3.1.20 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\) [20]

Optimal. Leaf size=114 \[ \frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {13 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a^3 \tan ^3(c+d x)}{3 d}+\frac {a^3 \tan ^5(c+d x)}{5 d} \]

[Out]

13/8*a^3*arctanh(sin(d*x+c))/d+4*a^3*tan(d*x+c)/d+13/8*a^3*sec(d*x+c)*tan(d*x+c)/d+3/4*a^3*sec(d*x+c)^3*tan(d*
x+c)/d+5/3*a^3*tan(d*x+c)^3/d+1/5*a^3*tan(d*x+c)^5/d

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Rubi [A]
time = 0.10, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3876, 3853, 3855, 3852} \begin {gather*} \frac {a^3 \tan ^5(c+d x)}{5 d}+\frac {5 a^3 \tan ^3(c+d x)}{3 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {13 a^3 \tan (c+d x) \sec (c+d x)}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a^3*Tan[c + d*x])/d + (13*a^3*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (3*
a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^3*Tan[c + d*x]^3)/(3*d) + (a^3*Tan[c + d*x]^5)/(5*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 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 \, dx &=\int \left (a^3 \sec ^3(c+d x)+3 a^3 \sec ^4(c+d x)+3 a^3 \sec ^5(c+d x)+a^3 \sec ^6(c+d x)\right ) \, dx\\ &=a^3 \int \sec ^3(c+d x) \, dx+a^3 \int \sec ^6(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^4(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^5(c+d x) \, dx\\ &=\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{2} a^3 \int \sec (c+d x) \, dx+\frac {1}{4} \left (9 a^3\right ) \int \sec ^3(c+d x) \, dx-\frac {a^3 \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {13 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a^3 \tan ^3(c+d x)}{3 d}+\frac {a^3 \tan ^5(c+d x)}{5 d}+\frac {1}{8} \left (9 a^3\right ) \int \sec (c+d x) \, dx\\ &=\frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {13 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a^3 \tan ^3(c+d x)}{3 d}+\frac {a^3 \tan ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(487\) vs. \(2(114)=228\).
time = 1.55, size = 487, normalized size = 4.27 \begin {gather*} -\frac {a^3 \sec (c) \sec ^5(c+d x) \left (975 \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 )+975 \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 )+195 \cos (4 c+5 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+195 \cos (6 c+5 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+1950 \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 )+1950 \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 )-975 \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 )-975 \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 )-195 \cos (4 c+5 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-195 \cos (6 c+5 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-4640 \sin (d x)+1440 \sin (2 c+d x)-1500 \sin (c+2 d x)-1500 \sin (3 c+2 d x)-3040 \sin (2 c+3 d x)-390 \sin (3 c+4 d x)-390 \sin (5 c+4 d x)-608 \sin (4 c+5 d x)\right )}{3840 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3840*(a^3*Sec[c]*Sec[c + d*x]^5*(975*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 975*Cos[4*
c + 3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 195*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]] + 195*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 1950*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]]) + 1950*Cos[2*c + d*x]*(Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 975*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]] - 975*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 195*Cos[4*c + 5*d*x]*Log[Cos[(
c + d*x)/2] + Sin[(c + d*x)/2]] - 195*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 4640*Sin[d*x
] + 1440*Sin[2*c + d*x] - 1500*Sin[c + 2*d*x] - 1500*Sin[3*c + 2*d*x] - 3040*Sin[2*c + 3*d*x] - 390*Sin[3*c +
4*d*x] - 390*Sin[5*c + 4*d*x] - 608*Sin[4*c + 5*d*x]))/d

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Maple [A]
time = 0.10, size = 146, normalized size = 1.28

method	result	size
risch	\(-\frac {i a^{3} \left (195 \,{\mathrm e}^{9 i \left (d x +c \right )}+750 \,{\mathrm e}^{7 i \left (d x +c \right )}-720 \,{\mathrm e}^{6 i \left (d x +c \right )}-2320 \,{\mathrm e}^{4 i \left (d x +c \right )}-750 \,{\mathrm e}^{3 i \left (d x +c \right )}-1520 \,{\mathrm e}^{2 i \left (d x +c \right )}-195 \,{\mathrm e}^{i \left (d x +c \right )}-304\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(145\)
derivativedivides	\(\frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+3 a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(146\)
default	\(\frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+3 a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(146\)
norman	\(\frac {-\frac {51 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {133 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {416 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {91 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {13 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {13 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {13 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-a^3*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+3*a^3*(-(-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)))-3*a^3*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a^3*(1/2*sec(d*x+c)*tan(d*x+c)
+1/2*ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [A]
time = 0.29, size = 179, normalized size = 1.57 \begin {gather*} \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - 45 \, a^{3} {\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 )} - 60 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))
*a^3 - 45*a^3*(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)) - 60*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + lo
g(sin(d*x + c) - 1)))/d

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Fricas [A]
time = 2.76, size = 124, normalized size = 1.09 \begin {gather*} \frac {195 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 195 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (304 \, a^{3} \cos \left (d x + c\right )^{4} + 195 \, a^{3} \cos \left (d x + c\right )^{3} + 152 \, a^{3} \cos \left (d x + c\right )^{2} + 90 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{240 \, 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)^3*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/240*(195*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 195*a^3*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(304*a
^3*cos(d*x + c)^4 + 195*a^3*cos(d*x + c)^3 + 152*a^3*cos(d*x + c)^2 + 90*a^3*cos(d*x + c) + 24*a^3)*sin(d*x +
c))/(d*cos(d*x + c)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(sec(c + d*x)**3, x) + Integral(3*sec(c + d*x)**4, x) + Integral(3*sec(c + d*x)**5, x) + Integra
l(sec(c + d*x)**6, x))

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Giac [A]
time = 0.47, size = 138, normalized size = 1.21 \begin {gather*} \frac {195 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 195 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (195 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 910 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1330 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, a^{3} \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 )}^{5}}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(195*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 195*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(195*a^3*ta
n(1/2*d*x + 1/2*c)^9 - 910*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*a^3*tan(1/2*d*x + 1/2*c)^5 - 1330*a^3*tan(1/2*d*x
+ 1/2*c)^3 + 765*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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Mupad [B]
time = 5.48, size = 170, normalized size = 1.49 \begin {gather*} \frac {13\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}-\frac {91\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {416\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {133\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {51\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{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 + a/cos(c + d*x))^3/cos(c + d*x)^3,x)

[Out]

(13*a^3*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((416*a^3*tan(c/2 + (d*x)/2)^5)/15 - (133*a^3*tan(c/2 + (d*x)/2)^3)
/6 - (91*a^3*tan(c/2 + (d*x)/2)^7)/6 + (13*a^3*tan(c/2 + (d*x)/2)^9)/4 + (51*a^3*tan(c/2 + (d*x)/2))/4)/(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))

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