3.1219 \(\int \frac{a+b \tan (e+f x)}{(c+d \tan (e+f x))^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.153337, 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} \[ \frac{b c-a d}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}+\frac{\left (2 a c d-b \left (c^2-d^2\right )\right ) \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right )^2}+\frac{x \left (a \left (c^2-d^2\right )+2 b c d\right )}{\left (c^2+d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 1.85308, size = 189, normalized size = 1.7 \[ \frac{(b c-a d) \left (\frac{2 d \left (\frac{c^2+d^2}{c+d \tan (e+f x)}-2 c \log (c+d \tan (e+f x))\right )}{\left (c^2+d^2\right )^2}+\frac{i \log (-\tan (e+f x)+i)}{(c+i d)^2}-\frac{i \log (\tan (e+f x)+i)}{(c-i d)^2}\right )+\frac{b ((-d-i c) \log (-\tan (e+f x)+i)+i (c+i d) \log (\tan (e+f x)+i)+2 d \log (c+d \tan (e+f x)))}{c^2+d^2}}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.033, size = 301, normalized size = 2.7 \begin{align*} -{\frac{a\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) b{c}^{2}}{2\,f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) b{d}^{2}}{2\,f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{a\arctan \left ( \tan \left ( fx+e \right ) \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{a\arctan \left ( \tan \left ( fx+e \right ) \right ){d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) bcd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{ad}{f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}+{\frac{bc}{f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}+2\,{\frac{a\ln \left ( c+d\tan \left ( fx+e \right ) \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) b{c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) b{d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.67479, size = 239, normalized size = 2.15 \begin{align*} \frac{\frac{2 \,{\left (a c^{2} + 2 \, b c d - a d^{2}\right )}{\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} - \frac{2 \,{\left (b c^{2} - 2 \, a c d - b d^{2}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac{{\left (b c^{2} - 2 \, a c d - b d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac{2 \,{\left (b c - a d\right )}}{c^{3} + c d^{2} +{\left (c^{2} d + d^{3}\right )} \tan \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.38805, size = 489, normalized size = 4.41 \begin{align*} \frac{2 \, b c d^{2} - 2 \, a d^{3} + 2 \,{\left (a c^{3} + 2 \, b c^{2} d - a c d^{2}\right )} f x -{\left (b c^{3} - 2 \, a c^{2} d - b c d^{2} +{\left (b c^{2} d - 2 \, a c d^{2} - b d^{3}\right )} \tan \left (f x + e\right )\right )} \log \left (\frac{d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \,{\left (b c^{2} d - a c d^{2} -{\left (a c^{2} d + 2 \, b c d^{2} - a d^{3}\right )} f x\right )} \tan \left (f x + e\right )}{2 \,{\left ({\left (c^{4} d + 2 \, c^{2} d^{3} + d^{5}\right )} f \tan \left (f x + e\right ) +{\left (c^{5} + 2 \, c^{3} d^{2} + c d^{4}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.30577, size = 325, normalized size = 2.93 \begin{align*} \frac{\frac{2 \,{\left (a c^{2} + 2 \, b c d - a d^{2}\right )}{\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac{{\left (b c^{2} - 2 \, a c d - b d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} - \frac{2 \,{\left (b c^{2} d - 2 \, a c d^{2} - b d^{3}\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{c^{4} d + 2 \, c^{2} d^{3} + d^{5}} + \frac{2 \,{\left (b c^{2} d \tan \left (f x + e\right ) - 2 \, a c d^{2} \tan \left (f x + e\right ) - b d^{3} \tan \left (f x + e\right ) + 2 \, b c^{3} - 3 \, a c^{2} d - a d^{3}\right )}}{{\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )}{\left (d \tan \left (f x + e\right ) + c\right )}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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