3.54 \(\int \frac{\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=66 \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{5 \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

ArcTanh[Sin[c + d*x]]/(a^2*d) - (5*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + Tan[c + d*x]/(3*d*(a + a*Sec[c
+ d*x])^2)

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Rubi [A]  time = 0.118135, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3799, 3998, 3770, 3794} \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{5 \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

ArcTanh[Sin[c + d*x]]/(a^2*d) - (5*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + Tan[c + d*x]/(3*d*(a + a*Sec[c
+ d*x])^2)

Rule 3799

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(
a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\frac{\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec (c+d x) (-2 a+3 a \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac{\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \sec (c+d x) \, dx}{a^2}-\frac{5 \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{5 \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.353587, size = 160, normalized size = 2.42 \[ -\frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (\tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+6 \cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+8 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

(-2*Cos[(c + d*x)/2]*Sec[c + d*x]^2*(6*Cos[(c + d*x)/2]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sin[(d*x)/2] + 8*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + Cos[(
c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Sec[c + d*x])^2)

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Maple [A]  time = 0.031, size = 77, normalized size = 1.2 \begin{align*} -{\frac{1}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{d{a}^{2}}\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(sec(d*x+c)^3/(a+a*sec(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.19073, size = 132, normalized size = 2. \begin{align*} -\frac{\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(
d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d

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Fricas [A]  time = 1.70834, size = 305, normalized size = 4.62 \begin{align*} \frac{3 \,{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(3*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(sin(d*x + c) + 1) - 3*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*l
og(-sin(d*x + c) + 1) - 2*(4*cos(d*x + c) + 5)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^
2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**3/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.46181, size = 104, normalized size = 1.58 \begin{align*} \frac{\frac{6 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(6*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - (a^4*tan(1/2*d*x +
1/2*c)^3 + 9*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d