3.3 \(\int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=87 \[ \frac {a \sec ^5(c+d x)}{5 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {2 a \sec ^3(c+d x)}{3 d}-\frac {a \sec ^2(c+d x)}{d}+\frac {a \sec (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

[Out]

-a*ln(cos(d*x+c))/d+a*sec(d*x+c)/d-a*sec(d*x+c)^2/d-2/3*a*sec(d*x+c)^3/d+1/4*a*sec(d*x+c)^4/d+1/5*a*sec(d*x+c)
^5/d

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Rubi [A]  time = 0.05, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3879, 88} \[ \frac {a \sec ^5(c+d x)}{5 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {2 a \sec ^3(c+d x)}{3 d}-\frac {a \sec ^2(c+d x)}{d}+\frac {a \sec (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Log[Cos[c + d*x]])/d) + (a*Sec[c + d*x])/d - (a*Sec[c + d*x]^2)/d - (2*a*Sec[c + d*x]^3)/(3*d) + (a*Sec[c
+ d*x]^4)/(4*d) + (a*Sec[c + d*x]^5)/(5*d)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(a-a x)^2 (a+a x)^3}{x^6} \, dx,x,\cos (c+d x)\right )}{a^4 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {a^5}{x^6}+\frac {a^5}{x^5}-\frac {2 a^5}{x^4}-\frac {2 a^5}{x^3}+\frac {a^5}{x^2}+\frac {a^5}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^4 d}\\ &=-\frac {a \log (\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}-\frac {a \sec ^2(c+d x)}{d}-\frac {2 a \sec ^3(c+d x)}{3 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {a \sec ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 82, normalized size = 0.94 \[ \frac {a \sec ^5(c+d x)}{5 d}-\frac {2 a \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x)}{d}-\frac {a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[c + d*x])/d - (2*a*Sec[c + d*x]^3)/(3*d) + (a*Sec[c + d*x]^5)/(5*d) - (a*(4*Log[Cos[c + d*x]] + 2*Tan[c
+ d*x]^2 - Tan[c + d*x]^4))/(4*d)

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fricas [A]  time = 0.56, size = 79, normalized size = 0.91 \[ -\frac {60 \, a \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 60 \, a \cos \left (d x + c\right )^{4} + 60 \, a \cos \left (d x + c\right )^{3} + 40 \, a \cos \left (d x + c\right )^{2} - 15 \, a \cos \left (d x + c\right ) - 12 \, a}{60 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a*cos(d*x + c)^5*log(-cos(d*x + c)) - 60*a*cos(d*x + c)^4 + 60*a*cos(d*x + c)^3 + 40*a*cos(d*x + c)^
2 - 15*a*cos(d*x + c) - 12*a)/(d*cos(d*x + c)^5)

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giac [B]  time = 3.09, size = 201, normalized size = 2.31 \[ \frac {60 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {201 \, a + \frac {1125 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2610 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1970 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {805 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {137 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + (201*a + 1125*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2610*a*(cos(d*x + c) - 1)^2/(cos(d*x +
c) + 1)^2 + 1970*a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 805*a*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^
4 + 137*a*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^5)/d

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maple [A]  time = 0.74, size = 161, normalized size = 1.85 \[ \frac {a \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}-\frac {a \left (\sin ^{6}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}+\frac {a \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )}+\frac {8 a \cos \left (d x +c \right )}{15 d}+\frac {a \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 a \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*tan(d*x+c)^4/d-1/2*a*tan(d*x+c)^2/d-a*ln(cos(d*x+c))/d+1/5/d*a*sin(d*x+c)^6/cos(d*x+c)^5-1/15/d*a*sin(d*
x+c)^6/cos(d*x+c)^3+1/5/d*a*sin(d*x+c)^6/cos(d*x+c)+8/15*a*cos(d*x+c)/d+1/5/d*a*cos(d*x+c)*sin(d*x+c)^4+4/15/d
*a*cos(d*x+c)*sin(d*x+c)^2

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maxima [A]  time = 0.61, size = 72, normalized size = 0.83 \[ -\frac {60 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac {60 \, a \cos \left (d x + c\right )^{4} - 60 \, a \cos \left (d x + c\right )^{3} - 40 \, a \cos \left (d x + c\right )^{2} + 15 \, a \cos \left (d x + c\right ) + 12 \, a}{\cos \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a*log(cos(d*x + c)) - (60*a*cos(d*x + c)^4 - 60*a*cos(d*x + c)^3 - 40*a*cos(d*x + c)^2 + 15*a*cos(d*
x + c) + 12*a)/cos(d*x + c)^5)/d

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mupad [B]  time = 5.73, size = 151, normalized size = 1.74 \[ \frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {62\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {22\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {16\,a}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)^2))/d - ((16*a)/15 - (22*a*tan(c/2 + (d*x)/2)^2)/3 + (62*a*tan(c/2 + (d*x)/2)^4)
/3 - 10*a*tan(c/2 + (d*x)/2)^6 + 2*a*tan(c/2 + (d*x)/2)^8)/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^
4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))

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sympy [A]  time = 3.06, size = 112, normalized size = 1.29 \[ \begin {cases} \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{5 d} + \frac {a \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {4 a \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 d} - \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {8 a \sec {\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right ) \tan ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*log(tan(c + d*x)**2 + 1)/(2*d) + a*tan(c + d*x)**4*sec(c + d*x)/(5*d) + a*tan(c + d*x)**4/(4*d) -
4*a*tan(c + d*x)**2*sec(c + d*x)/(15*d) - a*tan(c + d*x)**2/(2*d) + 8*a*sec(c + d*x)/(15*d), Ne(d, 0)), (x*(a
*sec(c) + a)*tan(c)**5, True))

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