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

Optimal. Leaf size=82 \[ \frac {a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {2 a b x}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3529, 3531, 3530} \[ \frac {a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {2 a b x}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.48, size = 181, normalized size = 2.21 \[ \frac {a \left (2 \left (\left (b^2-a^2\right ) \log (a+b \tan (c+d x))+a^2+b^2\right )+(a-i b)^2 \log (-\tan (c+d x)+i)+(a+i b)^2 \log (\tan (c+d x)+i)\right )+b \tan (c+d x) \left (2 \left (b^2-a^2\right ) \log (a+b \tan (c+d x))+(a-i b)^2 \log (-\tan (c+d x)+i)+(a+i b)^2 \log (\tan (c+d x)+i)\right )}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 157, normalized size = 1.91 \[ \frac {4 \, a^{2} b d x + 2 \, a b^{2} - {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (2 \, a b^{2} d x - a^{2} b\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.43, size = 173, normalized size = 2.11 \[ \frac {\frac {4 \, {\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac {2 \, {\left (a^{2} b \tan \left (d x + c\right ) - b^{3} \tan \left (d x + c\right ) + 2 \, a^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 162, normalized size = 1.98 \[ \frac {a}{\left (a^{2}+b^{2}\right ) d \left (a +b \tan \left (d x +c \right )\right )}-\frac {a^{2} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) b^{2}}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.56, size = 139, normalized size = 1.70 \[ \frac {\frac {4 \, {\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, a}{a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.17, size = 133, normalized size = 1.62 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{2\,d\,\left (a^2+a\,b\,2{}\mathrm {i}-b^2\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {1}{a^2+b^2}-\frac {2\,b^2}{{\left (a^2+b^2\right )}^2}\right )}{d}+\frac {a}{d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.70, size = 1482, normalized size = 18.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x/tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-d*x*tan(c + d*x)**2/(4*I*b**2*d*tan(c + d*x)**2 +
8*b**2*d*tan(c + d*x) - 4*I*b**2*d) + 2*I*d*x*tan(c + d*x)/(4*I*b**2*d*tan(c + d*x)**2 + 8*b**2*d*tan(c + d*x)
- 4*I*b**2*d) + d*x/(4*I*b**2*d*tan(c + d*x)**2 + 8*b**2*d*tan(c + d*x) - 4*I*b**2*d) - tan(c + d*x)/(4*I*b**
2*d*tan(c + d*x)**2 + 8*b**2*d*tan(c + d*x) - 4*I*b**2*d), Eq(a, -I*b)), (-d*x*tan(c + d*x)**2/(-4*I*b**2*d*ta
n(c + d*x)**2 + 8*b**2*d*tan(c + d*x) + 4*I*b**2*d) - 2*I*d*x*tan(c + d*x)/(-4*I*b**2*d*tan(c + d*x)**2 + 8*b*
*2*d*tan(c + d*x) + 4*I*b**2*d) + d*x/(-4*I*b**2*d*tan(c + d*x)**2 + 8*b**2*d*tan(c + d*x) + 4*I*b**2*d) - tan
(c + d*x)/(-4*I*b**2*d*tan(c + d*x)**2 + 8*b**2*d*tan(c + d*x) + 4*I*b**2*d), Eq(a, I*b)), (x*tan(c)/(a + b*ta
n(c))**2, Eq(d, 0)), (log(tan(c + d*x)**2 + 1)/(2*a**2*d), Eq(b, 0)), (-2*a**3*log(a/b + tan(c + d*x))/(2*a**5
*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)
) + a**3*log(tan(c + d*x)**2 + 1)/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c +
d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) + 2*a**3/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a*
*2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) + 4*a**2*b*d*x/(2*a**5*d + 2*a**4*b*d*tan(c + d*x
) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) - 2*a**2*b*log(a/b + tan(
c + d*x))*tan(c + d*x)/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*
b**4*d + 2*b**5*d*tan(c + d*x)) + a**2*b*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*d + 2*a**4*b*d*tan(c +
d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) + 4*a*b**2*d*x*tan(c +
d*x)/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d
*tan(c + d*x)) + 2*a*b**2*log(a/b + tan(c + d*x))/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a**2
*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) - a*b**2*log(tan(c + d*x)**2 + 1)/(2*a**5*d + 2*a**
4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) + 2*a*b*
*2/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*ta
n(c + d*x)) + 2*b**3*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d
+ 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) - b**3*log(tan(c + d*x)**2 + 1)*tan(c + d*x
)/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan
(c + d*x)), True))

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