3.2.25 \(\int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx\) [125]

Optimal. Leaf size=49 \[ \frac {A x}{a}+\frac {C \tanh ^{-1}(\sin (c+d x))}{a d}-\frac {(A+C) \tan (c+d x)}{a d (1+\sec (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4136, 3855, 4004, 3879} \begin {gather*} -\frac {(A+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 + 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 + 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 4136

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

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx &=\frac {\int \frac {a A-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-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+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. \(143\) vs. \(2(49)=98\).
time = 0.48, size = 143, normalized size = 2.92 \begin {gather*} -\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (C+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+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+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.38, size = 75, normalized size = 1.53

method	result	size
derivativedivides	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(75\)
default	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(75\)
risch	\(\frac {A x}{a}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}\)	\(97\)
norman	\(\frac {\frac {A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {A x}{a}-\frac {\left (A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+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)-C*tan(1/2*d*x+1/2*c)+C*ln(tan(1/2*d*x+1/2*c)+1)+2*A*arctan(tan(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. 125 vs. \(2 (49) = 98\).
time = 0.51, size = 125, normalized size = 2.55 \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 )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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))))/
d

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Fricas [A]
time = 4.28, size = 88, normalized size = 1.80 \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 + 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+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 + 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 {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+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

(Integral(A/(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 = 0.46, size = 80, normalized size = 1.63 \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 ) + 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+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) + C*tan(1/2*d*x + 1/2*c))/a)/d

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Mupad [B]
time = 2.74, size = 101, normalized size = 2.06 \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 )+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 + 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) + C*sin(c/2 + (d*x)/2))/(a*d*cos(c/2 + (d*x)/2))

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