3.66 \(\int \frac{\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=49 \[ -\frac{\tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{\tan (c+d x) (2-\sec (c+d x))}{2 a d}+\frac{x}{a} \]

[Out]

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

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Rubi [A]  time = 0.0771262, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3888, 3881, 3770} \[ -\frac{\tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{\tan (c+d x) (2-\sec (c+d x))}{2 a d}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + a*Sec[c + d*x]),x]

[Out]

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

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

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 \frac{\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx &=\frac{\int (-a+a \sec (c+d x)) \tan ^2(c+d x) \, dx}{a^2}\\ &=-\frac{(2-\sec (c+d x)) \tan (c+d x)}{2 a d}-\frac{\int (-2 a+a \sec (c+d x)) \, dx}{2 a^2}\\ &=\frac{x}{a}-\frac{(2-\sec (c+d x)) \tan (c+d x)}{2 a d}-\frac{\int \sec (c+d x) \, dx}{2 a}\\ &=\frac{x}{a}-\frac{\tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{(2-\sec (c+d x)) \tan (c+d x)}{2 a d}\\ \end{align*}

Mathematica [B]  time = 0.855145, size = 241, normalized size = 4.92 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (-\frac{4 \sin (d x)}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+4 x\right )}{2 a (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + a*Sec[c + d*x]),x]

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(4*x + (2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d - (2*Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]])/d + 1/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - 1/(d*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2])^2) - (4*Sin[d*x])/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]
)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(2*a*(1 + Sec[c + d*x]))

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Maple [B]  time = 0.063, size = 144, normalized size = 2.9 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{3}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a*arctan(tan(1/2*d*x+1/2*c))-1/2/a/d/(tan(1/2*d*x+1/2*c)+1)^2+3/2/a/d/(tan(1/2*d*x+1/2*c)+1)-1/2/a/d*ln(ta
n(1/2*d*x+1/2*c)+1)+1/2/a/d/(tan(1/2*d*x+1/2*c)-1)^2+3/2/a/d/(tan(1/2*d*x+1/2*c)-1)+1/2/a/d*ln(tan(1/2*d*x+1/2
*c)-1)

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Maxima [B]  time = 1.69882, size = 220, normalized size = 4.49 \begin{align*} -\frac{\frac{2 \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{4 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 4*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + log(
sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a)/d

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Fricas [A]  time = 1.19055, size = 224, normalized size = 4.57 \begin{align*} \frac{4 \, d x \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \, \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{4 \, a d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(c + d*x)**4/(sec(c + d*x) + 1), x)/a

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Giac [B]  time = 2.51276, size = 130, normalized size = 2.65 \begin{align*} \frac{\frac{2 \,{\left (d x + c\right )}}{a} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} + \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \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} a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

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