3.5.70 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [470]

Optimal. Leaf size=115 \[ \frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

A*x/a^3-1/5*(A-B+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(7*A-2*B-3*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-1/15*
(22*A-2*B-3*C)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]
time = 0.14, antiderivative size = 115, 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 = {4137, 4007, 4004, 3879} \begin {gather*} -\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {A x}{a^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \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])^3,x]

[Out]

(A*x)/a^3 - ((A - B + C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((7*A - 2*B - 3*C)*Tan[c + d*x])/(15*a*d
*(a + a*Sec[c + d*x])^2) - ((22*A - 2*B - 3*C)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[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 4007

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

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))^3} \, dx &=-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {-5 a A+a (2 A-2 B-3 C) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {15 a^2 A-a^2 (7 A-2 B-3 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(115)=230\).
time = 0.99, size = 289, normalized size = 2.51 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (150 A d x \cos \left (\frac {d x}{2}\right )+150 A d x \cos \left (c+\frac {d x}{2}\right )+75 A d x \cos \left (c+\frac {3 d x}{2}\right )+75 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-370 A \sin \left (\frac {d x}{2}\right )+80 B \sin \left (\frac {d x}{2}\right )+30 C \sin \left (\frac {d x}{2}\right )+270 A \sin \left (c+\frac {d x}{2}\right )-60 B \sin \left (c+\frac {d x}{2}\right )-30 C \sin \left (c+\frac {d x}{2}\right )-230 A \sin \left (c+\frac {3 d x}{2}\right )+40 B \sin \left (c+\frac {3 d x}{2}\right )+30 C \sin \left (c+\frac {3 d x}{2}\right )+90 A \sin \left (2 c+\frac {3 d x}{2}\right )-30 B \sin \left (2 c+\frac {3 d x}{2}\right )-64 A \sin \left (2 c+\frac {5 d x}{2}\right )+14 B \sin \left (2 c+\frac {5 d x}{2}\right )+6 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{480 a^3 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])^3,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*A*d*x*Cos[(d*x)/2] + 150*A*d*x*Cos[c + (d*x)/2] + 75*A*d*x*Cos[c + (3*d*x)/2
] + 75*A*d*x*Cos[2*c + (3*d*x)/2] + 15*A*d*x*Cos[2*c + (5*d*x)/2] + 15*A*d*x*Cos[3*c + (5*d*x)/2] - 370*A*Sin[
(d*x)/2] + 80*B*Sin[(d*x)/2] + 30*C*Sin[(d*x)/2] + 270*A*Sin[c + (d*x)/2] - 60*B*Sin[c + (d*x)/2] - 30*C*Sin[c
+ (d*x)/2] - 230*A*Sin[c + (3*d*x)/2] + 40*B*Sin[c + (3*d*x)/2] + 30*C*Sin[c + (3*d*x)/2] + 90*A*Sin[2*c + (3
*d*x)/2] - 30*B*Sin[2*c + (3*d*x)/2] - 64*A*Sin[2*c + (5*d*x)/2] + 14*B*Sin[2*c + (5*d*x)/2] + 6*C*Sin[2*c + (
5*d*x)/2]))/(480*a^3*d)

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Maple [A]
time = 0.66, size = 127, normalized size = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {4 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(127\)
default	\(\frac {-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {4 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(127\)
norman	\(\frac {\frac {A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {A x}{a}-\frac {\left (A -B +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (7 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (23 A -13 B +3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}-\frac {\left (25 A -5 B -3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}\)	\(155\)
risch	\(\frac {A x}{a^{3}}-\frac {2 i \left (45 A \,{\mathrm e}^{4 i \left (d x +c \right )}-15 B \,{\mathrm e}^{4 i \left (d x +c \right )}+135 A \,{\mathrm e}^{3 i \left (d x +c \right )}-30 B \,{\mathrm e}^{3 i \left (d x +c \right )}-15 C \,{\mathrm e}^{3 i \left (d x +c \right )}+185 A \,{\mathrm e}^{2 i \left (d x +c \right )}-40 B \,{\mathrm e}^{2 i \left (d x +c \right )}-15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+115 \,{\mathrm e}^{i \left (d x +c \right )} A -20 B \,{\mathrm e}^{i \left (d x +c \right )}-15 C \,{\mathrm e}^{i \left (d x +c \right )}+32 A -7 B -3 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(172\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^3*(-1/5*A*tan(1/2*d*x+1/2*c)^5+1/5*B*tan(1/2*d*x+1/2*c)^5-1/5*C*tan(1/2*d*x+1/2*c)^5+4/3*A*tan(1/2*d*x
+1/2*c)^3-2/3*B*tan(1/2*d*x+1/2*c)^3-7*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)+8*A*arct
an(tan(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.50, size = 205, normalized size = 1.78 \begin {gather*} -\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, 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))^3,x, algorithm="maxima")

[Out]

-1/60*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - B*(15*sin(d*x + c)/(cos(d*x + c) +
1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 3*C*(5*sin(d*x + c
)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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Fricas [A]
time = 2.78, size = 147, normalized size = 1.28 \begin {gather*} \frac {15 \, A d x \cos \left (d x + c\right )^{3} + 45 \, A d x \cos \left (d x + c\right )^{2} + 45 \, A d x \cos \left (d x + c\right ) + 15 \, A d x - {\left ({\left (32 \, A - 7 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 22 \, A - 2 \, B - 3 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/15*(15*A*d*x*cos(d*x + c)^3 + 45*A*d*x*cos(d*x + c)^2 + 45*A*d*x*cos(d*x + c) + 15*A*d*x - ((32*A - 7*B - 3*
C)*cos(d*x + c)^2 + 3*(17*A - 2*B - 3*C)*cos(d*x + c) + 22*A - 2*B - 3*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3
+ 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \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))**3,x)

[Out]

(Integral(A/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)/(sec(c +
d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c
+ d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

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Giac [A]
time = 0.49, size = 153, normalized size = 1.33 \begin {gather*} \frac {\frac {60 \, {\left (d x + c\right )} A}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, 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))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)*A/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1
/2*d*x + 1/2*c)^5 - 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d
*x + 1/2*c) - 15*B*a^12*tan(1/2*d*x + 1/2*c) - 15*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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Mupad [B]
time = 3.44, size = 158, normalized size = 1.37 \begin {gather*} \frac {A\,x}{a^3}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {7\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )+{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}\right )-\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}-\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \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))^3,x)

[Out]

(A*x)/a^3 + (cos(c/2 + (d*x)/2)^4*((B*sin(c/2 + (d*x)/2))/4 - (7*A*sin(c/2 + (d*x)/2))/4 + (C*sin(c/2 + (d*x)/
2))/4) + cos(c/2 + (d*x)/2)^2*((A*sin(c/2 + (d*x)/2)^3)/3 - (B*sin(c/2 + (d*x)/2)^3)/6) - (A*sin(c/2 + (d*x)/2
)^5)/20 + (B*sin(c/2 + (d*x)/2)^5)/20 - (C*sin(c/2 + (d*x)/2)^5)/20)/(a^3*d*cos(c/2 + (d*x)/2)^5)

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