3.1.54 \(\int \sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [54]

Optimal. Leaf size=122 \[ \frac {3 B \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d} \]

[Out]

3/8*B*arctanh(sin(d*x+c))/d+1/5*(5*A+4*C)*tan(d*x+c)/d+3/8*B*sec(d*x+c)*tan(d*x+c)/d+1/4*B*sec(d*x+c)^3*tan(d*
x+c)/d+1/5*C*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(5*A+4*C)*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 = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4132, 3853, 3855, 4131, 3852} \begin {gather*} \frac {(5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {3 B \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 B \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(3*B*ArcTanh[Sin[c + d*x]])/(8*d) + ((5*A + 4*C)*Tan[c + d*x])/(5*d) + (3*B*Sec[c + d*x]*Tan[c + d*x])/(8*d) +
(B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (C*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((5*A + 4*C)*Tan[c + d*x]^3)/
(15*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 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]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=B \int \sec ^5(c+d x) \, dx+\int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 B) \int \sec ^3(c+d x) \, dx+\frac {1}{5} (5 A+4 C) \int \sec ^4(c+d x) \, dx\\ &=\frac {3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} (3 B) \int \sec (c+d x) \, dx-\frac {(5 A+4 C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {3 B \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]
time = 0.42, size = 80, normalized size = 0.66 \begin {gather*} \frac {45 B \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (45 B \sec (c+d x)+30 B \sec ^3(c+d x)+8 \left (15 (A+C)+5 (A+2 C) \tan ^2(c+d x)+3 C \tan ^4(c+d x)\right )\right )}{120 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(45*B*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(45*B*Sec[c + d*x] + 30*B*Sec[c + d*x]^3 + 8*(15*(A + C) + 5*(A + 2
*C)*Tan[c + d*x]^2 + 3*C*Tan[c + d*x]^4)))/(120*d)

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Maple [A]
time = 0.54, size = 104, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {-A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(104\)
default	\(\frac {-A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(104\)
risch	\(-\frac {i \left (45 B \,{\mathrm e}^{9 i \left (d x +c \right )}+210 B \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,{\mathrm e}^{6 i \left (d x +c \right )}-560 A \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C \,{\mathrm e}^{4 i \left (d x +c \right )}-210 B \,{\mathrm e}^{3 i \left (d x +c \right )}-400 A \,{\mathrm e}^{2 i \left (d x +c \right )}-320 C \,{\mathrm e}^{2 i \left (d x +c \right )}-45 B \,{\mathrm e}^{i \left (d x +c \right )}-80 A -64 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\)	\(174\)
norman	\(\frac {-\frac {4 \left (25 A +29 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {\left (8 A -5 B +8 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (8 A +5 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (32 A -3 B +16 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (32 A +3 B +16 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {3 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 127, normalized size = 1.04 \begin {gather*} \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A + 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 )} C - 15 \, B {\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 )}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C
- 15*B*(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 = 3.03, size = 122, normalized size = 1.00 \begin {gather*} \frac {45 \, B \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, B \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} + 45 \, B \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 30 \, B \cos \left (d x + c\right ) + 24 \, C\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)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(45*B*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*B*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(5*A + 4
*C)*cos(d*x + c)^4 + 45*B*cos(d*x + c)^3 + 8*(5*A + 4*C)*cos(d*x + c)^2 + 30*B*cos(d*x + c) + 24*C)*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*} \int \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (110) = 220\).
time = 0.45, size = 246, normalized size = 2.02 \begin {gather*} \frac {45 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 320 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 320 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C \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)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(45*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*B*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*tan(1/2*d*
x + 1/2*c)^9 - 75*B*tan(1/2*d*x + 1/2*c)^9 + 120*C*tan(1/2*d*x + 1/2*c)^9 - 320*A*tan(1/2*d*x + 1/2*c)^7 + 30*
B*tan(1/2*d*x + 1/2*c)^7 - 160*C*tan(1/2*d*x + 1/2*c)^7 + 400*A*tan(1/2*d*x + 1/2*c)^5 + 464*C*tan(1/2*d*x + 1
/2*c)^5 - 320*A*tan(1/2*d*x + 1/2*c)^3 - 30*B*tan(1/2*d*x + 1/2*c)^3 - 160*C*tan(1/2*d*x + 1/2*c)^3 + 120*A*ta
n(1/2*d*x + 1/2*c) + 75*B*tan(1/2*d*x + 1/2*c) + 120*C*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 = 5.17, size = 197, normalized size = 1.61 \begin {gather*} \frac {3\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\left (2\,A-\frac {5\,B}{4}+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B}{2}-\frac {16\,A}{3}-\frac {8\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A}{3}+\frac {116\,C}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,A}{3}-\frac {B}{2}-\frac {8\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+\frac {5\,B}{4}+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{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/cos(c + d*x) + C/cos(c + d*x)^2)/cos(c + d*x)^4,x)

[Out]

(3*B*atanh(tan(c/2 + (d*x)/2)))/(4*d) - (tan(c/2 + (d*x)/2)^5*((20*A)/3 + (116*C)/15) + tan(c/2 + (d*x)/2)*(2*
A + (5*B)/4 + 2*C) + tan(c/2 + (d*x)/2)^9*(2*A - (5*B)/4 + 2*C) - tan(c/2 + (d*x)/2)^3*((16*A)/3 + B/2 + (8*C)
/3) - tan(c/2 + (d*x)/2)^7*((16*A)/3 - B/2 + (8*C)/3))/(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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