3.21 \(\int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=120 \[ \frac {a^2 \sec ^6(c+d x)}{6 d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}-\frac {a^2 \sec ^4(c+d x)}{4 d}-\frac {4 a^2 \sec ^3(c+d x)}{3 d}-\frac {a^2 \sec ^2(c+d x)}{2 d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \log (\cos (c+d x))}{d} \]

[Out]

-a^2*ln(cos(d*x+c))/d+2*a^2*sec(d*x+c)/d-1/2*a^2*sec(d*x+c)^2/d-4/3*a^2*sec(d*x+c)^3/d-1/4*a^2*sec(d*x+c)^4/d+
2/5*a^2*sec(d*x+c)^5/d+1/6*a^2*sec(d*x+c)^6/d

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3879, 88} \[ \frac {a^2 \sec ^6(c+d x)}{6 d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}-\frac {a^2 \sec ^4(c+d x)}{4 d}-\frac {4 a^2 \sec ^3(c+d x)}{3 d}-\frac {a^2 \sec ^2(c+d x)}{2 d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Log[Cos[c + d*x]])/d) + (2*a^2*Sec[c + d*x])/d - (a^2*Sec[c + d*x]^2)/(2*d) - (4*a^2*Sec[c + d*x]^3)/(3
*d) - (a^2*Sec[c + d*x]^4)/(4*d) + (2*a^2*Sec[c + d*x]^5)/(5*d) + (a^2*Sec[c + d*x]^6)/(6*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))^2 \tan ^5(c+d x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(a-a x)^2 (a+a x)^4}{x^7} \, dx,x,\cos (c+d x)\right )}{a^4 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {a^6}{x^7}+\frac {2 a^6}{x^6}-\frac {a^6}{x^5}-\frac {4 a^6}{x^4}-\frac {a^6}{x^3}+\frac {2 a^6}{x^2}+\frac {a^6}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^4 d}\\ &=-\frac {a^2 \log (\cos (c+d x))}{d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \sec ^2(c+d x)}{2 d}-\frac {4 a^2 \sec ^3(c+d x)}{3 d}-\frac {a^2 \sec ^4(c+d x)}{4 d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}+\frac {a^2 \sec ^6(c+d x)}{6 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.42, size = 125, normalized size = 1.04 \[ \frac {a^2 \sec ^6(c+d x) (312 \cos (c+d x)-5 (-28 \cos (3 (c+d x))+6 \cos (4 (c+d x))-12 \cos (5 (c+d x))+18 \cos (4 (c+d x)) \log (\cos (c+d x))+3 \cos (6 (c+d x)) \log (\cos (c+d x))+30 \log (\cos (c+d x))+9 \cos (2 (c+d x)) (5 \log (\cos (c+d x))+4)+14))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(312*Cos[c + d*x] - 5*(14 - 28*Cos[3*(c + d*x)] + 6*Cos[4*(c + d*x)] - 12*Cos[5*(c + d*x)] + 30*Log[Cos[c
+ d*x]] + 18*Cos[4*(c + d*x)]*Log[Cos[c + d*x]] + 3*Cos[6*(c + d*x)]*Log[Cos[c + d*x]] + 9*Cos[2*(c + d*x)]*(
4 + 5*Log[Cos[c + d*x]])))*Sec[c + d*x]^6)/(480*d)

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 104, normalized size = 0.87 \[ -\frac {60 \, a^{2} \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) - 120 \, a^{2} \cos \left (d x + c\right )^{5} + 30 \, a^{2} \cos \left (d x + c\right )^{4} + 80 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )^{2} - 24 \, a^{2} \cos \left (d x + c\right ) - 10 \, a^{2}}{60 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a^2*cos(d*x + c)^6*log(-cos(d*x + c)) - 120*a^2*cos(d*x + c)^5 + 30*a^2*cos(d*x + c)^4 + 80*a^2*cos(
d*x + c)^3 + 15*a^2*cos(d*x + c)^2 - 24*a^2*cos(d*x + c) - 10*a^2)/(d*cos(d*x + c)^6)

________________________________________________________________________________________

giac [B]  time = 3.35, size = 242, normalized size = 2.02 \[ \frac {60 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {275 \, a^{2} + \frac {1770 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {4845 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4780 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2925 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1002 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {147 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 1)) + (275*a^2 + 1770*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 4845*a^2*(cos(d*x + c) - 1)^2/
(cos(d*x + c) + 1)^2 + 4780*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 2925*a^2*(cos(d*x + c) - 1)^4/(cos
(d*x + c) + 1)^4 + 1002*a^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 147*a^2*(cos(d*x + c) - 1)^6/(cos(d*x
+ c) + 1)^6)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^6)/d

________________________________________________________________________________________

maple [A]  time = 0.74, size = 203, normalized size = 1.69 \[ \frac {a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {2 a^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}-\frac {2 a^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}+\frac {2 a^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )}+\frac {16 a^{2} \cos \left (d x +c \right )}{15 d}+\frac {2 a^{2} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}+\frac {8 a^{2} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}+\frac {a^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*tan(d*x+c)^4/d-1/2*a^2*tan(d*x+c)^2/d-a^2*ln(cos(d*x+c))/d+2/5/d*a^2*sin(d*x+c)^6/cos(d*x+c)^5-2/15/d*
a^2*sin(d*x+c)^6/cos(d*x+c)^3+2/5/d*a^2*sin(d*x+c)^6/cos(d*x+c)+16/15*a^2*cos(d*x+c)/d+2/5/d*a^2*cos(d*x+c)*si
n(d*x+c)^4+8/15/d*a^2*cos(d*x+c)*sin(d*x+c)^2+1/6/d*a^2*sin(d*x+c)^6/cos(d*x+c)^6

________________________________________________________________________________________

maxima [A]  time = 0.70, size = 97, normalized size = 0.81 \[ -\frac {60 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {120 \, a^{2} \cos \left (d x + c\right )^{5} - 30 \, a^{2} \cos \left (d x + c\right )^{4} - 80 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right )^{2} + 24 \, a^{2} \cos \left (d x + c\right ) + 10 \, a^{2}}{\cos \left (d x + c\right )^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a^2*log(cos(d*x + c)) - (120*a^2*cos(d*x + c)^5 - 30*a^2*cos(d*x + c)^4 - 80*a^2*cos(d*x + c)^3 - 15
*a^2*cos(d*x + c)^2 + 24*a^2*cos(d*x + c) + 10*a^2)/cos(d*x + c)^6)/d

________________________________________________________________________________________

mupad [B]  time = 5.05, size = 192, normalized size = 1.60 \[ \frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {92\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}-44\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {74\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-\frac {32\,a^2}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*atanh(tan(c/2 + (d*x)/2)^2))/d - ((74*a^2*tan(c/2 + (d*x)/2)^2)/5 - 44*a^2*tan(c/2 + (d*x)/2)^4 + (92*a
^2*tan(c/2 + (d*x)/2)^6)/3 - 12*a^2*tan(c/2 + (d*x)/2)^8 + 2*a^2*tan(c/2 + (d*x)/2)^10 - (32*a^2)/15)/(d*(15*t
an(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 +
(d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))

________________________________________________________________________________________

sympy [A]  time = 5.33, size = 189, normalized size = 1.58 \[ \begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{6 d} + \frac {2 a^{2} \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{5 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{6 d} - \frac {8 a^{2} \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {a^{2} \sec ^{2}{\left (c + d x \right )}}{6 d} + \frac {16 a^{2} \sec {\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right )^{2} \tan ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**2*tan(d*x+c)**5,x)

[Out]

Piecewise((a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**4*sec(c + d*x)**2/(6*d) + 2*a**2*tan(c + d
*x)**4*sec(c + d*x)/(5*d) + a**2*tan(c + d*x)**4/(4*d) - a**2*tan(c + d*x)**2*sec(c + d*x)**2/(6*d) - 8*a**2*t
an(c + d*x)**2*sec(c + d*x)/(15*d) - a**2*tan(c + d*x)**2/(2*d) + a**2*sec(c + d*x)**2/(6*d) + 16*a**2*sec(c +
d*x)/(15*d), Ne(d, 0)), (x*(a*sec(c) + a)**2*tan(c)**5, True))

________________________________________________________________________________________