3.1 \(\int (a+a \sec (c+d x)) \tan ^9(c+d x) \, dx\)

Optimal. Leaf size=151 \[ \frac {a \sec ^9(c+d x)}{9 d}+\frac {a \sec ^8(c+d x)}{8 d}-\frac {4 a \sec ^7(c+d x)}{7 d}-\frac {2 a \sec ^6(c+d x)}{3 d}+\frac {6 a \sec ^5(c+d x)}{5 d}+\frac {3 a \sec ^4(c+d x)}{2 d}-\frac {4 a \sec ^3(c+d x)}{3 d}-\frac {2 a \sec ^2(c+d x)}{d}+\frac {a \sec (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

[Out]

-a*ln(cos(d*x+c))/d+a*sec(d*x+c)/d-2*a*sec(d*x+c)^2/d-4/3*a*sec(d*x+c)^3/d+3/2*a*sec(d*x+c)^4/d+6/5*a*sec(d*x+
c)^5/d-2/3*a*sec(d*x+c)^6/d-4/7*a*sec(d*x+c)^7/d+1/8*a*sec(d*x+c)^8/d+1/9*a*sec(d*x+c)^9/d

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Rubi [A]  time = 0.07, antiderivative size = 151, 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} \[ \frac {a \sec ^9(c+d x)}{9 d}+\frac {a \sec ^8(c+d x)}{8 d}-\frac {4 a \sec ^7(c+d x)}{7 d}-\frac {2 a \sec ^6(c+d x)}{3 d}+\frac {6 a \sec ^5(c+d x)}{5 d}+\frac {3 a \sec ^4(c+d x)}{2 d}-\frac {4 a \sec ^3(c+d x)}{3 d}-\frac {2 a \sec ^2(c+d x)}{d}+\frac {a \sec (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])*Tan[c + d*x]^9,x]

[Out]

-((a*Log[Cos[c + d*x]])/d) + (a*Sec[c + d*x])/d - (2*a*Sec[c + d*x]^2)/d - (4*a*Sec[c + d*x]^3)/(3*d) + (3*a*S
ec[c + d*x]^4)/(2*d) + (6*a*Sec[c + d*x]^5)/(5*d) - (2*a*Sec[c + d*x]^6)/(3*d) - (4*a*Sec[c + d*x]^7)/(7*d) +
(a*Sec[c + d*x]^8)/(8*d) + (a*Sec[c + d*x]^9)/(9*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 ^9(c+d x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(a-a x)^4 (a+a x)^5}{x^{10}} \, dx,x,\cos (c+d x)\right )}{a^8 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {a^9}{x^{10}}+\frac {a^9}{x^9}-\frac {4 a^9}{x^8}-\frac {4 a^9}{x^7}+\frac {6 a^9}{x^6}+\frac {6 a^9}{x^5}-\frac {4 a^9}{x^4}-\frac {4 a^9}{x^3}+\frac {a^9}{x^2}+\frac {a^9}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^8 d}\\ &=-\frac {a \log (\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}-\frac {2 a \sec ^2(c+d x)}{d}-\frac {4 a \sec ^3(c+d x)}{3 d}+\frac {3 a \sec ^4(c+d x)}{2 d}+\frac {6 a \sec ^5(c+d x)}{5 d}-\frac {2 a \sec ^6(c+d x)}{3 d}-\frac {4 a \sec ^7(c+d x)}{7 d}+\frac {a \sec ^8(c+d x)}{8 d}+\frac {a \sec ^9(c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 134, normalized size = 0.89 \[ \frac {a \sec ^9(c+d x)}{9 d}-\frac {4 a \sec ^7(c+d x)}{7 d}+\frac {6 a \sec ^5(c+d x)}{5 d}-\frac {4 a \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x)}{d}-\frac {a \left (-3 \tan ^8(c+d x)+4 \tan ^6(c+d x)-6 \tan ^4(c+d x)+12 \tan ^2(c+d x)+24 \log (\cos (c+d x))\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])*Tan[c + d*x]^9,x]

[Out]

(a*Sec[c + d*x])/d - (4*a*Sec[c + d*x]^3)/(3*d) + (6*a*Sec[c + d*x]^5)/(5*d) - (4*a*Sec[c + d*x]^7)/(7*d) + (a
*Sec[c + d*x]^9)/(9*d) - (a*(24*Log[Cos[c + d*x]] + 12*Tan[c + d*x]^2 - 6*Tan[c + d*x]^4 + 4*Tan[c + d*x]^6 -
3*Tan[c + d*x]^8))/(24*d)

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fricas [A]  time = 0.71, size = 123, normalized size = 0.81 \[ -\frac {2520 \, a \cos \left (d x + c\right )^{9} \log \left (-\cos \left (d x + c\right )\right ) - 2520 \, a \cos \left (d x + c\right )^{8} + 5040 \, a \cos \left (d x + c\right )^{7} + 3360 \, a \cos \left (d x + c\right )^{6} - 3780 \, a \cos \left (d x + c\right )^{5} - 3024 \, a \cos \left (d x + c\right )^{4} + 1680 \, a \cos \left (d x + c\right )^{3} + 1440 \, a \cos \left (d x + c\right )^{2} - 315 \, a \cos \left (d x + c\right ) - 280 \, a}{2520 \, d \cos \left (d x + c\right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*tan(d*x+c)^9,x, algorithm="fricas")

[Out]

-1/2520*(2520*a*cos(d*x + c)^9*log(-cos(d*x + c)) - 2520*a*cos(d*x + c)^8 + 5040*a*cos(d*x + c)^7 + 3360*a*cos
(d*x + c)^6 - 3780*a*cos(d*x + c)^5 - 3024*a*cos(d*x + c)^4 + 1680*a*cos(d*x + c)^3 + 1440*a*cos(d*x + c)^2 -
315*a*cos(d*x + c) - 280*a)/(d*cos(d*x + c)^9)

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giac [B]  time = 24.07, size = 293, normalized size = 1.94 \[ \frac {2520 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 2520 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {9177 \, a + \frac {87633 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {375732 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {953988 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1594782 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1336734 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {781956 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {302004 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {69201 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {7129 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{9}}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*tan(d*x+c)^9,x, algorithm="giac")

[Out]

1/2520*(2520*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 2520*a*log(abs(-(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) - 1)) + (9177*a + 87633*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 375732*a*(cos(d*x + c) - 1)^2/
(cos(d*x + c) + 1)^2 + 953988*a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 1594782*a*(cos(d*x + c) - 1)^4/(co
s(d*x + c) + 1)^4 + 1336734*a*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 781956*a*(cos(d*x + c) - 1)^6/(cos(d
*x + c) + 1)^6 + 302004*a*(cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7 + 69201*a*(cos(d*x + c) - 1)^8/(cos(d*x +
c) + 1)^8 + 7129*a*(cos(d*x + c) - 1)^9/(cos(d*x + c) + 1)^9)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^9)/d

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maple [A]  time = 0.85, size = 273, normalized size = 1.81 \[ \frac {a \left (\tan ^{8}\left (d x +c \right )\right )}{8 d}-\frac {a \left (\tan ^{6}\left (d x +c \right )\right )}{6 d}+\frac {a \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a \left (\sin ^{10}\left (d x +c \right )\right )}{9 d \cos \left (d x +c \right )^{9}}-\frac {a \left (\sin ^{10}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )^{7}}+\frac {a \left (\sin ^{10}\left (d x +c \right )\right )}{105 d \cos \left (d x +c \right )^{5}}-\frac {a \left (\sin ^{10}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )^{3}}+\frac {a \left (\sin ^{10}\left (d x +c \right )\right )}{9 d \cos \left (d x +c \right )}+\frac {128 a \cos \left (d x +c \right )}{315 d}+\frac {\cos \left (d x +c \right ) \left (\sin ^{8}\left (d x +c \right )\right ) a}{9 d}+\frac {8 a \cos \left (d x +c \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{63 d}+\frac {16 a \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{105 d}+\frac {64 a \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{315 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))*tan(d*x+c)^9,x)

[Out]

1/8/d*a*tan(d*x+c)^8-1/6/d*a*tan(d*x+c)^6+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+1/9/d*a
*sin(d*x+c)^10/cos(d*x+c)^9-1/63/d*a*sin(d*x+c)^10/cos(d*x+c)^7+1/105/d*a*sin(d*x+c)^10/cos(d*x+c)^5-1/63/d*a*
sin(d*x+c)^10/cos(d*x+c)^3+1/9/d*a*sin(d*x+c)^10/cos(d*x+c)+128/315*a*cos(d*x+c)/d+1/9/d*cos(d*x+c)*sin(d*x+c)
^8*a+8/63/d*a*cos(d*x+c)*sin(d*x+c)^6+16/105/d*a*cos(d*x+c)*sin(d*x+c)^4+64/315/d*a*cos(d*x+c)*sin(d*x+c)^2

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maxima [A]  time = 0.80, size = 116, normalized size = 0.77 \[ -\frac {2520 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac {2520 \, a \cos \left (d x + c\right )^{8} - 5040 \, a \cos \left (d x + c\right )^{7} - 3360 \, a \cos \left (d x + c\right )^{6} + 3780 \, a \cos \left (d x + c\right )^{5} + 3024 \, a \cos \left (d x + c\right )^{4} - 1680 \, a \cos \left (d x + c\right )^{3} - 1440 \, a \cos \left (d x + c\right )^{2} + 315 \, a \cos \left (d x + c\right ) + 280 \, a}{\cos \left (d x + c\right )^{9}}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*tan(d*x+c)^9,x, algorithm="maxima")

[Out]

-1/2520*(2520*a*log(cos(d*x + c)) - (2520*a*cos(d*x + c)^8 - 5040*a*cos(d*x + c)^7 - 3360*a*cos(d*x + c)^6 + 3
780*a*cos(d*x + c)^5 + 3024*a*cos(d*x + c)^4 - 1680*a*cos(d*x + c)^3 - 1440*a*cos(d*x + c)^2 + 315*a*cos(d*x +
c) + 280*a)/cos(d*x + c)^9)/d

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mupad [B]  time = 5.13, size = 259, normalized size = 1.72 \[ \frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+\frac {218\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{3}-174\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {1382\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{5}-\frac {2114\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}+\frac {1654\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{35}-\frac {326\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}+\frac {256\,a}{315}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+126\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-126\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^9*(a + a/cos(c + d*x)),x)

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)^2))/d - ((256*a)/315 - (326*a*tan(c/2 + (d*x)/2)^2)/35 + (1654*a*tan(c/2 + (d*x)
/2)^4)/35 - (2114*a*tan(c/2 + (d*x)/2)^6)/15 + (1382*a*tan(c/2 + (d*x)/2)^8)/5 - 174*a*tan(c/2 + (d*x)/2)^10 +
(218*a*tan(c/2 + (d*x)/2)^12)/3 - 18*a*tan(c/2 + (d*x)/2)^14 + 2*a*tan(c/2 + (d*x)/2)^16)/(d*(9*tan(c/2 + (d*
x)/2)^2 - 36*tan(c/2 + (d*x)/2)^4 + 84*tan(c/2 + (d*x)/2)^6 - 126*tan(c/2 + (d*x)/2)^8 + 126*tan(c/2 + (d*x)/2
)^10 - 84*tan(c/2 + (d*x)/2)^12 + 36*tan(c/2 + (d*x)/2)^14 - 9*tan(c/2 + (d*x)/2)^16 + tan(c/2 + (d*x)/2)^18 -
1))

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sympy [A]  time = 21.44, size = 184, normalized size = 1.22 \[ \begin {cases} \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{8}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{9 d} + \frac {a \tan ^{8}{\left (c + d x \right )}}{8 d} - \frac {8 a \tan ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{63 d} - \frac {a \tan ^{6}{\left (c + d x \right )}}{6 d} + \frac {16 a \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{105 d} + \frac {a \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {64 a \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{315 d} - \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {128 a \sec {\left (c + d x \right )}}{315 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right ) \tan ^{9}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*tan(d*x+c)**9,x)

[Out]

Piecewise((a*log(tan(c + d*x)**2 + 1)/(2*d) + a*tan(c + d*x)**8*sec(c + d*x)/(9*d) + a*tan(c + d*x)**8/(8*d) -
8*a*tan(c + d*x)**6*sec(c + d*x)/(63*d) - a*tan(c + d*x)**6/(6*d) + 16*a*tan(c + d*x)**4*sec(c + d*x)/(105*d)
+ a*tan(c + d*x)**4/(4*d) - 64*a*tan(c + d*x)**2*sec(c + d*x)/(315*d) - a*tan(c + d*x)**2/(2*d) + 128*a*sec(c
+ d*x)/(315*d), Ne(d, 0)), (x*(a*sec(c) + a)*tan(c)**9, True))

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