3.1195 \(\int \frac{c+d \tan (e+f x)}{(a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=111 \[ -\frac{b c-a d}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{\left (a^2 (-d)+2 a b c+b^2 d\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^2}+\frac{x \left (a^2 c+2 a b d-b^2 c\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.147237, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3529, 3531, 3530} \[ -\frac{b c-a d}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{\left (a^2 (-d)+2 a b c+b^2 d\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^2}+\frac{x \left (a^2 c+2 a b d-b^2 c\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [C]  time = 1.91284, size = 190, normalized size = 1.71 \[ \frac{\frac{d ((-b-i a) \log (-\tan (e+f x)+i)+i (a+i b) \log (\tan (e+f x)+i)+2 b \log (a+b \tan (e+f x)))}{a^2+b^2}-(b c-a d) \left (\frac{2 b \left (\frac{a^2+b^2}{a+b \tan (e+f x)}-2 a \log (a+b \tan (e+f x))\right )}{\left (a^2+b^2\right )^2}+\frac{i \log (-\tan (e+f x)+i)}{(a+i b)^2}-\frac{i \log (\tan (e+f x)+i)}{(a-i b)^2}\right )}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.03, size = 301, normalized size = 2.7 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){a}^{2}d}{2\,f \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) abc}{f \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){b}^{2}d}{2\,f \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){a}^{2}c}{f \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) abd}{f \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}c}{f \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{ad}{f \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( fx+e \right ) \right ) }}-{\frac{bc}{f \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( fx+e \right ) \right ) }}-{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ){a}^{2}d}{f \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+2\,{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ) abc}{f \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ){b}^{2}d}{f \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.89206, size = 243, normalized size = 2.19 \begin{align*} \frac{\frac{2 \,{\left (2 \, a b d +{\left (a^{2} - b^{2}\right )} c\right )}{\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (2 \, a b c -{\left (a^{2} - b^{2}\right )} d\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (2 \, a b c -{\left (a^{2} - b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (b c - a d\right )}}{a^{3} + a b^{2} +{\left (a^{2} b + b^{3}\right )} \tan \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.39608, size = 490, normalized size = 4.41 \begin{align*} -\frac{2 \, b^{3} c - 2 \, a b^{2} d - 2 \,{\left (2 \, a^{2} b d +{\left (a^{3} - a b^{2}\right )} c\right )} f x -{\left (2 \, a^{2} b c -{\left (a^{3} - a b^{2}\right )} d +{\left (2 \, a b^{2} c -{\left (a^{2} b - b^{3}\right )} d\right )} \tan \left (f x + e\right )\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 b^{2} c - a^{2} b d +{\left (2 \, a b^{2} d +{\left (a^{2} b - b^{3}\right )} c\right )} f x\right )} \tan \left (f x + e\right )}{2 \,{\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} f \tan \left (f x + e\right ) +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.28911, size = 325, normalized size = 2.93 \begin{align*} \frac{\frac{2 \,{\left (a^{2} c - b^{2} c + 2 \, a b d\right )}{\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (2 \, a b c - a^{2} d + b^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (2 \, a b^{2} c - a^{2} b d + b^{3} d\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac{2 \,{\left (2 \, a b^{2} c \tan \left (f x + e\right ) - a^{2} b d \tan \left (f x + e\right ) + b^{3} d \tan \left (f x + e\right ) + 3 \, a^{2} b c + b^{3} c - 2 \, a^{3} d\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \tan \left (f x + e\right ) + a\right )}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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