3.31 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=111 \[ \frac{a^4 \tan ^5(c+d x)}{5 d}+\frac{8 a^4 \tan ^3(c+d x)}{3 d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{7 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^4 \tan (c+d x) \sec ^3(c+d x)}{d}+\frac{7 a^4 \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

(7*a^4*ArcTanh[Sin[c + d*x]])/(2*d) + (8*a^4*Tan[c + d*x])/d + (7*a^4*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^4*
Sec[c + d*x]^3*Tan[c + d*x])/d + (8*a^4*Tan[c + d*x]^3)/(3*d) + (a^4*Tan[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.135141, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3791, 3767, 8, 3768, 3770} \[ \frac{a^4 \tan ^5(c+d x)}{5 d}+\frac{8 a^4 \tan ^3(c+d x)}{3 d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{7 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^4 \tan (c+d x) \sec ^3(c+d x)}{d}+\frac{7 a^4 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^4,x]

[Out]

(7*a^4*ArcTanh[Sin[c + d*x]])/(2*d) + (8*a^4*Tan[c + d*x])/d + (7*a^4*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^4*
Sec[c + d*x]^3*Tan[c + d*x])/d + (8*a^4*Tan[c + d*x]^3)/(3*d) + (a^4*Tan[c + d*x]^5)/(5*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 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]

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 \sec ^2(c+d x) (a+a \sec (c+d x))^4 \, dx &=\int \left (a^4 \sec ^2(c+d x)+4 a^4 \sec ^3(c+d x)+6 a^4 \sec ^4(c+d x)+4 a^4 \sec ^5(c+d x)+a^4 \sec ^6(c+d x)\right ) \, dx\\ &=a^4 \int \sec ^2(c+d x) \, dx+a^4 \int \sec ^6(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^3(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^5(c+d x) \, dx+\left (6 a^4\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{2 a^4 \sec (c+d x) \tan (c+d x)}{d}+\frac{a^4 \sec ^3(c+d x) \tan (c+d x)}{d}+\left (2 a^4\right ) \int \sec (c+d x) \, dx+\left (3 a^4\right ) \int \sec ^3(c+d x) \, dx-\frac{a^4 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}-\frac{a^4 \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d}-\frac{\left (6 a^4\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{2 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{7 a^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^4 \sec ^3(c+d x) \tan (c+d x)}{d}+\frac{8 a^4 \tan ^3(c+d x)}{3 d}+\frac{a^4 \tan ^5(c+d x)}{5 d}+\frac{1}{2} \left (3 a^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{7 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{7 a^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^4 \sec ^3(c+d x) \tan (c+d x)}{d}+\frac{8 a^4 \tan ^3(c+d x)}{3 d}+\frac{a^4 \tan ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [B]  time = 1.52693, size = 498, normalized size = 4.49 \[ -\frac{a^4 \sec (c) \sec ^5(c+d x) \left (960 \sin (2 c+d x)-660 \sin (c+2 d x)-660 \sin (3 c+2 d x)-1600 \sin (2 c+3 d x)+60 \sin (4 c+3 d x)-210 \sin (3 c+4 d x)-210 \sin (5 c+4 d x)-332 \sin (4 c+5 d x)+525 \cos (2 c+3 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+525 \cos (4 c+3 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+105 \cos (4 c+5 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+105 \cos (6 c+5 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+1050 \cos (d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+1050 \cos (2 c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-525 \cos (2 c+3 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-525 \cos (4 c+3 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-105 \cos (4 c+5 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-105 \cos (6 c+5 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-2360 \sin (d x)\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^4,x]

[Out]

-(a^4*Sec[c]*Sec[c + d*x]^5*(525*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 525*Cos[4*c + 3*d
*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 105*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] +
105*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 1050*Cos[d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 1050*Cos[2*c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 525*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*
x)/2]] - 525*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 105*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x
)/2] + Sin[(c + d*x)/2]] - 105*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2360*Sin[d*x] + 960
*Sin[2*c + d*x] - 660*Sin[c + 2*d*x] - 660*Sin[3*c + 2*d*x] - 1600*Sin[2*c + 3*d*x] + 60*Sin[4*c + 3*d*x] - 21
0*Sin[3*c + 4*d*x] - 210*Sin[5*c + 4*d*x] - 332*Sin[4*c + 5*d*x]))/(960*d)

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Maple [A]  time = 0.039, size = 123, normalized size = 1.1 \begin{align*}{\frac{83\,{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{7\,{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{34\,{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{4} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^4,x)

[Out]

83/15*a^4*tan(d*x+c)/d+7/2*a^4*sec(d*x+c)*tan(d*x+c)/d+7/2/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+34/15/d*a^4*tan(d*x
+c)*sec(d*x+c)^2+a^4*sec(d*x+c)^3*tan(d*x+c)/d+1/5/d*a^4*tan(d*x+c)*sec(d*x+c)^4

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Maxima [A]  time = 1.16907, size = 257, normalized size = 2.32 \begin{align*} \frac{4 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{4} + 120 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} - 15 \, a^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a^{4}{\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 )} + 60 \, a^{4} \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/60*(4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^4 + 120*(tan(d*x + c)^3 + 3*tan(d*x + c))*a
^4 - 15*a^4*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c
) + 1) + 3*log(sin(d*x + c) - 1)) - 60*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(
sin(d*x + c) - 1)) + 60*a^4*tan(d*x + c))/d

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Fricas [A]  time = 1.75651, size = 325, normalized size = 2.93 \begin{align*} \frac{105 \, a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (166 \, a^{4} \cos \left (d x + c\right )^{4} + 105 \, a^{4} \cos \left (d x + c\right )^{3} + 68 \, a^{4} \cos \left (d x + c\right )^{2} + 30 \, a^{4} \cos \left (d x + c\right ) + 6 \, a^{4}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/60*(105*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 105*a^4*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(166*a^
4*cos(d*x + c)^4 + 105*a^4*cos(d*x + c)^3 + 68*a^4*cos(d*x + c)^2 + 30*a^4*cos(d*x + c) + 6*a^4)*sin(d*x + c))
/(d*cos(d*x + c)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\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)**2*(a+a*sec(d*x+c))**4,x)

[Out]

a**4*(Integral(sec(c + d*x)**2, x) + Integral(4*sec(c + d*x)**3, x) + Integral(6*sec(c + d*x)**4, x) + Integra
l(4*sec(c + d*x)**5, x) + Integral(sec(c + d*x)**6, x))

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Giac [A]  time = 1.36585, size = 186, normalized size = 1.68 \begin{align*} \frac{105 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (105 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 490 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 896 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 790 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 375 \, a^{4} \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 )}^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/30*(105*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(105*a^4*tan
(1/2*d*x + 1/2*c)^9 - 490*a^4*tan(1/2*d*x + 1/2*c)^7 + 896*a^4*tan(1/2*d*x + 1/2*c)^5 - 790*a^4*tan(1/2*d*x +
1/2*c)^3 + 375*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d