3.2.34 \(\int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [134]

Optimal. Leaf size=68 \[ \frac {A x}{a^2}-\frac {2 (2 A-C) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4138, 4004, 3879} \begin {gather*} -\frac {2 (2 A-C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {A x}{a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x)/a^2 - (2*(2*A - C)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A + C)*Tan[c + d*x])/(3*d*(a + a*Sec[c
+ d*x])^2)

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 4138

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(
-a)*(A + C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc
[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) - a*(A*(m + 1) - C*m)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A,
C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {-3 a A+a (A-2 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {A x}{a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 (2 A-C)) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}\\ &=\frac {A x}{a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {2 (2 A-C) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(141\) vs. \(2(68)=136\).
time = 0.54, size = 141, normalized size = 2.07 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (9 A d x \cos \left (\frac {d x}{2}\right )+9 A d x \cos \left (c+\frac {d x}{2}\right )+3 A d x \cos \left (c+\frac {3 d x}{2}\right )+3 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-18 A \sin \left (\frac {d x}{2}\right )+6 C \sin \left (\frac {d x}{2}\right )+12 A \sin \left (c+\frac {d x}{2}\right )-10 A \sin \left (c+\frac {3 d x}{2}\right )+2 C \sin \left (c+\frac {3 d x}{2}\right )\right )}{24 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(9*A*d*x*Cos[(d*x)/2] + 9*A*d*x*Cos[c + (d*x)/2] + 3*A*d*x*Cos[c + (3*d*x)/2] + 3
*A*d*x*Cos[2*c + (3*d*x)/2] - 18*A*Sin[(d*x)/2] + 6*C*Sin[(d*x)/2] + 12*A*Sin[c + (d*x)/2] - 10*A*Sin[c + (3*d
*x)/2] + 2*C*Sin[c + (3*d*x)/2]))/(24*a^2*d)

________________________________________________________________________________________

Maple [A]
time = 0.36, size = 74, normalized size = 1.09

method	result	size
risch	\(\frac {A x}{a^{2}}-\frac {2 i \left (6 A \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )} A -3 C \,{\mathrm e}^{i \left (d x +c \right )}+5 A -C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(73\)
derivativedivides	\(\frac {\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(74\)
default	\(\frac {\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(74\)
norman	\(\frac {\frac {A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {A x}{a}+\frac {\left (A +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (3 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (5 A -C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(117\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.52, size = 119, normalized size = 1.75 \begin {gather*} -\frac {A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, 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))^2,x, algorithm="maxima")

[Out]

-1/6*(A*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2) - C*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/
d

________________________________________________________________________________________

Fricas [A]
time = 2.55, size = 94, normalized size = 1.38 \begin {gather*} \frac {3 \, A d x \cos \left (d x + c\right )^{2} + 6 \, A d x \cos \left (d x + c\right ) + 3 \, A d x - {\left ({\left (5 \, A - C\right )} \cos \left (d x + c\right ) + 4 \, A - 2 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} 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))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.44, size = 84, normalized size = 1.24 \begin {gather*} \frac {\frac {6 \, {\left (d x + c\right )} A}{a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*A/a^2 + (A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x +
1/2*c) + 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

________________________________________________________________________________________

Mupad [B]
time = 2.58, size = 64, normalized size = 0.94 \begin {gather*} \frac {3\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,A\,d\,x}{6\,a^2\,d} \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))^2,x)

[Out]

(3*C*tan(c/2 + (d*x)/2) - 9*A*tan(c/2 + (d*x)/2) + A*tan(c/2 + (d*x)/2)^3 + C*tan(c/2 + (d*x)/2)^3 + 6*A*d*x)/
(6*a^2*d)

________________________________________________________________________________________