3.428 \(\int (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=39 \[ x \left (a^2-b^2\right )-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3477, 3475} \[ x \left (a^2-b^2\right )-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^2,x]

[Out]

(a^2 - b^2)*x - (2*a*b*Log[Cos[c + d*x]])/d + (b^2*Tan[c + d*x])/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 3477

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

Rubi steps

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

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Mathematica [C]  time = 0.11, size = 69, normalized size = 1.77 \[ \frac {2 b^2 \tan (c+d x)-i \left ((a+i b)^2 \log (-\tan (c+d x)+i)-(a-i b)^2 \log (\tan (c+d x)+i)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.46, size = 44, normalized size = 1.13 \[ \frac {{\left (a^{2} - b^{2}\right )} d x - a b \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + b^{2} \tan \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.96, size = 201, normalized size = 5.15 \[ \frac {a^{2} d x \tan \left (d x\right ) \tan \relax (c) - b^{2} d x \tan \left (d x\right ) \tan \relax (c) - a b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \left (d x\right ) \tan \relax (c) - a^{2} d x + b^{2} d x + a b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) - b^{2} \tan \left (d x\right ) - b^{2} \tan \relax (c)}{d \tan \left (d x\right ) \tan \relax (c) - d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 61, normalized size = 1.56 \[ \frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{d}-\frac {\arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{d}+\frac {a b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {b^{2} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.85, size = 41, normalized size = 1.05 \[ a^{2} x - \frac {{\left (d x + c - \tan \left (d x + c\right )\right )} b^{2}}{d} + \frac {2 \, a b \log \left (\sec \left (d x + c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.09, size = 136, normalized size = 3.49 \[ \frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )}{a^2-b^2}-\frac {b^2\,\mathrm {tan}\left (c+d\,x\right )}{a^2-b^2}\right )}{d}-\frac {b^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )}{a^2-b^2}-\frac {b^2\,\mathrm {tan}\left (c+d\,x\right )}{a^2-b^2}\right )}{d}+\frac {b^2\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {a\,b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 0.18, size = 48, normalized size = 1.23 \[ \begin {cases} a^{2} x + \frac {a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - b^{2} x + \frac {b^{2} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**2,x)

[Out]

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

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