3.458 \(\int \frac{\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=97 \[ \frac{a^4 \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )}+\frac{b \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac{a x}{a^2+b^2}-\frac{a \tan (c+d x)}{b^2 d}+\frac{\tan ^2(c+d x)}{2 b d} \]

[Out]

(a*x)/(a^2 + b^2) + (b*Log[Cos[c + d*x]])/((a^2 + b^2)*d) + (a^4*Log[a + b*Tan[c + d*x]])/(b^3*(a^2 + b^2)*d)
- (a*Tan[c + d*x])/(b^2*d) + Tan[c + d*x]^2/(2*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.211595, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3566, 3647, 3627, 3617, 31, 3475} \[ \frac{a^4 \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )}+\frac{b \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac{a x}{a^2+b^2}-\frac{a \tan (c+d x)}{b^2 d}+\frac{\tan ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + b*Tan[c + d*x]),x]

[Out]

(a*x)/(a^2 + b^2) + (b*Log[Cos[c + d*x]])/((a^2 + b^2)*d) + (a^4*Log[a + b*Tan[c + d*x]])/(b^3*(a^2 + b^2)*d)
- (a*Tan[c + d*x])/(b^2*d) + Tan[c + d*x]^2/(2*b*d)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3627

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

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 ^4(c+d x)}{a+b \tan (c+d x)} \, dx &=\frac{\tan ^2(c+d x)}{2 b d}+\frac{\int \frac{\tan (c+d x) \left (-2 a-2 b \tan (c+d x)-2 a \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b}\\ &=-\frac{a \tan (c+d x)}{b^2 d}+\frac{\tan ^2(c+d x)}{2 b d}+\frac{\int \frac{2 a^2+2 \left (a^2-b^2\right ) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{2 b^2}\\ &=\frac{a x}{a^2+b^2}-\frac{a \tan (c+d x)}{b^2 d}+\frac{\tan ^2(c+d x)}{2 b d}+\frac{a^4 \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^2 \left (a^2+b^2\right )}-\frac{b \int \tan (c+d x) \, dx}{a^2+b^2}\\ &=\frac{a x}{a^2+b^2}+\frac{b \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac{a \tan (c+d x)}{b^2 d}+\frac{\tan ^2(c+d x)}{2 b d}+\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 \left (a^2+b^2\right ) d}\\ &=\frac{a x}{a^2+b^2}+\frac{b \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac{a^4 \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right ) d}-\frac{a \tan (c+d x)}{b^2 d}+\frac{\tan ^2(c+d x)}{2 b d}\\ \end{align*}

Mathematica [C]  time = 0.402089, size = 107, normalized size = 1.1 \[ \frac{\frac{2 a^4 \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right )}-\frac{2 a \tan (c+d x)}{b^2}+\frac{\log (-\tan (c+d x)+i)}{-b+i a}-\frac{\log (\tan (c+d x)+i)}{b+i a}+\frac{\tan ^2(c+d x)}{b}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + b*Tan[c + d*x]),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.019, size = 110, normalized size = 1.1 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,bd}}-{\frac{a\tan \left ( dx+c \right ) }{{b}^{2}d}}-{\frac{b\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{a\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{{a}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{3} \left ({a}^{2}+{b}^{2} \right ) d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+b*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.50073, size = 134, normalized size = 1.38 \begin{align*} \frac{\frac{2 \, a^{4} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{3} + b^{5}} + \frac{2 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac{b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{b \tan \left (d x + c\right )^{2} - 2 \, a \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.2621, size = 306, normalized size = 3.15 \begin{align*} \frac{2 \, a b^{3} d x + a^{4} \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^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2} -{\left (a^{4} - b^{4}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{2 \,{\left (a^{2} b^{3} + b^{5}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 11.233, size = 677, normalized size = 6.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+b*tan(d*x+c)),x)

[Out]

Piecewise((zoo*x*tan(c)**3, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (3*I*d*x*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*
b*d) + 3*d*x/(-2*b*d*tan(c + d*x) + 2*I*b*d) + 2*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*b*d*tan(c + d*x) +
2*I*b*d) - 2*I*log(tan(c + d*x)**2 + 1)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - tan(c + d*x)**3/(-2*b*d*tan(c + d*x)
+ 2*I*b*d) - I*tan(c + d*x)**2/(-2*b*d*tan(c + d*x) + 2*I*b*d) - 3*I/(-2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, -
I*b)), (3*I*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) - 3*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - 2*log(tan
(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) - 2*I*log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d*
x) + 2*I*b*d) + tan(c + d*x)**3/(2*b*d*tan(c + d*x) + 2*I*b*d) - I*tan(c + d*x)**2/(2*b*d*tan(c + d*x) + 2*I*b
*d) - 3*I/(2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, I*b)), ((x + tan(c + d*x)**3/(3*d) - tan(c + d*x)/d)/a, Eq(b,
0)), (x*tan(c)**4/(a + b*tan(c)), Eq(d, 0)), (2*a**4*log(a/b + tan(c + d*x))/(2*a**2*b**3*d + 2*b**5*d) - 2*a*
*3*b*tan(c + d*x)/(2*a**2*b**3*d + 2*b**5*d) + a**2*b**2*tan(c + d*x)**2/(2*a**2*b**3*d + 2*b**5*d) + 2*a*b**3
*d*x/(2*a**2*b**3*d + 2*b**5*d) - 2*a*b**3*tan(c + d*x)/(2*a**2*b**3*d + 2*b**5*d) - b**4*log(tan(c + d*x)**2
+ 1)/(2*a**2*b**3*d + 2*b**5*d) + b**4*tan(c + d*x)**2/(2*a**2*b**3*d + 2*b**5*d), True))

________________________________________________________________________________________

Giac [A]  time = 2.32486, size = 135, normalized size = 1.39 \begin{align*} \frac{\frac{2 \, a^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{3} + b^{5}} + \frac{2 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac{b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{b \tan \left (d x + c\right )^{2} - 2 \, a \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

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