3.1.55 \(\int \sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [55]

Optimal. Leaf size=97 \[ \frac {(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x)}{d}+\frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {B \tan ^3(c+d x)}{3 d} \]

[Out]

1/8*(4*A+3*C)*arctanh(sin(d*x+c))/d+B*tan(d*x+c)/d+1/8*(4*A+3*C)*sec(d*x+c)*tan(d*x+c)/d+1/4*C*sec(d*x+c)^3*ta
n(d*x+c)/d+1/3*B*tan(d*x+c)^3/d

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Rubi [A]
time = 0.06, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4132, 3852, 4131, 3853, 3855} \begin {gather*} \frac {(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {B \tan ^3(c+d x)}{3 d}+\frac {B \tan (c+d x)}{d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (B*Tan[c + d*x])/d + ((4*A + 3*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d)
+ (C*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (B*Tan[c + d*x]^3)/(3*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 ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=B \int \sec ^4(c+d x) \, dx+\int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} (4 A+3 C) \int \sec ^3(c+d x) \, dx-\frac {B \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {B \tan (c+d x)}{d}+\frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {B \tan ^3(c+d x)}{3 d}+\frac {1}{8} (4 A+3 C) \int \sec (c+d x) \, dx\\ &=\frac {(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x)}{d}+\frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {B \tan ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 71, normalized size = 0.73 \begin {gather*} \frac {3 (4 A+3 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (4 A+3 C) \sec (c+d x)+6 C \sec ^3(c+d x)+8 B \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(3*(4*A + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(4*A + 3*C)*Sec[c + d*x] + 6*C*Sec[c + d*x]^3 + 8*B*(3
+ Tan[c + d*x]^2)))/(24*d)

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Maple [A]
time = 0.56, size = 106, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(106\)
default	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(106\)
norman	\(\frac {\frac {\left (4 A +5 C -8 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 A +5 C +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 A -9 C -40 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {\left (12 A -9 C +40 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(169\)
risch	\(-\frac {i \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+9 C \,{\mathrm e}^{7 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+33 C \,{\mathrm e}^{5 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-33 C \,{\mathrm e}^{3 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 \,{\mathrm e}^{i \left (d x +c \right )} A -9 C \,{\mathrm e}^{i \left (d x +c \right )}-16 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-B*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C*(-(-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))))

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Maxima [A]
time = 0.28, size = 139, normalized size = 1.43 \begin {gather*} \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B - 3 \, C {\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 )} - 12 \, A {\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 )}}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B - 3*C*(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)) - 12*A*(2*sin(d*x + c)/(sin(d*x + c)^
2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]
time = 3.40, size = 117, normalized size = 1.21 \begin {gather*} \frac {3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, C\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(3*(4*A + 3*C)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A + 3*C)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)
+ 2*(16*B*cos(d*x + c)^3 + 3*(4*A + 3*C)*cos(d*x + c)^2 + 8*B*cos(d*x + c) + 6*C)*sin(d*x + c))/(d*cos(d*x +
c)^4)

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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 ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(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)**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (89) = 178\).
time = 0.47, size = 230, normalized size = 2.37 \begin {gather*} \frac {3 \, {\left (4 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, 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 )}^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(4*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*
(12*A*tan(1/2*d*x + 1/2*c)^7 - 24*B*tan(1/2*d*x + 1/2*c)^7 + 15*C*tan(1/2*d*x + 1/2*c)^7 - 12*A*tan(1/2*d*x +
1/2*c)^5 + 40*B*tan(1/2*d*x + 1/2*c)^5 + 9*C*tan(1/2*d*x + 1/2*c)^5 - 12*A*tan(1/2*d*x + 1/2*c)^3 - 40*B*tan(1
/2*d*x + 1/2*c)^3 + 9*C*tan(1/2*d*x + 1/2*c)^3 + 12*A*tan(1/2*d*x + 1/2*c) + 24*B*tan(1/2*d*x + 1/2*c) + 15*C*
tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 5.03, size = 160, normalized size = 1.65 \begin {gather*} \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {3\,C}{4}\right )}{d}+\frac {\left (A-2\,B+\frac {5\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {10\,B}{3}-A+\frac {3\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,C}{4}-\frac {10\,B}{3}-A\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A+2\,B+\frac {5\,C}{4}\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 )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\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)^3,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*(A + (3*C)/4))/d + (tan(c/2 + (d*x)/2)*(A + 2*B + (5*C)/4) + tan(c/2 + (d*x)/2)^7*(
A - 2*B + (5*C)/4) - tan(c/2 + (d*x)/2)^3*(A + (10*B)/3 - (3*C)/4) + tan(c/2 + (d*x)/2)^5*((10*B)/3 - A + (3*C
)/4))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)
)

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