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

Optimal. Leaf size=96 \[ \frac {4 a^4 \tan ^3(c+d x)}{3 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {35 a^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {27 a^4 \tan (c+d x) \sec (c+d x)}{8 d} \]

[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

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Rubi [A]  time = 0.13, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 12, 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} \[ \frac {4 a^4 \tan ^3(c+d x)}{3 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {35 a^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {27 a^4 \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[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 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 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 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 (a+a \cos (c+d x))^4 \sec ^5(c+d x) \, dx &=\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 \tanh ^{-1}(\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 ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}-\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {4 a^4 \tanh ^{-1}(\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 \tanh ^{-1}(\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*}

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Mathematica [B]  time = 6.37, size = 797, normalized size = 8.30 \[ -\frac {35 (\cos (c+d x) a+a)^4 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{128 d}+\frac {35 (\cos (c+d x) a+a)^4 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{128 d}+\frac {5 (\cos (c+d x) a+a)^4 \sin \left (\frac {d x}{2}\right ) \sec ^8\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 )}+\frac {5 (\cos (c+d x) a+a)^4 \sin \left (\frac {d x}{2}\right ) \sec ^8\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 )}+\frac {(\cos (c+d x) a+a)^4 \left (97 \cos \left (\frac {c}{2}\right )-65 \sin \left (\frac {c}{2}\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{768 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)^4 \left (-97 \cos \left (\frac {c}{2}\right )-65 \sin \left (\frac {c}{2}\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{768 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)^4 \sin \left (\frac {d x}{2}\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{24 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)^4 \sin \left (\frac {d x}{2}\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{24 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)^4 \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{256 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)^4 \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right )}{256 d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*(a + a*Cos[c + d*x])^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8)/(128*d) + (35*(
a + a*Cos[c + d*x])^4*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8)/(128*d) + ((a + a*Cos
[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(256*d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^4) + ((a + a*Cos[c + d*x])
^4*Sec[c/2 + (d*x)/2]^8*Sin[(d*x)/2])/(24*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])^4*Sec[c/2 + (d*x)/2]^8*(97*Cos[c/2] - 65*Sin[c/2]))/(768*d*(Cos[c/2] - Sin[c/2])*(Cos
[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (5*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(d*x)/2])/(12*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])^4*Sec[c/2 + (d*x)/2]^
8)/(256*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^4) + ((a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(d*x
)/2])/(24*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])^4*Sec[c
/2 + (d*x)/2]^8*(-97*Cos[c/2] - 65*Sin[c/2]))/(768*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*
x)/2])^2) + (5*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(d*x)/2])/(12*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2
+ (d*x)/2] + Sin[c/2 + (d*x)/2]))

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fricas [A]  time = 2.00, size = 111, normalized size = 1.16 \[ \frac {105 \, a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (160 \, a^{4} \cos \left (d x + c\right )^{3} + 81 \, a^{4} \cos \left (d x + c\right )^{2} + 32 \, a^{4} \cos \left (d x + c\right ) + 6 \, a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.86, size = 122, normalized size = 1.27 \[ \frac {105 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 385 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 279 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.16, size = 102, normalized size = 1.06 \[ \frac {35 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {20 a^{4} \tan \left (d x +c \right )}{3 d}+\frac {27 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {4 a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+20/3*a^4*tan(d*x+c)/d+27/8*a^4*sec(d*x+c)*tan(d*x+c)/d+4/3/d*a^4*tan(d*x+
c)*sec(d*x+c)^2+1/4*a^4*sec(d*x+c)^3*tan(d*x+c)/d

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maxima [B]  time = 1.01, size = 182, normalized size = 1.90 \[ \frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} - 3 \, a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 192 \, a^{4} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 3.50, size = 141, normalized size = 1.47 \[ \frac {35\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {35\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {385\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {511\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}-\frac {93\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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