3.6.8 \(\int \sec ^5(c+d x) (a+b \tan (c+d x)) \, dx\) [508]

Optimal. Leaf size=74 \[ \frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \sec ^5(c+d x)}{5 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[Out]

3/8*a*arctanh(sin(d*x+c))/d+1/5*b*sec(d*x+c)^5/d+3/8*a*sec(d*x+c)*tan(d*x+c)/d+1/4*a*sec(d*x+c)^3*tan(d*x+c)/d

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Rubi [A]
time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3567, 3853, 3855} \begin {gather*} \frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b \sec ^5(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + (b*Sec[c + d*x]^5)/(5*d) + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*Sec[
c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a+b \tan (c+d x)) \, dx &=\frac {b \sec ^5(c+d x)}{5 d}+a \int \sec ^5(c+d x) \, dx\\ &=\frac {b \sec ^5(c+d x)}{5 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} (3 a) \int \sec ^3(c+d x) \, dx\\ &=\frac {b \sec ^5(c+d x)}{5 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} (3 a) \int \sec (c+d x) \, dx\\ &=\frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \sec ^5(c+d x)}{5 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 68, normalized size = 0.92 \begin {gather*} \frac {b \sec ^5(c+d x)}{5 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {3 a \left (\tanh ^{-1}(\sin (c+d x))+\sec (c+d x) \tan (c+d x)\right )}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sec[c + d*x]^5)/(5*d) + (a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*a*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*
Tan[c + d*x]))/(8*d)

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Maple [A]
time = 0.21, size = 63, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b}{5 \cos \left (d x +c \right )^{5}}}{d}\)	\(63\)
default	\(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b}{5 \cos \left (d x +c \right )^{5}}}{d}\)	\(63\)
risch	\(\frac {-15 i a \,{\mathrm e}^{9 i \left (d x +c \right )}-70 i a \,{\mathrm e}^{7 i \left (d x +c \right )}+128 b \,{\mathrm e}^{5 i \left (d x +c \right )}+70 i a \,{\mathrm e}^{3 i \left (d x +c \right )}+15 i a \,{\mathrm e}^{i \left (d x +c \right )}}{20 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+1/5*b/cos(d*x+c)^5)

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Maxima [A]
time = 0.27, size = 86, normalized size = 1.16 \begin {gather*} -\frac {5 \, a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {16 \, b}{\cos \left (d x + c\right )^{5}}}{80 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*(5*a*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c)
+ 1) + 3*log(sin(d*x + c) - 1)) - 16*b/cos(d*x + c)^5)/d

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Fricas [A]
time = 0.38, size = 88, normalized size = 1.19 \begin {gather*} \frac {15 \, a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 10 \, {\left (3 \, a \cos \left (d x + c\right )^{3} + 2 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 16 \, b}{80 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(15*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*a*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 10*(3*a*cos(d*x
+ c)^3 + 2*a*cos(d*x + c))*sin(d*x + c) + 16*b)/(d*cos(d*x + c)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(c + d*x))*sec(c + d*x)**5, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (66) = 132\).
time = 0.55, size = 141, normalized size = 1.91 \begin {gather*} \frac {15 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (25 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 40 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 25 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{40 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(15*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(25*a*tan(1/2*d*x
+ 1/2*c)^9 - 40*b*tan(1/2*d*x + 1/2*c)^8 - 10*a*tan(1/2*d*x + 1/2*c)^7 - 80*b*tan(1/2*d*x + 1/2*c)^4 + 10*a*ta
n(1/2*d*x + 1/2*c)^3 - 25*a*tan(1/2*d*x + 1/2*c) - 8*b)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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Mupad [B]
time = 7.07, size = 175, normalized size = 2.36 \begin {gather*} \frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {2\,b}{5}}{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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((2*b)/5 + (5*a*tan(c/2 + (d*x)/2))/4 - (a*tan(c/2 + (d*x)/2)^3)/2 + (
a*tan(c/2 + (d*x)/2)^7)/2 - (5*a*tan(c/2 + (d*x)/2)^9)/4 + 4*b*tan(c/2 + (d*x)/2)^4 + 2*b*tan(c/2 + (d*x)/2)^8
)/(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 + ta
n(c/2 + (d*x)/2)^10 - 1))

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