3.1.42 \(\int \frac {\sec ^5(c+d x)}{a+a \sec (c+d x)} \, dx\) [42]

Optimal. Leaf size=103 \[ -\frac {3 \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {4 \tan (c+d x)}{a d}-\frac {3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {4 \tan ^3(c+d x)}{3 a d} \]

[Out]

-3/2*arctanh(sin(d*x+c))/a/d+4*tan(d*x+c)/a/d-3/2*sec(d*x+c)*tan(d*x+c)/a/d-sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec
(d*x+c))+4/3*tan(d*x+c)^3/a/d

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Rubi [A]
time = 0.07, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3903, 3872, 3853, 3855, 3852} \begin {gather*} \frac {4 \tan ^3(c+d x)}{3 a d}+\frac {4 \tan (c+d x)}{a d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {\tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}-\frac {3 \tan (c+d x) \sec (c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[Sin[c + d*x]])/(2*a*d) + (4*Tan[c + d*x])/(a*d) - (3*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (Sec[c +
d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) + (4*Tan[c + d*x]^3)/(3*a*d)

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/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]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3903

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[d^2*Cot[e +
f*x]*((d*Csc[e + f*x])^(n - 2)/(f*(a + b*Csc[e + f*x]))), x] - Dist[d^2/(a*b), Int[(d*Csc[e + f*x])^(n - 2)*(
b*(n - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{a+a \sec (c+d x)} \, dx &=-\frac {\sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {\int \sec ^3(c+d x) (3 a-4 a \sec (c+d x)) \, dx}{a^2}\\ &=-\frac {\sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {3 \int \sec ^3(c+d x) \, dx}{a}+\frac {4 \int \sec ^4(c+d x) \, dx}{a}\\ &=-\frac {3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {3 \int \sec (c+d x) \, dx}{2 a}-\frac {4 \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a d}\\ &=-\frac {3 \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {4 \tan (c+d x)}{a d}-\frac {3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {4 \tan ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(374\) vs. \(2(103)=206\).
time = 3.35, size = 374, normalized size = 3.63 \begin {gather*} \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (6 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\frac {1}{8} \cos \left (\frac {1}{2} (c+d x)\right ) \sec (c) \sec ^3(c+d x) \left (9 \cos (2 c+3 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 \cos (4 c+3 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+27 \cos (d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+27 \cos (2 c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-9 \cos (2 c+3 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 \cos (4 c+3 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+48 \sin (d x)-12 \sin (2 c+d x)-6 \sin (c+2 d x)-6 \sin (3 c+2 d x)+20 \sin (2 c+3 d x)\right )\right )}{3 a d (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]*(6*Sec[c/2]*Sin[(d*x)/2] + (Cos[(c + d*x)/2]*Sec[c]*Sec[c + d*x]^3*(9*Cos[2*c +
3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
+ 27*Cos[d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 27*Cos[
2*c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 9*Cos[2*c +
3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 9*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]
+ 48*Sin[d*x] - 12*Sin[2*c + d*x] - 6*Sin[c + 2*d*x] - 6*Sin[3*c + 2*d*x] + 20*Sin[2*c + 3*d*x]))/8))/(3*a*d*
(1 + Sec[c + d*x]))

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Maple [A]
time = 0.09, size = 134, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {5}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}}{d a}\)	\(134\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {5}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}}{d a}\)	\(134\)
risch	\(\frac {i \left (9 \,{\mathrm e}^{6 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}+24 \,{\mathrm e}^{4 i \left (d x +c \right )}+24 \,{\mathrm e}^{3 i \left (d x +c \right )}+39 \,{\mathrm e}^{2 i \left (d x +c \right )}+7 \,{\mathrm e}^{i \left (d x +c \right )}+16\right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}\)	\(147\)
norman	\(\frac {\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {37 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {49 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {9 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (97) = 194\).
time = 0.28, size = 205, normalized size = 1.99 \begin {gather*} \frac {\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*sin(
d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d*x
+ c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]
time = 4.57, size = 124, normalized size = 1.20 \begin {gather*} -\frac {9 \, {\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, \cos \left (d x + c\right )^{3} + 7 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(9*(cos(d*x + c)^4 + cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 9*(cos(d*x + c)^4 + cos(d*x + c)^3)*log(-si
n(d*x + c) + 1) - 2*(16*cos(d*x + c)^3 + 7*cos(d*x + c)^2 - cos(d*x + c) + 2)*sin(d*x + c))/(a*d*cos(d*x + c)^
4 + a*d*cos(d*x + c)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**5/(sec(c + d*x) + 1), x)/a

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Giac [A]
time = 0.44, size = 114, normalized size = 1.11 \begin {gather*} -\frac {\frac {9 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {9 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(9*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 9*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 6*tan(1/2*d*x + 1/2*c)
/a + 2*(15*tan(1/2*d*x + 1/2*c)^5 - 16*tan(1/2*d*x + 1/2*c)^3 + 9*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)
^2 - 1)^3*a))/d

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Mupad [B]
time = 0.93, size = 96, normalized size = 0.93 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d}-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + a/cos(c + d*x))),x)

[Out]

tan(c/2 + (d*x)/2)/(a*d) - (3*atanh(tan(c/2 + (d*x)/2)))/(a*d) - (3*tan(c/2 + (d*x)/2) - (16*tan(c/2 + (d*x)/2
)^3)/3 + 5*tan(c/2 + (d*x)/2)^5)/(a*d*(tan(c/2 + (d*x)/2)^2 - 1)^3)

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