3.412 \(\int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=93 \[ \frac{a \tan ^4(c+d x)}{4 d}-\frac{a \tan ^2(c+d x)}{2 d}-\frac{a \log (\cos (c+d x))}{d}+\frac{b \tan ^5(c+d x)}{5 d}-\frac{b \tan ^3(c+d x)}{3 d}+\frac{b \tan (c+d x)}{d}-b x \]

[Out]

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

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Rubi [A]  time = 0.098645, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3528, 3525, 3475} \[ \frac{a \tan ^4(c+d x)}{4 d}-\frac{a \tan ^2(c+d x)}{2 d}-\frac{a \log (\cos (c+d x))}{d}+\frac{b \tan ^5(c+d x)}{5 d}-\frac{b \tan ^3(c+d x)}{3 d}+\frac{b \tan (c+d x)}{d}-b x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

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

Mathematica [A]  time = 0.286726, size = 95, normalized size = 1.02 \[ -\frac{a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d}+\frac{b \tan ^5(c+d x)}{5 d}-\frac{b \tan ^3(c+d x)}{3 d}-\frac{b \tan ^{-1}(\tan (c+d x))}{d}+\frac{b \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 99, normalized size = 1.1 \begin{align*}{\frac{b \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{b \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{b\tan \left ( dx+c \right ) }{d}}+{\frac{a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d}}-{\frac{b\arctan \left ( \tan \left ( dx+c \right ) \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]

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

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Maxima [A]  time = 1.72754, size = 109, normalized size = 1.17 \begin{align*} \frac{12 \, b \tan \left (d x + c\right )^{5} + 15 \, a \tan \left (d x + c\right )^{4} - 20 \, b \tan \left (d x + c\right )^{3} - 30 \, a \tan \left (d x + c\right )^{2} - 60 \,{\left (d x + c\right )} b + 30 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 60 \, b \tan \left (d x + c\right )}{60 \, 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]

1/60*(12*b*tan(d*x + c)^5 + 15*a*tan(d*x + c)^4 - 20*b*tan(d*x + c)^3 - 30*a*tan(d*x + c)^2 - 60*(d*x + c)*b +
30*a*log(tan(d*x + c)^2 + 1) + 60*b*tan(d*x + c))/d

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Fricas [A]  time = 1.65255, size = 219, normalized size = 2.35 \begin{align*} \frac{12 \, b \tan \left (d x + c\right )^{5} + 15 \, a \tan \left (d x + c\right )^{4} - 20 \, b \tan \left (d x + c\right )^{3} - 60 \, b d x - 30 \, a \tan \left (d x + c\right )^{2} - 30 \, a \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 60 \, b \tan \left (d x + c\right )}{60 \, 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]

1/60*(12*b*tan(d*x + c)^5 + 15*a*tan(d*x + c)^4 - 20*b*tan(d*x + c)^3 - 60*b*d*x - 30*a*tan(d*x + c)^2 - 30*a*
log(1/(tan(d*x + c)^2 + 1)) + 60*b*tan(d*x + c))/d

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Sympy [A]  time = 0.778538, size = 97, normalized size = 1.04 \begin{align*} \begin{cases} \frac{a \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{a \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac{a \tan ^{2}{\left (c + d x \right )}}{2 d} - b x + \frac{b \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac{b \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac{b \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right ) \tan ^{5}{\left (c \right )} & \text{otherwise} \end{cases} \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]

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

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Giac [B]  time = 5.97245, size = 1278, normalized size = 13.74 \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)^5*(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(60*b*d*x*tan(d*x)^5*tan(c)^5 + 30*a*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + t
an(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^5*tan(c)^5 - 300*b*d*x*tan(d*x)^4*tan(c)^4
+ 45*a*tan(d*x)^5*tan(c)^5 - 150*a*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^
2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^4*tan(c)^4 + 60*b*tan(d*x)^5*tan(c)^4 + 60*b*tan(d*
x)^4*tan(c)^5 + 600*b*d*x*tan(d*x)^3*tan(c)^3 + 30*a*tan(d*x)^5*tan(c)^3 - 165*a*tan(d*x)^4*tan(c)^4 + 30*a*ta
n(d*x)^3*tan(c)^5 - 20*b*tan(d*x)^5*tan(c)^2 + 300*a*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*
tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^3*tan(c)^3 - 300*b*tan(d*x)^4*tan
(c)^3 - 300*b*tan(d*x)^3*tan(c)^4 - 20*b*tan(d*x)^2*tan(c)^5 - 15*a*tan(d*x)^5*tan(c) - 600*b*d*x*tan(d*x)^2*t
an(c)^2 - 150*a*tan(d*x)^4*tan(c)^2 + 180*a*tan(d*x)^3*tan(c)^3 - 150*a*tan(d*x)^2*tan(c)^4 - 15*a*tan(d*x)*ta
n(c)^5 + 12*b*tan(d*x)^5 + 100*b*tan(d*x)^4*tan(c) - 300*a*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d
*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^2 + 600*b*tan(d*x)
^3*tan(c)^2 + 600*b*tan(d*x)^2*tan(c)^3 + 100*b*tan(d*x)*tan(c)^4 + 12*b*tan(c)^5 + 15*a*tan(d*x)^4 + 300*b*d*
x*tan(d*x)*tan(c) + 150*a*tan(d*x)^3*tan(c) - 180*a*tan(d*x)^2*tan(c)^2 + 150*a*tan(d*x)*tan(c)^3 + 15*a*tan(c
)^4 - 20*b*tan(d*x)^3 + 150*a*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan
(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)*tan(c) - 300*b*tan(d*x)^2*tan(c) - 300*b*tan(d*x)*tan(c)
^2 - 20*b*tan(c)^3 - 60*b*d*x - 30*a*tan(d*x)^2 + 165*a*tan(d*x)*tan(c) - 30*a*tan(c)^2 - 30*a*log(4*(tan(c)^2
+ 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))
+ 60*b*tan(d*x) + 60*b*tan(c) - 45*a)/(d*tan(d*x)^5*tan(c)^5 - 5*d*tan(d*x)^4*tan(c)^4 + 10*d*tan(d*x)^3*tan(c
)^3 - 10*d*tan(d*x)^2*tan(c)^2 + 5*d*tan(d*x)*tan(c) - d)