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

Optimal. Leaf size=58 \[ -\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {(a+b \tan (c+d x))^2}{2 d}+\frac {a b \tan (c+d x)}{d}-2 a b x \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3528, 3525, 3475} \[ -\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {(a+b \tan (c+d x))^2}{2 d}+\frac {a b \tan (c+d x)}{d}-2 a b x \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]*(a + b*Tan[c + d*x])^2,x]

[Out]

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

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Mathematica [C]  time = 0.28, size = 74, normalized size = 1.28 \[ \frac {4 a b \tan (c+d x)+(a-i b)^2 \log (\tan (c+d x)+i)+(a+i b)^2 \log (-\tan (c+d x)+i)+b^2 \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.47, size = 58, normalized size = 1.00 \[ -\frac {4 \, a b d x - b^{2} \tan \left (d x + c\right )^{2} - 4 \, a b \tan \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.15, size = 554, normalized size = 9.55 \[ -\frac {4 \, a b d x \tan \left (d x\right )^{2} \tan \relax (c)^{2} + a^{2} \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} - b^{2} \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} - 8 \, a b d x \tan \left (d x\right ) \tan \relax (c) - b^{2} \tan \left (d x\right )^{2} \tan \relax (c)^{2} - 2 \, a^{2} \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) + 2 \, b^{2} \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) + 4 \, a b \tan \left (d x\right )^{2} \tan \relax (c) + 4 \, a b \tan \left (d x\right ) \tan \relax (c)^{2} + 4 \, a b d x - b^{2} \tan \left (d x\right )^{2} - b^{2} \tan \relax (c)^{2} + a^{2} \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} \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 ) - 4 \, a b \tan \left (d x\right ) - 4 \, a b \tan \relax (c) - b^{2}}{2 \, {\left (d \tan \left (d x\right )^{2} \tan \relax (c)^{2} - 2 \, 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)*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(4*a*b*d*x*tan(d*x)^2*tan(c)^2 + a^2*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 - b^2*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*t
an(c)^2 - 8*a*b*d*x*tan(d*x)*tan(c) - b^2*tan(d*x)^2*tan(c)^2 - 2*a^2*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) + 2*b^2*l
og(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)/(t
an(c)^2 + 1))*tan(d*x)*tan(c) + 4*a*b*tan(d*x)^2*tan(c) + 4*a*b*tan(d*x)*tan(c)^2 + 4*a*b*d*x - b^2*tan(d*x)^2
- b^2*tan(c)^2 + a^2*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*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)) - 4*a*b*tan(d*x) - 4*a*b*tan(c) - b^2)/(d*tan(d*x)^2
*tan(c)^2 - 2*d*tan(d*x)*tan(c) + d)

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maple [A]  time = 0.02, size = 83, normalized size = 1.43 \[ \frac {b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {2 a b \tan \left (d x +c \right )}{d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2}}{2 d}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2}}{2 d}-\frac {2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.85, size = 58, normalized size = 1.00 \[ \frac {b^{2} \tan \left (d x + c\right )^{2} - 4 \, {\left (d x + c\right )} a b + 4 \, a b \tan \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.02, size = 57, normalized size = 0.98 \[ \frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )-2\,a\,b\,d\,x}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.25, size = 85, normalized size = 1.47 \[ \begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 2 a b x + \frac {2 a b \tan {\left (c + d x \right )}}{d} - \frac {b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right )^{2} \tan {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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