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

Optimal. Leaf size=103 \[ \frac{b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}-\frac{4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+x \left (-6 a^2 b^2+a^4+b^4\right )+\frac{b (a+b \tan (c+d x))^3}{3 d}+\frac{a b (a+b \tan (c+d x))^2}{d} \]

[Out]

(a^4 - 6*a^2*b^2 + b^4)*x - (4*a*b*(a^2 - b^2)*Log[Cos[c + d*x]])/d + (b^2*(3*a^2 - b^2)*Tan[c + d*x])/d + (a*
b*(a + b*Tan[c + d*x])^2)/d + (b*(a + b*Tan[c + d*x])^3)/(3*d)

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Rubi [A]  time = 0.105084, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3482, 3528, 3525, 3475} \[ \frac{b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}-\frac{4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+x \left (-6 a^2 b^2+a^4+b^4\right )+\frac{b (a+b \tan (c+d x))^3}{3 d}+\frac{a b (a+b \tan (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^4,x]

[Out]

(a^4 - 6*a^2*b^2 + b^4)*x - (4*a*b*(a^2 - b^2)*Log[Cos[c + d*x]])/d + (b^2*(3*a^2 - b^2)*Tan[c + d*x])/d + (a*
b*(a + b*Tan[c + d*x])^2)/d + (b*(a + b*Tan[c + d*x])^3)/(3*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 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 (a+b \tan (c+d x))^4 \, dx &=\frac{b (a+b \tan (c+d x))^3}{3 d}+\int (a+b \tan (c+d x))^2 \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=\frac{a b (a+b \tan (c+d x))^2}{d}+\frac{b (a+b \tan (c+d x))^3}{3 d}+\int (a+b \tan (c+d x)) \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\left (a^4-6 a^2 b^2+b^4\right ) x+\frac{b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+\frac{a b (a+b \tan (c+d x))^2}{d}+\frac{b (a+b \tan (c+d x))^3}{3 d}+\left (4 a b \left (a^2-b^2\right )\right ) \int \tan (c+d x) \, dx\\ &=\left (a^4-6 a^2 b^2+b^4\right ) x-\frac{4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac{b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+\frac{a b (a+b \tan (c+d x))^2}{d}+\frac{b (a+b \tan (c+d x))^3}{3 d}\\ \end{align*}

Mathematica [C]  time = 0.355845, size = 105, normalized size = 1.02 \[ \frac{-6 b^2 \left (b^2-6 a^2\right ) \tan (c+d x)+12 a b^3 \tan ^2(c+d x)+3 i (a-i b)^4 \log (\tan (c+d x)+i)-3 i (a+i b)^4 \log (-\tan (c+d x)+i)+2 b^4 \tan ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^4,x]

[Out]

((-3*I)*(a + I*b)^4*Log[I - Tan[c + d*x]] + (3*I)*(a - I*b)^4*Log[I + Tan[c + d*x]] - 6*b^2*(-6*a^2 + b^2)*Tan
[c + d*x] + 12*a*b^3*Tan[c + d*x]^2 + 2*b^4*Tan[c + d*x]^3)/(6*d)

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Maple [A]  time = 0.003, size = 152, normalized size = 1.5 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{b}^{4}}{3\,d}}+2\,{\frac{{b}^{3}a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+6\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}-{\frac{{b}^{4}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{3}b}{d}}-2\,{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) a{b}^{3}}{d}}+{\frac{{a}^{4}\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}}-6\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}{b}^{2}}{d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{4}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.5532, size = 153, normalized size = 1.49 \begin{align*} a^{4} x - \frac{6 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b^{2}}{d} + \frac{{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} b^{4}}{3 \, d} - \frac{2 \, a 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 )}}{d} + \frac{4 \, a^{3} 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))^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.00038, size = 227, normalized size = 2.2 \begin{align*} \frac{b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 3 \,{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x - 6 \,{\left (a^{3} b - a b^{3}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 3 \,{\left (6 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.487494, size = 131, normalized size = 1.27 \begin{align*} \begin{cases} a^{4} x + \frac{2 a^{3} b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - 6 a^{2} b^{2} x + \frac{6 a^{2} b^{2} \tan{\left (c + d x \right )}}{d} - \frac{2 a b^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac{2 a b^{3} \tan ^{2}{\left (c + d x \right )}}{d} + b^{4} x + \frac{b^{4} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac{b^{4} \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right )^{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x + 2*a**3*b*log(tan(c + d*x)**2 + 1)/d - 6*a**2*b**2*x + 6*a**2*b**2*tan(c + d*x)/d - 2*a*b**
3*log(tan(c + d*x)**2 + 1)/d + 2*a*b**3*tan(c + d*x)**2/d + b**4*x + b**4*tan(c + d*x)**3/(3*d) - b**4*tan(c +
d*x)/d, Ne(d, 0)), (x*(a + b*tan(c))**4, True))

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Giac [B]  time = 2.27849, size = 1446, normalized size = 14.04 \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))^4,x, algorithm="giac")

[Out]

1/3*(3*a^4*d*x*tan(d*x)^3*tan(c)^3 - 18*a^2*b^2*d*x*tan(d*x)^3*tan(c)^3 + 3*b^4*d*x*tan(d*x)^3*tan(c)^3 - 6*a^
3*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)^3*tan(c)^3 + 6*a*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)^3*tan(c)^3 - 9*a^4*d*x*tan(d*x)^2*tan
(c)^2 + 54*a^2*b^2*d*x*tan(d*x)^2*tan(c)^2 - 9*b^4*d*x*tan(d*x)^2*tan(c)^2 + 6*a*b^3*tan(d*x)^3*tan(c)^3 + 18*
a^3*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*t
an(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^2 - 18*a*b^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*t
an(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 - 18*a^2*b^2*tan(d*x)^3
*tan(c)^2 + 3*b^4*tan(d*x)^3*tan(c)^2 - 18*a^2*b^2*tan(d*x)^2*tan(c)^3 + 3*b^4*tan(d*x)^2*tan(c)^3 + 9*a^4*d*x
*tan(d*x)*tan(c) - 54*a^2*b^2*d*x*tan(d*x)*tan(c) + 9*b^4*d*x*tan(d*x)*tan(c) + 6*a*b^3*tan(d*x)^3*tan(c) - 6*
a*b^3*tan(d*x)^2*tan(c)^2 + 6*a*b^3*tan(d*x)*tan(c)^3 - b^4*tan(d*x)^3 - 18*a^3*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) + 18*a*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) + 36*a^2*b^2*tan(d*x)^2*tan(c) - 9*b^4*tan(d*x)^2*tan(c) + 36*a^2
*b^2*tan(d*x)*tan(c)^2 - 9*b^4*tan(d*x)*tan(c)^2 - b^4*tan(c)^3 - 3*a^4*d*x + 18*a^2*b^2*d*x - 3*b^4*d*x - 6*a
*b^3*tan(d*x)^2 + 6*a*b^3*tan(d*x)*tan(c) - 6*a*b^3*tan(c)^2 + 6*a^3*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)) - 6*a*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))
- 18*a^2*b^2*tan(d*x) + 3*b^4*tan(d*x) - 18*a^2*b^2*tan(c) + 3*b^4*tan(c) - 6*a*b^3)/(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)