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

Optimal. Leaf size=101 \[ -\frac{a^2-b^2}{e \left (a^2+b^2\right ) (a \tan (d+e x)+b)}+\frac{b \left (3 a^2-b^2\right ) \log (a \sin (d+e x)+b \cos (d+e x))}{e \left (a^2+b^2\right )^2}-\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.257506, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3708, 3529, 3531, 3530} \[ -\frac{a^2-b^2}{e \left (a^2+b^2\right ) (a \tan (d+e x)+b)}+\frac{b \left (3 a^2-b^2\right ) \log (a \sin (d+e x)+b \cos (d+e x))}{e \left (a^2+b^2\right )^2}-\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^2} \]

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),x]

[Out]

-((a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^2) + (b*(3*a^2 - b^2)*Log[b*Cos[d + e*x] + a*Sin[d + e*x]])/((a^2 + b^2)^2*e
) - (a^2 - b^2)/((a^2 + b^2)*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)}{b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)} \, dx &=\left (4 a^2\right ) \int \frac{a+b \tan (d+e x)}{\left (2 a b+2 a^2 \tan (d+e x)\right )^2} \, dx\\ &=-\frac{a^2-b^2}{\left (a^2+b^2\right ) e (b+a \tan (d+e x))}+\frac{\int \frac{4 a^2 b-2 a \left (a^2-b^2\right ) \tan (d+e x)}{2 a b+2 a^2 \tan (d+e x)} \, dx}{a^2+b^2}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^2}-\frac{a^2-b^2}{\left (a^2+b^2\right ) e (b+a \tan (d+e x))}+\frac{\left (b \left (3 a^2-b^2\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 )^2}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^2}+\frac{b \left (3 a^2-b^2\right ) \log (b \cos (d+e x)+a \sin (d+e x))}{\left (a^2+b^2\right )^2 e}-\frac{a^2-b^2}{\left (a^2+b^2\right ) e (b+a \tan (d+e x))}\\ \end{align*}

Mathematica [C]  time = 2.0414, size = 187, normalized size = 1.85 \[ \frac{\frac{b (-(a+i b) \log (-\tan (d+e x)+i)-(a-i b) \log (\tan (d+e x)+i)+2 a \log (a \tan (d+e x)+b))}{a^2+b^2}+(a-b) (a+b) \left (\frac{2 a \left (2 b \log (a \tan (d+e x)+b)-\frac{a^2+b^2}{a \tan (d+e x)+b}\right )}{\left (a^2+b^2\right )^2}+\frac{i \log (-\tan (d+e x)+i)}{(a-i b)^2}-\frac{i \log (\tan (d+e x)+i)}{(a+i b)^2}\right )}{2 a e} \]

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),x]

[Out]

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

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Maple [B]  time = 0.05, size = 222, normalized size = 2.2 \begin{align*} -{\frac{{a}^{2}}{e \left ({a}^{2}+{b}^{2} \right ) \left ( b+a\tan \left ( ex+d \right ) \right ) }}+{\frac{{b}^{2}}{e \left ({a}^{2}+{b}^{2} \right ) \left ( b+a\tan \left ( ex+d \right ) \right ) }}+3\,{\frac{b\ln \left ( b+a\tan \left ( ex+d \right ) \right ){a}^{2}}{e \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{b}^{3}\ln \left ( b+a\tan \left ( ex+d \right ) \right ) }{e \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){a}^{2}b}{2\,e \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){b}^{3}}{2\,e \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ){a}^{3}}{e \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+3\,{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ) a{b}^{2}}{e \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \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),x)

[Out]

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

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Maxima [A]  time = 1.50309, size = 217, normalized size = 2.15 \begin{align*} -\frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (e x + d\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )} \log \left (a \tan \left (e x + d\right ) + b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (a^{2} - b^{2}\right )}}{a^{2} b + b^{3} +{\left (a^{3} + a b^{2}\right )} \tan \left (e x + d\right )}}{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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.73982, size = 417, normalized size = 4.13 \begin{align*} -\frac{2 \, a^{4} - 2 \, a^{2} b^{2} + 2 \,{\left (a^{3} b - 3 \, a b^{3}\right )} e x -{\left (3 \, a^{2} b^{2} - b^{4} +{\left (3 \, a^{3} b - a b^{3}\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 ) - 2 \,{\left (a^{3} b - a b^{3} -{\left (a^{4} - 3 \, a^{2} b^{2}\right )} e x\right )} \tan \left (e x + d\right )}{2 \,{\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} e \tan \left (e x + d\right ) +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\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),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.57471, size = 275, normalized size = 2.72 \begin{align*} -\frac{1}{2} \,{\left (\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (x e + d\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (x e + d\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (3 \, a^{3} b - a b^{3}\right )} \log \left ({\left | a \tan \left (x e + d\right ) + b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac{2 \,{\left (3 \, a^{3} b \tan \left (x e + d\right ) - a b^{3} \tan \left (x e + d\right ) + a^{4} + 3 \, a^{2} b^{2} - 2 \, b^{4}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a \tan \left (x e + d\right ) + b\right )}}\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),x, algorithm="giac")

[Out]

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