3.1.2 \(\int \sec ^5(c+d x) (A+C \sec ^2(c+d x)) \, dx\) [2]

Optimal. Leaf size=98 \[ \frac {(6 A+5 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {(6 A+5 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(6 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {C \sec ^5(c+d x) \tan (c+d x)}{6 d} \]

[Out]

1/16*(6*A+5*C)*arctanh(sin(d*x+c))/d+1/16*(6*A+5*C)*sec(d*x+c)*tan(d*x+c)/d+1/24*(6*A+5*C)*sec(d*x+c)^3*tan(d*
x+c)/d+1/6*C*sec(d*x+c)^5*tan(d*x+c)/d

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Rubi [A]
time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4131, 3853, 3855} \begin {gather*} \frac {(6 A+5 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {(6 A+5 C) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {(6 A+5 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {C \tan (c+d x) \sec ^5(c+d x)}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]

[Out]

((6*A + 5*C)*ArcTanh[Sin[c + d*x]])/(16*d) + ((6*A + 5*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((6*A + 5*C)*Sec
[c + d*x]^3*Tan[c + d*x])/(24*d) + (C*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

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]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} (6 A+5 C) \int \sec ^5(c+d x) \, dx\\ &=\frac {(6 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {C \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{8} (6 A+5 C) \int \sec ^3(c+d x) \, dx\\ &=\frac {(6 A+5 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(6 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {C \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{16} (6 A+5 C) \int \sec (c+d x) \, dx\\ &=\frac {(6 A+5 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {(6 A+5 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(6 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {C \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 75, normalized size = 0.77 \begin {gather*} \frac {3 (6 A+5 C) \tanh ^{-1}(\sin (c+d x))+\sec (c+d x) \left (3 (6 A+5 C)+2 (6 A+5 C) \sec ^2(c+d x)+8 C \sec ^4(c+d x)\right ) \tan (c+d x)}{48 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]

[Out]

(3*(6*A + 5*C)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(3*(6*A + 5*C) + 2*(6*A + 5*C)*Sec[c + d*x]^2 + 8*C*Sec[c
+ d*x]^4)*Tan[c + d*x])/(48*d)

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Maple [A]
time = 0.45, size = 108, normalized size = 1.10

method	result	size
derivativedivides	\(\frac {A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(108\)
default	\(\frac {A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(108\)
norman	\(\frac {\frac {\left (2 A +15 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (2 A +15 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (10 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (10 A +11 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {\left (42 A -5 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {\left (42 A -5 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}-\frac {\left (6 A +5 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {\left (6 A +5 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\)	\(203\)
risch	\(-\frac {i \left (18 A \,{\mathrm e}^{11 i \left (d x +c \right )}+15 C \,{\mathrm e}^{11 i \left (d x +c \right )}+102 A \,{\mathrm e}^{9 i \left (d x +c \right )}+85 C \,{\mathrm e}^{9 i \left (d x +c \right )}+84 A \,{\mathrm e}^{7 i \left (d x +c \right )}+198 C \,{\mathrm e}^{7 i \left (d x +c \right )}-84 A \,{\mathrm e}^{5 i \left (d x +c \right )}-198 C \,{\mathrm e}^{5 i \left (d x +c \right )}-102 A \,{\mathrm e}^{3 i \left (d x +c \right )}-85 C \,{\mathrm e}^{3 i \left (d x +c \right )}-18 \,{\mathrm e}^{i \left (d x +c \right )} A -15 C \,{\mathrm e}^{i \left (d x +c \right )}\right )}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{16 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{16 d}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),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)))+C*(-(-1/6*sec(d*x+c)^5-5
/24*sec(d*x+c)^3-5/16*sec(d*x+c))*tan(d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [A]
time = 0.29, size = 126, normalized size = 1.29 \begin {gather*} \frac {3 \, {\left (6 \, A + 5 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (6 \, A + 5 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )^{5} - 8 \, {\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )^{3} + 3 \, {\left (10 \, A + 11 \, C\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(3*(6*A + 5*C)*log(sin(d*x + c) + 1) - 3*(6*A + 5*C)*log(sin(d*x + c) - 1) - 2*(3*(6*A + 5*C)*sin(d*x + c
)^5 - 8*(6*A + 5*C)*sin(d*x + c)^3 + 3*(10*A + 11*C)*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(
d*x + c)^2 - 1))/d

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Fricas [A]
time = 3.77, size = 114, normalized size = 1.16 \begin {gather*} \frac {3 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, C\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/96*(3*(6*A + 5*C)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(6*A + 5*C)*cos(d*x + c)^6*log(-sin(d*x + c) + 1)
+ 2*(3*(6*A + 5*C)*cos(d*x + c)^4 + 2*(6*A + 5*C)*cos(d*x + c)^2 + 8*C)*sin(d*x + c))/(d*cos(d*x + c)^6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + C \sec ^{2}{\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+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**5, x)

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Giac [A]
time = 0.47, size = 121, normalized size = 1.23 \begin {gather*} \frac {3 \, {\left (6 \, A + 5 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (6 \, A + 5 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, A \sin \left (d x + c\right )^{5} + 15 \, C \sin \left (d x + c\right )^{5} - 48 \, A \sin \left (d x + c\right )^{3} - 40 \, C \sin \left (d x + c\right )^{3} + 30 \, A \sin \left (d x + c\right ) + 33 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/96*(3*(6*A + 5*C)*log(abs(sin(d*x + c) + 1)) - 3*(6*A + 5*C)*log(abs(sin(d*x + c) - 1)) - 2*(18*A*sin(d*x +
c)^5 + 15*C*sin(d*x + c)^5 - 48*A*sin(d*x + c)^3 - 40*C*sin(d*x + c)^3 + 30*A*sin(d*x + c) + 33*C*sin(d*x + c)
)/(sin(d*x + c)^2 - 1)^3)/d

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Mupad [B]
time = 2.48, size = 102, normalized size = 1.04 \begin {gather*} \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {3\,A}{8}+\frac {5\,C}{16}\right )}{d}-\frac {\left (\frac {3\,A}{8}+\frac {5\,C}{16}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-A-\frac {5\,C}{6}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {5\,A}{8}+\frac {11\,C}{16}\right )\,\sin \left (c+d\,x\right )}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\sin \left (c+d\,x\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/cos(c + d*x)^5,x)

[Out]

(atanh(sin(c + d*x))*((3*A)/8 + (5*C)/16))/d - (sin(c + d*x)*((5*A)/8 + (11*C)/16) - sin(c + d*x)^3*(A + (5*C)
/6) + sin(c + d*x)^5*((3*A)/8 + (5*C)/16))/(d*(3*sin(c + d*x)^2 - 3*sin(c + d*x)^4 + sin(c + d*x)^6 - 1))

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