3.39 \(\int (a+a \cos (c+d x))^4 \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=73 \[ \frac{a^4 \tan ^3(c+d x)}{3 d}+\frac{7 a^4 \tan (c+d x)}{d}+\frac{6 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a^4 \tan (c+d x) \sec (c+d x)}{d}+a^4 x \]

[Out]

a^4*x + (6*a^4*ArcTanh[Sin[c + d*x]])/d + (7*a^4*Tan[c + d*x])/d + (2*a^4*Sec[c + d*x]*Tan[c + d*x])/d + (a^4*
Tan[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.0956729, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2757, 3770, 3767, 8, 3768} \[ \frac{a^4 \tan ^3(c+d x)}{3 d}+\frac{7 a^4 \tan (c+d x)}{d}+\frac{6 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a^4 \tan (c+d x) \sec (c+d x)}{d}+a^4 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*x + (6*a^4*ArcTanh[Sin[c + d*x]])/d + (7*a^4*Tan[c + d*x])/d + (2*a^4*Sec[c + d*x]*Tan[c + d*x])/d + (a^4*
Tan[c + d*x]^3)/(3*d)

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, 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 (a+a \cos (c+d x))^4 \sec ^4(c+d x) \, dx &=\int \left (a^4+4 a^4 \sec (c+d x)+6 a^4 \sec ^2(c+d x)+4 a^4 \sec ^3(c+d x)+a^4 \sec ^4(c+d x)\right ) \, dx\\ &=a^4 x+a^4 \int \sec ^4(c+d x) \, dx+\left (4 a^4\right ) \int \sec (c+d x) \, dx+\left (4 a^4\right ) \int \sec ^3(c+d x) \, dx+\left (6 a^4\right ) \int \sec ^2(c+d x) \, dx\\ &=a^4 x+\frac{4 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a^4 \sec (c+d x) \tan (c+d x)}{d}+\left (2 a^4\right ) \int \sec (c+d x) \, dx-\frac{a^4 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}-\frac{\left (6 a^4\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=a^4 x+\frac{6 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{7 a^4 \tan (c+d x)}{d}+\frac{2 a^4 \sec (c+d x) \tan (c+d x)}{d}+\frac{a^4 \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.0327334, size = 61, normalized size = 0.84 \[ a^4 \left (\frac{\tan ^3(c+d x)}{3 d}+\frac{7 \tan (c+d x)}{d}+\frac{6 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 \tan (c+d x) \sec (c+d x)}{d}+x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*(x + (6*ArcTanh[Sin[c + d*x]])/d + (7*Tan[c + d*x])/d + (2*Sec[c + d*x]*Tan[c + d*x])/d + Tan[c + d*x]^3/(
3*d))

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Maple [A]  time = 0.09, size = 93, normalized size = 1.3 \begin{align*}{a}^{4}x+{\frac{{a}^{4}c}{d}}+6\,{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{20\,{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+2\,{\frac{{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}\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((a+cos(d*x+c)*a)^4*sec(d*x+c)^4,x)

[Out]

a^4*x+1/d*a^4*c+6/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+20/3*a^4*tan(d*x+c)/d+2*a^4*sec(d*x+c)*tan(d*x+c)/d+1/3/d*a^
4*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 1.13387, size = 162, normalized size = 2.22 \begin{align*} \frac{{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} + 3 \,{\left (d x + c\right )} a^{4} - 3 \, 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 )} + 6 \, a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, a^{4} \tan \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((tan(d*x + c)^3 + 3*tan(d*x + c))*a^4 + 3*(d*x + c)*a^4 - 3*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - lo
g(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*a^4*
tan(d*x + c))/d

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Fricas [A]  time = 2.013, size = 281, normalized size = 3.85 \begin{align*} \frac{3 \, a^{4} d x \cos \left (d x + c\right )^{3} + 9 \, a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (20 \, a^{4} \cos \left (d x + c\right )^{2} + 6 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^4*d*x*cos(d*x + c)^3 + 9*a^4*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 9*a^4*cos(d*x + c)^3*log(-sin(d*x
+ c) + 1) + (20*a^4*cos(d*x + c)^2 + 6*a^4*cos(d*x + c) + a^4)*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.45785, size = 157, normalized size = 2.15 \begin{align*} \frac{3 \,{\left (d x + c\right )} a^{4} + 18 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 38 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, 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 )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*a^4 + 18*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(15*a^4*tan(1/2*d*x + 1/2*c)^5 - 38*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^3)/d