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

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

[Out]

-a*(a^2-3*b^2)*x/(a^2+b^2)^2+b*(3*a^2-b^2)*ln(b*cos(d*x+c)+a*sin(d*x+c))/(a^2+b^2)^2/d+(-a^2+b^2)/(a^2+b^2)/d/
(b+a*tan(d*x+c))

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Rubi [A]  time = 0.13, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3529, 3531, 3530} \[ -\frac {a^2-b^2}{d \left (a^2+b^2\right ) (a \tan (c+d x)+b)}+\frac {b \left (3 a^2-b^2\right ) \log (a \sin (c+d x)+b \cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])/(b + a*Tan[c + d*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[c + d*x] + a*Sin[c + d*x]])/((a^2 + b^2)^2*d
) - (a^2 - b^2)/((a^2 + b^2)*d*(b + a*Tan[c + d*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]

Rubi steps

\begin {align*} \int \frac {a+b \tan (c+d x)}{(b+a \tan (c+d x))^2} \, dx &=-\frac {a^2-b^2}{\left (a^2+b^2\right ) d (b+a \tan (c+d x))}+\frac {\int \frac {2 a b-\left (a^2-b^2\right ) \tan (c+d x)}{b+a \tan (c+d 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 ) d (b+a \tan (c+d x))}+\frac {\left (b \left (3 a^2-b^2\right )\right ) \int \frac {a-b \tan (c+d x)}{b+a \tan (c+d 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 (c+d x)+a \sin (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a^2-b^2}{\left (a^2+b^2\right ) d (b+a \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 2.30, size = 187, normalized size = 1.85 \[ \frac {\frac {b (-((a+i b) \log (-\tan (c+d x)+i))-(a-i b) \log (\tan (c+d x)+i)+2 a \log (a \tan (c+d x)+b))}{a^2+b^2}+(a-b) (a+b) \left (\frac {2 a \left (2 b \log (a \tan (c+d x)+b)-\frac {a^2+b^2}{a \tan (c+d x)+b}\right )}{\left (a^2+b^2\right )^2}+\frac {i \log (-\tan (c+d x)+i)}{(a-i b)^2}-\frac {i \log (\tan (c+d x)+i)}{(a+i b)^2}\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.01, size = 191, normalized size = 1.89 \[ -\frac {2 \, a^{4} - 2 \, a^{2} b^{2} + 2 \, {\left (a^{3} b - 3 \, a b^{3}\right )} d x - {\left (3 \, a^{2} b^{2} - b^{4} + {\left (3 \, a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {a^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + b^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (a^{3} b - a b^{3} - {\left (a^{4} - 3 \, a^{2} b^{2}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \tan \left (d x + c\right ) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.54, size = 199, normalized size = 1.97 \[ -\frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (3 \, a^{3} b - a b^{3}\right )} \log \left ({\left | a \tan \left (d x + c\right ) + b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac {2 \, {\left (3 \, a^{3} b \tan \left (d x + c\right ) - a b^{3} \tan \left (d x + c\right ) + a^{4} + 3 \, a^{2} b^{2} - 2 \, b^{4}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (d x + c\right ) + b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.26, size = 222, normalized size = 2.20 \[ -\frac {a^{2}}{d \left (a^{2}+b^{2}\right ) \left (b +a \tan \left (d x +c \right )\right )}+\frac {b^{2}}{d \left (a^{2}+b^{2}\right ) \left (b +a \tan \left (d x +c \right )\right )}+\frac {3 b \ln \left (b +a \tan \left (d x +c \right )\right ) a^{2}}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {b^{3} \ln \left (b +a \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{3}}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {3 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 161, normalized size = 1.59 \[ -\frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (a \tan \left (d x + c\right ) + b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (a^{2} - b^{2}\right )}}{a^{2} b + b^{3} + {\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.63, size = 152, normalized size = 1.50 \[ \frac {b\,\ln \left (b+a\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a^2-b^2\right )}{d\,{\left (a^2+b^2\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (a-b\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {a^2-b^2}{d\,\left (a^2+b^2\right )\,\left (b+a\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,d\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.57, size = 1346, normalized size = 13.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (1/(2*b*d*tan(c + d*x)**2 + 4*I*b*d*tan(c + d*x) - 2
*b*d), Eq(a, -I*b)), (1/(2*b*d*tan(c + d*x)**2 - 4*I*b*d*tan(c + d*x) - 2*b*d), Eq(a, I*b)), (x*(a + b*tan(c))
/(a*tan(c) + b)**2, Eq(d, 0)), (log(tan(c + d*x)**2 + 1)/(2*b*d), Eq(a, 0)), (-2*a**4*d*x*tan(c + d*x)/(2*a**5
*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d
) - 2*a**4/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c
+ d*x) + 2*b**5*d) - 2*a**3*b*d*x/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b
**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 6*a**3*b*log(tan(c + d*x) + b/a)*tan(c + d*x)/(2*a**5*d*tan(c +
d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) - 3*a**3*
b*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a
**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 6*a**2*b**2*d*x*tan(c + d*x)/(2*a**5*d*tan(c + d*x) + 2*a**
4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 6*a**2*b**2*log(tan
(c + d*x) + b/a)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d
*tan(c + d*x) + 2*b**5*d) - 3*a**2*b**2*log(tan(c + d*x)**2 + 1)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*
b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 6*a*b**3*d*x/(2*a**5*d*tan(c + d*x
) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) - 2*a*b**3*l
og(tan(c + d*x) + b/a)*tan(c + d*x)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*
b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + a*b**3*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*d*tan(c +
d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) - 2*b**4*
log(tan(c + d*x) + b/a)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a
*b**4*d*tan(c + d*x) + 2*b**5*d) + b**4*log(tan(c + d*x)**2 + 1)/(2*a**5*d*tan(c + d*x) + 2*a**4*b*d + 4*a**3*
b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d) + 2*b**4/(2*a**5*d*tan(c + d*x) + 2*
a**4*b*d + 4*a**3*b**2*d*tan(c + d*x) + 4*a**2*b**3*d + 2*a*b**4*d*tan(c + d*x) + 2*b**5*d), True))

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