3.13 \(\int (a+a \sec (c+d x)) \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=45 \[ -\frac{a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\tan (c+d x) (a \sec (c+d x)+2 a)}{2 d}-a x \]

[Out]

-(a*x) - (a*ArcTanh[Sin[c + d*x]])/(2*d) + ((2*a + a*Sec[c + d*x])*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.0339243, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3881, 3770} \[ -\frac{a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\tan (c+d x) (a \sec (c+d x)+2 a)}{2 d}-a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x) - (a*ArcTanh[Sin[c + d*x]])/(2*d) + ((2*a + a*Sec[c + d*x])*Tan[c + d*x])/(2*d)

Rule 3881

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(e*(e*Cot[
c + d*x])^(m - 1)*(a*m + b*(m - 1)*Csc[c + d*x]))/(d*m*(m - 1)), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m - 2)
*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3770

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

Rubi steps

\begin{align*} \int (a+a \sec (c+d x)) \tan ^2(c+d x) \, dx &=\frac{(2 a+a \sec (c+d x)) \tan (c+d x)}{2 d}-\frac{1}{2} \int (2 a+a \sec (c+d x)) \, dx\\ &=-a x+\frac{(2 a+a \sec (c+d x)) \tan (c+d x)}{2 d}-\frac{1}{2} a \int \sec (c+d x) \, dx\\ &=-a x-\frac{a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(2 a+a \sec (c+d x)) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0293039, size = 60, normalized size = 1.33 \[ -\frac{a \tan ^{-1}(\tan (c+d x))}{d}+\frac{a \tan (c+d x)}{d}-\frac{a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*ArcTan[Tan[c + d*x]])/d) - (a*ArcTanh[Sin[c + d*x]])/(2*d) + (a*Tan[c + d*x])/d + (a*Sec[c + d*x]*Tan[c +
d*x])/(2*d)

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Maple [A]  time = 0.032, size = 78, normalized size = 1.7 \begin{align*} -ax+{\frac{a\tan \left ( dx+c \right ) }{d}}-{\frac{ac}{d}}+{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{a\sin \left ( dx+c \right ) }{2\,d}}-{\frac{a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.72673, size = 88, normalized size = 1.96 \begin{align*} -\frac{4 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a + a{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.904682, size = 234, normalized size = 5.2 \begin{align*} -\frac{4 \, a d x \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*a*d*x*cos(d*x + c)^2 + a*cos(d*x + c)^2*log(sin(d*x + c) + 1) - a*cos(d*x + c)^2*log(-sin(d*x + c) + 1
) - 2*(2*a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \tan ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(tan(c + d*x)**2*sec(c + d*x), x) + Integral(tan(c + d*x)**2, x))

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Giac [B]  time = 1.75812, size = 119, normalized size = 2.64 \begin{align*} -\frac{2 \,{\left (d x + c\right )} a + a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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