3.22 \(\int (a+a \cos (c+d x))^2 \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=96 \[ \frac{2 a^2 \tan ^3(c+d x)}{3 d}+\frac{2 a^2 \tan (c+d x)}{d}+\frac{7 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{7 a^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.10797, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2757, 3768, 3770, 3767} \[ \frac{2 a^2 \tan ^3(c+d x)}{3 d}+\frac{2 a^2 \tan (c+d x)}{d}+\frac{7 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{7 a^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^2*Sec[c + d*x]^5,x]

[Out]

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

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]

Rubi steps

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

Mathematica [B]  time = 6.37008, size = 797, normalized size = 8.3 \[ -\frac{7 (\cos (c+d x) a+a)^2 \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{32 d}+\frac{7 (\cos (c+d x) a+a)^2 \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{32 d}+\frac{(\cos (c+d x) a+a)^2 \sin \left (\frac{d x}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{(\cos (c+d x) a+a)^2 \sin \left (\frac{d x}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{(\cos (c+d x) a+a)^2 \left (29 \cos \left (\frac{c}{2}\right )-13 \sin \left (\frac{c}{2}\right )\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{192 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}+\frac{(\cos (c+d x) a+a)^2 \left (-29 \cos \left (\frac{c}{2}\right )-13 \sin \left (\frac{c}{2}\right )\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{192 d \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}+\frac{(\cos (c+d x) a+a)^2 \sin \left (\frac{d x}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{12 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3}+\frac{(\cos (c+d x) a+a)^2 \sin \left (\frac{d x}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{12 d \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3}+\frac{(\cos (c+d x) a+a)^2 \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{64 d \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^4}-\frac{(\cos (c+d x) a+a)^2 \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{64 d \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Cos[c + d*x])^2*Sec[c + d*x]^5,x]

[Out]

(-7*(a + a*Cos[c + d*x])^2*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4)/(32*d) + (7*(a +
a*Cos[c + d*x])^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4)/(32*d) + ((a + a*Cos[c +
d*x])^2*Sec[c/2 + (d*x)/2]^4)/(64*d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^4) + ((a + a*Cos[c + d*x])^2*Se
c[c/2 + (d*x)/2]^4*Sin[(d*x)/2])/(12*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + ((
a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(29*Cos[c/2] - 13*Sin[c/2]))/(192*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2
+ (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + ((a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*Sin[(d*x)/2])/(3*d*(Cos[c/2
] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) - ((a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4)/(64*d
*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^4) + ((a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*Sin[(d*x)/2])/(12
*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + ((a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x
)/2]^4*(-29*Cos[c/2] - 13*Sin[c/2]))/(192*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2)
+ ((a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*Sin[(d*x)/2])/(3*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] +
Sin[c/2 + (d*x)/2]))

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Maple [A]  time = 0.082, size = 102, normalized size = 1.1 \begin{align*}{\frac{7\,{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{7\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{4\,{a}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^2*sec(d*x+c)^5,x)

[Out]

7/8/d*a^2*sec(d*x+c)*tan(d*x+c)+7/8/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+4/3*a^2*tan(d*x+c)/d+2/3*a^2*sec(d*x+c)^2*
tan(d*x+c)/d+1/4*a^2*sec(d*x+c)^3*tan(d*x+c)/d

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Maxima [A]  time = 1.12332, size = 196, normalized size = 2.04 \begin{align*} \frac{32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 3 \, a^{2}{\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 )} - 12 \, a^{2}{\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 )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 - 3*a^2*(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)) - 12*a^2*(2*sin(d*x + c)/(sin(d*x
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 1.72642, size = 288, normalized size = 3. \begin{align*} \frac{21 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (32 \, a^{2} \cos \left (d x + c\right )^{3} + 21 \, a^{2} \cos \left (d x + c\right )^{2} + 16 \, a^{2} \cos \left (d x + c\right ) + 6 \, a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(21*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 21*a^2*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(32*a^2*c
os(d*x + c)^3 + 21*a^2*cos(d*x + c)^2 + 16*a^2*cos(d*x + c) + 6*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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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))**2*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.46997, size = 165, normalized size = 1.72 \begin{align*} \frac{21 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 21 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (21 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 77 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 83 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 75 \, a^{2} \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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(21*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 21*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(21*a^2*tan(1/
2*d*x + 1/2*c)^7 - 77*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*a^2*tan(1/2*d*x + 1/2*c)^3 - 75*a^2*tan(1/2*d*x + 1/2*c)
)/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d