\(\int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx\) [40]

Optimal result

Integrand size = 21, antiderivative size = 96 \[ \int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {35 a^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {27 a^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d} \]

[Out]

35/8*a^4*arctanh(sin(d*x+c))/d+8*a^4*tan(d*x+c)/d+27/8*a^4*sec(d*x+c)*tan(d*x+c)/d+1/4*a^4*sec(d*x+c)^3*tan(d*
x+c)/d+4/3*a^4*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2836, 3855, 3852, 8, 3853} \[ \int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {35 a^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {a^4 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {27 a^4 \tan (c+d x) \sec (c+d x)}{8 d} \]

[In]

Int[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^5,x]

[Out]

(35*a^4*ArcTanh[Sin[c + d*x]])/(8*d) + (8*a^4*Tan[c + d*x])/d + (27*a^4*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^
4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (4*a^4*Tan[c + d*x]^3)/(3*d)

Rule 8

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

Rule 2836

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 3852

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 3853

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a^4 \sec (c+d x)+4 a^4 \sec ^2(c+d x)+6 a^4 \sec ^3(c+d x)+4 a^4 \sec ^4(c+d x)+a^4 \sec ^5(c+d x)\right ) \, dx \\ & = a^4 \int \sec (c+d x) \, dx+a^4 \int \sec ^5(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^2(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^4(c+d x) \, dx+\left (6 a^4\right ) \int \sec ^3(c+d x) \, dx \\ & = \frac {a^4 \text {arctanh}(\sin (c+d x))}{d}+\frac {3 a^4 \sec (c+d x) \tan (c+d x)}{d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \left (3 a^4\right ) \int \sec ^3(c+d x) \, dx+\left (3 a^4\right ) \int \sec (c+d x) \, dx-\frac {\left (4 a^4\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}-\frac {\left (4 a^4\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {4 a^4 \text {arctanh}(\sin (c+d x))}{d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {27 a^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d}+\frac {1}{8} \left (3 a^4\right ) \int \sec (c+d x) \, dx \\ & = \frac {35 a^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {27 a^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {35 a^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {27 a^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^5,x]

[Out]

(35*a^4*ArcTanh[Sin[c + d*x]])/(8*d) + (8*a^4*Tan[c + d*x])/d + (27*a^4*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^
4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (4*a^4*Tan[c + d*x]^3)/(3*d)

Maple [A] (verified)

Time = 3.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.41

method	result	size
parts	\(\frac {a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {4 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {3 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {4 a^{4} \tan \left (d x +c \right )}{d}\)	\(135\)
derivativedivides	\(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} \tan \left (d x +c \right )+6 a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-4 a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(142\)
default	\(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} \tan \left (d x +c \right )+6 a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-4 a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(142\)
risch	\(-\frac {i a^{4} \left (81 \,{\mathrm e}^{7 i \left (d x +c \right )}-96 \,{\mathrm e}^{6 i \left (d x +c \right )}+105 \,{\mathrm e}^{5 i \left (d x +c \right )}-480 \,{\mathrm e}^{4 i \left (d x +c \right )}-105 \,{\mathrm e}^{3 i \left (d x +c \right )}-544 \,{\mathrm e}^{2 i \left (d x +c \right )}-81 \,{\mathrm e}^{i \left (d x +c \right )}-160\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\)	\(145\)
parallelrisch	\(\frac {a^{4} \left (105 \left (-\cos \left (4 d x +4 c \right )-4 \cos \left (2 d x +2 c \right )-3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+105 \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+160 \sin \left (4 d x +4 c \right )+210 \sin \left (d x +c \right )+448 \sin \left (2 d x +2 c \right )+162 \sin \left (3 d x +3 c \right )\right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(149\)
norman	\(\frac {\frac {93 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {605 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {5 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {515 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {125 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {133 a^{4} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {35 a^{4} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {35 a^{4} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {35 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {35 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(224\)

[In]

int((a+cos(d*x+c)*a)^4*sec(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.16 \[ \int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {105 \, a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (160 \, a^{4} \cos \left (d x + c\right )^{3} + 81 \, a^{4} \cos \left (d x + c\right )^{2} + 32 \, a^{4} \cos \left (d x + c\right ) + 6 \, a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(105*a^4*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 105*a^4*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(160*a^
4*cos(d*x + c)^3 + 81*a^4*cos(d*x + c)^2 + 32*a^4*cos(d*x + c) + 6*a^4)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**4*sec(d*x+c)**5,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (88) = 176\).

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.90 \[ \int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} - 3 \, a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 192 \, a^{4} \tan \left (d x + c\right )}{48 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(64*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^4 - 3*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)) - 72*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)) + 24*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x +
c) - 1)) + 192*a^4*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {105 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 385 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 279 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(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)^7 - 385*a^4*tan(1/2*d*x + 1/2*c)^5 + 511*a^4*tan(1/2*d*x + 1/2*c)^3 - 279*a^4*tan(1/2*d*x +
1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 17.75 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.47 \[ \int (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {35\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {35\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {385\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {511\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}-\frac {93\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int((a + a*cos(c + d*x))^4/cos(c + d*x)^5,x)

[Out]

(35*a^4*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((511*a^4*tan(c/2 + (d*x)/2)^3)/12 - (385*a^4*tan(c/2 + (d*x)/2)^5)
/12 + (35*a^4*tan(c/2 + (d*x)/2)^7)/4 - (93*a^4*tan(c/2 + (d*x)/2))/4)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2
+ (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))