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

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

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Rubi [A]  time = 0.54, antiderivative size = 197, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 5.41, 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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fricas [B]  time = 2.02, size = 580, normalized size = 2.94 \[ -\frac {2 \, a^{8} + 7 \, a^{6} b^{2} + 66 \, a^{4} b^{4} - 27 \, a^{2} b^{6} + {\left (21 \, a^{7} b - 56 \, a^{5} b^{3} + 11 \, a^{3} b^{5} - 6 \, {\left (a^{8} - 10 \, a^{6} b^{2} + 5 \, a^{4} b^{4}\right )} e x\right )} \tan \left (e x + d\right )^{3} - 6 \, {\left (a^{5} b^{3} - 10 \, a^{3} b^{5} + 5 \, a b^{7}\right )} e x - 3 \, {\left (2 \, a^{8} - 31 \, a^{6} b^{2} + 46 \, a^{4} b^{4} - 9 \, a^{2} b^{6} + 6 \, {\left (a^{7} b - 10 \, a^{5} b^{3} + 5 \, a^{3} b^{5}\right )} e x\right )} \tan \left (e x + d\right )^{2} + 3 \, {\left (5 \, a^{4} b^{4} - 10 \, a^{2} b^{6} + b^{8} + {\left (5 \, a^{7} b - 10 \, a^{5} b^{3} + a^{3} b^{5}\right )} \tan \left (e x + d\right )^{3} + 3 \, {\left (5 \, a^{6} b^{2} - 10 \, a^{4} b^{4} + a^{2} b^{6}\right )} \tan \left (e x + d\right )^{2} + 3 \, {\left (5 \, a^{5} b^{3} - 10 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (e x + d\right )\right )} \log \left (\frac {a^{2} \tan \left (e x + d\right )^{2} + 2 \, a b \tan \left (e x + d\right ) + b^{2}}{\tan \left (e x + d\right )^{2} + 1}\right ) - 3 \, {\left (a^{7} b - 46 \, a^{5} b^{3} + 35 \, a^{3} b^{5} - 6 \, a b^{7} + 6 \, {\left (a^{6} b^{2} - 10 \, a^{4} b^{4} + 5 \, a^{2} b^{6}\right )} e x\right )} \tan \left (e x + d\right )}{6 \, {\left ({\left (a^{11} + 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} + 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} e \tan \left (e x + d\right )^{3} + 3 \, {\left (a^{10} b + 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} + 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} e \tan \left (e x + d\right )^{2} + 3 \, {\left (a^{9} b^{2} + 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} + 4 \, a^{3} b^{8} + a b^{10}\right )} e \tan \left (e x + d\right ) + {\left (a^{8} b^{3} + 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} + 4 \, a^{2} b^{9} + b^{11}\right )} e\right )}} \]

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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giac [B]  time = 1.94, size = 451, normalized size = 2.29 \[ \frac {1}{6} \, {\left (\frac {6 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} {\left (x e + d\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {3 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (x e + d\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {6 \, {\left (5 \, a^{5} b - 10 \, a^{3} b^{3} + a b^{5}\right )} \log \left ({\left | a \tan \left (x e + d\right ) + b \right |}\right )}{a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}} + \frac {55 \, a^{7} b \tan \left (x e + d\right )^{3} - 110 \, a^{5} b^{3} \tan \left (x e + d\right )^{3} + 11 \, a^{3} b^{5} \tan \left (x e + d\right )^{3} + 6 \, a^{8} \tan \left (x e + d\right )^{2} + 135 \, a^{6} b^{2} \tan \left (x e + d\right )^{2} - 360 \, a^{4} b^{4} \tan \left (x e + d\right )^{2} + 39 \, a^{2} b^{6} \tan \left (x e + d\right )^{2} + 3 \, a^{7} b \tan \left (x e + d\right ) + 90 \, a^{5} b^{3} \tan \left (x e + d\right ) - 393 \, a^{3} b^{5} \tan \left (x e + d\right ) + 48 \, a b^{7} \tan \left (x e + d\right ) - 2 \, a^{8} - 7 \, a^{6} b^{2} + 10 \, a^{4} b^{4} - 139 \, a^{2} b^{6} + 22 \, b^{8}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (a \tan \left (x e + d\right ) + b\right )}^{3}}\right )} e^{\left (-1\right )} \]

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)

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maple [B]  time = 0.18, size = 458, normalized size = 2.32 \[ -\frac {a^{2}}{3 e \left (a^{2}+b^{2}\right ) \left (b +a \tan \left (e x +d \right )\right )^{3}}+\frac {b^{2}}{3 e \left (a^{2}+b^{2}\right ) \left (b +a \tan \left (e x +d \right )\right )^{3}}-\frac {3 b \,a^{2}}{2 e \left (a^{2}+b^{2}\right )^{2} \left (b +a \tan \left (e x +d \right )\right )^{2}}+\frac {b^{3}}{2 e \left (a^{2}+b^{2}\right )^{2} \left (b +a \tan \left (e x +d \right )\right )^{2}}+\frac {a^{4}}{e \left (a^{2}+b^{2}\right )^{3} \left (b +a \tan \left (e x +d \right )\right )}-\frac {6 a^{2} b^{2}}{e \left (a^{2}+b^{2}\right )^{3} \left (b +a \tan \left (e x +d \right )\right )}+\frac {b^{4}}{e \left (a^{2}+b^{2}\right )^{3} \left (b +a \tan \left (e x +d \right )\right )}-\frac {5 b \ln \left (b +a \tan \left (e x +d \right )\right ) a^{4}}{e \left (a^{2}+b^{2}\right )^{4}}+\frac {10 b^{3} \ln \left (b +a \tan \left (e x +d \right )\right ) a^{2}}{e \left (a^{2}+b^{2}\right )^{4}}-\frac {b^{5} \ln \left (b +a \tan \left (e x +d \right )\right )}{e \left (a^{2}+b^{2}\right )^{4}}+\frac {5 \ln \left (1+\tan ^{2}\left (e x +d \right )\right ) a^{4} b}{2 e \left (a^{2}+b^{2}\right )^{4}}-\frac {5 \ln \left (1+\tan ^{2}\left (e x +d \right )\right ) a^{2} b^{3}}{e \left (a^{2}+b^{2}\right )^{4}}+\frac {\ln \left (1+\tan ^{2}\left (e x +d \right )\right ) b^{5}}{2 e \left (a^{2}+b^{2}\right )^{4}}+\frac {\arctan \left (\tan \left (e x +d \right )\right ) a^{5}}{e \left (a^{2}+b^{2}\right )^{4}}-\frac {10 \arctan \left (\tan \left (e x +d \right )\right ) a^{3} b^{2}}{e \left (a^{2}+b^{2}\right )^{4}}+\frac {5 \arctan \left (\tan \left (e x +d \right )\right ) a \,b^{4}}{e \left (a^{2}+b^{2}\right )^{4}} \]

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 = 0.43, size = 419, normalized size = 2.13 \[ \frac {\frac {6 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} {\left (e x + d\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {6 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \tan \left (e x + d\right ) + b\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {3 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {2 \, a^{6} + 5 \, a^{4} b^{2} + 40 \, a^{2} b^{4} - 11 \, b^{6} - 6 \, {\left (a^{6} - 6 \, a^{4} b^{2} + a^{2} b^{4}\right )} \tan \left (e x + d\right )^{2} - 3 \, {\left (a^{5} b - 26 \, a^{3} b^{3} + 5 \, a b^{5}\right )} \tan \left (e x + d\right )}{a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9} + {\left (a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6}\right )} \tan \left (e x + d\right )^{3} + 3 \, {\left (a^{8} b + 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} + a^{2} b^{7}\right )} \tan \left (e x + d\right )^{2} + 3 \, {\left (a^{7} b^{2} + 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (e x + d\right )}}{6 \, e} \]

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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mupad [B]  time = 3.26, size = 388, normalized size = 1.97 \[ \frac {\frac {{\mathrm {tan}\left (d+e\,x\right )}^2\,\left (a^6-6\,a^4\,b^2+a^2\,b^4\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}-\frac {2\,a^6+5\,a^4\,b^2+40\,a^2\,b^4-11\,b^6}{6\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {\mathrm {tan}\left (d+e\,x\right )\,\left (a^5\,b-26\,a^3\,b^3+5\,a\,b^5\right )}{2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}}{e\,\left (a^3\,{\mathrm {tan}\left (d+e\,x\right )}^3+3\,a^2\,b\,{\mathrm {tan}\left (d+e\,x\right )}^2+3\,a\,b^2\,\mathrm {tan}\left (d+e\,x\right )+b^3\right )}-\frac {\ln \left (b+a\,\mathrm {tan}\left (d+e\,x\right )\right )\,\left (\frac {5\,b}{{\left (a^2+b^2\right )}^2}-\frac {20\,b^3}{{\left (a^2+b^2\right )}^3}+\frac {16\,b^5}{{\left (a^2+b^2\right )}^4}\right )}{e}+\frac {\ln \left (\mathrm {tan}\left (d+e\,x\right )-\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,e\,\left (a^4\,1{}\mathrm {i}+4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}-4\,a\,b^3+b^4\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (d+e\,x\right )+1{}\mathrm {i}\right )\,\left (a-b\,1{}\mathrm {i}\right )}{2\,e\,\left (a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(d + e*x)^2*(a^6 + a^2*b^4 - 6*a^4*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (2*a^6 - 11*b^6 + 40*a^2*b
^4 + 5*a^4*b^2)/(6*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (tan(d + e*x)*(5*a*b^5 + a^5*b - 26*a^3*b^3))/(2*(a^
6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))/(e*(b^3 + a^3*tan(d + e*x)^3 + 3*a^2*b*tan(d + e*x)^2 + 3*a*b^2*tan(d + e*x
))) - (log(b + a*tan(d + e*x))*((5*b)/(a^2 + b^2)^2 - (20*b^3)/(a^2 + b^2)^3 + (16*b^5)/(a^2 + b^2)^4))/e + (l
og(tan(d + e*x) - 1i)*(a + b*1i))/(2*e*(4*a^3*b - 4*a*b^3 + a^4*1i + b^4*1i - a^2*b^2*6i)) - (log(tan(d + e*x)
+ 1i)*(a - b*1i))/(2*e*(4*a*b^3 - 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i))

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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((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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