3.413 \(\int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=77 \[ \frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+a x+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}-\frac {b \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3528, 3525, 3475} \[ \frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+a x+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}-\frac {b \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3475

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 79, normalized size = 1.03 \[ \frac {a \tan ^{-1}(\tan (c+d x))}{d}+\frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}-\frac {b \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 69, normalized size = 0.90 \[ \frac {3 \, b \tan \left (d x + c\right )^{4} + 4 \, a \tan \left (d x + c\right )^{3} + 12 \, a d x - 6 \, b \tan \left (d x + c\right )^{2} - 6 \, b \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 12 \, a \tan \left (d x + c\right )}{12 \, d} \]

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]

1/12*(3*b*tan(d*x + c)^4 + 4*a*tan(d*x + c)^3 + 12*a*d*x - 6*b*tan(d*x + c)^2 - 6*b*log(1/(tan(d*x + c)^2 + 1)
) - 12*a*tan(d*x + c))/d

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giac [B]  time = 6.46, size = 716, normalized size = 9.30 \[ \text {result too large to display} \]

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]

1/12*(12*a*d*x*tan(d*x)^4*tan(c)^4 - 6*b*log(4*(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(c)^2 + 1))*tan(d*x)^4*tan(c)^4 - 48*a*d*x*tan(d*x)^3*tan(c)^3 - 9
*b*tan(d*x)^4*tan(c)^4 + 24*b*log(4*(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(c)^2 + 1))*tan(d*x)^3*tan(c)^3 + 12*a*tan(d*x)^4*tan(c)^3 + 12*a*tan(d*x)^3*
tan(c)^4 + 72*a*d*x*tan(d*x)^2*tan(c)^2 - 6*b*tan(d*x)^4*tan(c)^2 + 24*b*tan(d*x)^3*tan(c)^3 - 6*b*tan(d*x)^2*
tan(c)^4 - 4*a*tan(d*x)^4*tan(c) - 36*b*log(4*(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(c)^2 + 1))*tan(d*x)^2*tan(c)^2 - 48*a*tan(d*x)^3*tan(c)^2 - 48*a*t
an(d*x)^2*tan(c)^3 - 4*a*tan(d*x)*tan(c)^4 + 3*b*tan(d*x)^4 - 48*a*d*x*tan(d*x)*tan(c) + 24*b*tan(d*x)^3*tan(c
) - 12*b*tan(d*x)^2*tan(c)^2 + 24*b*tan(d*x)*tan(c)^3 + 3*b*tan(c)^4 + 4*a*tan(d*x)^3 + 24*b*log(4*(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(c)^2 + 1))*ta
n(d*x)*tan(c) + 48*a*tan(d*x)^2*tan(c) + 48*a*tan(d*x)*tan(c)^2 + 4*a*tan(c)^3 + 12*a*d*x - 6*b*tan(d*x)^2 + 2
4*b*tan(d*x)*tan(c) - 6*b*tan(c)^2 - 6*b*log(4*(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(c)^2 + 1)) - 12*a*tan(d*x) - 12*a*tan(c) - 9*b)/(d*tan(d*x)^4*tan
(c)^4 - 4*d*tan(d*x)^3*tan(c)^3 + 6*d*tan(d*x)^2*tan(c)^2 - 4*d*tan(d*x)*tan(c) + d)

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maple [A]  time = 0.02, size = 85, normalized size = 1.10 \[ \frac {b \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {b \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \tan \left (d x +c \right )}{d}+\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

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]

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

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maxima [A]  time = 0.61, size = 70, normalized size = 0.91 \[ \frac {3 \, b \tan \left (d x + c\right )^{4} + 4 \, a \tan \left (d x + c\right )^{3} - 6 \, b \tan \left (d x + c\right )^{2} + 12 \, {\left (d x + c\right )} a + 6 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 12 \, a \tan \left (d x + c\right )}{12 \, d} \]

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]

1/12*(3*b*tan(d*x + c)^4 + 4*a*tan(d*x + c)^3 - 6*b*tan(d*x + c)^2 + 12*(d*x + c)*a + 6*b*log(tan(d*x + c)^2 +
1) - 12*a*tan(d*x + c))/d

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mupad [B]  time = 4.01, size = 65, normalized size = 0.84 \[ \frac {\frac {b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}-a\,\mathrm {tan}\left (c+d\,x\right )+\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}-\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}+a\,d\,x}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.43, size = 83, normalized size = 1.08 \[ \begin {cases} a x + \frac {a \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {a \tan {\left (c + d x \right )}}{d} + \frac {b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {b \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right ) \tan ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

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

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