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

Optimal. Leaf size=84 \[ \frac {\tan ^4(c+d x) (5 a+4 b \sec (c+d x))}{20 d}-\frac {\tan ^2(c+d x) (15 a+8 b \sec (c+d x))}{30 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {8 b \sec (c+d x)}{15 d} \]

[Out]

-a*ln(cos(d*x+c))/d+8/15*b*sec(d*x+c)/d-1/30*(15*a+8*b*sec(d*x+c))*tan(d*x+c)^2/d+1/20*(5*a+4*b*sec(d*x+c))*ta
n(d*x+c)^4/d

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3881, 3884, 3475, 2606, 8} \[ \frac {\tan ^4(c+d x) (5 a+4 b \sec (c+d x))}{20 d}-\frac {\tan ^2(c+d x) (15 a+8 b \sec (c+d x))}{30 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {8 b \sec (c+d x)}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Log[Cos[c + d*x]])/d) + (8*b*Sec[c + d*x])/(15*d) - ((15*a + 8*b*Sec[c + d*x])*Tan[c + d*x]^2)/(30*d) + (
(5*a + 4*b*Sec[c + d*x])*Tan[c + d*x]^4)/(20*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3881

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(e*(e*Cot[
c + d*x])^(m - 1)*(a*m + b*(m - 1)*Csc[c + d*x]))/(d*m*(m - 1)), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m - 2)
*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x)) \tan ^5(c+d x) \, dx &=\frac {(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}-\frac {1}{5} \int (5 a+4 b \sec (c+d x)) \tan ^3(c+d x) \, dx\\ &=-\frac {(15 a+8 b \sec (c+d x)) \tan ^2(c+d x)}{30 d}+\frac {(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}+\frac {1}{15} \int (15 a+8 b \sec (c+d x)) \tan (c+d x) \, dx\\ &=-\frac {(15 a+8 b \sec (c+d x)) \tan ^2(c+d x)}{30 d}+\frac {(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}+a \int \tan (c+d x) \, dx+\frac {1}{15} (8 b) \int \sec (c+d x) \tan (c+d x) \, dx\\ &=-\frac {a \log (\cos (c+d x))}{d}-\frac {(15 a+8 b \sec (c+d x)) \tan ^2(c+d x)}{30 d}+\frac {(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}+\frac {(8 b) \operatorname {Subst}(\int 1 \, dx,x,\sec (c+d x))}{15 d}\\ &=-\frac {a \log (\cos (c+d x))}{d}+\frac {8 b \sec (c+d x)}{15 d}-\frac {(15 a+8 b \sec (c+d x)) \tan ^2(c+d x)}{30 d}+\frac {(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 82, normalized size = 0.98 \[ -\frac {a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d}+\frac {b \sec ^5(c+d x)}{5 d}-\frac {2 b \sec ^3(c+d x)}{3 d}+\frac {b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sec[c + d*x])/d - (2*b*Sec[c + d*x]^3)/(3*d) + (b*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 = 1.13, size = 79, normalized size = 0.94 \[ -\frac {60 \, a \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 60 \, b \cos \left (d x + c\right )^{4} + 60 \, a \cos \left (d x + c\right )^{3} + 40 \, b \cos \left (d x + c\right )^{2} - 15 \, a \cos \left (d x + c\right ) - 12 \, b}{60 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [B]  time = 3.01, size = 248, normalized size = 2.95 \[ \frac {60 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {137 \, a + 64 \, b + \frac {805 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {320 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {1970 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {640 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1970 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {805 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {137 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*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)) + (137*a + 64*b + 805*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 320*b*(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) + 1970*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 640*b*(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 [B]  time = 0.64, size = 161, normalized size = 1.92 \[ \frac {a \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {b \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}-\frac {b \left (\sin ^{6}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}+\frac {b \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )}+\frac {8 b \cos \left (d x +c \right )}{15 d}+\frac {b \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 b \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.50, size = 72, normalized size = 0.86 \[ -\frac {60 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac {60 \, b \cos \left (d x + c\right )^{4} - 60 \, a \cos \left (d x + c\right )^{3} - 40 \, b \cos \left (d x + c\right )^{2} + 15 \, a \cos \left (d x + c\right ) + 12 \, b}{\cos \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.85, size = 162, normalized size = 1.93 \[ \frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (10\,a+\frac {32\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-2\,a-\frac {16\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {16\,b}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)^2))/d - ((16*b)/15 - tan(c/2 + (d*x)/2)^2*(2*a + (16*b)/3) + tan(c/2 + (d*x)/2)^
4*(10*a + (32*b)/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*ta
n(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 = 3.13, size = 112, normalized size = 1.33 \[ \begin {cases} \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {b \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{5 d} - \frac {4 b \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 d} + \frac {8 b \sec {\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a + b \sec {\relax (c )}\right ) \tan ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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