3.457 \(\int \frac{\tan ^5(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=125 \[ \frac{\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac{a^5 \log (a+b \tan (c+d x))}{b^4 d \left (a^2+b^2\right )}-\frac{a \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac{b x}{a^2+b^2}-\frac{a \tan ^2(c+d x)}{2 b^2 d}+\frac{\tan ^3(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.375735, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3566, 3647, 3648, 3626, 3617, 31, 3475} \[ \frac{\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac{a^5 \log (a+b \tan (c+d x))}{b^4 d \left (a^2+b^2\right )}-\frac{a \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac{b x}{a^2+b^2}-\frac{a \tan ^2(c+d x)}{2 b^2 d}+\frac{\tan ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + b*Tan[c + d*x]),x]

[Out]

(b*x)/(a^2 + b^2) - (a*Log[Cos[c + d*x]])/((a^2 + b^2)*d) - (a^5*Log[a + b*Tan[c + d*x]])/(b^4*(a^2 + b^2)*d)
+ ((a^2 - b^2)*Tan[c + d*x])/(b^3*d) - (a*Tan[c + d*x]^2)/(2*b^2*d) + Tan[c + d*x]^3/(3*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 3648

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (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 - b*C)*(m + n + 1)*Tan[e + f*x] - C*m*(b*c - a*d)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

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 ^5(c+d x)}{a+b \tan (c+d x)} \, dx &=\frac{\tan ^3(c+d x)}{3 b d}+\frac{\int \frac{\tan ^2(c+d x) \left (-3 a-3 b \tan (c+d x)-3 a \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{3 b}\\ &=-\frac{a \tan ^2(c+d x)}{2 b^2 d}+\frac{\tan ^3(c+d x)}{3 b d}+\frac{\int \frac{\tan (c+d x) \left (6 a^2+6 \left (a^2-b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{6 b^2}\\ &=\frac{\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac{a \tan ^2(c+d x)}{2 b^2 d}+\frac{\tan ^3(c+d x)}{3 b d}+\frac{\int \frac{-6 a \left (a^2-b^2\right )+6 b^3 \tan (c+d x)-6 a \left (a^2-b^2\right ) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{6 b^3}\\ &=\frac{b x}{a^2+b^2}+\frac{\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac{a \tan ^2(c+d x)}{2 b^2 d}+\frac{\tan ^3(c+d x)}{3 b d}+\frac{a \int \tan (c+d x) \, dx}{a^2+b^2}-\frac{a^5 \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^3 \left (a^2+b^2\right )}\\ &=\frac{b x}{a^2+b^2}-\frac{a \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac{\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac{a \tan ^2(c+d x)}{2 b^2 d}+\frac{\tan ^3(c+d x)}{3 b d}-\frac{a^5 \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^4 \left (a^2+b^2\right ) d}\\ &=\frac{b x}{a^2+b^2}-\frac{a \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac{a^5 \log (a+b \tan (c+d x))}{b^4 \left (a^2+b^2\right ) d}+\frac{\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac{a \tan ^2(c+d x)}{2 b^2 d}+\frac{\tan ^3(c+d x)}{3 b d}\\ \end{align*}

Mathematica [C]  time = 0.594649, size = 155, normalized size = 1.24 \[ \frac{a^2 \tan (c+d x)}{b^3 d}-\frac{a^5 \log (a+b \tan (c+d x))}{b^4 d \left (a^2+b^2\right )}-\frac{a \tan ^2(c+d x)}{2 b^2 d}+\frac{\log (-\tan (c+d x)+i)}{2 d (a+i b)}+\frac{\log (\tan (c+d x)+i)}{2 d (a-i b)}+\frac{\tan ^3(c+d x)}{3 b d}-\frac{\tan (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + b*Tan[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.02, size = 143, normalized size = 1.1 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,bd}}-{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,{b}^{2}d}}+{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d{b}^{3}}}-{\frac{\tan \left ( dx+c \right ) }{bd}}+{\frac{a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{b\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{{a}^{5}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{4} \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)^5/(a+b*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.5288, size = 166, normalized size = 1.33 \begin{align*} -\frac{\frac{6 \, a^{5} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{4} + b^{6}} - \frac{6 \,{\left (d x + c\right )} b}{a^{2} + b^{2}} - \frac{3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right )^{2} + 6 \,{\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{b^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.32172, size = 362, normalized size = 2.9 \begin{align*} \frac{6 \, b^{5} d x - 3 \, a^{5} \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 (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )^{3} - 3 \,{\left (a^{3} b^{2} + a b^{4}\right )} \tan \left (d x + c\right )^{2} + 3 \,{\left (a^{5} - a b^{4}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \,{\left (a^{4} b - b^{5}\right )} \tan \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} + b^{6}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+b*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 3.44017, size = 174, normalized size = 1.39 \begin{align*} -\frac{\frac{6 \, a^{5} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{4} + b^{6}} - \frac{6 \,{\left (d x + c\right )} b}{a^{2} + b^{2}} - \frac{3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right )^{2} + 6 \, a^{2} \tan \left (d x + c\right ) - 6 \, b^{2} \tan \left (d x + c\right )}{b^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

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