3.482 \(\int \frac{\tan (c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=129 \[ \frac{a}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a^2-b^2}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{a \left (a^2-3 b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.165085, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3529, 3531, 3530} \[ \frac{a}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a^2-b^2}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{a \left (a^2-3 b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + b*Tan[c + d*x])^3,x]

[Out]

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

Mathematica [C]  time = 3.44904, size = 234, normalized size = 1.81 \[ \frac{-\frac{2 b}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{4 a b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2}+a \left (\frac{b \left (\frac{\left (a^2+b^2\right ) \left (5 a^2+4 a b \tan (c+d x)+b^2\right )}{(a+b \tan (c+d x))^2}+\left (2 b^2-6 a^2\right ) \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^3}+\frac{i \log (-\tan (c+d x)+i)}{(a+i b)^3}-\frac{\log (\tan (c+d x)+i)}{(b+i a)^3}\right )-\frac{i \log (-\tan (c+d x)+i)}{(a+i b)^2}+\frac{i \log (\tan (c+d x)+i)}{(a-i b)^2}}{2 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + b*Tan[c + d*x])^3,x]

[Out]

(((-I)*Log[I - Tan[c + d*x]])/(a + I*b)^2 + (I*Log[I + Tan[c + d*x]])/(a - I*b)^2 + (4*a*b*Log[a + b*Tan[c + d
*x]])/(a^2 + b^2)^2 - (2*b)/((a^2 + b^2)*(a + b*Tan[c + d*x])) + a*((I*Log[I - Tan[c + d*x]])/(a + I*b)^3 - Lo
g[I + Tan[c + d*x]]/(I*a + b)^3 + (b*((-6*a^2 + 2*b^2)*Log[a + b*Tan[c + d*x]] + ((a^2 + b^2)*(5*a^2 + b^2 + 4
*a*b*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2))/(a^2 + b^2)^3))/(2*b*d)

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Maple [A]  time = 0.033, size = 249, normalized size = 1.9 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) a{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{a}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{a\ln \left ( a+b\tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+b*tan(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.60783, size = 342, normalized size = 2.65 \begin{align*} \frac{\frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{3 \, a^{3} - a b^{2} + 2 \,{\left (a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} +{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(2*(3*a^2*b - b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(a^3 - 3*a*b^2)*log(b*tan(d*x + c) +
a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4
+ b^6) + (3*a^3 - a*b^2 + 2*(a^2*b - b^3)*tan(d*x + c))/(a^6 + 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 + 2*a^2*b^4 + b^
6)*tan(d*x + c)^2 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*tan(d*x + c)))/d

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Fricas [B]  time = 1.65872, size = 703, normalized size = 5.45 \begin{align*} \frac{5 \, a^{3} b^{2} - a b^{4} + 2 \,{\left (3 \, a^{4} b - a^{2} b^{3}\right )} d x -{\left (3 \, a^{3} b^{2} - 3 \, a b^{4} - 2 \,{\left (3 \, a^{2} b^{3} - b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} -{\left (a^{5} - 3 \, a^{3} b^{2} +{\left (a^{3} b^{2} - 3 \, a b^{4}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b - 3 \, a^{2} b^{3}\right )} \tan \left (d x + c\right )\right )} \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 (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5} - 2 \,{\left (3 \, a^{3} b^{2} - a b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) +{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(5*a^3*b^2 - a*b^4 + 2*(3*a^4*b - a^2*b^3)*d*x - (3*a^3*b^2 - 3*a*b^4 - 2*(3*a^2*b^3 - b^5)*d*x)*tan(d*x +
c)^2 - (a^5 - 3*a^3*b^2 + (a^3*b^2 - 3*a*b^4)*tan(d*x + c)^2 + 2*(a^4*b - 3*a^2*b^3)*tan(d*x + c))*log((b^2*t
an(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) - 2*(2*a^4*b - 3*a^2*b^3 + b^5 - 2*(3*a^3*b^2
- a*b^4)*d*x)*tan(d*x + c))/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*tan(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 +
3*a^3*b^5 + a*b^7)*d*tan(d*x + c) + (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*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(tan(d*x+c)/(a+b*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.30328, size = 371, normalized size = 2.88 \begin{align*} \frac{\frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (a^{3} b - 3 \, a b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} + \frac{3 \, a^{3} b^{2} \tan \left (d x + c\right )^{2} - 9 \, a b^{4} \tan \left (d x + c\right )^{2} + 8 \, a^{4} b \tan \left (d x + c\right ) - 18 \, a^{2} b^{3} \tan \left (d x + c\right ) - 2 \, b^{5} \tan \left (d x + c\right ) + 6 \, a^{5} - 7 \, a^{3} b^{2} - a b^{4}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(2*(3*a^2*b - b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1)
/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(a^3*b - 3*a*b^3)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3
*a^2*b^5 + b^7) + (3*a^3*b^2*tan(d*x + c)^2 - 9*a*b^4*tan(d*x + c)^2 + 8*a^4*b*tan(d*x + c) - 18*a^2*b^3*tan(d
*x + c) - 2*b^5*tan(d*x + c) + 6*a^5 - 7*a^3*b^2 - a*b^4)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(b*tan(d*x + c)
+ a)^2))/d