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

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

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 103, normalized size of antiderivative = 1.00, 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 3475

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

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 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 (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*}

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Mathematica [C]  time = 0.37, 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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fricas [A]  time = 0.43, size = 100, normalized size = 0.97 \[ \frac {b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x - 6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 3 \, {\left (6 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right )}{3 \, d} \]

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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giac [B]  time = 3.33, size = 1071, normalized size = 10.40 \[ \text {result too large to display} \]

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(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 + 6*a*b^3*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 - 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(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 - 18*a*b^3*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 - 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(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*a*b^3*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) + 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(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*a*b^3*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))
- 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)

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maple [A]  time = 0.02, size = 152, normalized size = 1.48 \[ \frac {b^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 a \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {6 a^{2} b^{2} \tan \left (d x +c \right )}{d}-\frac {b^{4} \tan \left (d x +c \right )}{d}+\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} b}{d}-\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a \,b^{3}}{d}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{4}}{d}-\frac {6 \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b^{2}}{d}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) b^{4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^4*tan(d*x+c)^3/d+2*a*b^3*tan(d*x+c)^2/d+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*arctan(tan(d*x+c))*a^4-6/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 = 0.76, size = 113, normalized size = 1.10 \[ a^{4} x - \frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b^{2}}{d} + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} b^{4}}{3 \, d} - \frac {2 \, a b^{3} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{d} + \frac {4 \, a^{3} b \log \left (\sec \left (d x + c\right )\right )}{d} \]

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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mupad [B]  time = 3.89, size = 167, normalized size = 1.62 \[ \frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (b^4-6\,a^2\,b^2\right )}{d}-\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (2\,a\,b^3-2\,a^3\,b\right )}{d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-a^2+2\,a\,b+b^2\right )\,\left (a^2+2\,a\,b-b^2\right )}{a^4-6\,a^2\,b^2+b^4}\right )\,\left (-a^2+2\,a\,b+b^2\right )\,\left (a^2+2\,a\,b-b^2\right )}{d}+\frac {2\,a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^4,x)

[Out]

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

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sympy [A]  time = 0.39, size = 131, normalized size = 1.27 \[ \begin {cases} a^{4} x + \frac {2 a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - 6 a^{2} b^{2} x + \frac {6 a^{2} b^{2} \tan {\left (c + d x \right )}}{d} - \frac {2 a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 a b^{3} \tan ^{2}{\left (c + d x \right )}}{d} + b^{4} x + \frac {b^{4} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{4} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right )^{4} & \text {otherwise} \end {cases} \]

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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