3.40 \(\int (a+a \sec (c+d x))^3 \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=99 \[ \frac {a^3 \sec ^5(c+d x)}{5 d}+\frac {3 a^3 \sec ^4(c+d x)}{4 d}+\frac {2 a^3 \sec ^3(c+d x)}{3 d}-\frac {a^3 \sec ^2(c+d x)}{d}-\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 99, 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, 75} \[ \frac {a^3 \sec ^5(c+d x)}{5 d}+\frac {3 a^3 \sec ^4(c+d x)}{4 d}+\frac {2 a^3 \sec ^3(c+d x)}{3 d}-\frac {a^3 \sec ^2(c+d x)}{d}-\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^3,x]

[Out]

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

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

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

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Mathematica [A]  time = 0.30, size = 92, normalized size = 0.93 \[ -\frac {a^3 \sec ^5(c+d x) (280 \cos (2 (c+d x))+90 \cos (4 (c+d x))+\cos (3 (c+d x)) (60-75 \log (\cos (c+d x)))-150 \cos (c+d x) \log (\cos (c+d x))-15 \cos (5 (c+d x)) \log (\cos (c+d x))+142)}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^3,x]

[Out]

-1/240*(a^3*(142 + 280*Cos[2*(c + d*x)] + 90*Cos[4*(c + d*x)] + Cos[3*(c + d*x)]*(60 - 75*Log[Cos[c + d*x]]) -
150*Cos[c + d*x]*Log[Cos[c + d*x]] - 15*Cos[5*(c + d*x)]*Log[Cos[c + d*x]])*Sec[c + d*x]^5)/d

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fricas [A]  time = 0.49, size = 91, normalized size = 0.92 \[ \frac {60 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 180 \, a^{3} \cos \left (d x + c\right )^{4} - 60 \, a^{3} \cos \left (d x + c\right )^{3} + 40 \, a^{3} \cos \left (d x + c\right )^{2} + 45 \, a^{3} \cos \left (d x + c\right ) + 12 \, a^{3}}{60 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

1/60*(60*a^3*cos(d*x + c)^5*log(-cos(d*x + c)) - 180*a^3*cos(d*x + c)^4 - 60*a^3*cos(d*x + c)^3 + 40*a^3*cos(d
*x + c)^2 + 45*a^3*cos(d*x + c) + 12*a^3)/(d*cos(d*x + c)^5)

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giac [B]  time = 1.52, size = 217, normalized size = 2.19 \[ -\frac {60 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {393 \, a^{3} + \frac {2085 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2610 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1970 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {805 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {137 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^3,x, algorithm="giac")

[Out]

-1/60*(60*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d
*x + c) + 1) - 1)) + (393*a^3 + 2085*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2610*a^3*(cos(d*x + c) - 1)^2
/(cos(d*x + c) + 1)^2 + 1970*a^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 805*a^3*(cos(d*x + c) - 1)^4/(cos
(d*x + c) + 1)^4 + 137*a^3*(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 = 0.78, size = 164, normalized size = 1.66 \[ \frac {a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {16 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}-\frac {16 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )}-\frac {16 a^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}-\frac {32 a^{3} \cos \left (d x +c \right )}{15 d}+\frac {3 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*tan(d*x+c)^3,x)

[Out]

1/2*a^3*tan(d*x+c)^2/d+a^3*ln(cos(d*x+c))/d+16/15/d*a^3*sin(d*x+c)^4/cos(d*x+c)^3-16/15/d*a^3*sin(d*x+c)^4/cos
(d*x+c)-16/15/d*a^3*cos(d*x+c)*sin(d*x+c)^2-32/15*a^3*cos(d*x+c)/d+3/4/d*a^3*sin(d*x+c)^4/cos(d*x+c)^4+1/5/d*a
^3*sin(d*x+c)^4/cos(d*x+c)^5

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maxima [A]  time = 0.47, size = 84, normalized size = 0.85 \[ \frac {60 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) - \frac {180 \, a^{3} \cos \left (d x + c\right )^{4} + 60 \, a^{3} \cos \left (d x + c\right )^{3} - 40 \, a^{3} \cos \left (d x + c\right )^{2} - 45 \, a^{3} \cos \left (d x + c\right ) - 12 \, a^{3}}{\cos \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

1/60*(60*a^3*log(cos(d*x + c)) - (180*a^3*cos(d*x + c)^4 + 60*a^3*cos(d*x + c)^3 - 40*a^3*cos(d*x + c)^2 - 45*
a^3*cos(d*x + c) - 12*a^3)/cos(d*x + c)^5)/d

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mupad [B]  time = 5.55, size = 162, normalized size = 1.64 \[ \frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {62\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {70\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {64\,a^3}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a^3\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + a/cos(c + d*x))^3,x)

[Out]

((62*a^3*tan(c/2 + (d*x)/2)^4)/3 - (70*a^3*tan(c/2 + (d*x)/2)^2)/3 - 10*a^3*tan(c/2 + (d*x)/2)^6 + 2*a^3*tan(c
/2 + (d*x)/2)^8 + (64*a^3)/15)/(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)) - (2*a^3*atanh(tan(c/2 + (d*x)/2)^2))/d

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sympy [A]  time = 3.24, size = 165, normalized size = 1.67 \[ \begin {cases} - \frac {a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{3} \tan ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{5 d} + \frac {3 a^{3} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{4 d} + \frac {a^{3} \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{d} + \frac {a^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} - \frac {2 a^{3} \sec ^{3}{\left (c + d x \right )}}{15 d} - \frac {3 a^{3} \sec ^{2}{\left (c + d x \right )}}{4 d} - \frac {2 a^{3} \sec {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right )^{3} \tan ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**3*tan(d*x+c)**3,x)

[Out]

Piecewise((-a**3*log(tan(c + d*x)**2 + 1)/(2*d) + a**3*tan(c + d*x)**2*sec(c + d*x)**3/(5*d) + 3*a**3*tan(c +
d*x)**2*sec(c + d*x)**2/(4*d) + a**3*tan(c + d*x)**2*sec(c + d*x)/d + a**3*tan(c + d*x)**2/(2*d) - 2*a**3*sec(
c + d*x)**3/(15*d) - 3*a**3*sec(c + d*x)**2/(4*d) - 2*a**3*sec(c + d*x)/d, Ne(d, 0)), (x*(a*sec(c) + a)**3*tan
(c)**3, True))

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