3.414 \(\int \tan ^3(c+d x) (a+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=60 \[ \frac {a \tan ^2(c+d x)}{2 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {b \tan ^3(c+d x)}{3 d}-\frac {b \tan (c+d x)}{d}+b x \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3528, 3525, 3475} \[ \frac {a \tan ^2(c+d x)}{2 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {b \tan ^3(c+d x)}{3 d}-\frac {b \tan (c+d x)}{d}+b x \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(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)

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 ^3(c+d x) (a+b \tan (c+d x)) \, dx &=\frac {b \tan ^3(c+d x)}{3 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}+\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}-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}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 67, normalized size = 1.12 \[ \frac {a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d}+\frac {b \tan ^{-1}(\tan (c+d x))}{d}+\frac {b \tan ^3(c+d x)}{3 d}-\frac {b \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.49, size = 58, normalized size = 0.97 \[ \frac {2 \, b \tan \left (d x + c\right )^{3} + 6 \, b d x + 3 \, a \tan \left (d x + c\right )^{2} + 3 \, a \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 6 \, b \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 3.06, size = 515, normalized size = 8.58 \[ \frac {6 \, b d x \tan \left (d x\right )^{3} \tan \relax (c)^{3} + 3 \, a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \left (d x\right )^{3} \tan \relax (c)^{3} - 18 \, b d x \tan \left (d x\right )^{2} \tan \relax (c)^{2} + 3 \, a \tan \left (d x\right )^{3} \tan \relax (c)^{3} - 9 \, a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \left (d x\right )^{2} \tan \relax (c)^{2} + 6 \, b \tan \left (d x\right )^{3} \tan \relax (c)^{2} + 6 \, b \tan \left (d x\right )^{2} \tan \relax (c)^{3} + 18 \, b d x \tan \left (d x\right ) \tan \relax (c) + 3 \, a \tan \left (d x\right )^{3} \tan \relax (c) - 3 \, a \tan \left (d x\right )^{2} \tan \relax (c)^{2} + 3 \, a \tan \left (d x\right ) \tan \relax (c)^{3} - 2 \, b \tan \left (d x\right )^{3} + 9 \, a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \left (d x\right ) \tan \relax (c) - 18 \, b \tan \left (d x\right )^{2} \tan \relax (c) - 18 \, b \tan \left (d x\right ) \tan \relax (c)^{2} - 2 \, b \tan \relax (c)^{3} - 6 \, b d x - 3 \, a \tan \left (d x\right )^{2} + 3 \, a \tan \left (d x\right ) \tan \relax (c) - 3 \, a \tan \relax (c)^{2} - 3 \, a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) + 6 \, b \tan \left (d x\right ) + 6 \, b \tan \relax (c) - 3 \, a}{6 \, {\left (d \tan \left (d x\right )^{3} \tan \relax (c)^{3} - 3 \, d \tan \left (d x\right )^{2} \tan \relax (c)^{2} + 3 \, d \tan \left (d x\right ) \tan \relax (c) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*b*d*x*tan(d*x)^3*tan(c)^3 + 3*a*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 - 18*b*d*x*tan(d*x)^2*tan(c)^2 + 3*a
*tan(d*x)^3*tan(c)^3 - 9*a*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 + 6*b*tan(d*x)^3*tan(c)^2 + 6*b*tan(d*x)^2*tan(c
)^3 + 18*b*d*x*tan(d*x)*tan(c) + 3*a*tan(d*x)^3*tan(c) - 3*a*tan(d*x)^2*tan(c)^2 + 3*a*tan(d*x)*tan(c)^3 - 2*b
*tan(d*x)^3 + 9*a*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)*tan(c) - 18*b*tan(d*x)^2*tan(c) - 18*b*tan(d*x)*tan(c)^2 - 2*b*tan(c
)^3 - 6*b*d*x - 3*a*tan(d*x)^2 + 3*a*tan(d*x)*tan(c) - 3*a*tan(c)^2 - 3*a*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)) + 6*b*tan(d*x) + 6*b*
tan(c) - 3*a)/(d*tan(d*x)^3*tan(c)^3 - 3*d*tan(d*x)^2*tan(c)^2 + 3*d*tan(d*x)*tan(c) - d)

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maple [A]  time = 0.02, size = 71, normalized size = 1.18 \[ \frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b \tan \left (d x +c \right )}{d}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {b \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 = 0.92, size = 59, normalized size = 0.98 \[ \frac {2 \, b \tan \left (d x + c\right )^{3} + 3 \, a \tan \left (d x + c\right )^{2} + 6 \, {\left (d x + c\right )} b - 3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, b \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.04, size = 54, normalized size = 0.90 \[ \frac {\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}-\frac {a\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}-b\,\mathrm {tan}\left (c+d\,x\right )+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+b\,d\,x}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.31, size = 70, normalized size = 1.17 \[ \begin {cases} - \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} + b x + \frac {b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right ) \tan ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(a+b*tan(d*x+c)),x)

[Out]

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

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