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

Optimal. Leaf size=101 \[ -\frac{a^2-b^2}{d \left (a^2+b^2\right ) (a \tan (c+d x)+b)}+\frac{b \left (3 a^2-b^2\right ) \log (a \sin (c+d x)+b \cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^2} \]

[Out]

-((a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^2) + (b*(3*a^2 - b^2)*Log[b*Cos[c + d*x] + a*Sin[c + d*x]])/((a^2 + b^2)^2*d
) - (a^2 - b^2)/((a^2 + b^2)*d*(b + a*Tan[c + d*x]))

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Rubi [A]  time = 0.127962, antiderivative size = 101, 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} \[ -\frac{a^2-b^2}{d \left (a^2+b^2\right ) (a \tan (c+d x)+b)}+\frac{b \left (3 a^2-b^2\right ) \log (a \sin (c+d x)+b \cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^2) + (b*(3*a^2 - b^2)*Log[b*Cos[c + d*x] + a*Sin[c + d*x]])/((a^2 + b^2)^2*d
) - (a^2 - b^2)/((a^2 + b^2)*d*(b + a*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 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 (c+d x)}{(b+a \tan (c+d x))^2} \, dx &=-\frac{a^2-b^2}{\left (a^2+b^2\right ) d (b+a \tan (c+d x))}+\frac{\int \frac{2 a b-\left (a^2-b^2\right ) \tan (c+d x)}{b+a \tan (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^2}-\frac{a^2-b^2}{\left (a^2+b^2\right ) d (b+a \tan (c+d x))}+\frac{\left (b \left (3 a^2-b^2\right )\right ) \int \frac{a-b \tan (c+d x)}{b+a \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^2}+\frac{b \left (3 a^2-b^2\right ) \log (b \cos (c+d x)+a \sin (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac{a^2-b^2}{\left (a^2+b^2\right ) d (b+a \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 1.94609, size = 187, normalized size = 1.85 \[ \frac{\frac{b (-(a+i b) \log (-\tan (c+d x)+i)-(a-i b) \log (\tan (c+d x)+i)+2 a \log (a \tan (c+d x)+b))}{a^2+b^2}+(a-b) (a+b) \left (\frac{2 a \left (2 b \log (a \tan (c+d x)+b)-\frac{a^2+b^2}{a \tan (c+d x)+b}\right )}{\left (a^2+b^2\right )^2}+\frac{i \log (-\tan (c+d x)+i)}{(a-i b)^2}-\frac{i \log (\tan (c+d x)+i)}{(a+i b)^2}\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.039, size = 222, normalized size = 2.2 \begin{align*} -{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) b{a}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){b}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) \left ( b+a\tan \left ( dx+c \right ) \right ) }}+{\frac{{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) \left ( b+a\tan \left ( dx+c \right ) \right ) }}+3\,{\frac{b\ln \left ( b+a\tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{b}^{3}\ln \left ( b+a\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2/d/(a^2+b^2)^2*ln(1+tan(d*x+c)^2)*b*a^2+1/2/d/(a^2+b^2)^2*ln(1+tan(d*x+c)^2)*b^3-1/d/(a^2+b^2)^2*arctan(ta
n(d*x+c))*a^3+3/d/(a^2+b^2)^2*arctan(tan(d*x+c))*a*b^2-1/d/(a^2+b^2)/(b+a*tan(d*x+c))*a^2+1/d/(a^2+b^2)/(b+a*t
an(d*x+c))*b^2+3/d*b/(a^2+b^2)^2*ln(b+a*tan(d*x+c))*a^2-1/d*b^3/(a^2+b^2)^2*ln(b+a*tan(d*x+c))

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Maxima [A]  time = 1.7985, size = 217, normalized size = 2.15 \begin{align*} -\frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )} \log \left (a \tan \left (d x + c\right ) + b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (3 \, a^{2} b - b^{3}\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^{2}\right )}}{a^{2} b + b^{3} +{\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.71119, size = 417, normalized size = 4.13 \begin{align*} -\frac{2 \, a^{4} - 2 \, a^{2} b^{2} + 2 \,{\left (a^{3} b - 3 \, a b^{3}\right )} d x -{\left (3 \, a^{2} b^{2} - b^{4} +{\left (3 \, a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{a^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + b^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (a^{3} b - a b^{3} -{\left (a^{4} - 3 \, a^{2} b^{2}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \tan \left (d x + c\right ) +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.23514, size = 269, normalized size = 2.66 \begin{align*} -\frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (3 \, a^{3} b - a b^{3}\right )} \log \left ({\left | a \tan \left (d x + c\right ) + b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac{2 \,{\left (3 \, a^{3} b \tan \left (d x + c\right ) - a b^{3} \tan \left (d x + c\right ) + a^{4} + 3 \, a^{2} b^{2} - 2 \, b^{4}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a \tan \left (d x + c\right ) + b\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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