3.1.57 \(\int \sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [57]

Optimal. Leaf size=51 \[ \frac {(2 A+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

1/2*(2*A+C)*arctanh(sin(d*x+c))/d+B*tan(d*x+c)/d+1/2*C*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]
time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4132, 3852, 8, 4131, 3855} \begin {gather*} \frac {(2 A+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \tan (c+d x) \sec (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A + C)*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Tan[c + d*x])/d + (C*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

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Mathematica [A]
time = 0.02, size = 59, normalized size = 1.16 \begin {gather*} \frac {A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*ArcTanh[Sin[c + d*x]])/d + (C*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Tan[c + d*x])/d + (C*Sec[c + d*x]*Tan[c + d
*x])/(2*d)

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Maple [A]
time = 0.38, size = 63, normalized size = 1.24

method	result	size
derivativedivides	\(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(63\)
default	\(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(63\)
norman	\(\frac {\frac {\left (2 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (2 B -C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {\left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(104\)
risch	\(-\frac {i \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(A*ln(sec(d*x+c)+tan(d*x+c))+B*tan(d*x+c)+C*(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 = 0.28, size = 75, normalized size = 1.47 \begin {gather*} -\frac {C {\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 )} - 4 \, A \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, B \tan \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(C*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*A*log(sec(d*
x + c) + tan(d*x + c)) - 4*B*tan(d*x + c))/d

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Fricas [A]
time = 2.19, size = 82, normalized size = 1.61 \begin {gather*} \frac {{\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + C\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*A + C)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*A + C)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*B
*cos(d*x + c) + C)*sin(d*x + c))/(d*cos(d*x + c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (47) = 94\).
time = 0.44, size = 115, normalized size = 2.25 \begin {gather*} \frac {{\left (2 \, A + C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A + C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 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 )}^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.14, size = 89, normalized size = 1.75 \begin {gather*} \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A+C\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B+C\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\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),x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*(2*A + C))/d - (tan(c/2 + (d*x)/2)^3*(2*B - C) - tan(c/2 + (d*x)/2)*(2*B + C))/(d*(
tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1))

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