3.470 \(\int \frac{\tan ^3(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=114 \[ \frac{a^3}{b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{a^2 \left (a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2}+\frac{\left (a^2-b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{2 a b x}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.161869, antiderivative size = 121, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3565, 3626, 3617, 31, 3475} \[ -\frac{a^2 \tan (c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{a^2 \left (a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2}+\frac{\left (a^2-b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{2 a b x}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + b*Tan[c + d*x])^2,x]

[Out]

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

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\tan ^3(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=-\frac{a^2 \tan (c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{a^2-a b \tan (c+d x)+\left (a^2+b^2\right ) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{2 a b x}{\left (a^2+b^2\right )^2}-\frac{a^2 \tan (c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\left (a^2-b^2\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^2}+\frac{\left (a^2 \left (a^2+3 b^2\right )\right ) \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )^2}\\ &=-\frac{2 a b x}{\left (a^2+b^2\right )^2}+\frac{\left (a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac{a^2 \tan (c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (a^2 \left (a^2+3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^2 \left (a^2+b^2\right )^2 d}\\ &=-\frac{2 a b x}{\left (a^2+b^2\right )^2}+\frac{\left (a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac{a^2 \left (a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan (c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 1.03153, size = 251, normalized size = 2.2 \[ \frac{-2 i a^2 \left (a^2+3 b^2\right ) \tan ^{-1}(\tan (c+d x)) (a+b \tan (c+d x))+a \left (-2 \left (a^2+b^2\right )^2 \log (\cos (c+d x))+a \left (a \left (a^2+3 b^2\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+2 (2 b+i a) (a+i b)^2 (c+d x)\right )\right )+b \tan (c+d x) \left (-2 \left (a^2+b^2\right )^2 \log (\cos (c+d x))+a \left (a \left (a^2+3 b^2\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+2 i \left (a^3 (c+d x+i)+a b^2 (3 c+3 d x+i)+2 i b^3 (c+d x)\right )\right )\right )}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + b*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.024, size = 170, normalized size = 1.5 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){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}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-2\,{\frac{ab\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{{a}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{2}d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+3\,{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{{a}^{3}}{ \left ({a}^{2}+{b}^{2} \right ){b}^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3/(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.50756, size = 209, normalized size = 1.83 \begin{align*} \frac{\frac{2 \, a^{3}}{a^{3} b^{2} + a b^{4} +{\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )} - \frac{4 \,{\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (a^{4} + 3 \, a^{2} b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} - \frac{{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.33256, size = 510, normalized size = 4.47 \begin{align*} -\frac{4 \, a^{2} b^{3} d x - 2 \, a^{3} b^{2} -{\left (a^{5} + 3 \, a^{3} b^{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 ) +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \,{\left (2 \, a b^{4} d x + a^{4} b\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} d \tan \left (d x + c\right ) +{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.7659, size = 244, normalized size = 2.14 \begin{align*} -\frac{\frac{4 \,{\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\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^{4} + 3 \, a^{2} b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (a^{4} \tan \left (d x + c\right ) + 3 \, a^{2} b^{2} \tan \left (d x + c\right ) + 2 \, a^{3} b\right )}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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