3.5.53 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx\) [453]

Optimal. Leaf size=52 \[ \frac {A x}{a}+\frac {C \tanh ^{-1}(\sin (c+d x))}{a d}-\frac {(A-B+C) \tan (c+d x)}{a d (1+\sec (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4135, 3855, 4004, 3879} \begin {gather*} -\frac {(A-B+C) \tan (c+d x)}{a d (\sec (c+d x)+1)}+\frac {A x}{a}+\frac {C \tanh ^{-1}(\sin (c+d x))}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x)/a + (C*ArcTanh[Sin[c + d*x]])/(a*d) - ((A - B + C)*Tan[c + d*x])/(a*d*(1 + Sec[c + d*x]))

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4135

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

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(163\) vs. \(2(52)=104\).
time = 0.55, size = 163, normalized size = 3.13 \begin {gather*} \frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (A d x-C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-(A-B+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )\right )}{a d (1+\cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*(Cos[(c + d*x)/2]*(A*d*x - C*Log[Cos[(c + d*x)/2]
- Sin[(c + d*x)/2]] + C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - (A - B + C)*Sec[c/2]*Sin[(d*x)/2]))/(a*d*(
1 + Cos[c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]
time = 0.53, size = 86, normalized size = 1.65

method	result	size
derivativedivides	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(86\)
default	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(86\)
risch	\(\frac {A x}{a}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}\)	\(120\)
norman	\(\frac {\frac {A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {A x}{a}-\frac {\left (A -B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a*(-A*tan(1/2*d*x+1/2*c)+B*tan(1/2*d*x+1/2*c)-C*tan(1/2*d*x+1/2*c)+C*ln(tan(1/2*d*x+1/2*c)+1)+2*A*arctan(t
an(1/2*d*x+1/2*c))-C*ln(tan(1/2*d*x+1/2*c)-1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (52) = 104\).
time = 0.48, size = 146, normalized size = 2.81 \begin {gather*} \frac {A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + C {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {B \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.07, size = 91, normalized size = 1.75 \begin {gather*} \frac {2 \, A d x \cos \left (d x + c\right ) + 2 \, A d x + {\left (C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A - B + C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*d*x*cos(d*x + c) + 2*A*d*x + (C*cos(d*x + c) + C)*log(sin(d*x + c) + 1) - (C*cos(d*x + c) + C)*log(-s
in(d*x + c) + 1) - 2*(A - B + C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A/(sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)*
*2/(sec(c + d*x) + 1), x))/a

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Giac [A]
time = 0.49, size = 92, normalized size = 1.77 \begin {gather*} \frac {\frac {{\left (d x + c\right )} A}{a} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 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 )}{a}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.42, size = 113, normalized size = 2.17 \begin {gather*} \frac {2\,A\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,C\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\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)/(a + a/cos(c + d*x)),x)

[Out]

(2*A*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*C*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a*d) - (
A*sin(c/2 + (d*x)/2) - B*sin(c/2 + (d*x)/2) + C*sin(c/2 + (d*x)/2))/(a*d*cos(c/2 + (d*x)/2))

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