3.12.94 \(\int \frac {c+d \tan (e+f x)}{a+b \tan (e+f x)} \, dx\) [1194]

Optimal. Leaf size=58 \[ \frac {(a c+b d) x}{a^2+b^2}+\frac {(b c-a d) \log (a \cos (e+f x)+b \sin (e+f x))}{\left (a^2+b^2\right ) f} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3612, 3611} \begin {gather*} \frac {(b c-a d) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )}+\frac {x (a c+b d)}{a^2+b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])/(a + b*Tan[e + f*x]),x]

[Out]

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

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]

Rule 3612

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

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Mathematica [A]
time = 0.13, size = 66, normalized size = 1.14 \begin {gather*} \frac {2 (a c+b d) \text {ArcTan}(\tan (e+f x))-(b c-a d) \left (\log \left (\sec ^2(e+f x)\right )-2 \log (a+b \tan (e+f x))\right )}{2 \left (a^2+b^2\right ) f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])/(a + b*Tan[e + f*x]),x]

[Out]

(2*(a*c + b*d)*ArcTan[Tan[e + f*x]] - (b*c - a*d)*(Log[Sec[e + f*x]^2] - 2*Log[a + b*Tan[e + f*x]]))/(2*(a^2 +
b^2)*f)

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Maple [A]
time = 0.17, size = 83, normalized size = 1.43

method	result	size
derivativedivides	\(\frac {-\frac {\left (a d -b c \right ) \ln \left (a +b \tan \left (f x +e \right )\right )}{a^{2}+b^{2}}+\frac {\frac {\left (a d -b c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a c +b d \right ) \arctan \left (\tan \left (f x +e \right )\right )}{a^{2}+b^{2}}}{f}\)	\(83\)
default	\(\frac {-\frac {\left (a d -b c \right ) \ln \left (a +b \tan \left (f x +e \right )\right )}{a^{2}+b^{2}}+\frac {\frac {\left (a d -b c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a c +b d \right ) \arctan \left (\tan \left (f x +e \right )\right )}{a^{2}+b^{2}}}{f}\)	\(83\)
norman	\(\frac {\left (a c +b d \right ) x}{a^{2}+b^{2}}+\frac {\left (a d -b c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (a^{2}+b^{2}\right )}-\frac {\left (a d -b c \right ) \ln \left (a +b \tan \left (f x +e \right )\right )}{f \left (a^{2}+b^{2}\right )}\)	\(86\)
risch	\(\frac {i x d}{i b -a}-\frac {x c}{i b -a}+\frac {2 i a d x}{a^{2}+b^{2}}-\frac {2 i b c x}{a^{2}+b^{2}}+\frac {2 i a d e}{f \left (a^{2}+b^{2}\right )}-\frac {2 i b c e}{f \left (a^{2}+b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i b +a}{i b -a}\right ) a d}{f \left (a^{2}+b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i b +a}{i b -a}\right ) b c}{f \left (a^{2}+b^{2}\right )}\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 92, normalized size = 1.59 \begin {gather*} \frac {\frac {2 \, {\left (a c + b d\right )} {\left (f x + e\right )}}{a^{2} + b^{2}} + \frac {2 \, {\left (b c - a d\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{2} + b^{2}} - \frac {{\left (b c - a d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.12, size = 78, normalized size = 1.34 \begin {gather*} \frac {2 \, {\left (a c + b d\right )} f x + {\left (b c - a d\right )} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a^{2} + b^{2}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.47, size = 524, normalized size = 9.03 \begin {gather*} \begin {cases} \frac {\tilde {\infty } x \left (c + d \tan {\left (e \right )}\right )}{\tan {\left (e \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {c x + \frac {d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f}}{a} & \text {for}\: b = 0 \\\frac {i c f x \tan {\left (e + f x \right )}}{2 b f \tan {\left (e + f x \right )} - 2 i b f} + \frac {c f x}{2 b f \tan {\left (e + f x \right )} - 2 i b f} + \frac {i c}{2 b f \tan {\left (e + f x \right )} - 2 i b f} + \frac {d f x \tan {\left (e + f x \right )}}{2 b f \tan {\left (e + f x \right )} - 2 i b f} - \frac {i d f x}{2 b f \tan {\left (e + f x \right )} - 2 i b f} - \frac {d}{2 b f \tan {\left (e + f x \right )} - 2 i b f} & \text {for}\: a = - i b \\- \frac {i c f x \tan {\left (e + f x \right )}}{2 b f \tan {\left (e + f x \right )} + 2 i b f} + \frac {c f x}{2 b f \tan {\left (e + f x \right )} + 2 i b f} - \frac {i c}{2 b f \tan {\left (e + f x \right )} + 2 i b f} + \frac {d f x \tan {\left (e + f x \right )}}{2 b f \tan {\left (e + f x \right )} + 2 i b f} + \frac {i d f x}{2 b f \tan {\left (e + f x \right )} + 2 i b f} - \frac {d}{2 b f \tan {\left (e + f x \right )} + 2 i b f} & \text {for}\: a = i b \\\frac {x \left (c + d \tan {\left (e \right )}\right )}{a + b \tan {\left (e \right )}} & \text {for}\: f = 0 \\\frac {2 a c f x}{2 a^{2} f + 2 b^{2} f} - \frac {2 a d \log {\left (\frac {a}{b} + \tan {\left (e + f x \right )} \right )}}{2 a^{2} f + 2 b^{2} f} + \frac {a d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a^{2} f + 2 b^{2} f} + \frac {2 b c \log {\left (\frac {a}{b} + \tan {\left (e + f x \right )} \right )}}{2 a^{2} f + 2 b^{2} f} - \frac {b c \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a^{2} f + 2 b^{2} f} + \frac {2 b d f x}{2 a^{2} f + 2 b^{2} f} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.48, size = 98, normalized size = 1.69 \begin {gather*} \frac {\frac {2 \, {\left (a c + b d\right )} {\left (f x + e\right )}}{a^{2} + b^{2}} - \frac {{\left (b c - a d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (b^{2} c - a b d\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.68, size = 94, normalized size = 1.62 \begin {gather*} -\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-d+c\,1{}\mathrm {i}\right )}{2\,f\,\left (a+b\,1{}\mathrm {i}\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a\,d-b\,c\right )}{f\,\left (a^2+b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (c-d\,1{}\mathrm {i}\right )}{2\,f\,\left (b+a\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*tan(e + f*x))/(a + b*tan(e + f*x)),x)

[Out]

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

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