \tan \left (x e + d\right ) - 393 \, a^{3} b^{5} \tan \left (x e + d\right ) + 48 \, a b^{7} \tan \left (x e + d\right ) - 2 \, a^{8} - 7 \, a^{6} b^{2} + 10 \, a^{4} b^{4} - 139 \, a^{2} b^{6} + 22 \, b^{8}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )}{\left (a \tan \left (x e + d\right ) + b\right )}^{3}}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(a^5 - 10*a^3*b^2 + 5*a*b^4)*(x*e + d)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 3*(5*a^4*b - 1
0*a^2*b^3 + b^5)*log(tan(x*e + d)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 6*(5*a^5*b - 10*a^3
*b^3 + a*b^5)*log(abs(a*tan(x*e + d) + b))/(a^9 + 4*a^7*b^2 + 6*a^5*b^4 + 4*a^3*b^6 + a*b^8) + (55*a^7*b*tan(x
*e + d)^3 - 110*a^5*b^3*tan(x*e + d)^3 + 11*a^3*b^5*tan(x*e + d)^3 + 6*a^8*tan(x*e + d)^2 + 135*a^6*b^2*tan(x*
e + d)^2 - 360*a^4*b^4*tan(x*e + d)^2 + 39*a^2*b^6*tan(x*e + d)^2 + 3*a^7*b*tan(x*e + d) + 90*a^5*b^3*tan(x*e
+ d) - 393*a^3*b^5*tan(x*e + d) + 48*a*b^7*tan(x*e + d) - 2*a^8 - 7*a^6*b^2 + 10*a^4*b^4 - 139*a^2*b^6 + 22*b^
8)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(a*tan(x*e + d) + b)^3))*e^(-1)