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

Optimal. Leaf size=72 \[ -\frac{b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+a x \left (a^2-3 b^2\right )+\frac{2 a b^2 \tan (c+d x)}{d}+\frac{b (a+b \tan (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0513432, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3482, 3525, 3475} \[ -\frac{b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+a x \left (a^2-3 b^2\right )+\frac{2 a b^2 \tan (c+d x)}{d}+\frac{b (a+b \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^3,x]

[Out]

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

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

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

Mathematica [C]  time = 0.244122, size = 79, normalized size = 1.1 \[ \frac{6 a b^2 \tan (c+d x)+(-b+i a)^3 \log (-\tan (c+d x)+i)-(b+i a)^3 \log (\tan (c+d x)+i)+b^3 \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^3,x]

[Out]

((I*a - b)^3*Log[I - Tan[c + d*x]] - (I*a + b)^3*Log[I + Tan[c + d*x]] + 6*a*b^2*Tan[c + d*x] + b^3*Tan[c + d*
x]^2)/(2*d)

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Maple [A]  time = 0.003, size = 102, normalized size = 1.4 \begin{align*}{\frac{{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) b{a}^{2}}{2\,d}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){b}^{3}}{2\,d}}+{\frac{{a}^{3}\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.63881, size = 105, normalized size = 1.46 \begin{align*} a^{3} x - \frac{3 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a b^{2}}{d} - \frac{b^{3}{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{2 \, d} + \frac{3 \, a^{2} b \log \left (\sec \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.7461, size = 163, normalized size = 2.26 \begin{align*} \frac{b^{3} \tan \left (d x + c\right )^{2} + 6 \, a b^{2} \tan \left (d x + c\right ) + 2 \,{\left (a^{3} - 3 \, a b^{2}\right )} d x -{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.29789, size = 94, normalized size = 1.31 \begin{align*} \begin{cases} a^{3} x + \frac{3 a^{2} b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 3 a b^{2} x + \frac{3 a b^{2} \tan{\left (c + d x \right )}}{d} - \frac{b^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{b^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.76174, size = 814, normalized size = 11.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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