3.2.2 \(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [102]

Optimal. Leaf size=102 \[ -\frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(3 A-8 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(3 A+7 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-1/5*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(3*A-8*B)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2+1/15*(3*A+7*B)*tan
(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]
time = 0.14, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4093, 4085, 3879} \begin {gather*} \frac {(3 A+7 B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(3 A-8 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

-1/5*((A - B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + ((3*A - 8*B)*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x
])^2) + ((3*A + 7*B)*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 4085

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

Rule 4093

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

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec (c+d x) (-3 a (A-B)-5 a B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(3 A-8 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(3 A+7 B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=-\frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(3 A-8 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(3 A+7 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 96, normalized size = 0.94 \begin {gather*} \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (5 (3 A+4 B) \sin \left (\frac {d x}{2}\right )-15 A \sin \left (c+\frac {d x}{2}\right )+(3 A+2 B) \left (5 \sin \left (c+\frac {3 d x}{2}\right )+\sin \left (2 c+\frac {5 d x}{2}\right )\right )\right )}{30 a^3 d (1+\cos (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(5*(3*A + 4*B)*Sin[(d*x)/2] - 15*A*Sin[c + (d*x)/2] + (3*A + 2*B)*(5*Sin[c + (3*d*x
)/2] + Sin[2*c + (5*d*x)/2])))/(30*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]
time = 0.25, size = 64, normalized size = 0.63

method	result	size
derivativedivides	\(\frac {\frac {\left (-A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(64\)
default	\(\frac {\frac {\left (-A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(64\)
risch	\(\frac {2 i \left (15 A \,{\mathrm e}^{3 i \left (d x +c \right )}+15 A \,{\mathrm e}^{2 i \left (d x +c \right )}+20 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )} A +10 B \,{\mathrm e}^{i \left (d x +c \right )}+3 A +2 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(90\)
norman	\(\frac {-\frac {\left (A -B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (3 A +2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (3 A +2 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}+\frac {\left (6 A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a^{2}}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^3*(1/5*(-A+B)*tan(1/2*d*x+1/2*c)^5+2/3*B*tan(1/2*d*x+1/2*c)^3+A*tan(1/2*d*x+1/2*c)+B*tan(1/2*d*x+1/2*c
))

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Maxima [A]
time = 0.27, size = 115, normalized size = 1.13 \begin {gather*} \frac {\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 \, A {\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(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(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*A*(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 = 5.45, size = 93, normalized size = 0.91 \begin {gather*} \frac {{\left ({\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right ) + 3 \, A + 7 \, B\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(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*((3*A + 2*B)*cos(d*x + c)^2 + 3*(3*A + 2*B)*cos(d*x + c) + 3*A + 7*B)*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 ^{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 + \int \frac {B \sec ^{3}{\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(sec(d*x+c)**2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c
+ d*x)**3/(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 = 0.50, size = 75, normalized size = 0.74 \begin {gather*} -\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.92, size = 66, normalized size = 0.65 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+15\,B-3\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+10\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^2*(a + a/cos(c + d*x))^3),x)

[Out]

(tan(c/2 + (d*x)/2)*(15*A + 15*B - 3*A*tan(c/2 + (d*x)/2)^4 + 10*B*tan(c/2 + (d*x)/2)^2 + 3*B*tan(c/2 + (d*x)/
2)^4))/(60*a^3*d)

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