3.5.62 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [462]

Optimal. Leaf size=74 \[ \frac {A x}{a^2}-\frac {(4 A-B-2 C) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4137, 4004, 3879} \begin {gather*} -\frac {(4 A-B-2 C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {A x}{a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \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])^2,x]

[Out]

(A*x)/a^2 - ((4*A - B - 2*C)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B + 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 4137

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

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A-B+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {-3 a A+a (A-B-2 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {A x}{a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-B-2 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}\\ &=\frac {A x}{a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-B-2 C) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(175\) vs. \(2(74)=148\).
time = 0.54, size = 175, normalized size = 2.36 \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 B \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 )-6 B \sin \left (c+\frac {d x}{2}\right )-10 A \sin \left (c+\frac {3 d x}{2}\right )+4 B \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 + B*Sec[c + d*x] + 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*B*Sin[(d*x)/2] + 6*C*Sin[(d*x)/2] + 12*A*Sin[c + (d*x)/2]
- 6*B*Sin[c + (d*x)/2] - 10*A*Sin[c + (3*d*x)/2] + 4*B*Sin[c + (3*d*x)/2] + 2*C*Sin[c + (3*d*x)/2]))/(24*a^2*d
)

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Maple [A]
time = 0.54, size = 99, normalized size = 1.34

method	result	size
risch	\(\frac {A x}{a^{2}}+\frac {2 i \left (-6 A \,{\mathrm e}^{2 i \left (d x +c \right )}+3 B \,{\mathrm e}^{2 i \left (d x +c \right )}-9 \,{\mathrm e}^{i \left (d x +c \right )} A +3 B \,{\mathrm e}^{i \left (d x +c \right )}+3 C \,{\mathrm e}^{i \left (d x +c \right )}-5 A +2 B +C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(98\)
derivativedivides	\(\frac {\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(99\)
default	\(\frac {\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(99\)
norman	\(\frac {\frac {A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {A x}{a}+\frac {\left (A -B +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (3 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (5 A -2 B -C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(126\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (70) = 140\).
time = 0.51, size = 164, normalized size = 2.22 \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}} - \frac {B {\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+B*sec(d*x+c)+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 -
B*(3*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

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Fricas [A]
time = 2.83, size = 100, normalized size = 1.35 \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 - 2 \, B - C\right )} \cos \left (d x + c\right ) + 4 \, A - B - 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+B*sec(d*x+c)+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 - 2*B - C)*cos(d*x + c) + 4*A - B - 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)

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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 {B \sec {\left (c + d x \right )}}{\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+B*sec(d*x+c)+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(B*sec(c + d*x)/(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

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Giac [A]
time = 0.48, size = 116, normalized size = 1.57 \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} - B 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 \, B 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+B*sec(d*x+c)+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 - B*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*B*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

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Mupad [B]
time = 3.48, size = 113, normalized size = 1.53 \begin {gather*} \frac {B\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )-\frac {5\,A\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{2}-\frac {3\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {3\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {C\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{2}+\frac {9\,A\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (c+d\,x\right )}{2}+\frac {3\,A\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (c+d\,x\right )}{2}}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \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))^2,x)

[Out]

(B*sin((3*c)/2 + (3*d*x)/2) - (5*A*sin((3*c)/2 + (3*d*x)/2))/2 - (3*A*sin(c/2 + (d*x)/2))/2 + (3*C*sin(c/2 + (
d*x)/2))/2 + (C*sin((3*c)/2 + (3*d*x)/2))/2 + (9*A*cos(c/2 + (d*x)/2)*(c + d*x))/2 + (3*A*cos((3*c)/2 + (3*d*x
)/2)*(c + d*x))/2)/(6*a^2*d*cos(c/2 + (d*x)/2)^3)

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