3.20 \(\int (a+a \sec (c+d x))^2 \tan ^7(c+d x) \, dx\)

Optimal. Leaf size=132 \[ \frac {a^2 \sec ^8(c+d x)}{8 d}+\frac {2 a^2 \sec ^7(c+d x)}{7 d}-\frac {a^2 \sec ^6(c+d x)}{3 d}-\frac {6 a^2 \sec ^5(c+d x)}{5 d}+\frac {2 a^2 \sec ^3(c+d x)}{d}+\frac {a^2 \sec ^2(c+d x)}{d}-\frac {2 a^2 \sec (c+d x)}{d}+\frac {a^2 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 132, 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} \[ \frac {a^2 \sec ^8(c+d x)}{8 d}+\frac {2 a^2 \sec ^7(c+d x)}{7 d}-\frac {a^2 \sec ^6(c+d x)}{3 d}-\frac {6 a^2 \sec ^5(c+d x)}{5 d}+\frac {2 a^2 \sec ^3(c+d x)}{d}+\frac {a^2 \sec ^2(c+d x)}{d}-\frac {2 a^2 \sec (c+d x)}{d}+\frac {a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

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Mathematica [A]  time = 0.31, size = 110, normalized size = 0.83 \[ \frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (105 \sec ^8(c+d x)+240 \sec ^7(c+d x)-280 \sec ^6(c+d x)-1008 \sec ^5(c+d x)+1680 \sec ^3(c+d x)+840 \sec ^2(c+d x)-1680 \sec (c+d x)+840 \log (\cos (c+d x))\right )}{3360 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^7,x]

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(840*Log[Cos[c + d*x]] - 1680*Sec[c + d*x] + 840*Sec[c + d*x]^2 +
1680*Sec[c + d*x]^3 - 1008*Sec[c + d*x]^5 - 280*Sec[c + d*x]^6 + 240*Sec[c + d*x]^7 + 105*Sec[c + d*x]^8))/(3
360*d)

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fricas [A]  time = 0.73, size = 117, normalized size = 0.89 \[ \frac {840 \, a^{2} \cos \left (d x + c\right )^{8} \log \left (-\cos \left (d x + c\right )\right ) - 1680 \, a^{2} \cos \left (d x + c\right )^{7} + 840 \, a^{2} \cos \left (d x + c\right )^{6} + 1680 \, a^{2} \cos \left (d x + c\right )^{5} - 1008 \, a^{2} \cos \left (d x + c\right )^{3} - 280 \, a^{2} \cos \left (d x + c\right )^{2} + 240 \, a^{2} \cos \left (d x + c\right ) + 105 \, a^{2}}{840 \, d \cos \left (d x + c\right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^7,x, algorithm="fricas")

[Out]

1/840*(840*a^2*cos(d*x + c)^8*log(-cos(d*x + c)) - 1680*a^2*cos(d*x + c)^7 + 840*a^2*cos(d*x + c)^6 + 1680*a^2
*cos(d*x + c)^5 - 1008*a^2*cos(d*x + c)^3 - 280*a^2*cos(d*x + c)^2 + 240*a^2*cos(d*x + c) + 105*a^2)/(d*cos(d*
x + c)^8)

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giac [B]  time = 11.00, size = 292, normalized size = 2.21 \[ -\frac {840 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 840 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {3819 \, a^{2} + \frac {32232 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {120372 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {261464 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {258370 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {175448 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {77364 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {19944 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {2283 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{8}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^7,x, algorithm="giac")

[Out]

-1/840*(840*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 840*a^2*log(abs(-(cos(d*x + c) - 1)/(co
s(d*x + c) + 1) - 1)) + (3819*a^2 + 32232*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 120372*a^2*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2 + 261464*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 258370*a^2*(cos(d*x + c)
- 1)^4/(cos(d*x + c) + 1)^4 + 175448*a^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 77364*a^2*(cos(d*x + c)
- 1)^6/(cos(d*x + c) + 1)^6 + 19944*a^2*(cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7 + 2283*a^2*(cos(d*x + c) - 1
)^8/(cos(d*x + c) + 1)^8)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^8)/d

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maple [B]  time = 0.78, size = 264, normalized size = 2.00 \[ \frac {a^{2} \left (\tan ^{6}\left (d x +c \right )\right )}{6 d}-\frac {a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {2 a^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{7}}-\frac {2 a^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )^{5}}+\frac {2 a^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )^{3}}-\frac {2 a^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )}-\frac {32 a^{2} \cos \left (d x +c \right )}{35 d}-\frac {2 a^{2} \cos \left (d x +c \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{7 d}-\frac {12 a^{2} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{35 d}-\frac {16 a^{2} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{35 d}+\frac {a^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*tan(d*x+c)^7,x)

[Out]

1/6/d*a^2*tan(d*x+c)^6-1/4*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+2/7/d*a^2*sin(d*x+c)
^8/cos(d*x+c)^7-2/35/d*a^2*sin(d*x+c)^8/cos(d*x+c)^5+2/35/d*a^2*sin(d*x+c)^8/cos(d*x+c)^3-2/7/d*a^2*sin(d*x+c)
^8/cos(d*x+c)-32/35*a^2*cos(d*x+c)/d-2/7/d*a^2*cos(d*x+c)*sin(d*x+c)^6-12/35/d*a^2*cos(d*x+c)*sin(d*x+c)^4-16/
35/d*a^2*cos(d*x+c)*sin(d*x+c)^2+1/8/d*a^2*sin(d*x+c)^8/cos(d*x+c)^8

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maxima [A]  time = 0.43, size = 110, normalized size = 0.83 \[ \frac {840 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {1680 \, a^{2} \cos \left (d x + c\right )^{7} - 840 \, a^{2} \cos \left (d x + c\right )^{6} - 1680 \, a^{2} \cos \left (d x + c\right )^{5} + 1008 \, a^{2} \cos \left (d x + c\right )^{3} + 280 \, a^{2} \cos \left (d x + c\right )^{2} - 240 \, a^{2} \cos \left (d x + c\right ) - 105 \, a^{2}}{\cos \left (d x + c\right )^{8}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^7,x, algorithm="maxima")

[Out]

1/840*(840*a^2*log(cos(d*x + c)) - (1680*a^2*cos(d*x + c)^7 - 840*a^2*cos(d*x + c)^6 - 1680*a^2*cos(d*x + c)^5
+ 1008*a^2*cos(d*x + c)^3 + 280*a^2*cos(d*x + c)^2 - 240*a^2*cos(d*x + c) - 105*a^2)/cos(d*x + c)^8)/d

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mupad [B]  time = 4.82, size = 249, normalized size = 1.89 \[ \frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\frac {170\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{3}-\frac {352\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {2386\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}-\frac {336\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {582\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}-\frac {64\,a^2}{35}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^7*(a + a/cos(c + d*x))^2,x)

[Out]

((582*a^2*tan(c/2 + (d*x)/2)^2)/35 - (336*a^2*tan(c/2 + (d*x)/2)^4)/5 + (2386*a^2*tan(c/2 + (d*x)/2)^6)/15 - (
352*a^2*tan(c/2 + (d*x)/2)^8)/3 + (170*a^2*tan(c/2 + (d*x)/2)^10)/3 - 16*a^2*tan(c/2 + (d*x)/2)^12 + 2*a^2*tan
(c/2 + (d*x)/2)^14 - (64*a^2)/35)/(d*(28*tan(c/2 + (d*x)/2)^4 - 8*tan(c/2 + (d*x)/2)^2 - 56*tan(c/2 + (d*x)/2)
^6 + 70*tan(c/2 + (d*x)/2)^8 - 56*tan(c/2 + (d*x)/2)^10 + 28*tan(c/2 + (d*x)/2)^12 - 8*tan(c/2 + (d*x)/2)^14 +
tan(c/2 + (d*x)/2)^16 + 1)) - (2*a^2*atanh(tan(c/2 + (d*x)/2)^2))/d

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sympy [A]  time = 14.18, size = 252, normalized size = 1.91 \[ \begin {cases} - \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{6}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{8 d} + \frac {2 a^{2} \tan ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{7 d} + \frac {a^{2} \tan ^{6}{\left (c + d x \right )}}{6 d} - \frac {a^{2} \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{8 d} - \frac {12 a^{2} \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{35 d} - \frac {a^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} + \frac {a^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{8 d} + \frac {16 a^{2} \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{35 d} + \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} - \frac {a^{2} \sec ^{2}{\left (c + d x \right )}}{8 d} - \frac {32 a^{2} \sec {\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right )^{2} \tan ^{7}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**2*tan(d*x+c)**7,x)

[Out]

Piecewise((-a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**6*sec(c + d*x)**2/(8*d) + 2*a**2*tan(c +
d*x)**6*sec(c + d*x)/(7*d) + a**2*tan(c + d*x)**6/(6*d) - a**2*tan(c + d*x)**4*sec(c + d*x)**2/(8*d) - 12*a**2
*tan(c + d*x)**4*sec(c + d*x)/(35*d) - a**2*tan(c + d*x)**4/(4*d) + a**2*tan(c + d*x)**2*sec(c + d*x)**2/(8*d)
+ 16*a**2*tan(c + d*x)**2*sec(c + d*x)/(35*d) + a**2*tan(c + d*x)**2/(2*d) - a**2*sec(c + d*x)**2/(8*d) - 32*
a**2*sec(c + d*x)/(35*d), Ne(d, 0)), (x*(a*sec(c) + a)**2*tan(c)**7, True))

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