3.5.62 \(\int \frac {1}{a+b \tan (c+d x)} \, dx\) [462]

Optimal. Leaf size=45 \[ \frac {a x}{a^2+b^2}+\frac {b \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right ) d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3565, 3611} \begin {gather*} \frac {b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac {a x}{a^2+b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-1),x]

[Out]

(a*x)/(a^2 + b^2) + (b*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)*d)

Rule 3565

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[a*(x/(a^2 + b^2)), x] + Dist[b/(a^2 + b^2),
Int[(b - a*Tan[c + d*x])/(a + b*Tan[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[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] && EqQ[a*c + b*d, 0]

Rubi steps

\begin {align*} \int \frac {1}{a+b \tan (c+d x)} \, dx &=\frac {a x}{a^2+b^2}+\frac {b \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=\frac {a x}{a^2+b^2}+\frac {b \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right ) d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.06, size = 76, normalized size = 1.69 \begin {gather*} \frac {(-i a-b) \log (i-\tan (c+d x))+i (a+i b) \log (i+\tan (c+d x))+2 b \log (a+b \tan (c+d x))}{2 \left (a^2+b^2\right ) d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(-1),x]

[Out]

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

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Maple [A]
time = 0.09, size = 62, normalized size = 1.38

method	result	size
derivativedivides	\(\frac {\frac {-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}+\frac {b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(62\)
default	\(\frac {\frac {-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}+\frac {b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(62\)
norman	\(\frac {a x}{a^{2}+b^{2}}+\frac {b \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )}-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}\)	\(65\)
risch	\(-\frac {x}{i b -a}-\frac {2 i b x}{a^{2}+b^{2}}-\frac {2 i b c}{d \left (a^{2}+b^{2}\right )}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{2}+b^{2}\right )}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.52, size = 69, normalized size = 1.53 \begin {gather*} \frac {\frac {2 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac {2 \, b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.27, size = 62, normalized size = 1.38 \begin {gather*} \frac {2 \, a d x + b \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 (a^{2} + b^{2}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.41, size = 241, normalized size = 5.36 \begin {gather*} \begin {cases} \frac {\tilde {\infty } x}{\tan {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {i d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {i}{2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = - i b \\- \frac {i d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {i}{2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = i b \\\frac {x}{a + b \tan {\left (c \right )}} & \text {for}\: d = 0 \\\frac {x}{a} & \text {for}\: b = 0 \\\frac {2 a d x}{2 a^{2} d + 2 b^{2} d} + \frac {2 b \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} - \frac {b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)),x)

[Out]

Piecewise((zoo*x/tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (I*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) - 2*I*b*d) +
d*x/(2*b*d*tan(c + d*x) - 2*I*b*d) + I/(2*b*d*tan(c + d*x) - 2*I*b*d), Eq(a, -I*b)), (-I*d*x*tan(c + d*x)/(2*
b*d*tan(c + d*x) + 2*I*b*d) + d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - I/(2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, I*b
)), (x/(a + b*tan(c)), Eq(d, 0)), (x/a, Eq(b, 0)), (2*a*d*x/(2*a**2*d + 2*b**2*d) + 2*b*log(a/b + tan(c + d*x)
)/(2*a**2*d + 2*b**2*d) - b*log(tan(c + d*x)**2 + 1)/(2*a**2*d + 2*b**2*d), True))

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Giac [A]
time = 0.59, size = 74, normalized size = 1.64 \begin {gather*} \frac {\frac {2 \, b^{2} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}} + \frac {2 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.18, size = 73, normalized size = 1.62 \begin {gather*} \frac {b\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,\left (a^2+b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(c + d*x)),x)

[Out]

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

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