3.22 \(\int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=72 \[ \frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{4 a^3 \tan (c+d x)}{d}+\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

(5*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (4*a^3*Tan[c + d*x])/d + (3*a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^3*
Tan[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.0752543, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3791, 3770, 3767, 8, 3768} \[ \frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{4 a^3 \tan (c+d x)}{d}+\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (4*a^3*Tan[c + d*x])/d + (3*a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^3*
Tan[c + d*x]^3)/(3*d)

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rubi steps

\begin{align*} \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx &=\int \left (a^3 \sec (c+d x)+3 a^3 \sec ^2(c+d x)+3 a^3 \sec ^3(c+d x)+a^3 \sec ^4(c+d x)\right ) \, dx\\ &=a^3 \int \sec (c+d x) \, dx+a^3 \int \sec ^4(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (3 a^3\right ) \int \sec (c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}-\frac{\left (3 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{4 a^3 \tan (c+d x)}{d}+\frac{3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^3 \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 5.94417, size = 154, normalized size = 2.14 \[ -\frac{a^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^3 \left (-4 \tan (c) \cos (c+d x)-\sec (c) (-20 \sin (2 c+d x)+9 \sin (c+2 d x)+9 \sin (3 c+2 d x)+22 \sin (2 c+3 d x)+50 \sin (d x))+60 \cos ^3(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(60*Cos[c + d*x]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - L
og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(50*Sin[d*x] - 20*Sin[2*c + d*x] + 9*Sin[c + 2*d*x] + 9*Sin[
3*c + 2*d*x] + 22*Sin[2*c + 3*d*x]) - 4*Cos[c + d*x]*Tan[c]))/(192*d)

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Maple [A]  time = 0.036, size = 80, normalized size = 1.1 \begin{align*}{\frac{5\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{11\,{a}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{3\,{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sec(d*x+c))^3,x)

[Out]

5/2/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+11/3*a^3*tan(d*x+c)/d+3/2*a^3*sec(d*x+c)*tan(d*x+c)/d+1/3/d*a^3*tan(d*x+c)
*sec(d*x+c)^2

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Maxima [A]  time = 1.08559, size = 140, normalized size = 1.94 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - 9 \, a^{3}{\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 )} + 12 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, a^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - 9*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) + 12*a^3*log(sec(d*x + c) + tan(d*x + c)) + 36*a^3*tan(d*x + c))/d

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Fricas [A]  time = 1.70999, size = 254, normalized size = 3.53 \begin{align*} \frac{15 \, a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (22 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(15*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 15*a^3*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(22*a^3*c
os(d*x + c)^2 + 9*a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \sec{\left (c + d x \right )}\, dx + \int 3 \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 \sec ^{3}{\left (c + d x \right )}\, dx + \int \sec ^{4}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(sec(c + d*x), x) + Integral(3*sec(c + d*x)**2, x) + Integral(3*sec(c + d*x)**3, x) + Integral(s
ec(c + d*x)**4, x))

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Giac [A]  time = 1.36951, size = 143, normalized size = 1.99 \begin{align*} \frac{15 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 40 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 33 \, a^{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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(15*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*a^3*tan(1/2
*d*x + 1/2*c)^5 - 40*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)
/d