3.42 \(\int (a+a \cos (c+d x))^4 \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=136 \[ \frac{4 a^4 \tan ^5(c+d x)}{5 d}+\frac{4 a^4 \tan ^3(c+d x)}{d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{49 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac{41 a^4 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{49 a^4 \tan (c+d x) \sec (c+d x)}{16 d} \]

[Out]

(49*a^4*ArcTanh[Sin[c + d*x]])/(16*d) + (8*a^4*Tan[c + d*x])/d + (49*a^4*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (
41*a^4*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (a^4*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) + (4*a^4*Tan[c + d*x]^3)/
d + (4*a^4*Tan[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.182264, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2757, 3768, 3770, 3767} \[ \frac{4 a^4 \tan ^5(c+d x)}{5 d}+\frac{4 a^4 \tan ^3(c+d x)}{d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{49 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac{41 a^4 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{49 a^4 \tan (c+d x) \sec (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^7,x]

[Out]

(49*a^4*ArcTanh[Sin[c + d*x]])/(16*d) + (8*a^4*Tan[c + d*x])/d + (49*a^4*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (
41*a^4*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (a^4*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) + (4*a^4*Tan[c + d*x]^3)/
d + (4*a^4*Tan[c + d*x]^5)/(5*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))^4 \sec ^7(c+d x) \, dx &=\int \left (a^4 \sec ^3(c+d x)+4 a^4 \sec ^4(c+d x)+6 a^4 \sec ^5(c+d x)+4 a^4 \sec ^6(c+d x)+a^4 \sec ^7(c+d x)\right ) \, dx\\ &=a^4 \int \sec ^3(c+d x) \, dx+a^4 \int \sec ^7(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^4(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^6(c+d x) \, dx+\left (6 a^4\right ) \int \sec ^5(c+d x) \, dx\\ &=\frac{a^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{3 a^4 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac{a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{2} a^4 \int \sec (c+d x) \, dx+\frac{1}{6} \left (5 a^4\right ) \int \sec ^5(c+d x) \, dx+\frac{1}{2} \left (9 a^4\right ) \int \sec ^3(c+d x) \, dx-\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}-\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{11 a^4 \sec (c+d x) \tan (c+d x)}{4 d}+\frac{41 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{4 a^4 \tan ^3(c+d x)}{d}+\frac{4 a^4 \tan ^5(c+d x)}{5 d}+\frac{1}{8} \left (5 a^4\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{4} \left (9 a^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{11 a^4 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{49 a^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{41 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{4 a^4 \tan ^3(c+d x)}{d}+\frac{4 a^4 \tan ^5(c+d x)}{5 d}+\frac{1}{16} \left (5 a^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{49 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{8 a^4 \tan (c+d x)}{d}+\frac{49 a^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{41 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{4 a^4 \tan ^3(c+d x)}{d}+\frac{4 a^4 \tan ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.781411, size = 211, normalized size = 1.55 \[ -\frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (23520 \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (3750 \sin (2 c+d x)+15360 \sin (c+2 d x)-1920 \sin (3 c+2 d x)+3845 \sin (2 c+3 d x)+3845 \sin (4 c+3 d x)+6912 \sin (3 c+4 d x)+735 \sin (4 c+5 d x)+735 \sin (6 c+5 d x)+1152 \sin (5 c+6 d x)-11520 \sin (c)+3750 \sin (d x))\right )}{122880 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^7,x]

[Out]

-(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*Sec[c + d*x]^6*(23520*Cos[c + d*x]^6*(Log[Cos[(c + d*x)/2] - Sin
[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-11520*Sin[c] + 3750*Sin[d*x] + 3750*Sin[
2*c + d*x] + 15360*Sin[c + 2*d*x] - 1920*Sin[3*c + 2*d*x] + 3845*Sin[2*c + 3*d*x] + 3845*Sin[4*c + 3*d*x] + 69
12*Sin[3*c + 4*d*x] + 735*Sin[4*c + 5*d*x] + 735*Sin[6*c + 5*d*x] + 1152*Sin[5*c + 6*d*x])))/(122880*d)

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Maple [A]  time = 0.093, size = 146, normalized size = 1.1 \begin{align*}{\frac{49\,{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{49\,{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{24\,{a}^{4}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{12\,{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{41\,{a}^{4} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{24\,d}}+{\frac{4\,{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{4} \left ( \sec \left ( dx+c \right ) \right ) ^{5}\tan \left ( dx+c \right ) }{6\,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)^7,x)

[Out]

49/16*a^4*sec(d*x+c)*tan(d*x+c)/d+49/16/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+24/5*a^4*tan(d*x+c)/d+12/5/d*a^4*tan(d
*x+c)*sec(d*x+c)^2+41/24*a^4*sec(d*x+c)^3*tan(d*x+c)/d+4/5/d*a^4*tan(d*x+c)*sec(d*x+c)^4+1/6*a^4*sec(d*x+c)^5*
tan(d*x+c)/d

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Maxima [B]  time = 1.13769, size = 365, normalized size = 2.68 \begin{align*} \frac{128 \,{\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} + 640 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} - 5 \, a^{4}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, 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 )} - 120 \, 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 )}}{480 \, 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)^7,x, algorithm="maxima")

[Out]

1/480*(128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^4 + 640*(tan(d*x + c)^3 + 3*tan(d*x + c)
)*a^4 - 5*a^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4
+ 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 180*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))
- 120*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)))/d

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Fricas [A]  time = 1.70632, size = 366, normalized size = 2.69 \begin{align*} \frac{735 \, a^{4} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 735 \, a^{4} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (1152 \, a^{4} \cos \left (d x + c\right )^{5} + 735 \, a^{4} \cos \left (d x + c\right )^{4} + 576 \, a^{4} \cos \left (d x + c\right )^{3} + 410 \, a^{4} \cos \left (d x + c\right )^{2} + 192 \, a^{4} \cos \left (d x + c\right ) + 40 \, a^{4}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \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)^7,x, algorithm="fricas")

[Out]

1/480*(735*a^4*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 735*a^4*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(1152*
a^4*cos(d*x + c)^5 + 735*a^4*cos(d*x + c)^4 + 576*a^4*cos(d*x + c)^3 + 410*a^4*cos(d*x + c)^2 + 192*a^4*cos(d*
x + c) + 40*a^4)*sin(d*x + c))/(d*cos(d*x + c)^6)

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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)**7,x)

[Out]

Timed out

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Giac [A]  time = 1.50627, size = 208, normalized size = 1.53 \begin{align*} \frac{735 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 735 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (735 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 4165 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 9702 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 11802 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 7355 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3105 \, 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 )}^{6}}}{240 \, 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)^7,x, algorithm="giac")

[Out]

1/240*(735*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 735*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(735*a^4*ta
n(1/2*d*x + 1/2*c)^11 - 4165*a^4*tan(1/2*d*x + 1/2*c)^9 + 9702*a^4*tan(1/2*d*x + 1/2*c)^7 - 11802*a^4*tan(1/2*
d*x + 1/2*c)^5 + 7355*a^4*tan(1/2*d*x + 1/2*c)^3 - 3105*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)
^6)/d