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)*ln(b*cos(e*x+d)+a*sin(e*x+d))/(a^2+b^2)^2/e+(-a^2+b^2)/(a^2+b^2)/e/
(b+a*tan(e*x+d))

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Rubi [A]  time = 0.26, antiderivative size = 101, normalized size of antiderivative = 1.00, 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 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)}{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*}

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Mathematica [C]  time = 2.34, 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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fricas [A]  time = 0.72, size = 191, normalized size = 1.89 \[ -\frac {2 \, a^{4} - 2 \, a^{2} b^{2} + 2 \, {\left (a^{3} b - 3 \, a b^{3}\right )} e x - {\left (3 \, a^{2} b^{2} - b^{4} + {\left (3 \, a^{3} b - a b^{3}\right )} \tan \left (e x + d\right )\right )} \log \left (\frac {a^{2} \tan \left (e x + d\right )^{2} + 2 \, a b \tan \left (e x + d\right ) + b^{2}}{\tan \left (e x + d\right )^{2} + 1}\right ) - 2 \, {\left (a^{3} b - a b^{3} - {\left (a^{4} - 3 \, a^{2} b^{2}\right )} e x\right )} \tan \left (e x + d\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} e \tan \left (e x + d\right ) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} e\right )}} \]

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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giac [B]  time = 0.89, size = 204, normalized size = 2.02 \[ -\frac {1}{2} \, {\left (\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (x e + d\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (x e + d\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (3 \, a^{3} b - a b^{3}\right )} \log \left ({\left | a \tan \left (x e + d\right ) + b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac {2 \, {\left (3 \, a^{3} b \tan \left (x e + d\right ) - a b^{3} \tan \left (x e + d\right ) + a^{4} + 3 \, a^{2} b^{2} - 2 \, b^{4}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (x e + d\right ) + b\right )}}\right )} e^{\left (-1\right )} \]

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)

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maple [B]  time = 0.16, size = 222, normalized size = 2.20 \[ -\frac {a^{2}}{e \left (a^{2}+b^{2}\right ) \left (b +a \tan \left (e x +d \right )\right )}+\frac {b^{2}}{e \left (a^{2}+b^{2}\right ) \left (b +a \tan \left (e x +d \right )\right )}+\frac {3 b \ln \left (b +a \tan \left (e x +d \right )\right ) a^{2}}{e \left (a^{2}+b^{2}\right )^{2}}-\frac {b^{3} \ln \left (b +a \tan \left (e x +d \right )\right )}{e \left (a^{2}+b^{2}\right )^{2}}-\frac {3 \ln \left (1+\tan ^{2}\left (e x +d \right )\right ) a^{2} b}{2 e \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (e x +d \right )\right ) b^{3}}{2 e \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \left (e x +d \right )\right ) a^{3}}{e \left (a^{2}+b^{2}\right )^{2}}+\frac {3 \arctan \left (\tan \left (e x +d \right )\right ) a \,b^{2}}{e \left (a^{2}+b^{2}\right )^{2}} \]

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 = 0.41, size = 161, normalized size = 1.59 \[ -\frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (e x + d\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (a \tan \left (e x + d\right ) + b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (a^{2} - b^{2}\right )}}{a^{2} b + b^{3} + {\left (a^{3} + a b^{2}\right )} \tan \left (e x + d\right )}}{2 \, e} \]

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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mupad [B]  time = 3.09, size = 152, normalized size = 1.50 \[ \frac {b\,\ln \left (b+a\,\mathrm {tan}\left (d+e\,x\right )\right )\,\left (3\,a^2-b^2\right )}{e\,{\left (a^2+b^2\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (d+e\,x\right )+1{}\mathrm {i}\right )\,\left (a-b\,1{}\mathrm {i}\right )}{2\,e\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {a^2-b^2}{e\,\left (a^2+b^2\right )\,\left (b+a\,\mathrm {tan}\left (d+e\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (d+e\,x\right )-\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,e\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )} \]

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)),x)

[Out]

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

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sympy [A]  time = 1.62, size = 1358, normalized size = 13.45 \[ \text {result too large to display} \]

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]

Piecewise((zoo*x*tan(d), Eq(a, 0) & Eq(b, 0) & Eq(e, 0)), (log(tan(d + e*x)**2 + 1)/(2*b*e), Eq(a, 0)), (1/(2*
b*e*tan(d + e*x)**2 + 4*I*b*e*tan(d + e*x) - 2*b*e), Eq(a, -I*b)), (1/(2*b*e*tan(d + e*x)**2 - 4*I*b*e*tan(d +
e*x) - 2*b*e), Eq(a, I*b)), (x*(a + b*tan(d))/(a**2*tan(d)**2 + 2*a*b*tan(d) + b**2), Eq(e, 0)), (-2*a**4*e*x
*tan(d + e*x)/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*ta
n(d + e*x) + 2*b**5*e) - 2*a**4/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3
*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e) - 2*a**3*b*e*x/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*ta
n(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e) + 6*a**3*b*log(tan(d + e*x) + b/a)*tan(d + e*
x)/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x)
+ 2*b**5*e) - 3*a**3*b*log(tan(d + e*x)**2 + 1)*tan(d + e*x)/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2
*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e) + 6*a**2*b**2*e*x*tan(d + e*x)/(2*a**5*e
*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e)
+ 6*a**2*b**2*log(tan(d + e*x) + b/a)/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**
2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e) - 3*a**2*b**2*log(tan(d + e*x)**2 + 1)/(2*a**5*e*tan(d + e*x) +
2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e) + 6*a*b**3*e*x/
(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2
*b**5*e) - 2*a*b**3*log(tan(d + e*x) + b/a)*tan(d + e*x)/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*t
an(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e) + a*b**3*log(tan(d + e*x)**2 + 1)*tan(d + e*
x)/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x)
+ 2*b**5*e) - 2*b**4*log(tan(d + e*x) + b/a)/(2*a**5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x)
+ 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e) + b**4*log(tan(d + e*x)**2 + 1)/(2*a**5*e*tan(d + e*x) +
2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*e) + 2*b**4/(2*a**
5*e*tan(d + e*x) + 2*a**4*b*e + 4*a**3*b**2*e*tan(d + e*x) + 4*a**2*b**3*e + 2*a*b**4*e*tan(d + e*x) + 2*b**5*
e), True))

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