3.517 \(\int \frac{a+b \tan (d+e x)}{(b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x))^{3/2}} \, dx\)

Optimal. Leaf size=316 \[ -\frac{\left (a^2-b^2\right ) (a \tan (d+e x)+b)}{2 e \left (a^2+b^2\right ) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}-\frac{4 b x \left (a^2-b^2\right ) \left (a^2 \tan (d+e x)+a b\right )^3}{a^2 \left (a^2+b^2\right )^3 \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}-\frac{b \left (3 a^2-b^2\right ) \left (a^2 \tan (d+e x)+a b\right )^3}{e \left (a^2+b^2\right )^2 \left (a^3 b+a^4 \tan (d+e x)\right ) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}-\frac{\left (-6 a^2 b^2+a^4+b^4\right ) (a \tan (d+e x)+b)^3 \log (a \sin (d+e x)+b \cos (d+e x))}{e \left (a^2+b^2\right )^3 \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.401571, antiderivative size = 316, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {3710, 3529, 3531, 3530} \[ -\frac{\left (a^2-b^2\right ) (a \tan (d+e x)+b)}{2 e \left (a^2+b^2\right ) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}-\frac{4 b x \left (a^2-b^2\right ) \left (a^2 \tan (d+e x)+a b\right )^3}{a^2 \left (a^2+b^2\right )^3 \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}-\frac{b \left (3 a^2-b^2\right ) \left (a^2 \tan (d+e x)+a b\right )^3}{e \left (a^2+b^2\right )^2 \left (a^3 b+a^4 \tan (d+e x)\right ) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}-\frac{\left (-6 a^2 b^2+a^4+b^4\right ) (a \tan (d+e x)+b)^3 \log (a \sin (d+e x)+b \cos (d+e x))}{e \left (a^2+b^2\right )^3 \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}} \]

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)^(3/2),x]

[Out]

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

Rule 3710

Int[((A_) + (B_.)*tan[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*tan[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[(a + b*Tan[d + e*x] + c*Tan[d + e*x]^2)^n/(b + 2*c*Tan[d + e*x])^(2*n), Int[(A +
B*Tan[d + e*x])*(b + 2*c*Tan[d + e*x])^(2*n), x], x] /; FreeQ[{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 )^{3/2}} \, dx &=\frac{\left (2 a b+2 a^2 \tan (d+e x)\right )^3 \int \frac{a+b \tan (d+e x)}{\left (2 a b+2 a^2 \tan (d+e x)\right )^3} \, dx}{\left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}\\ &=-\frac{\left (a^2-b^2\right ) (b+a \tan (d+e x))}{2 \left (a^2+b^2\right ) e \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}+\frac{\left (2 a b+2 a^2 \tan (d+e x)\right )^3 \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 )^2} \, dx}{4 a^2 \left (a^2+b^2\right ) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}\\ &=-\frac{\left (a^2-b^2\right ) (b+a \tan (d+e x))}{2 \left (a^2+b^2\right ) e \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}-\frac{b \left (3 a^2-b^2\right ) (b+a \tan (d+e x))^2}{\left (a^2+b^2\right )^2 e \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}+\frac{\left (2 a b+2 a^2 \tan (d+e x)\right )^3 \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)}{2 a b+2 a^2 \tan (d+e x)} \, dx}{16 a^4 \left (a^2+b^2\right )^2 \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}\\ &=-\frac{\left (a^2-b^2\right ) (b+a \tan (d+e x))}{2 \left (a^2+b^2\right ) e \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}-\frac{b \left (3 a^2-b^2\right ) (b+a \tan (d+e x))^2}{\left (a^2+b^2\right )^2 e \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}-\frac{4 b \left (a^2-b^2\right ) x \left (a b+a^2 \tan (d+e x)\right )^3}{a^2 \left (a^2+b^2\right )^3 \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}-\frac{\left (\left (a^4-6 a^2 b^2+b^4\right ) \left (2 a b+2 a^2 \tan (d+e x)\right )^3\right ) \int \frac{2 a^2-2 a b \tan (d+e x)}{2 a b+2 a^2 \tan (d+e x)} \, dx}{8 a^3 \left (a^2+b^2\right )^3 \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}\\ &=-\frac{\left (a^2-b^2\right ) (b+a \tan (d+e x))}{2 \left (a^2+b^2\right ) e \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}-\frac{b \left (3 a^2-b^2\right ) (b+a \tan (d+e x))^2}{\left (a^2+b^2\right )^2 e \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}-\frac{\left (a^4-6 a^2 b^2+b^4\right ) \log (b \cos (d+e x)+a \sin (d+e x)) (b+a \tan (d+e x))^3}{\left (a^2+b^2\right )^3 e \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}-\frac{4 b \left (a^2-b^2\right ) x \left (a b+a^2 \tan (d+e x)\right )^3}{a^2 \left (a^2+b^2\right )^3 \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 3.18623, size = 268, normalized size = 0.85 \[ \frac{(a \tan (d+e x)+b)^3 \left (b \left (\frac{2 a \left (2 b \log (a \tan (d+e x)+b)-\frac{a^2+b^2}{a \tan (d+e x)+b}\right )}{\left (a^2+b^2\right )^2}+\frac{i \log (-\tan (d+e x)+i)}{(a-i b)^2}-\frac{i \log (\tan (d+e x)+i)}{(a+i b)^2}\right )+(a-b) (a+b) \left (\frac{a \left (-\frac{\left (a^2+b^2\right ) \left (a^2+4 a b \tan (d+e x)+5 b^2\right )}{(a \tan (d+e x)+b)^2}-2 \left (a^2-3 b^2\right ) \log (a \tan (d+e x)+b)\right )}{\left (a^2+b^2\right )^3}+\frac{\log (-\tan (d+e x)+i)}{(a-i b)^3}+\frac{\log (\tan (d+e x)+i)}{(a+i b)^3}\right )\right )}{2 a e \left ((a \tan (d+e x)+b)^2\right )^{3/2}} \]

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)^(3/2),x]

[Out]

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

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Maple [B]  time = 0.094, size = 622, normalized size = 2. \begin{align*} \text{result too large to display} \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)^(3/2),x)

[Out]

-1/2/e*(a^6+2*ln(b+a*tan(e*x+d))*b^6-ln(1+tan(e*x+d)^2)*b^6+3*a^2*b^4+7*a^4*b^2-ln(1+tan(e*x+d)^2)*tan(e*x+d)^
2*a^6-3*b^6-12*ln(b+a*tan(e*x+d))*a^2*b^4-ln(1+tan(e*x+d)^2)*a^4*b^2+6*ln(1+tan(e*x+d)^2)*a^2*b^4+8*arctan(tan
(e*x+d))*a^3*b^3-8*arctan(tan(e*x+d))*a*b^5+2*ln(b+a*tan(e*x+d))*tan(e*x+d)^2*a^6-2*tan(e*x+d)*a*b^5+6*tan(e*x
+d)*a^5*b+4*tan(e*x+d)*a^3*b^3+2*ln(b+a*tan(e*x+d))*a^4*b^2+4*ln(b+a*tan(e*x+d))*tan(e*x+d)*a^5*b-24*ln(b+a*ta
n(e*x+d))*tan(e*x+d)*a^3*b^3+4*ln(b+a*tan(e*x+d))*tan(e*x+d)*a*b^5-2*ln(1+tan(e*x+d)^2)*tan(e*x+d)*a^5*b+12*ln
(1+tan(e*x+d)^2)*tan(e*x+d)*a^3*b^3-2*ln(1+tan(e*x+d)^2)*tan(e*x+d)*a*b^5+16*arctan(tan(e*x+d))*tan(e*x+d)*a^4
*b^2-16*arctan(tan(e*x+d))*tan(e*x+d)*a^2*b^4-12*ln(b+a*tan(e*x+d))*tan(e*x+d)^2*a^4*b^2+2*ln(b+a*tan(e*x+d))*
tan(e*x+d)^2*a^2*b^4+6*ln(1+tan(e*x+d)^2)*tan(e*x+d)^2*a^4*b^2-ln(1+tan(e*x+d)^2)*tan(e*x+d)^2*a^2*b^4+8*arcta
n(tan(e*x+d))*tan(e*x+d)^2*a^5*b-8*arctan(tan(e*x+d))*tan(e*x+d)^2*a^3*b^3)*(b+a*tan(e*x+d))/(a^2+b^2)^3/((b+a
*tan(e*x+d))^2)^(3/2)

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Maxima [A]  time = 1.54713, size = 672, normalized size = 2.13 \begin{align*} -\frac{{\left (\frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (e x + d\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (a \tan \left (e x + d\right ) + b\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{4 \, a^{2} b \tan \left (e x + d\right ) + a^{3} + 5 \, a b^{2}}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6} +{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \tan \left (e x + d\right )^{2} + 2 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (e x + d\right )}\right )} a +{\left (\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (e x + d\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )} \log \left (a \tan \left (e x + d\right ) + b\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{a^{2} b - 3 \, b^{3} + 2 \,{\left (a^{3} - a b^{2}\right )} \tan \left (e x + d\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6} +{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \tan \left (e x + d\right )^{2} + 2 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (e x + d\right )}\right )} b}{2 \, 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)^(3/2),x, algorithm="maxima")

[Out]

-1/2*((2*(3*a^2*b - b^3)*(e*x + d)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(a^3 - 3*a*b^2)*log(a*tan(e*x + d)
+ b)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (a^3 - 3*a*b^2)*log(tan(e*x + d)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^
4 + b^6) + (4*a^2*b*tan(e*x + d) + a^3 + 5*a*b^2)/(a^4*b^2 + 2*a^2*b^4 + b^6 + (a^6 + 2*a^4*b^2 + a^2*b^4)*tan
(e*x + d)^2 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*tan(e*x + d)))*a + (2*(a^3 - 3*a*b^2)*(e*x + d)/(a^6 + 3*a^4*b^2 +
3*a^2*b^4 + b^6) - 2*(3*a^2*b - b^3)*log(a*tan(e*x + d) + b)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (3*a^2*b -
b^3)*log(tan(e*x + d)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^2*b - 3*b^3 + 2*(a^3 - a*b^2)*tan(e*x +
d))/(a^4*b^2 + 2*a^2*b^4 + b^6 + (a^6 + 2*a^4*b^2 + a^2*b^4)*tan(e*x + d)^2 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*t
an(e*x + d)))*b)/e

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Fricas [A]  time = 1.73784, size = 764, normalized size = 2.42 \begin{align*} -\frac{a^{6} + 8 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 8 \,{\left (a^{3} b^{3} - a b^{5}\right )} e x +{\left (a^{6} - 8 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + 8 \,{\left (a^{5} b - a^{3} b^{3}\right )} e x\right )} \tan \left (e x + d\right )^{2} +{\left (a^{4} b^{2} - 6 \, a^{2} b^{4} + b^{6} +{\left (a^{6} - 6 \, a^{4} b^{2} + a^{2} b^{4}\right )} \tan \left (e x + d\right )^{2} + 2 \,{\left (a^{5} b - 6 \, a^{3} b^{3} + a b^{5}\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 ) + 4 \,{\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} + 4 \,{\left (a^{4} b^{2} - a^{2} b^{4}\right )} e x\right )} \tan \left (e x + d\right )}{2 \,{\left ({\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} e \tan \left (e x + d\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} e \tan \left (e x + d\right ) +{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \tan{\left (d + e x \right )}}{\left (\left (a \tan{\left (d + e x \right )} + b\right )^{2}\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral((a + b*tan(d + e*x))/((a*tan(d + e*x) + b)**2)**(3/2), x)

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Giac [B]  time = 3.00569, size = 2121, normalized size = 6.71 \begin{align*} \text{result too large to display} \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)^(3/2),x, algorithm="giac")

[Out]

1/2*(4*(a^3*b - a*b^3)*(pi*sgn(tan(1/2*x*e + 1/2*d)) + 2*arctan(1/2*(tan(1/2*x*e + 1/2*d)^2 - 1)/tan(1/2*x*e +
1/2*d)))/(a^6*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*t
an(1/2*x*e + 1/2*d) - b) + 3*a^4*b^2*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*
x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b) + 3*a^2*b^4*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1
/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b) + b^6*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a
*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b)) - (a^4 - 6*a^2*b^2 + b^4
)*log((1/tan(1/2*x*e + 1/2*d) - tan(1/2*x*e + 1/2*d))^2 + 4)/(a^6*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*
x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b) + 3*a^4*b^2*sgn(-b*tan(1/2*x*e + 1
/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b) + 3*a^2*b^4*
sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/
2*d) - b) + b^6*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*
tan(1/2*x*e + 1/2*d) - b)) + 2*(a^4*b - 6*a^2*b^3 + b^5)*log(abs(-b*(1/tan(1/2*x*e + 1/2*d) - tan(1/2*x*e + 1/
2*d)) - 2*a))/(a^6*b*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 -
2*a*tan(1/2*x*e + 1/2*d) - b) + 3*a^4*b^3*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*ta
n(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b) + 3*a^2*b^5*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x
*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b) + b^7*sgn(-b*tan(1/2*x*e + 1/2*d)^4
+ 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b)) - (3*a^4*b^4*(1/ta
n(1/2*x*e + 1/2*d) - tan(1/2*x*e + 1/2*d))^2 - 18*a^2*b^6*(1/tan(1/2*x*e + 1/2*d) - tan(1/2*x*e + 1/2*d))^2 +
3*b^8*(1/tan(1/2*x*e + 1/2*d) - tan(1/2*x*e + 1/2*d))^2 + 4*a^7*b*(1/tan(1/2*x*e + 1/2*d) - tan(1/2*x*e + 1/2*
d)) + 28*a^5*b^3*(1/tan(1/2*x*e + 1/2*d) - tan(1/2*x*e + 1/2*d)) - 68*a^3*b^5*(1/tan(1/2*x*e + 1/2*d) - tan(1/
2*x*e + 1/2*d)) + 4*a*b^7*(1/tan(1/2*x*e + 1/2*d) - tan(1/2*x*e + 1/2*d)) + 4*a^8 + 40*a^6*b^2 - 60*a^4*b^4)/(
(a^6*b^2*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2
*x*e + 1/2*d) - b) + 3*a^4*b^4*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e +
1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b) + 3*a^2*b^6*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1/2*x*e + 1/2*d)^
3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b) + b^8*sgn(-b*tan(1/2*x*e + 1/2*d)^4 + 2*a*tan(1
/2*x*e + 1/2*d)^3 + 2*b*tan(1/2*x*e + 1/2*d)^2 - 2*a*tan(1/2*x*e + 1/2*d) - b))*(b*(1/tan(1/2*x*e + 1/2*d) - t
an(1/2*x*e + 1/2*d)) + 2*a)^2))*e^(-1)