3.1.10 \(\int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx\) [10]

Optimal. Leaf size=122 \[ \frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {9 a^2 \tan (c+d x)}{5 d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {3 a^2 \tan ^3(c+d x)}{5 d} \]

[Out]

3/4*a^2*arctanh(sin(d*x+c))/d+9/5*a^2*tan(d*x+c)/d+3/4*a^2*sec(d*x+c)*tan(d*x+c)/d+1/2*a^2*sec(d*x+c)^3*tan(d*
x+c)/d+1/5*a^2*sec(d*x+c)^4*tan(d*x+c)/d+3/5*a^2*tan(d*x+c)^3/d

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Rubi [A]
time = 0.07, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3873, 3853, 3855, 4131, 3852} \begin {gather*} \frac {3 a^2 \tan ^3(c+d x)}{5 d}+\frac {9 a^2 \tan (c+d x)}{5 d}+\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^2,x]

[Out]

(3*a^2*ArcTanh[Sin[c + d*x]])/(4*d) + (9*a^2*Tan[c + d*x])/(5*d) + (3*a^2*Sec[c + d*x]*Tan[c + d*x])/(4*d) + (
a^2*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) + (a^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (3*a^2*Tan[c + d*x]^3)/(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 3873

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

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx &=\left (2 a^2\right ) \int \sec ^5(c+d x) \, dx+\int \sec ^4(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} \left (3 a^2\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (9 a^2\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (3 a^2\right ) \int \sec (c+d x) \, dx-\frac {\left (9 a^2\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {9 a^2 \tan (c+d x)}{5 d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {3 a^2 \tan ^3(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. \(2(122)=244\).
time = 1.62, size = 487, normalized size = 3.99 \begin {gather*} -\frac {a^2 \sec (c) \sec ^5(c+d x) \left (75 \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 )+75 \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 )+15 \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 )+15 \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 )+150 \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 )+150 \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 )-75 \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 )-75 \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 )-15 \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 )-15 \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 )-400 \sin (d x)+80 \sin (2 c+d x)-140 \sin (c+2 d x)-140 \sin (3 c+2 d x)-240 \sin (2 c+3 d x)-30 \sin (3 c+4 d x)-30 \sin (5 c+4 d x)-48 \sin (4 c+5 d x)\right )}{640 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^2,x]

[Out]

-1/640*(a^2*Sec[c]*Sec[c + d*x]^5*(75*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 75*Cos[4*c +
3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 15*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]
] + 15*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 150*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]]) + 150*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]]) - 75*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]] - 75*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 15*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] - 15*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 400*Sin[d*x] + 80*Sin[2
*c + d*x] - 140*Sin[c + 2*d*x] - 140*Sin[3*c + 2*d*x] - 240*Sin[2*c + 3*d*x] - 30*Sin[3*c + 4*d*x] - 30*Sin[5*
c + 4*d*x] - 48*Sin[4*c + 5*d*x]))/d

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Maple [A]
time = 0.12, size = 111, normalized size = 0.91

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+2 a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(111\)
default	\(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+2 a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(111\)
risch	\(-\frac {i a^{2} \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}+70 \,{\mathrm e}^{7 i \left (d x +c \right )}-40 \,{\mathrm e}^{6 i \left (d x +c \right )}-200 \,{\mathrm e}^{4 i \left (d x +c \right )}-70 \,{\mathrm e}^{3 i \left (d x +c \right )}-120 \,{\mathrm e}^{2 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}-24\right )}{10 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}\)	\(145\)
norman	\(\frac {-\frac {13 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {9 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {72 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {7 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {3 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {3 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 133, normalized size = 1.09 \begin {gather*} \frac {8 \, {\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^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 15 \, a^{2} {\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 )}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/120*(8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*a
^2 - 15*a^2*(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)))/d

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Fricas [A]
time = 2.10, size = 124, normalized size = 1.02 \begin {gather*} \frac {15 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, a^{2} \cos \left (d x + c\right )^{4} + 15 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 10 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )} \sin \left (d x + c\right )}{40 \, 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)^4*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/40*(15*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*a^2*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(24*a^2*c
os(d*x + c)^4 + 15*a^2*cos(d*x + c)^3 + 12*a^2*cos(d*x + c)^2 + 10*a^2*cos(d*x + c) + 4*a^2)*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^{2} \left (\int \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 \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)**4*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(sec(c + d*x)**4, x) + Integral(2*sec(c + d*x)**5, x) + Integral(sec(c + d*x)**6, x))

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Giac [A]
time = 0.45, size = 138, normalized size = 1.13 \begin {gather*} \frac {15 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 144 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 65 \, a^{2} \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}}}{20 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/20*(15*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*a^2*tan(1/
2*d*x + 1/2*c)^9 - 70*a^2*tan(1/2*d*x + 1/2*c)^7 + 144*a^2*tan(1/2*d*x + 1/2*c)^5 - 90*a^2*tan(1/2*d*x + 1/2*c
)^3 + 65*a^2*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 = 5.70, size = 170, normalized size = 1.39 \begin {gather*} \frac {3\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}-7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}-9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {13\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{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))^2/cos(c + d*x)^4,x)

[Out]

(3*a^2*atanh(tan(c/2 + (d*x)/2)))/(2*d) - ((72*a^2*tan(c/2 + (d*x)/2)^5)/5 - 9*a^2*tan(c/2 + (d*x)/2)^3 - 7*a^
2*tan(c/2 + (d*x)/2)^7 + (3*a^2*tan(c/2 + (d*x)/2)^9)/2 + (13*a^2*tan(c/2 + (d*x)/2))/2)/(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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